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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
771
|
A
|
Bear and Friendship Condition
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"dsu",
"graphs"
] | null | null |
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
|
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
|
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
|
[
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] |
[
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] |
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.
| 250
|
[
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
"input": "15 42\n8 1\n3 14\n7 14\n12 3\n7 9\n6 7\n6 12\n14 12\n3 10\n10 14\n6 3\n3 13\n13 10\n7 12\n7 2\n6 10\n11 4\n9 3\n8 4\n7 3\n2 3\n2 10\n9 13\n2 14\n6 14\n13 2\n1 4\n13 6\n7 10\n13 14\n12 10\n13 7\n12 2\n9 10\n13 12\n2 6\n9 14\n6 9\n12 9\n11 1\n2 9\n11 8",
"output": "YES"
},
{
"input": "20 80\n17 4\n10 1\n11 10\n17 7\n15 10\n14 15\n13 1\n18 13\n3 13\n12 7\n9 13\n10 12\n14 12\n18 11\n4 7\n10 13\n11 3\n19 8\n14 7\n10 17\n14 3\n7 11\n11 14\n19 5\n10 14\n15 17\n3 1\n9 10\n11 1\n4 1\n11 4\n9 1\n12 3\n13 7\n1 14\n11 12\n7 1\n9 12\n18 15\n17 3\n7 15\n4 10\n7 18\n7 9\n12 17\n14 18\n3 18\n18 17\n9 15\n14 4\n14 9\n9 18\n12 4\n7 10\n15 4\n4 18\n15 13\n1 12\n7 3\n13 11\n4 13\n5 8\n12 18\n12 15\n17 9\n11 15\n3 10\n18 10\n4 3\n15 3\n13 12\n9 4\n9 11\n14 17\n13 17\n3 9\n13 14\n1 17\n15 1\n17 11",
"output": "NO"
},
{
"input": "99 26\n64 17\n48 70\n71 50\n3 50\n9 60\n61 64\n53 50\n25 12\n3 71\n71 53\n3 53\n65 70\n9 25\n9 12\n59 56\n39 60\n64 69\n65 94\n70 94\n25 60\n60 12\n94 48\n17 69\n61 17\n65 48\n61 69",
"output": "NO"
},
{
"input": "3 1\n1 2",
"output": "YES"
},
{
"input": "3 2\n3 2\n1 3",
"output": "NO"
},
{
"input": "3 3\n2 3\n1 2\n1 3",
"output": "YES"
},
{
"input": "4 2\n4 1\n2 1",
"output": "NO"
},
{
"input": "4 3\n3 1\n2 1\n3 2",
"output": "YES"
},
{
"input": "5 9\n1 2\n5 1\n3 1\n1 4\n2 4\n5 3\n5 4\n2 3\n5 2",
"output": "NO"
},
{
"input": "10 5\n9 5\n1 2\n6 8\n6 3\n10 6",
"output": "NO"
},
{
"input": "10 8\n10 7\n9 7\n5 7\n6 8\n3 5\n8 10\n3 4\n7 8",
"output": "NO"
},
{
"input": "10 20\n8 2\n8 3\n1 8\n9 5\n2 4\n10 1\n10 5\n7 5\n7 8\n10 7\n6 5\n3 7\n1 9\n9 8\n7 2\n2 10\n2 1\n6 4\n9 7\n4 3",
"output": "NO"
},
{
"input": "150000 10\n62562 50190\n48849 60549\n139470 18456\n21436 25159\n66845 120884\n99972 114453\n11631 99153\n62951 134848\n78114 146050\n136760 131762",
"output": "YES"
},
{
"input": "150000 0",
"output": "YES"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n1 4",
"output": "NO"
},
{
"input": "30 73\n25 2\n2 16\n20 12\n16 20\n7 18\n11 15\n13 11\n30 29\n16 12\n12 25\n2 1\n18 14\n9 8\n28 16\n2 9\n22 21\n1 25\n12 28\n14 7\n4 9\n26 7\n14 27\n12 2\n29 22\n1 9\n13 15\n3 10\n1 12\n8 20\n30 24\n25 20\n4 1\n4 12\n20 1\n8 4\n2 28\n25 16\n16 8\n20 4\n9 12\n21 30\n23 11\n19 6\n28 4\n29 21\n9 28\n30 10\n22 24\n25 8\n27 26\n25 4\n28 20\n9 25\n24 29\n20 9\n18 26\n1 28\n30 22\n23 15\n28 27\n8 2\n23 13\n12 8\n14 26\n16 4\n28 25\n8 1\n4 2\n9 16\n20 2\n18 27\n28 8\n27 7",
"output": "NO"
},
{
"input": "5 4\n1 2\n2 5\n3 4\n4 5",
"output": "NO"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n4 1",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 4\n4 3\n1 5\n5 6\n6 3",
"output": "NO"
},
{
"input": "3 2\n1 2\n1 3",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n1 6",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n2 4\n3 4",
"output": "NO"
},
{
"input": "6 9\n1 4\n1 5\n1 6\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n3 4",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n2 4",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 1",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "NO"
},
{
"input": "150000 3\n150000 149999\n149998 149999\n149998 150000",
"output": "YES"
},
{
"input": "10 15\n1 2\n1 3\n2 4\n2 5\n3 4\n3 5\n4 5\n1 6\n6 7\n6 8\n7 9\n7 10\n8 9\n8 10\n9 10",
"output": "NO"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "NO"
},
{
"input": "5 4\n1 2\n1 3\n1 4\n1 5",
"output": "NO"
},
{
"input": "9 9\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 1",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n3 4",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n4 2\n4 3",
"output": "NO"
},
{
"input": "5 5\n1 4\n5 3\n5 1\n2 3\n4 5",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n2 3\n1 4\n2 4",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n2 4",
"output": "NO"
},
{
"input": "6 12\n1 2\n1 4\n1 5\n1 6\n2 3\n2 5\n2 6\n3 4\n3 5\n3 6\n4 5\n4 6",
"output": "NO"
},
{
"input": "1000 4\n100 132\n100 164\n196 132\n196 164",
"output": "NO"
}
] | 1,496,423,092
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
# -*- coding: utf-8 -*-
def dfs(u, cnt_vert, cnt_edje):
vis[u-1] = True
cnt_vert += 1
cnt_edje += len(l[u-1])
for i in l[u-1]:
if not vis[i-1]:
dfs(i,cnt_vert,cnt_edje)
n,m = map(int, input().split(' '))
v = [i for i in range(n)]
l = [[] for i in range(n)]
for i in range(m):
x,y = map(int, input().split(' '))
l[x-1].append(y)
l[y-1].append(x)
vis = [False for i in range(n)]
b = True
for i in range(1, n+1):
if not vis[i]:
cnt_vert, cnt_edje = 0,0
dfs(i,cnt_vert,cnt_edje)
if cnt_edje != cnt_vert*(cnt_vert-1):
print('NO')
b = False
if b:
print('YES')
|
Title: Bear and Friendship Condition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.
|
```python
# -*- coding: utf-8 -*-
def dfs(u, cnt_vert, cnt_edje):
vis[u-1] = True
cnt_vert += 1
cnt_edje += len(l[u-1])
for i in l[u-1]:
if not vis[i-1]:
dfs(i,cnt_vert,cnt_edje)
n,m = map(int, input().split(' '))
v = [i for i in range(n)]
l = [[] for i in range(n)]
for i in range(m):
x,y = map(int, input().split(' '))
l[x-1].append(y)
l[y-1].append(x)
vis = [False for i in range(n)]
b = True
for i in range(1, n+1):
if not vis[i]:
cnt_vert, cnt_edje = 0,0
dfs(i,cnt_vert,cnt_edje)
if cnt_edje != cnt_vert*(cnt_vert-1):
print('NO')
b = False
if b:
print('YES')
```
| -1
|
|
867
|
A
|
Between the Offices
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
|
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
|
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
|
[
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] |
[
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] |
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
| 500
|
[
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFFFFSFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "20\nSSFSFSFSFSFSFSFSSFSF",
"output": "YES"
},
{
"input": "20\nSSSSFSFSSFSFSSSSSSFS",
"output": "NO"
},
{
"input": "100\nFFFSFSFSFSSFSFFSSFFFFFSSSSFSSFFFFSFFFFFSFFFSSFSSSFFFFSSFFSSFSFFSSFSSSFSFFSFSFFSFSFFSSFFSFSSSSFSFSFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFSFFFFFFFFFSFSSFFFFFFFFFFFFFFFFFFFFFFSFFSFFFFFSFFFFFFFFSFFFFFFFFFFFFFSFFFFFFFFSFFFFFFFSF",
"output": "NO"
},
{
"input": "100\nSFFSSFFFFFFSSFFFSSFSFFFFFSSFFFSFFFFFFSFSSSFSFSFFFFSFSSFFFFFFFFSFFFFFSFFFFFSSFFFSFFSFSFFFFSFFSFFFFFFF",
"output": "YES"
},
{
"input": "100\nFFFFSSSSSFFSSSFFFSFFFFFSFSSFSFFSFFSSFFSSFSFFFFFSFSFSFSFFFFFFFFFSFSFFSFFFFSFSFFFFFFFFFFFFSFSSFFSSSSFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFSSFFFFSFSFFFSFSSSFSSSSSFSSSSFFSSFFFSFSFSSFFFSSSFFSFSFSSFSFSSFSFFFSFFFFFSSFSFFFSSSFSSSFFS",
"output": "NO"
},
{
"input": "100\nFFFSSSFSFSSSSFSSFSFFSSSFFSSFSSFFSSFFSFSSSSFFFSFFFSFSFSSSFSSFSFSFSFFSSSSSFSSSFSFSFFSSFSFSSFFSSFSFFSFS",
"output": "NO"
},
{
"input": "100\nFFSSSSFSSSFSSSSFSSSFFSFSSFFSSFSSSFSSSFFSFFSSSSSSSSSSSSFSSFSSSSFSFFFSSFFFFFFSFSFSSSSSSFSSSFSFSSFSSFSS",
"output": "NO"
},
{
"input": "100\nSSSFFFSSSSFFSSSSSFSSSSFSSSFSSSSSFSSSSSSSSFSFFSSSFFSSFSSSSFFSSSSSSFFSSSSFSSSSSSFSSSFSSSSSSSFSSSSFSSSS",
"output": "NO"
},
{
"input": "100\nFSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSSSSSSFSSSSSSSSSSSSSFSSFSSSSSFSSFSSSSSSSSSFFSSSSSFSFSSSFFSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSFSSSSSSSSSSSSFSSSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSFS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFSFFFFFFFFFFFSFSFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFSSFFFFSFFFFFFFFFFFFFFFFFFFSFFFSSFFFFSFSFFFSFFFFFFFFFFFFFFFSSFFFFFFFFSSFFFFFFFFFFFFFFSFF",
"output": "YES"
},
{
"input": "100\nSFFSSSFFSFSFSFFFFSSFFFFSFFFFFFFFSFSFFFSFFFSFFFSFFFFSFSFFFFFFFSFFFFFFFFFFSFFSSSFFSSFFFFSFFFFSFFFFSFFF",
"output": "YES"
},
{
"input": "100\nSFFFSFFFFSFFFSSFFFSFSFFFSFFFSSFSFFFFFSFFFFFFFFSFSFSFFSFFFSFSSFSFFFSFSFFSSFSFSSSFFFFFFSSFSFFSFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSSSSFFFFSFFFFFFFSFFFFSFSFFFFSSFFFFFFFFFSFFSSFFFFFFSFSFSSFSSSFFFFFFFSFSFFFSSSFFFFFFFSFFFSSFFFFSSFFFSF",
"output": "YES"
},
{
"input": "100\nSSSFSSFFFSFSSSSFSSFSSSSFSSFFFFFSFFSSSSFFSSSFSSSFSSSSFSSSSFSSSSSSSFSFSSFFFSSFFSFFSSSSFSSFFSFSSFSFFFSF",
"output": "YES"
},
{
"input": "100\nSFFSFSSSSSSSFFSSSFSSSSFSFSSFFFSSSSSSFSSSSFSSFSSSFSSSSSSSFSSFSFFFSSFSSFSFSFSSSSSSSSSSSSFFFFSSSSSFSFFF",
"output": "YES"
},
{
"input": "100\nSSSFSFFSFSFFSSSSSFSSSFSSSFFFSSSSSSSSSFSFSFSSSSFSFSSFFFFFSSSSSSSSSSSSSSSSSSSFFSSSSSFSFSSSSFFSSSSFSSSF",
"output": "YES"
},
{
"input": "100\nSSSFSSSSSSSSSSFSSSSFSSSSSSFSSSSSSFSSSSSSSSSSSSSSFSSSFSSSFSSSSSSSSSSSFSSSSSSFSFSSSSFSSSSSSFSSSSSSSSFF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFFSSSSSSSSSFSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "2\nSS",
"output": "NO"
}
] | 1,586,086,177
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 109
| 0
|
c = int(input())
s = input()
if s[0] == "S" and s[c-1] == "F":
print("YES")
else:
print("NO")
|
Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
|
```python
c = int(input())
s = input()
if s[0] == "S" and s[c-1] == "F":
print("YES")
else:
print("NO")
```
| 3
|
|
669
|
A
|
Little Artem and Presents
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
|
Print the maximum possible number of times Artem can give presents to Masha.
|
[
"1\n",
"2\n",
"3\n",
"4\n"
] |
[
"1\n",
"1\n",
"2\n",
"3\n"
] |
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
| 500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"
},
{
"input": "1000000000",
"output": "666666667"
},
{
"input": "5",
"output": "3"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "666666666"
},
{
"input": "999999998",
"output": "666666665"
},
{
"input": "999999997",
"output": "666666665"
},
{
"input": "999999996",
"output": "666666664"
},
{
"input": "999999995",
"output": "666666663"
},
{
"input": "999999994",
"output": "666666663"
},
{
"input": "999999993",
"output": "666666662"
},
{
"input": "999999992",
"output": "666666661"
},
{
"input": "999999991",
"output": "666666661"
},
{
"input": "1000",
"output": "667"
},
{
"input": "10000",
"output": "6667"
},
{
"input": "100000",
"output": "66667"
},
{
"input": "1000000",
"output": "666667"
},
{
"input": "10000000",
"output": "6666667"
},
{
"input": "100000000",
"output": "66666667"
},
{
"input": "7",
"output": "5"
}
] | 1,655,928,614
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
a = int(input())
if a == 0:
print (0)
elif a - 1 == 0 or a - 1 == 1:
print (1)
elif a%4 == 0:
if a%4 == 1:
r = (a - 1) / 4
s = 2 * r + 1
print (s)
elif a%4 == 2:
r = (a - 2) / 4
s = 2 * r + 1
print (s)
elif a%4 == 3:
r = (a - 3) / 4
s = 2 * r + 1
print (s)
elif a == 4:
print (3)
else:
r = a / 4
s = r * 2
print (s)
else:
if a%3 == 0:
r = a / 3
s = 2 * r
print (s)
elif a%3 == 1:
r = (a - 1) / 3
s = 2 * r + 1
print (s)
else:
r = (a - 2) / 3
s = 2 * r + 1
print (s)
|
Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
|
```python
a = int(input())
if a == 0:
print (0)
elif a - 1 == 0 or a - 1 == 1:
print (1)
elif a%4 == 0:
if a%4 == 1:
r = (a - 1) / 4
s = 2 * r + 1
print (s)
elif a%4 == 2:
r = (a - 2) / 4
s = 2 * r + 1
print (s)
elif a%4 == 3:
r = (a - 3) / 4
s = 2 * r + 1
print (s)
elif a == 4:
print (3)
else:
r = a / 4
s = r * 2
print (s)
else:
if a%3 == 0:
r = a / 3
s = 2 * r
print (s)
elif a%3 == 1:
r = (a - 1) / 3
s = 2 * r + 1
print (s)
else:
r = (a - 2) / 3
s = 2 * r + 1
print (s)
```
| 0
|
|
507
|
B
|
Amr and Pins
|
PROGRAMMING
| 1,400
|
[
"geometry",
"math"
] | null | null |
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
|
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=≤<=*r*<=≤<=105, <=-<=105<=≤<=*x*,<=*y*,<=*x*',<=*y*'<=≤<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
|
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
|
[
"2 0 0 0 4\n",
"1 1 1 4 4\n",
"4 5 6 5 6\n"
] |
[
"1\n",
"3\n",
"0\n"
] |
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "2 0 0 0 4",
"output": "1"
},
{
"input": "1 1 1 4 4",
"output": "3"
},
{
"input": "4 5 6 5 6",
"output": "0"
},
{
"input": "10 20 0 40 0",
"output": "1"
},
{
"input": "9 20 0 40 0",
"output": "2"
},
{
"input": "5 -1 -6 -5 1",
"output": "1"
},
{
"input": "99125 26876 -21414 14176 17443",
"output": "1"
},
{
"input": "8066 7339 19155 -90534 -60666",
"output": "8"
},
{
"input": "100000 -100000 -100000 100000 100000",
"output": "2"
},
{
"input": "10 20 0 41 0",
"output": "2"
},
{
"input": "25 -64 -6 -56 64",
"output": "2"
},
{
"input": "125 455 450 439 721",
"output": "2"
},
{
"input": "5 6 3 7 2",
"output": "1"
},
{
"input": "24 130 14786 3147 2140",
"output": "271"
},
{
"input": "125 -363 176 93 330",
"output": "2"
},
{
"input": "1 14 30 30 14",
"output": "12"
},
{
"input": "25 96 13 7 2",
"output": "2"
},
{
"input": "4 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "1 3 4 2 5",
"output": "1"
},
{
"input": "1 -3 3 2 6",
"output": "3"
},
{
"input": "2 7 20 13 -5",
"output": "7"
},
{
"input": "1 1 1 1 4",
"output": "2"
},
{
"input": "249 -54242 -30537 -45023 -89682",
"output": "121"
},
{
"input": "4 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "97741 23818 78751 97583 26933",
"output": "1"
},
{
"input": "56767 -29030 51625 79823 -56297",
"output": "2"
},
{
"input": "98260 13729 74998 23701 9253",
"output": "1"
},
{
"input": "67377 -80131 -90254 -57320 14102",
"output": "1"
},
{
"input": "1 100000 100000 100000 -100000",
"output": "100000"
},
{
"input": "19312 19470 82059 58064 62231",
"output": "2"
},
{
"input": "67398 -68747 -79056 -34193 29400",
"output": "1"
},
{
"input": "91099 37184 -71137 75650 -3655",
"output": "1"
},
{
"input": "46456 -2621 -23623 -98302 -99305",
"output": "2"
},
{
"input": "100 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "1 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "8 0 0 0 32",
"output": "2"
},
{
"input": "100000 100000 1 -100000 0",
"output": "2"
}
] | 1,697,050,080
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 62
| 0
|
import math
r, x, y, a, b = map(int, input().split())
d = math.sqrt((a - x)**2 + (b - y)**2)
s = math.ceil(d / (2 * r))
print(s)
|
Title: Amr and Pins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input Specification:
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=≤<=*r*<=≤<=105, <=-<=105<=≤<=*x*,<=*y*,<=*x*',<=*y*'<=≤<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output Specification:
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Demo Input:
['2 0 0 0 4\n', '1 1 1 4 4\n', '4 5 6 5 6\n']
Demo Output:
['1\n', '3\n', '0\n']
Note:
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
import math
r, x, y, a, b = map(int, input().split())
d = math.sqrt((a - x)**2 + (b - y)**2)
s = math.ceil(d / (2 * r))
print(s)
```
| 3
|
|
937
|
A
|
Olympiad
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
|
Print a single integer — the desired number of ways.
|
[
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] |
[
"3\n",
"1\n",
"1\n"
] |
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
| 500
|
[
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
"input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600",
"output": "94"
},
{
"input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600",
"output": "1"
},
{
"input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12",
"output": "13"
},
{
"input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391",
"output": "33"
},
{
"input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3",
"output": "3"
},
{
"input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55",
"output": "93"
},
{
"input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13",
"output": "13"
},
{
"input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1",
"output": "1"
},
{
"input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37",
"output": "61"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 600",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "2"
},
{
"input": "4\n0 0 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "2\n0 5",
"output": "1"
},
{
"input": "5\n1 0 0 1 2",
"output": "2"
}
] | 1,618,256,593
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 21
| 108
| 0
|
n = int(input())
a = set([i for i in map(int,input().split()) if i!=0])
print(len(a))
|
Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
|
```python
n = int(input())
a = set([i for i in map(int,input().split()) if i!=0])
print(len(a))
```
| 3
|
|
651
|
B
|
Beautiful Paintings
|
PROGRAMMING
| 1,200
|
[
"greedy",
"sortings"
] | null | null |
There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.
We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=><=*a**i*.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting.
|
Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=><=*a**i*, after the optimal rearrangement.
|
[
"5\n20 30 10 50 40\n",
"4\n200 100 100 200\n"
] |
[
"4\n",
"2\n"
] |
In the first sample, the optimal order is: 10, 20, 30, 40, 50.
In the second sample, the optimal order is: 100, 200, 100, 200.
| 1,000
|
[
{
"input": "5\n20 30 10 50 40",
"output": "4"
},
{
"input": "4\n200 100 100 200",
"output": "2"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0"
},
{
"input": "1\n1000",
"output": "0"
},
{
"input": "2\n444 333",
"output": "1"
},
{
"input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14",
"output": "95"
},
{
"input": "1\n995",
"output": "0"
},
{
"input": "10\n103 101 103 103 101 102 100 100 101 104",
"output": "7"
},
{
"input": "20\n102 100 102 104 102 101 104 103 100 103 105 105 100 105 100 100 101 105 105 102",
"output": "15"
},
{
"input": "20\n990 994 996 999 997 994 990 992 990 993 992 990 999 999 992 994 997 990 993 998",
"output": "15"
},
{
"input": "100\n1 8 3 8 10 8 5 3 10 3 5 8 4 5 5 5 10 3 6 6 6 6 6 7 2 7 2 4 7 8 3 8 7 2 5 6 1 5 5 7 9 7 6 9 1 8 1 3 6 5 1 3 6 9 5 6 8 4 8 6 10 9 2 9 3 8 7 5 2 10 2 10 3 6 5 5 3 5 10 2 3 7 10 8 8 4 3 4 9 6 10 7 6 6 6 4 9 9 8 9",
"output": "84"
}
] | 1,623,781,480
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 109
| 1,843,200
|
"""
Bismillahir Rahmanir Rahim
A |-| |\| A F
"""
from collections import defaultdict
n = int(input())
ar = list(map(int, input().split()))
ar.sort()
chk = defaultdict(lambda : 0)
ans = []
for i in range(0, n):
if not chk[i]:
ans.append(ar[i])
cur = ar[i]
chk[i] = 1
for j in range(i+1, n):
if not chk[j] and ar[j] > cur:
ans.append(ar[j])
chk[j] = 1
cur = ar[j]
cur, anss = ar[0], 0
for i in range(1, n):
if ans[i] > cur:
anss += 1
cur = ans[i]
else: cur = ans[i]
print(anss)
|
Title: Beautiful Paintings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.
We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=><=*a**i*.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting.
Output Specification:
Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=><=*a**i*, after the optimal rearrangement.
Demo Input:
['5\n20 30 10 50 40\n', '4\n200 100 100 200\n']
Demo Output:
['4\n', '2\n']
Note:
In the first sample, the optimal order is: 10, 20, 30, 40, 50.
In the second sample, the optimal order is: 100, 200, 100, 200.
|
```python
"""
Bismillahir Rahmanir Rahim
A |-| |\| A F
"""
from collections import defaultdict
n = int(input())
ar = list(map(int, input().split()))
ar.sort()
chk = defaultdict(lambda : 0)
ans = []
for i in range(0, n):
if not chk[i]:
ans.append(ar[i])
cur = ar[i]
chk[i] = 1
for j in range(i+1, n):
if not chk[j] and ar[j] > cur:
ans.append(ar[j])
chk[j] = 1
cur = ar[j]
cur, anss = ar[0], 0
for i in range(1, n):
if ans[i] > cur:
anss += 1
cur = ans[i]
else: cur = ans[i]
print(anss)
```
| 3
|
|
975
|
B
|
Mancala
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.
Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction.
Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole.
After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.
Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.
|
The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole.
It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board.
|
Output one integer, the maximum possible score after one move.
|
[
"0 1 1 0 0 0 0 0 0 7 0 0 0 0\n",
"5 1 1 1 1 0 0 0 0 0 0 0 0 0\n"
] |
[
"4\n",
"8\n"
] |
In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$.
| 1,000
|
[
{
"input": "0 1 1 0 0 0 0 0 0 7 0 0 0 0",
"output": "4"
},
{
"input": "5 1 1 1 1 0 0 0 0 0 0 0 0 0",
"output": "8"
},
{
"input": "10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 1",
"output": "54294"
},
{
"input": "0 0 0 0 0 0 0 0 0 0 0 0 0 15",
"output": "2"
},
{
"input": "1 0 0 0 0 1 0 0 0 0 1 0 0 0",
"output": "0"
},
{
"input": "5 5 1 1 1 3 3 3 5 7 5 3 7 5",
"output": "38"
},
{
"input": "787 393 649 463 803 365 81 961 989 531 303 407 579 915",
"output": "7588"
},
{
"input": "8789651 4466447 1218733 6728667 1796977 6198853 8263135 6309291 8242907 7136751 3071237 5397369 6780785 9420869",
"output": "81063456"
},
{
"input": "0 0 0 0 0 0 0 0 0 0 0 0 0 29",
"output": "26"
},
{
"input": "282019717 109496191 150951267 609856495 953855615 569750143 6317733 255875779 645191029 572053369 290936613 338480779 879775193 177172893",
"output": "5841732816"
},
{
"input": "105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505",
"output": "120472578"
},
{
"input": "404418821 993626161 346204297 122439813 461187221 628048227 625919459 628611733 938993057 701270099 398043779 684205961 630975553 575964835",
"output": "8139909016"
},
{
"input": "170651077 730658441 824213789 583764177 129437345 717005779 675398017 314979709 380861369 265878463 746564659 797260041 506575735 335169317",
"output": "6770880638"
},
{
"input": "622585025 48249287 678950449 891575125 637411965 457739735 829353393 235216425 284006447 875591469 492839209 296444305 513776057 810057753",
"output": "7673796644"
},
{
"input": "475989857 930834747 786217439 927967137 489188151 869354161 276693267 56154399 131055697 509249443 143116853 426254423 44465165 105798821",
"output": "6172339560"
},
{
"input": "360122921 409370351 226220005 604004145 85173909 600403773 624052991 138163383 729239967 189036661 619842883 270087537 749500483 243727913",
"output": "5848946922"
},
{
"input": "997102881 755715147 273805839 436713689 547411799 72470207 522269145 647688957 137422311 422612659 197751751 679663349 821420227 387967237",
"output": "6900015198"
},
{
"input": "690518849 754551537 652949719 760695679 491633619 477564457 11669279 700467439 470069297 782338983 718169393 884421719 24619427 215745577",
"output": "7635414974"
},
{
"input": "248332749 486342237 662201929 917696895 555278549 252122023 850296207 463343655 832574345 954281071 168282553 825538865 996753493 461254663",
"output": "6400166934"
},
{
"input": "590789361 636464947 404477303 337309187 476703809 426863069 120608741 703406277 645444697 761482231 996635839 33459441 677458865 483861751",
"output": "7294965518"
},
{
"input": "297857621 238127103 749085829 139033277 597985489 202617713 982184715 183932743 278551059 297781685 330124279 338959601 682874531 187519685",
"output": "5201808164"
},
{
"input": "1 1 1 1 1 0 0 0 0 0 0 0 0 0",
"output": "2"
},
{
"input": "1 1 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "2"
},
{
"input": "1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "1 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "2"
},
{
"input": "0 0 0 0 0 0 0 0 0 0 0 0 1 1",
"output": "2"
}
] | 1,525,185,138
| 1,638
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 2
| 77
| 7,065,600
|
m = list(map(int, input().split()))
mx = 0
for i in range(14):
if m[i] == 0:
continue
e = m[i] // 14
k = m[i] % 14
g = [x + e for x in m]
g[i] -= m[i]
for u in range(i + 1, min(14, k)):
g[u] += 1
for u in range(0, k + i - 13):
g[u] += 1
s = 0
for d in g:
if not d % 2:
s += d
mx = max(mx, s)
print(mx)
|
Title: Mancala
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.
Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction.
Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole.
After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.
Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.
Input Specification:
The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole.
It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board.
Output Specification:
Output one integer, the maximum possible score after one move.
Demo Input:
['0 1 1 0 0 0 0 0 0 7 0 0 0 0\n', '5 1 1 1 1 0 0 0 0 0 0 0 0 0\n']
Demo Output:
['4\n', '8\n']
Note:
In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$.
|
```python
m = list(map(int, input().split()))
mx = 0
for i in range(14):
if m[i] == 0:
continue
e = m[i] // 14
k = m[i] % 14
g = [x + e for x in m]
g[i] -= m[i]
for u in range(i + 1, min(14, k)):
g[u] += 1
for u in range(0, k + i - 13):
g[u] += 1
s = 0
for d in g:
if not d % 2:
s += d
mx = max(mx, s)
print(mx)
```
| 0
|
|
937
|
A
|
Olympiad
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
|
Print a single integer — the desired number of ways.
|
[
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] |
[
"3\n",
"1\n",
"1\n"
] |
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
| 500
|
[
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
"input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600",
"output": "94"
},
{
"input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600",
"output": "1"
},
{
"input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12",
"output": "13"
},
{
"input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391",
"output": "33"
},
{
"input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3",
"output": "3"
},
{
"input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55",
"output": "93"
},
{
"input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13",
"output": "13"
},
{
"input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1",
"output": "1"
},
{
"input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37",
"output": "61"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 600",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "2"
},
{
"input": "4\n0 0 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "2\n0 5",
"output": "1"
},
{
"input": "5\n1 0 0 1 2",
"output": "2"
}
] | 1,565,364,688
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 140
| 0
|
input()
a = sorted(set(map(int, input().split())))
print(len(a) if a[0] != 0 else len(a) - 1)
|
Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
|
```python
input()
a = sorted(set(map(int, input().split())))
print(len(a) if a[0] != 0 else len(a) - 1)
```
| 3
|
|
681
|
C
|
Heap Operations
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"data structures",
"greedy"
] | null | null |
Petya has recently learned data structure named "Binary heap".
The heap he is now operating with allows the following operations:
- put the given number into the heap; - get the value of the minimum element in the heap; - extract the minimum element from the heap;
Thus, at any moment of time the heap contains several integers (possibly none), some of them might be equal.
In order to better learn this data structure Petya took an empty heap and applied some operations above to it. Also, he carefully wrote down all the operations and their results to his event log, following the format:
- insert *x* — put the element with value *x* in the heap; - getMin *x* — the value of the minimum element contained in the heap was equal to *x*; - removeMin — the minimum element was extracted from the heap (only one instance, if there were many).
All the operations were correct, i.e. there was at least one element in the heap each time getMin or removeMin operations were applied.
While Petya was away for a lunch, his little brother Vova came to the room, took away some of the pages from Petya's log and used them to make paper boats.
Now Vova is worried, if he made Petya's sequence of operations inconsistent. For example, if one apply operations one-by-one in the order they are written in the event log, results of getMin operations might differ from the results recorded by Petya, and some of getMin or removeMin operations may be incorrect, as the heap is empty at the moment they are applied.
Now Vova wants to add some new operation records to the event log in order to make the resulting sequence of operations correct. That is, the result of each getMin operation is equal to the result in the record, and the heap is non-empty when getMin ad removeMin are applied. Vova wants to complete this as fast as possible, as the Petya may get back at any moment. He asks you to add the least possible number of operation records to the current log. Note that arbitrary number of operations may be added at the beginning, between any two other operations, or at the end of the log.
|
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of the records left in Petya's journal.
Each of the following *n* lines describe the records in the current log in the order they are applied. Format described in the statement is used. All numbers in the input are integers not exceeding 109 by their absolute value.
|
The first line of the output should contain a single integer *m* — the minimum possible number of records in the modified sequence of operations.
Next *m* lines should contain the corrected sequence of records following the format of the input (described in the statement), one per line and in the order they are applied. All the numbers in the output should be integers not exceeding 109 by their absolute value.
Note that the input sequence of operations must be the subsequence of the output sequence.
It's guaranteed that there exists the correct answer consisting of no more than 1<=000<=000 operations.
|
[
"2\ninsert 3\ngetMin 4\n",
"4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2\n"
] |
[
"4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4\n",
"6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2\n"
] |
In the first sample, after number 3 is inserted into the heap, the minimum number is 3. To make the result of the first getMin equal to 4 one should firstly remove number 3 from the heap and then add number 4 into the heap.
In the second sample case number 1 is inserted two times, so should be similarly removed twice.
| 1,500
|
[
{
"input": "2\ninsert 3\ngetMin 4",
"output": "4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4"
},
{
"input": "4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2",
"output": "6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2"
},
{
"input": "1\ninsert 1",
"output": "1\ninsert 1"
},
{
"input": "1\ngetMin 31",
"output": "2\ninsert 31\ngetMin 31"
},
{
"input": "1\nremoveMin",
"output": "2\ninsert 0\nremoveMin"
},
{
"input": "2\ninsert 2\ngetMin 2",
"output": "2\ninsert 2\ngetMin 2"
},
{
"input": "2\ninsert 31\nremoveMin",
"output": "2\ninsert 31\nremoveMin"
},
{
"input": "2\ngetMin 31\nremoveMin",
"output": "3\ninsert 31\ngetMin 31\nremoveMin"
},
{
"input": "2\nremoveMin\ngetMin 31",
"output": "4\ninsert 0\nremoveMin\ninsert 31\ngetMin 31"
},
{
"input": "8\ninsert 219147240\nremoveMin\ngetMin 923854124\nremoveMin\ngetMin -876779400\nremoveMin\ninsert 387686853\ngetMin 749998368",
"output": "12\ninsert 219147240\nremoveMin\ninsert 923854124\ngetMin 923854124\nremoveMin\ninsert -876779400\ngetMin -876779400\nremoveMin\ninsert 387686853\nremoveMin\ninsert 749998368\ngetMin 749998368"
},
{
"input": "2\nremoveMin\ninsert 450653162",
"output": "3\ninsert 0\nremoveMin\ninsert 450653162"
},
{
"input": "6\ninsert -799688192\ngetMin 491561656\nremoveMin\ninsert -805250162\ninsert -945439443\nremoveMin",
"output": "8\ninsert -799688192\nremoveMin\ninsert 491561656\ngetMin 491561656\nremoveMin\ninsert -805250162\ninsert -945439443\nremoveMin"
},
{
"input": "30\ninsert 62350949\ngetMin -928976719\nremoveMin\ngetMin 766590157\ngetMin -276914351\ninsert 858958907\ngetMin -794653029\ngetMin 505812710\ngetMin -181182543\ninsert -805198995\nremoveMin\ninsert -200361579\nremoveMin\ninsert 988531216\ninsert -474257426\ninsert 579296921\nremoveMin\ninsert -410043658\ngetMin 716684155\nremoveMin\ngetMin -850837161\ngetMin 368670814\ninsert 579000842\nremoveMin\ngetMin -169833018\ninsert 313148949\nremoveMin\nremoveMin\ngetMin 228901059\ngetMin 599172503",
"output": "52\ninsert 62350949\ninsert -928976719\ngetMin -928976719\nremoveMin\nremoveMin\ninsert 766590157\ngetMin 766590157\ninsert -276914351\ngetMin -276914351\ninsert 858958907\ninsert -794653029\ngetMin -794653029\nremoveMin\nremoveMin\ninsert 505812710\ngetMin 505812710\ninsert -181182543\ngetMin -181182543\ninsert -805198995\nremoveMin\ninsert -200361579\nremoveMin\ninsert 988531216\ninsert -474257426\ninsert 579296921\nremoveMin\ninsert -410043658\nremoveMin\nremoveMin\nremoveMin\nremoveMin\ninsert 71668415..."
},
{
"input": "9\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ngetMin 3\nremoveMin\ngetMin 4\nremoveMin\ngetMin 5",
"output": "10\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ninsert 3\ngetMin 3\nremoveMin\ngetMin 4\nremoveMin\ngetMin 5"
},
{
"input": "9\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ngetMin 5\nremoveMin\ngetMin 4\nremoveMin\ngetMin 3",
"output": "12\ninsert 3\ninsert 4\ninsert 5\nremoveMin\nremoveMin\ngetMin 5\nremoveMin\ninsert 4\ngetMin 4\nremoveMin\ninsert 3\ngetMin 3"
},
{
"input": "1\ngetMin 0",
"output": "2\ninsert 0\ngetMin 0"
},
{
"input": "3\ninsert 3\ninsert 4\ngetMin 4",
"output": "4\ninsert 3\ninsert 4\nremoveMin\ngetMin 4"
},
{
"input": "3\ninsert 1\ninsert 2\ngetMin 2",
"output": "4\ninsert 1\ninsert 2\nremoveMin\ngetMin 2"
},
{
"input": "3\ninsert -1\ninsert 0\ngetMin 0",
"output": "4\ninsert -1\ninsert 0\nremoveMin\ngetMin 0"
},
{
"input": "2\ninsert 0\ngetMin 0",
"output": "2\ninsert 0\ngetMin 0"
},
{
"input": "1\ninsert -1",
"output": "1\ninsert -1"
},
{
"input": "3\ninsert 1\ninsert 0\ngetMin 1",
"output": "4\ninsert 1\ninsert 0\nremoveMin\ngetMin 1"
}
] | 1,650,823,762
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
command_count = int( input() )
heap = []
command_list = []
def insert(_heap, value):
_heap.append(value)
_heap.sort()
return _heap
for i in range(command_count):
_input = input()
try:
command, value = _input.split()
except:
command = _input
if command == 'insert':
heap = insert(heap, value)
command_list.append( f'insert {value}' )
elif command == 'getMin':
curr_min = heap[0] if len(heap) > 0 else None
if curr_min == None: # Lista vazia
command_list.extend( [f'insert {value}', f'getMin {value}'] )
else: # Lista tem elementos
if curr_min == value: # Elemento já estava na heap como minimo
command_list.extend( [f'getMin {value}'] )
elif curr_min > value: # O minimo atual é maior -> basta inserir o novo
command_list.extend( [f'insert {value}', f'getMin {value}'] )
else: # O minimo atual é menor -> tem que tirar tudo até esse novo
final = []
for elem in heap:
if elem < value:
command_list.append( 'removeMin' )
else:
final.append(elem)
heap = insert(final, value)
command_list.extend( [f'insert {value}', f'getMin {value}'] )
elif command == 'removeMin':
heap = heap[1:]
command_list.append( 'removeMin' )
# Output
print(len(command_list))
for elem in command_list:
print(elem)
|
Title: Heap Operations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has recently learned data structure named "Binary heap".
The heap he is now operating with allows the following operations:
- put the given number into the heap; - get the value of the minimum element in the heap; - extract the minimum element from the heap;
Thus, at any moment of time the heap contains several integers (possibly none), some of them might be equal.
In order to better learn this data structure Petya took an empty heap and applied some operations above to it. Also, he carefully wrote down all the operations and their results to his event log, following the format:
- insert *x* — put the element with value *x* in the heap; - getMin *x* — the value of the minimum element contained in the heap was equal to *x*; - removeMin — the minimum element was extracted from the heap (only one instance, if there were many).
All the operations were correct, i.e. there was at least one element in the heap each time getMin or removeMin operations were applied.
While Petya was away for a lunch, his little brother Vova came to the room, took away some of the pages from Petya's log and used them to make paper boats.
Now Vova is worried, if he made Petya's sequence of operations inconsistent. For example, if one apply operations one-by-one in the order they are written in the event log, results of getMin operations might differ from the results recorded by Petya, and some of getMin or removeMin operations may be incorrect, as the heap is empty at the moment they are applied.
Now Vova wants to add some new operation records to the event log in order to make the resulting sequence of operations correct. That is, the result of each getMin operation is equal to the result in the record, and the heap is non-empty when getMin ad removeMin are applied. Vova wants to complete this as fast as possible, as the Petya may get back at any moment. He asks you to add the least possible number of operation records to the current log. Note that arbitrary number of operations may be added at the beginning, between any two other operations, or at the end of the log.
Input Specification:
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of the records left in Petya's journal.
Each of the following *n* lines describe the records in the current log in the order they are applied. Format described in the statement is used. All numbers in the input are integers not exceeding 109 by their absolute value.
Output Specification:
The first line of the output should contain a single integer *m* — the minimum possible number of records in the modified sequence of operations.
Next *m* lines should contain the corrected sequence of records following the format of the input (described in the statement), one per line and in the order they are applied. All the numbers in the output should be integers not exceeding 109 by their absolute value.
Note that the input sequence of operations must be the subsequence of the output sequence.
It's guaranteed that there exists the correct answer consisting of no more than 1<=000<=000 operations.
Demo Input:
['2\ninsert 3\ngetMin 4\n', '4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2\n']
Demo Output:
['4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4\n', '6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2\n']
Note:
In the first sample, after number 3 is inserted into the heap, the minimum number is 3. To make the result of the first getMin equal to 4 one should firstly remove number 3 from the heap and then add number 4 into the heap.
In the second sample case number 1 is inserted two times, so should be similarly removed twice.
|
```python
command_count = int( input() )
heap = []
command_list = []
def insert(_heap, value):
_heap.append(value)
_heap.sort()
return _heap
for i in range(command_count):
_input = input()
try:
command, value = _input.split()
except:
command = _input
if command == 'insert':
heap = insert(heap, value)
command_list.append( f'insert {value}' )
elif command == 'getMin':
curr_min = heap[0] if len(heap) > 0 else None
if curr_min == None: # Lista vazia
command_list.extend( [f'insert {value}', f'getMin {value}'] )
else: # Lista tem elementos
if curr_min == value: # Elemento já estava na heap como minimo
command_list.extend( [f'getMin {value}'] )
elif curr_min > value: # O minimo atual é maior -> basta inserir o novo
command_list.extend( [f'insert {value}', f'getMin {value}'] )
else: # O minimo atual é menor -> tem que tirar tudo até esse novo
final = []
for elem in heap:
if elem < value:
command_list.append( 'removeMin' )
else:
final.append(elem)
heap = insert(final, value)
command_list.extend( [f'insert {value}', f'getMin {value}'] )
elif command == 'removeMin':
heap = heap[1:]
command_list.append( 'removeMin' )
# Output
print(len(command_list))
for elem in command_list:
print(elem)
```
| 0
|
|
136
|
A
|
Presents
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
|
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
|
[
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input": "5\n5 4 3 2 1",
"output": "5 4 3 2 1"
},
{
"input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19"
},
{
"input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19",
"output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19"
},
{
"input": "10\n3 4 5 6 7 8 9 10 1 2",
"output": "9 10 1 2 3 4 5 6 7 8"
},
{
"input": "8\n1 5 3 7 2 6 4 8",
"output": "1 5 3 7 2 6 4 8"
},
{
"input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47",
"output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19"
},
{
"input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17",
"output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25"
},
{
"input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85",
"output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30"
},
{
"input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49",
"output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49"
},
{
"input": "12\n3 8 7 4 6 5 2 1 11 9 10 12",
"output": "8 7 1 4 6 5 3 2 10 11 9 12"
},
{
"input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76",
"output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4"
},
{
"input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43",
"output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1"
},
{
"input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27",
"output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3"
},
{
"input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22",
"output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53"
},
{
"input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13",
"output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24"
},
{
"input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21",
"output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61"
},
{
"input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33",
"output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14"
},
{
"input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23",
"output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21"
},
{
"input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12",
"output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9"
},
{
"input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31",
"output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13"
},
{
"input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3",
"output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16"
},
{
"input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8",
"output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6"
},
{
"input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71",
"output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90"
},
{
"input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9",
"output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9"
},
{
"input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44",
"output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20"
},
{
"input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17",
"output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28"
},
{
"input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7",
"output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5"
},
{
"input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16",
"output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43"
},
{
"input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65",
"output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43"
},
{
"input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16",
"output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45"
},
{
"input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51",
"output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64"
},
{
"input": "12\n12 3 1 5 11 6 7 10 2 8 9 4",
"output": "3 9 2 12 4 6 7 10 11 8 5 1"
},
{
"input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17",
"output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44"
},
{
"input": "8\n1 3 5 2 4 8 6 7",
"output": "1 4 2 5 3 7 8 6"
},
{
"input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14",
"output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12"
},
{
"input": "10\n2 9 4 6 10 1 7 5 3 8",
"output": "6 1 9 3 8 4 7 10 2 5"
},
{
"input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3",
"output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7"
},
{
"input": "10\n2 4 9 3 6 8 10 5 1 7",
"output": "9 1 4 2 8 5 10 6 3 7"
},
{
"input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15",
"output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23"
},
{
"input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1",
"output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27"
},
{
"input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30",
"output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23"
},
{
"input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52",
"output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50"
},
{
"input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18",
"output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98"
},
{
"input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16",
"output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25"
},
{
"input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37",
"output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54"
},
{
"input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71",
"output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71"
},
{
"input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22",
"output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8"
},
{
"input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12",
"output": "31 9 72 77 48 17 78 84 65 2 20 96 75 69 41 40 50 87 25 57 35 71 92 76 28 58 10 86 34 45 18 30 52 95 89 90 24 5 32 60 16 21 27 14 70 4 67 22 83 49 23 29 37 73 61 79 68 11 15 54 59 88 33 56 36 93 55 74 8 82 47 66 46 38 91 19 7 1 13 12 80 3 44 85 94 53 26 62 81 43 6 39 64 63 42 51"
},
{
"input": "7\n2 1 5 7 3 4 6",
"output": "2 1 5 6 3 7 4"
},
{
"input": "51\n8 33 37 2 16 22 24 30 4 9 5 15 27 3 18 39 31 26 10 17 46 41 25 14 6 1 29 48 36 20 51 49 21 43 19 13 38 50 47 34 11 23 28 12 42 7 32 40 44 45 35",
"output": "26 4 14 9 11 25 46 1 10 19 41 44 36 24 12 5 20 15 35 30 33 6 42 7 23 18 13 43 27 8 17 47 2 40 51 29 3 37 16 48 22 45 34 49 50 21 39 28 32 38 31"
},
{
"input": "27\n12 14 7 3 20 21 25 13 22 15 23 4 2 24 10 17 19 8 26 11 27 18 9 5 6 1 16",
"output": "26 13 4 12 24 25 3 18 23 15 20 1 8 2 10 27 16 22 17 5 6 9 11 14 7 19 21"
},
{
"input": "71\n51 13 20 48 54 23 24 64 14 62 71 67 57 53 3 30 55 43 33 25 39 40 66 6 46 18 5 19 61 16 32 68 70 41 60 44 29 49 27 69 50 38 10 17 45 56 9 21 26 63 28 35 7 59 1 65 2 15 8 11 12 34 37 47 58 22 31 4 36 42 52",
"output": "55 57 15 68 27 24 53 59 47 43 60 61 2 9 58 30 44 26 28 3 48 66 6 7 20 49 39 51 37 16 67 31 19 62 52 69 63 42 21 22 34 70 18 36 45 25 64 4 38 41 1 71 14 5 17 46 13 65 54 35 29 10 50 8 56 23 12 32 40 33 11"
},
{
"input": "9\n8 5 2 6 1 9 4 7 3",
"output": "5 3 9 7 2 4 8 1 6"
},
{
"input": "29\n10 24 11 5 26 25 2 9 22 15 8 14 29 21 4 1 23 17 3 12 13 16 18 28 19 20 7 6 27",
"output": "16 7 19 15 4 28 27 11 8 1 3 20 21 12 10 22 18 23 25 26 14 9 17 2 6 5 29 24 13"
},
{
"input": "60\n39 25 42 4 55 60 16 18 47 1 11 40 7 50 19 35 49 54 12 3 30 38 2 58 17 26 45 6 33 43 37 32 52 36 15 23 27 59 24 20 28 14 8 9 13 29 44 46 41 21 5 48 51 22 31 56 57 53 10 34",
"output": "10 23 20 4 51 28 13 43 44 59 11 19 45 42 35 7 25 8 15 40 50 54 36 39 2 26 37 41 46 21 55 32 29 60 16 34 31 22 1 12 49 3 30 47 27 48 9 52 17 14 53 33 58 18 5 56 57 24 38 6"
},
{
"input": "50\n37 45 22 5 12 21 28 24 18 47 20 25 8 50 14 2 34 43 11 16 49 41 48 1 19 31 39 46 32 23 15 42 3 35 38 30 44 26 10 9 40 36 7 17 33 4 27 6 13 29",
"output": "24 16 33 46 4 48 43 13 40 39 19 5 49 15 31 20 44 9 25 11 6 3 30 8 12 38 47 7 50 36 26 29 45 17 34 42 1 35 27 41 22 32 18 37 2 28 10 23 21 14"
},
{
"input": "30\n8 29 28 16 17 25 27 15 21 11 6 20 2 13 1 30 5 4 24 10 14 3 23 18 26 9 12 22 19 7",
"output": "15 13 22 18 17 11 30 1 26 20 10 27 14 21 8 4 5 24 29 12 9 28 23 19 6 25 7 3 2 16"
},
{
"input": "46\n15 2 44 43 38 19 31 42 4 37 29 30 24 45 27 41 8 20 33 7 35 3 18 46 36 26 1 28 21 40 16 22 32 11 14 13 12 9 25 39 10 6 23 17 5 34",
"output": "27 2 22 9 45 42 20 17 38 41 34 37 36 35 1 31 44 23 6 18 29 32 43 13 39 26 15 28 11 12 7 33 19 46 21 25 10 5 40 30 16 8 4 3 14 24"
},
{
"input": "9\n4 8 6 5 3 9 2 7 1",
"output": "9 7 5 1 4 3 8 2 6"
},
{
"input": "46\n31 30 33 23 45 7 36 8 11 3 32 39 41 20 1 28 6 27 18 24 17 5 16 37 26 13 22 14 2 38 15 46 9 4 19 21 12 44 10 35 25 34 42 43 40 29",
"output": "15 29 10 34 22 17 6 8 33 39 9 37 26 28 31 23 21 19 35 14 36 27 4 20 41 25 18 16 46 2 1 11 3 42 40 7 24 30 12 45 13 43 44 38 5 32"
},
{
"input": "66\n27 12 37 48 46 21 34 58 38 28 66 2 64 32 44 31 13 36 40 15 19 11 22 5 30 29 6 7 61 39 20 42 23 54 51 33 50 9 60 8 57 45 49 10 62 41 59 3 55 63 52 24 25 26 43 56 65 4 16 14 1 35 18 17 53 47",
"output": "61 12 48 58 24 27 28 40 38 44 22 2 17 60 20 59 64 63 21 31 6 23 33 52 53 54 1 10 26 25 16 14 36 7 62 18 3 9 30 19 46 32 55 15 42 5 66 4 43 37 35 51 65 34 49 56 41 8 47 39 29 45 50 13 57 11"
},
{
"input": "13\n3 12 9 2 8 5 13 4 11 1 10 7 6",
"output": "10 4 1 8 6 13 12 5 3 11 9 2 7"
},
{
"input": "80\n21 25 56 50 20 61 7 74 51 69 8 2 46 57 45 71 14 52 17 43 9 30 70 78 31 10 38 13 23 15 37 79 6 16 77 73 80 4 49 48 18 28 26 58 33 41 64 22 54 72 59 60 40 63 53 27 1 5 75 67 62 34 19 39 68 65 44 55 3 32 11 42 76 12 35 47 66 36 24 29",
"output": "57 12 69 38 58 33 7 11 21 26 71 74 28 17 30 34 19 41 63 5 1 48 29 79 2 43 56 42 80 22 25 70 45 62 75 78 31 27 64 53 46 72 20 67 15 13 76 40 39 4 9 18 55 49 68 3 14 44 51 52 6 61 54 47 66 77 60 65 10 23 16 50 36 8 59 73 35 24 32 37"
},
{
"input": "63\n9 49 53 25 40 46 43 51 54 22 58 16 23 26 10 47 5 27 2 8 61 59 19 35 63 56 28 20 34 4 62 38 6 55 36 31 57 15 29 33 1 48 50 37 7 30 18 42 32 52 12 41 14 21 45 11 24 17 39 13 44 60 3",
"output": "41 19 63 30 17 33 45 20 1 15 56 51 60 53 38 12 58 47 23 28 54 10 13 57 4 14 18 27 39 46 36 49 40 29 24 35 44 32 59 5 52 48 7 61 55 6 16 42 2 43 8 50 3 9 34 26 37 11 22 62 21 31 25"
},
{
"input": "26\n11 4 19 13 17 9 2 24 6 5 22 23 14 15 3 25 16 8 18 10 21 1 12 26 7 20",
"output": "22 7 15 2 10 9 25 18 6 20 1 23 4 13 14 17 5 19 3 26 21 11 12 8 16 24"
},
{
"input": "69\n40 22 11 66 4 27 31 29 64 53 37 55 51 2 7 36 18 52 6 1 30 21 17 20 14 9 59 62 49 68 3 50 65 57 44 5 67 46 33 13 34 15 24 48 63 58 38 25 41 35 16 54 32 10 60 61 39 12 69 8 23 45 26 47 56 43 28 19 42",
"output": "20 14 31 5 36 19 15 60 26 54 3 58 40 25 42 51 23 17 68 24 22 2 61 43 48 63 6 67 8 21 7 53 39 41 50 16 11 47 57 1 49 69 66 35 62 38 64 44 29 32 13 18 10 52 12 65 34 46 27 55 56 28 45 9 33 4 37 30 59"
},
{
"input": "6\n4 3 6 5 1 2",
"output": "5 6 2 1 4 3"
},
{
"input": "9\n7 8 5 3 1 4 2 9 6",
"output": "5 7 4 6 3 9 1 2 8"
},
{
"input": "41\n27 24 16 30 25 8 32 2 26 20 39 33 41 22 40 14 36 9 28 4 34 11 31 23 19 18 17 35 3 10 6 13 5 15 29 38 7 21 1 12 37",
"output": "39 8 29 20 33 31 37 6 18 30 22 40 32 16 34 3 27 26 25 10 38 14 24 2 5 9 1 19 35 4 23 7 12 21 28 17 41 36 11 15 13"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "20\n2 6 4 18 7 10 17 13 16 8 14 9 20 5 19 12 1 3 15 11",
"output": "17 1 18 3 14 2 5 10 12 6 20 16 8 11 19 9 7 4 15 13"
},
{
"input": "2\n2 1",
"output": "2 1"
},
{
"input": "60\n2 4 31 51 11 7 34 20 3 14 18 23 48 54 15 36 38 60 49 40 5 33 41 26 55 58 10 8 13 9 27 30 37 1 21 59 44 57 35 19 46 43 42 45 12 22 39 32 24 16 6 56 53 52 25 17 47 29 50 28",
"output": "34 1 9 2 21 51 6 28 30 27 5 45 29 10 15 50 56 11 40 8 35 46 12 49 55 24 31 60 58 32 3 48 22 7 39 16 33 17 47 20 23 43 42 37 44 41 57 13 19 59 4 54 53 14 25 52 38 26 36 18"
},
{
"input": "14\n14 6 3 12 11 2 7 1 10 9 8 5 4 13",
"output": "8 6 3 13 12 2 7 11 10 9 5 4 14 1"
},
{
"input": "81\n13 43 79 8 7 21 73 46 63 4 62 78 56 11 70 68 61 53 60 49 16 27 59 47 69 5 22 44 77 57 52 48 1 9 72 81 28 55 58 33 51 18 31 17 41 20 42 3 32 54 19 2 75 34 64 10 65 50 30 29 67 12 71 66 74 15 26 23 6 38 25 35 37 24 80 76 40 45 39 36 14",
"output": "33 52 48 10 26 69 5 4 34 56 14 62 1 81 66 21 44 42 51 46 6 27 68 74 71 67 22 37 60 59 43 49 40 54 72 80 73 70 79 77 45 47 2 28 78 8 24 32 20 58 41 31 18 50 38 13 30 39 23 19 17 11 9 55 57 64 61 16 25 15 63 35 7 65 53 76 29 12 3 75 36"
},
{
"input": "42\n41 11 10 8 21 37 32 19 31 25 1 15 36 5 6 27 4 3 13 7 16 17 2 23 34 24 38 28 12 20 30 42 18 26 39 35 33 40 9 14 22 29",
"output": "11 23 18 17 14 15 20 4 39 3 2 29 19 40 12 21 22 33 8 30 5 41 24 26 10 34 16 28 42 31 9 7 37 25 36 13 6 27 35 38 1 32"
},
{
"input": "97\n20 6 76 42 4 18 35 59 39 63 27 7 66 47 61 52 15 36 88 93 19 33 10 92 1 34 46 86 78 57 51 94 77 29 26 73 41 2 58 97 43 65 17 74 21 49 25 3 91 82 95 12 96 13 84 90 69 24 72 37 16 55 54 71 64 62 48 89 11 70 80 67 30 40 44 85 53 83 79 9 56 45 75 87 22 14 81 68 8 38 60 50 28 23 31 32 5",
"output": "25 38 48 5 97 2 12 89 80 23 69 52 54 86 17 61 43 6 21 1 45 85 94 58 47 35 11 93 34 73 95 96 22 26 7 18 60 90 9 74 37 4 41 75 82 27 14 67 46 92 31 16 77 63 62 81 30 39 8 91 15 66 10 65 42 13 72 88 57 70 64 59 36 44 83 3 33 29 79 71 87 50 78 55 76 28 84 19 68 56 49 24 20 32 51 53 40"
},
{
"input": "62\n15 27 46 6 8 51 14 56 23 48 42 49 52 22 20 31 29 12 47 3 62 34 37 35 32 57 19 25 5 60 61 38 18 10 11 55 45 53 17 30 9 36 4 50 41 16 44 28 40 59 24 1 13 39 26 7 33 58 2 43 21 54",
"output": "52 59 20 43 29 4 56 5 41 34 35 18 53 7 1 46 39 33 27 15 61 14 9 51 28 55 2 48 17 40 16 25 57 22 24 42 23 32 54 49 45 11 60 47 37 3 19 10 12 44 6 13 38 62 36 8 26 58 50 30 31 21"
},
{
"input": "61\n35 27 4 61 52 32 41 46 14 37 17 54 55 31 11 26 44 49 15 30 9 50 45 39 7 38 53 3 58 40 13 56 18 19 28 6 43 5 21 42 20 34 2 25 36 12 33 57 16 60 1 8 59 10 22 23 24 48 51 47 29",
"output": "51 43 28 3 38 36 25 52 21 54 15 46 31 9 19 49 11 33 34 41 39 55 56 57 44 16 2 35 61 20 14 6 47 42 1 45 10 26 24 30 7 40 37 17 23 8 60 58 18 22 59 5 27 12 13 32 48 29 53 50 4"
},
{
"input": "59\n31 26 36 15 17 19 10 53 11 34 13 46 55 9 44 7 8 37 32 52 47 25 51 22 35 39 41 4 43 24 5 27 20 57 6 38 3 28 21 40 50 18 14 56 33 45 12 2 49 59 54 29 16 48 42 58 1 30 23",
"output": "57 48 37 28 31 35 16 17 14 7 9 47 11 43 4 53 5 42 6 33 39 24 59 30 22 2 32 38 52 58 1 19 45 10 25 3 18 36 26 40 27 55 29 15 46 12 21 54 49 41 23 20 8 51 13 44 34 56 50"
},
{
"input": "10\n2 10 7 4 1 5 8 6 3 9",
"output": "5 1 9 4 6 8 3 7 10 2"
},
{
"input": "14\n14 2 1 8 6 12 11 10 9 7 3 4 5 13",
"output": "3 2 11 12 13 5 10 4 9 8 7 6 14 1"
},
{
"input": "43\n28 38 15 14 31 42 27 30 19 33 43 26 22 29 18 32 3 13 1 8 35 34 4 12 11 17 41 21 5 25 39 37 20 23 7 24 16 10 40 9 6 36 2",
"output": "19 43 17 23 29 41 35 20 40 38 25 24 18 4 3 37 26 15 9 33 28 13 34 36 30 12 7 1 14 8 5 16 10 22 21 42 32 2 31 39 27 6 11"
},
{
"input": "86\n39 11 20 31 28 76 29 64 35 21 41 71 12 82 5 37 80 73 38 26 79 75 23 15 59 45 47 6 3 62 50 49 51 22 2 65 86 60 70 42 74 17 1 30 55 44 8 66 81 27 57 77 43 13 54 32 72 46 48 56 14 34 78 52 36 85 24 19 69 83 25 61 7 4 84 33 63 58 18 40 68 10 67 9 16 53",
"output": "43 35 29 74 15 28 73 47 84 82 2 13 54 61 24 85 42 79 68 3 10 34 23 67 71 20 50 5 7 44 4 56 76 62 9 65 16 19 1 80 11 40 53 46 26 58 27 59 32 31 33 64 86 55 45 60 51 78 25 38 72 30 77 8 36 48 83 81 69 39 12 57 18 41 22 6 52 63 21 17 49 14 70 75 66 37"
},
{
"input": "99\n65 78 56 98 33 24 61 40 29 93 1 64 57 22 25 52 67 95 50 3 31 15 90 68 71 83 38 36 6 46 89 26 4 87 14 88 72 37 23 43 63 12 80 96 5 34 73 86 9 48 92 62 99 10 16 20 66 27 28 2 82 70 30 94 49 8 84 69 18 60 58 59 44 39 21 7 91 76 54 19 75 85 74 47 55 32 97 77 51 13 35 79 45 42 11 41 17 81 53",
"output": "11 60 20 33 45 29 76 66 49 54 95 42 90 35 22 55 97 69 80 56 75 14 39 6 15 32 58 59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53"
},
{
"input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64",
"output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49"
},
{
"input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 61 26 82 96 4 98 25 40 31 44 24 47 30 52 14 16 39 27 9 29 45 18 67 63 37 43 90 66 19 69 88 22 92 77 34 42 73 80 56 64 20 35 15 33 86",
"output": "47 33 48 59 11 9 24 1 73 36 10 16 21 69 97 70 27 76 83 95 51 86 35 65 61 56 72 42 74 67 63 37 98 89 96 28 79 31 71 62 14 90 80 64 75 17 66 12 15 40 46 68 45 43 29 93 7 8 20 6 55 26 78 94 13 82 77 2 84 22 5 18 91 23 4 30 88 54 32 92 50 57 41 25 34 99 39 85 52 81 44 87 53 3 19 58 49 60 38"
},
{
"input": "99\n12 99 88 13 7 19 74 47 23 90 16 29 26 11 58 60 64 98 37 18 82 67 72 46 51 85 17 92 87 20 77 36 78 71 57 35 80 54 73 15 14 62 97 45 31 79 94 56 76 96 28 63 8 44 38 86 49 2 52 66 61 59 10 43 55 50 22 34 83 53 95 40 81 21 30 42 27 3 5 41 1 70 69 25 93 48 65 6 24 89 91 33 39 68 9 4 32 84 75",
"output": "81 58 78 96 79 88 5 53 95 63 14 1 4 41 40 11 27 20 6 30 74 67 9 89 84 13 77 51 12 75 45 97 92 68 36 32 19 55 93 72 80 76 64 54 44 24 8 86 57 66 25 59 70 38 65 48 35 15 62 16 61 42 52 17 87 60 22 94 83 82 34 23 39 7 99 49 31 33 46 37 73 21 69 98 26 56 29 3 90 10 91 28 85 47 71 50 43 18 2"
},
{
"input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53",
"output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5"
},
{
"input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39",
"output": "4 82 16 9 13 28 42 93 3 92 43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12"
},
{
"input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87",
"output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82"
},
{
"input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25",
"output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97"
},
{
"input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61",
"output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96"
},
{
"input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92",
"output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96"
},
{
"input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6",
"output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36"
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1"
}
] | 1,681,655,647
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
n = int(input())
l = list(map(int,input().split()))
if n<=3:
for i in l:
print(i,end=" ")
else:
ans = []
ans.append(l[-2])
ans.append(l[-1])
for i in range(n-2):
ans.append(l[i])
for i in ans:
print(i,end=" ")
|
Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none
|
```python
n = int(input())
l = list(map(int,input().split()))
if n<=3:
for i in l:
print(i,end=" ")
else:
ans = []
ans.append(l[-2])
ans.append(l[-1])
for i in range(n-2):
ans.append(l[i])
for i in ans:
print(i,end=" ")
```
| 0
|
|
808
|
B
|
Average Sleep Time
|
PROGRAMMING
| 1,300
|
[
"data structures",
"implementation",
"math"
] | null | null |
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
|
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
|
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
|
[
"3 2\n3 4 7\n",
"1 1\n10\n",
"8 2\n1 2 4 100000 123 456 789 1\n"
] |
[
"9.0000000000\n",
"10.0000000000\n",
"28964.2857142857\n"
] |
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
| 0
|
[
{
"input": "3 2\n3 4 7",
"output": "9.0000000000"
},
{
"input": "1 1\n10",
"output": "10.0000000000"
},
{
"input": "8 2\n1 2 4 100000 123 456 789 1",
"output": "28964.2857142857"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n100000",
"output": "100000.0000000000"
},
{
"input": "3 1\n1 2 3",
"output": "2.0000000000"
},
{
"input": "10 4\n11 3 5 20 12 7 9 2 2 20",
"output": "36.2857142857"
},
{
"input": "10 5\n15 9 3 2 17 10 9 18 4 19",
"output": "50.3333333333"
},
{
"input": "10 6\n19 3 20 16 14 10 1 13 7 3",
"output": "65.8000000000"
},
{
"input": "10 7\n8 16 2 13 15 9 5 13 9 2",
"output": "68.2500000000"
},
{
"input": "10 4\n127 1459 718 1183 880 1044 1857 1340 725 1496",
"output": "4574.4285714286"
},
{
"input": "10 5\n1384 1129 1780 1960 1567 1928 12 1523 1165 344",
"output": "6931.3333333333"
}
] | 1,599,369,350
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 29
| 264
| 17,203,200
|
from sys import stdin
a,b=map(int,stdin.readline().split())
c=list(map(int,stdin.readline().split()))
s=sum(c[:b]);k=s
for i in range(1,a-b+1):k=k-c[i-1]+c[i+b-1];s+=k
print((s)/(a-b+1))
|
Title: Average Sleep Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
Input Specification:
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
Demo Input:
['3 2\n3 4 7\n', '1 1\n10\n', '8 2\n1 2 4 100000 123 456 789 1\n']
Demo Output:
['9.0000000000\n', '10.0000000000\n', '28964.2857142857\n']
Note:
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
|
```python
from sys import stdin
a,b=map(int,stdin.readline().split())
c=list(map(int,stdin.readline().split()))
s=sum(c[:b]);k=s
for i in range(1,a-b+1):k=k-c[i-1]+c[i+b-1];s+=k
print((s)/(a-b+1))
```
| 3
|
|
352
|
A
|
Jeff and Digits
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation",
"math"
] | null | null |
Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?
Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card.
|
In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.
|
[
"4\n5 0 5 0\n",
"11\n5 5 5 5 5 5 5 5 0 5 5\n"
] |
[
"0\n",
"5555555550\n"
] |
In the first test you can make only one number that is a multiple of 90 — 0.
In the second test you can make number 5555555550, it is a multiple of 90.
| 500
|
[
{
"input": "4\n5 0 5 0",
"output": "0"
},
{
"input": "11\n5 5 5 5 5 5 5 5 0 5 5",
"output": "5555555550"
},
{
"input": "7\n5 5 5 5 5 5 5",
"output": "-1"
},
{
"input": "1\n5",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "11\n5 0 5 5 5 0 0 5 5 5 5",
"output": "0"
},
{
"input": "23\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0",
"output": "55555555555555555500000"
},
{
"input": "9\n5 5 5 5 5 5 5 5 5",
"output": "-1"
},
{
"input": "24\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0",
"output": "55555555555555555500000"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "10\n5 5 5 5 5 0 0 5 0 5",
"output": "0"
},
{
"input": "3\n5 5 0",
"output": "0"
},
{
"input": "5\n5 5 0 5 5",
"output": "0"
},
{
"input": "14\n0 5 5 0 0 0 0 0 0 5 5 5 5 5",
"output": "0"
},
{
"input": "3\n5 5 5",
"output": "-1"
},
{
"input": "3\n0 5 5",
"output": "0"
},
{
"input": "13\n0 0 5 0 5 0 5 5 0 0 0 0 0",
"output": "0"
},
{
"input": "9\n5 5 0 5 5 5 5 5 5",
"output": "0"
},
{
"input": "8\n0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "101\n5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 5 0 0 0 0 0 5 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 5 0 0 5 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 5 0 0",
"output": "5555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "214\n5 0 5 0 5 0 0 0 5 5 0 5 0 5 5 0 5 0 0 0 0 5 5 0 0 5 5 0 0 0 0 5 5 5 5 0 5 0 0 0 0 0 0 5 0 0 0 5 0 0 5 0 0 5 5 0 0 5 5 0 0 0 0 0 5 0 5 0 5 5 0 5 0 0 5 5 5 0 5 0 5 0 5 5 0 5 0 0 0 5 5 0 5 0 5 5 5 5 5 0 0 0 0 0 0 5 0 5 5 0 5 0 5 0 5 5 0 0 0 0 5 0 5 0 5 0 0 5 0 0 5 5 5 5 5 0 0 5 0 0 5 0 0 5 0 0 5 0 0 5 0 5 0 0 0 5 0 0 5 5 5 0 0 5 5 5 0 0 5 5 0 0 0 5 0 0 5 5 5 5 5 5 0 5 0 0 5 5 5 5 0 5 5 0 0 0 5 5 5 5 0 0 0 0 5 0 0 5 0 0 5 5 0 0",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "80\n0 0 0 0 5 0 5 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 0 0 5 5 0 5 0 0 0 0 0 0 5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 5 5 0 0 0",
"output": "555555555000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "0"
},
{
"input": "4\n5 5 5 5",
"output": "-1"
},
{
"input": "2\n0 5",
"output": "0"
},
{
"input": "14\n5 5 5 5 5 5 5 5 5 5 5 5 5 0",
"output": "5555555550"
},
{
"input": "18\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "-1"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 0",
"output": "5555555550"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 5",
"output": "-1"
},
{
"input": "20\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "-1"
}
] | 1,650,954,398
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 7
| 92
| 0
|
n1 = int(input())
list1 = []
for i in input().split():
list1.append(int(i))
flag = 0
no_five = len([i for i, x in enumerate(list1) if x == 5])
no_zero = len([i for i, x in enumerate(list1) if x == 0])
if no_five < 9:
if no_zero == 0:
print(-1)
else:
print(0)
else:
str1 = (no_five // 9) * 9 * "5" + no_zero * "0"
print(str1)
|
Title: Jeff and Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?
Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card.
Output Specification:
In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.
Demo Input:
['4\n5 0 5 0\n', '11\n5 5 5 5 5 5 5 5 0 5 5\n']
Demo Output:
['0\n', '5555555550\n']
Note:
In the first test you can make only one number that is a multiple of 90 — 0.
In the second test you can make number 5555555550, it is a multiple of 90.
|
```python
n1 = int(input())
list1 = []
for i in input().split():
list1.append(int(i))
flag = 0
no_five = len([i for i, x in enumerate(list1) if x == 5])
no_zero = len([i for i, x in enumerate(list1) if x == 0])
if no_five < 9:
if no_zero == 0:
print(-1)
else:
print(0)
else:
str1 = (no_five // 9) * 9 * "5" + no_zero * "0"
print(str1)
```
| 0
|
|
113
|
A
|
Grammar Lessons
|
PROGRAMMING
| 1,600
|
[
"implementation",
"strings"
] |
A. Grammar Lessons
|
5
|
256
|
Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of speech: the adjective, the noun, the verb. Each word in his language is an adjective, noun or verb. - There are two genders: masculine and feminine. Each word in his language has gender either masculine or feminine. - Masculine adjectives end with -lios, and feminine adjectives end with -liala. - Masculine nouns end with -etr, and feminime nouns end with -etra. - Masculine verbs end with -initis, and feminime verbs end with -inites. - Thus, each word in the Petya's language has one of the six endings, given above. There are no other endings in Petya's language. - It is accepted that the whole word consists of an ending. That is, words "lios", "liala", "etr" and so on belong to the Petya's language. - There aren't any punctuation marks, grammatical tenses, singular/plural forms or other language complications. - A sentence is either exactly one valid language word or exactly one statement.
Statement is any sequence of the Petya's language, that satisfy both conditions:
- Words in statement follow in the following order (from the left to the right): zero or more adjectives followed by exactly one noun followed by zero or more verbs. - All words in the statement should have the same gender.
After Petya's friend Vasya wrote instant messenger (an instant messaging program) that supported the Petya's language, Petya wanted to add spelling and grammar checking to the program. As Vasya was in the country and Petya didn't feel like waiting, he asked you to help him with this problem. Your task is to define by a given sequence of words, whether it is true that the given text represents exactly one sentence in Petya's language.
|
The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible that given words do not belong to the Petya's language.
|
If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes).
|
[
"petr\n",
"etis atis animatis etis atis amatis\n",
"nataliala kataliala vetra feinites\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "petr",
"output": "YES"
},
{
"input": "etis atis animatis etis atis amatis",
"output": "NO"
},
{
"input": "nataliala kataliala vetra feinites",
"output": "YES"
},
{
"input": "qweasbvflios",
"output": "YES"
},
{
"input": "lios lios petr initis qwe",
"output": "NO"
},
{
"input": "lios initis",
"output": "NO"
},
{
"input": "petr initis lios",
"output": "NO"
},
{
"input": "petra petra petra",
"output": "NO"
},
{
"input": "in",
"output": "NO"
},
{
"input": "liala petra initis",
"output": "NO"
},
{
"input": "liala petra inites",
"output": "YES"
},
{
"input": "liala initis",
"output": "NO"
},
{
"input": "liala petra petr inites",
"output": "NO"
},
{
"input": "liala petr inites",
"output": "NO"
},
{
"input": "llilitos",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzentsphxqhdungbylhnikwviuhccbstghhxlmvcjznnkjqkugsdysjbedwpmsmxmgxlrlxctnebtbwrsvgjktkrosffwymovxvsgfmmqwfflpvbumozikroxrdgwjrnstngstxbiyyuxehrhviteptedlmyetr",
"output": "YES"
},
{
"input": "i i i i i i i i i i i i i i i a a a a a a v v v v v v v v v v v",
"output": "NO"
},
{
"input": "fbvzqonvdlqdanwliolaqfj sbauorbinites xkbfnfinitespjy phbexglblzpobtqpisyijycmtliola aosinites lbpjiwcjoqyuhglthloiniteswb mjtxhoofohzzgefvhsywojcuxtetxmojrlktodhbgyrkeejgjzxkzyvrxwmyaqkeoqnvusnlrsfffrzeoqjdfumolhksqkrtzwhnforgpenziokrxlnhcapbbupctlmuetrani pigxerwetupjbkvlmgnjhdfjliolanz tqhaidxbqmdaeincxjuliola",
"output": "NO"
},
{
"input": "mfrmqetr",
"output": "YES"
},
{
"input": "hnwvfllholxfialiola cknjtxpliola daliola gqfapnhmmworliola qhetra qrisbexsrefcwzoxqwxrevinites wwldqkqhvrgwplqinites nqdpoauitczttxoinites fgbmdfpxkhahkinites",
"output": "NO"
},
{
"input": "kcymcpgqdxkudadewddualeemhixhsdazudnjdmuvxvrlrbrpsdpxpagmrogplltnifrtomdtahxwadguvetxaqkvsvnoyhowirnluhmyewzapirnpfdisvhtbenxmfezahqoflkjrfqjubwdfktnpeirodwubftzlcczzavfiooihzvnqincndisudihvbcaxptrwovekmhiiwsgzgbxydvuldlnktxtltrlajjzietkxbnhetra",
"output": "YES"
},
{
"input": "dosiydnwxemojaavfdvlwsyhzqywqjutovygtlcleklhybczhjqfzxwdmlwqwcqqyfjkzhsizlmdarrfronxqkcknwpkvhdlgatdyjisjoopvngpjggldxjfxaauoxmqirkuphydyweoixftstlozaoywnxgriscudwlokncbmaebpssccmmmfjennyjaryqlzjknnklqketra",
"output": "YES"
},
{
"input": "etretra linites",
"output": "YES"
},
{
"input": "petretra petr",
"output": "NO"
},
{
"input": "lialalios petraveryfunnypetr",
"output": "YES"
},
{
"input": "petropetrapetr petra",
"output": "NO"
},
{
"input": "lios petrnonono",
"output": "NO"
},
{
"input": "lios petr initisandinitisandliala petrainitis",
"output": "NO"
},
{
"input": "petro",
"output": "NO"
},
{
"input": "petr initesinitis",
"output": "YES"
},
{
"input": "lios initis",
"output": "NO"
},
{
"input": "liala initespetra",
"output": "YES"
},
{
"input": "lios petrapetr",
"output": "YES"
},
{
"input": "initis petr",
"output": "NO"
},
{
"input": "lioslialapetrpetrainitisinitesliosliala initesinitislioslialapetrpetrainitisinitetra",
"output": "YES"
},
{
"input": "veryfunnyprefixpetr",
"output": "YES"
},
{
"input": "veryfunnyprefixpetra",
"output": "YES"
},
{
"input": "veryfunnyprefixinitis",
"output": "YES"
},
{
"input": "veryfunnyprefixinites",
"output": "YES"
},
{
"input": "veryfunnyprefixliala",
"output": "YES"
},
{
"input": "veryfunnyprefixlios",
"output": "YES"
},
{
"input": "veryfunnyprefixlialas",
"output": "NO"
},
{
"input": "veryfunnyprefixliala veryfunnyprefixpetretra",
"output": "YES"
},
{
"input": "veryfunnyprefixlios veryfunnyprefixinitisetr",
"output": "YES"
},
{
"input": "veryfunnyprefixlios aabbinitis",
"output": "NO"
},
{
"input": "veryfunnyprefixlios inites",
"output": "NO"
},
{
"input": "lios petr initis",
"output": "YES"
},
{
"input": "liala etra inites",
"output": "YES"
},
{
"input": "lios",
"output": "YES"
},
{
"input": "liala",
"output": "YES"
},
{
"input": "initis",
"output": "YES"
},
{
"input": "inites",
"output": "YES"
},
{
"input": "tes",
"output": "NO"
},
{
"input": "tr",
"output": "NO"
},
{
"input": "a",
"output": "NO"
},
{
"input": "lios lios",
"output": "NO"
},
{
"input": "lios",
"output": "YES"
},
{
"input": "liala",
"output": "YES"
},
{
"input": "petr",
"output": "YES"
},
{
"input": "petra",
"output": "YES"
},
{
"input": "pinitis",
"output": "YES"
},
{
"input": "pinites",
"output": "YES"
},
{
"input": "plios pliala",
"output": "NO"
},
{
"input": "plios petr",
"output": "YES"
},
{
"input": "plios petra",
"output": "NO"
},
{
"input": "plios plios",
"output": "NO"
},
{
"input": "plios initis",
"output": "NO"
},
{
"input": "plios pinites",
"output": "NO"
},
{
"input": "pliala plios",
"output": "NO"
},
{
"input": "pliala ppliala",
"output": "NO"
},
{
"input": "pliala petr",
"output": "NO"
},
{
"input": "pliala petra",
"output": "YES"
},
{
"input": "pliala pinitis",
"output": "NO"
},
{
"input": "pliala pinites",
"output": "NO"
},
{
"input": "petr plios",
"output": "NO"
},
{
"input": "petr pliala",
"output": "NO"
},
{
"input": "petr petr",
"output": "NO"
},
{
"input": "petr petra",
"output": "NO"
},
{
"input": "petr pinitis",
"output": "YES"
},
{
"input": "petr pinites",
"output": "NO"
},
{
"input": "petra lios",
"output": "NO"
},
{
"input": "petra liala",
"output": "NO"
},
{
"input": "petra petr",
"output": "NO"
},
{
"input": "petra petra",
"output": "NO"
},
{
"input": "petra initis",
"output": "NO"
},
{
"input": "petra inites",
"output": "YES"
},
{
"input": "initis lios",
"output": "NO"
},
{
"input": "initis liala",
"output": "NO"
},
{
"input": "initis petr",
"output": "NO"
},
{
"input": "initis petra",
"output": "NO"
},
{
"input": "initis initis",
"output": "NO"
},
{
"input": "initis inites",
"output": "NO"
},
{
"input": "inites lios",
"output": "NO"
},
{
"input": "inites liala",
"output": "NO"
},
{
"input": "inites petr",
"output": "NO"
},
{
"input": "inites petra",
"output": "NO"
},
{
"input": "inites initis",
"output": "NO"
},
{
"input": "inites inites",
"output": "NO"
},
{
"input": "lios lios lios",
"output": "NO"
},
{
"input": "lios lios liala",
"output": "NO"
},
{
"input": "lios lios etr",
"output": "YES"
},
{
"input": "lios lios etra",
"output": "NO"
},
{
"input": "lios lios initis",
"output": "NO"
},
{
"input": "lios lios inites",
"output": "NO"
},
{
"input": "lios liala lios",
"output": "NO"
},
{
"input": "lios liala liala",
"output": "NO"
},
{
"input": "lios liala etr",
"output": "NO"
},
{
"input": "lios liala etra",
"output": "NO"
},
{
"input": "lios liala initis",
"output": "NO"
},
{
"input": "lios liala inites",
"output": "NO"
},
{
"input": "lios etr lios",
"output": "NO"
},
{
"input": "lios etr liala",
"output": "NO"
},
{
"input": "lios etr etr",
"output": "NO"
},
{
"input": "lios etr etra",
"output": "NO"
},
{
"input": "lios etr initis",
"output": "YES"
},
{
"input": "lios etr inites",
"output": "NO"
},
{
"input": "lios etra lios",
"output": "NO"
},
{
"input": "lios etra liala",
"output": "NO"
},
{
"input": "lios etra etr",
"output": "NO"
},
{
"input": "lios etra etra",
"output": "NO"
},
{
"input": "lios etra initis",
"output": "NO"
},
{
"input": "lios etra inites",
"output": "NO"
},
{
"input": "lios initis lios",
"output": "NO"
},
{
"input": "lios initis liala",
"output": "NO"
},
{
"input": "lios initis etr",
"output": "NO"
},
{
"input": "lios initis etra",
"output": "NO"
},
{
"input": "lios initis initis",
"output": "NO"
},
{
"input": "lios initis inites",
"output": "NO"
},
{
"input": "lios inites lios",
"output": "NO"
},
{
"input": "lios inites liala",
"output": "NO"
},
{
"input": "lios inites etr",
"output": "NO"
},
{
"input": "lios inites etra",
"output": "NO"
},
{
"input": "lios inites initis",
"output": "NO"
},
{
"input": "lios inites inites",
"output": "NO"
},
{
"input": "liala lios lios",
"output": "NO"
},
{
"input": "liala lios liala",
"output": "NO"
},
{
"input": "liala lios etr",
"output": "NO"
},
{
"input": "liala lios etra",
"output": "NO"
},
{
"input": "liala lios initis",
"output": "NO"
},
{
"input": "liala lios inites",
"output": "NO"
},
{
"input": "liala liala lios",
"output": "NO"
},
{
"input": "liala liala liala",
"output": "NO"
},
{
"input": "liala liala etr",
"output": "NO"
},
{
"input": "liala liala etra",
"output": "YES"
},
{
"input": "liala liala initis",
"output": "NO"
},
{
"input": "liala liala inites",
"output": "NO"
},
{
"input": "liala etr lios",
"output": "NO"
},
{
"input": "liala etr liala",
"output": "NO"
},
{
"input": "liala etr etr",
"output": "NO"
},
{
"input": "liala etr etra",
"output": "NO"
},
{
"input": "liala etr initis",
"output": "NO"
},
{
"input": "liala etr inites",
"output": "NO"
},
{
"input": "liala etra lios",
"output": "NO"
},
{
"input": "liala etra liala",
"output": "NO"
},
{
"input": "liala etra etr",
"output": "NO"
},
{
"input": "liala etra etra",
"output": "NO"
},
{
"input": "liala etra initis",
"output": "NO"
},
{
"input": "liala etra inites",
"output": "YES"
},
{
"input": "liala initis lios",
"output": "NO"
},
{
"input": "liala initis liala",
"output": "NO"
},
{
"input": "liala initis etr",
"output": "NO"
},
{
"input": "liala initis etra",
"output": "NO"
},
{
"input": "liala initis initis",
"output": "NO"
},
{
"input": "liala initis inites",
"output": "NO"
},
{
"input": "liala inites lios",
"output": "NO"
},
{
"input": "liala inites liala",
"output": "NO"
},
{
"input": "liala inites etr",
"output": "NO"
},
{
"input": "liala inites etra",
"output": "NO"
},
{
"input": "liala inites initis",
"output": "NO"
},
{
"input": "liala inites inites",
"output": "NO"
},
{
"input": "etr lios lios",
"output": "NO"
},
{
"input": "etr lios liala",
"output": "NO"
},
{
"input": "etr lios etr",
"output": "NO"
},
{
"input": "etr lios etra",
"output": "NO"
},
{
"input": "etr lios initis",
"output": "NO"
},
{
"input": "etr lios inites",
"output": "NO"
},
{
"input": "etr liala lios",
"output": "NO"
},
{
"input": "etr liala liala",
"output": "NO"
},
{
"input": "etr liala etr",
"output": "NO"
},
{
"input": "etr liala etra",
"output": "NO"
},
{
"input": "etr liala initis",
"output": "NO"
},
{
"input": "etr liala inites",
"output": "NO"
},
{
"input": "etr etr lios",
"output": "NO"
},
{
"input": "etr etr liala",
"output": "NO"
},
{
"input": "etr etr etr",
"output": "NO"
},
{
"input": "etr etr etra",
"output": "NO"
},
{
"input": "etr etr initis",
"output": "NO"
},
{
"input": "etr etr inites",
"output": "NO"
},
{
"input": "etr etra lios",
"output": "NO"
},
{
"input": "etr etra liala",
"output": "NO"
},
{
"input": "etr etra etr",
"output": "NO"
},
{
"input": "etr etra etra",
"output": "NO"
},
{
"input": "etr etra initis",
"output": "NO"
},
{
"input": "etr etra inites",
"output": "NO"
},
{
"input": "etr initis lios",
"output": "NO"
},
{
"input": "etr initis liala",
"output": "NO"
},
{
"input": "etr initis etr",
"output": "NO"
},
{
"input": "etr initis etra",
"output": "NO"
},
{
"input": "etr initis initis",
"output": "YES"
},
{
"input": "etr initis inites",
"output": "NO"
},
{
"input": "etr inites lios",
"output": "NO"
},
{
"input": "etr inites liala",
"output": "NO"
},
{
"input": "etr inites etr",
"output": "NO"
},
{
"input": "etr inites etra",
"output": "NO"
},
{
"input": "etr inites initis",
"output": "NO"
},
{
"input": "etr inites inites",
"output": "NO"
},
{
"input": "etra lios lios",
"output": "NO"
},
{
"input": "etra lios liala",
"output": "NO"
},
{
"input": "etra lios etr",
"output": "NO"
},
{
"input": "etra lios etra",
"output": "NO"
},
{
"input": "etra lios initis",
"output": "NO"
},
{
"input": "etra lios inites",
"output": "NO"
},
{
"input": "etra liala lios",
"output": "NO"
},
{
"input": "etra liala liala",
"output": "NO"
},
{
"input": "etra liala etr",
"output": "NO"
},
{
"input": "etra liala etra",
"output": "NO"
},
{
"input": "etra liala initis",
"output": "NO"
},
{
"input": "etra liala inites",
"output": "NO"
},
{
"input": "etra etr lios",
"output": "NO"
},
{
"input": "etra etr liala",
"output": "NO"
},
{
"input": "etra etr etr",
"output": "NO"
},
{
"input": "etra etr etra",
"output": "NO"
},
{
"input": "etra etr initis",
"output": "NO"
},
{
"input": "etra etr inites",
"output": "NO"
},
{
"input": "etra etra lios",
"output": "NO"
},
{
"input": "etra etra liala",
"output": "NO"
},
{
"input": "etra etra etr",
"output": "NO"
},
{
"input": "etra etra etra",
"output": "NO"
},
{
"input": "etra etra initis",
"output": "NO"
},
{
"input": "etra etra inites",
"output": "NO"
},
{
"input": "etra initis lios",
"output": "NO"
},
{
"input": "etra initis liala",
"output": "NO"
},
{
"input": "etra initis etr",
"output": "NO"
},
{
"input": "etra initis etra",
"output": "NO"
},
{
"input": "etra initis initis",
"output": "NO"
},
{
"input": "etra initis inites",
"output": "NO"
},
{
"input": "etra inites lios",
"output": "NO"
},
{
"input": "etra inites liala",
"output": "NO"
},
{
"input": "etra inites etr",
"output": "NO"
},
{
"input": "etra inites etra",
"output": "NO"
},
{
"input": "etra inites initis",
"output": "NO"
},
{
"input": "etra inites inites",
"output": "YES"
},
{
"input": "initis lios lios",
"output": "NO"
},
{
"input": "initis lios liala",
"output": "NO"
},
{
"input": "initis lios etr",
"output": "NO"
},
{
"input": "initis lios etra",
"output": "NO"
},
{
"input": "initis lios initis",
"output": "NO"
},
{
"input": "initis lios inites",
"output": "NO"
},
{
"input": "initis liala lios",
"output": "NO"
},
{
"input": "initis liala liala",
"output": "NO"
},
{
"input": "initis liala etr",
"output": "NO"
},
{
"input": "initis liala etra",
"output": "NO"
},
{
"input": "initis liala initis",
"output": "NO"
},
{
"input": "initis liala inites",
"output": "NO"
},
{
"input": "initis etr lios",
"output": "NO"
},
{
"input": "initis etr liala",
"output": "NO"
},
{
"input": "initis etr etr",
"output": "NO"
},
{
"input": "initis etr etra",
"output": "NO"
},
{
"input": "initis etr initis",
"output": "NO"
},
{
"input": "initis etr inites",
"output": "NO"
},
{
"input": "initis etra lios",
"output": "NO"
},
{
"input": "initis etra liala",
"output": "NO"
},
{
"input": "initis etra etr",
"output": "NO"
},
{
"input": "initis etra etra",
"output": "NO"
},
{
"input": "initis etra initis",
"output": "NO"
},
{
"input": "initis etra inites",
"output": "NO"
},
{
"input": "initis initis lios",
"output": "NO"
},
{
"input": "initis initis liala",
"output": "NO"
},
{
"input": "initis initis etr",
"output": "NO"
},
{
"input": "initis initis etra",
"output": "NO"
},
{
"input": "initis initis initis",
"output": "NO"
},
{
"input": "initis initis inites",
"output": "NO"
},
{
"input": "initis inites lios",
"output": "NO"
},
{
"input": "initis inites liala",
"output": "NO"
},
{
"input": "initis inites etr",
"output": "NO"
},
{
"input": "initis inites etra",
"output": "NO"
},
{
"input": "initis inites initis",
"output": "NO"
},
{
"input": "initis inites inites",
"output": "NO"
},
{
"input": "inites lios lios",
"output": "NO"
},
{
"input": "inites lios liala",
"output": "NO"
},
{
"input": "inites lios etr",
"output": "NO"
},
{
"input": "inites lios etra",
"output": "NO"
},
{
"input": "inites lios initis",
"output": "NO"
},
{
"input": "inites lios inites",
"output": "NO"
},
{
"input": "inites liala lios",
"output": "NO"
},
{
"input": "inites liala liala",
"output": "NO"
},
{
"input": "inites liala etr",
"output": "NO"
},
{
"input": "inites liala etra",
"output": "NO"
},
{
"input": "inites liala initis",
"output": "NO"
},
{
"input": "inites liala inites",
"output": "NO"
},
{
"input": "inites etr lios",
"output": "NO"
},
{
"input": "inites etr liala",
"output": "NO"
},
{
"input": "inites etr etr",
"output": "NO"
},
{
"input": "inites etr etra",
"output": "NO"
},
{
"input": "inites etr initis",
"output": "NO"
},
{
"input": "inites etr inites",
"output": "NO"
},
{
"input": "inites etra lios",
"output": "NO"
},
{
"input": "inites etra liala",
"output": "NO"
},
{
"input": "inites etra etr",
"output": "NO"
},
{
"input": "inites etra etra",
"output": "NO"
},
{
"input": "inites etra initis",
"output": "NO"
},
{
"input": "inites etra inites",
"output": "NO"
},
{
"input": "inites initis lios",
"output": "NO"
},
{
"input": "inites initis liala",
"output": "NO"
},
{
"input": "inites initis etr",
"output": "NO"
},
{
"input": "inites initis etra",
"output": "NO"
},
{
"input": "inites initis initis",
"output": "NO"
},
{
"input": "inites initis inites",
"output": "NO"
},
{
"input": "inites inites lios",
"output": "NO"
},
{
"input": "inites inites liala",
"output": "NO"
},
{
"input": "inites inites etr",
"output": "NO"
},
{
"input": "inites inites etra",
"output": "NO"
},
{
"input": "inites inites initis",
"output": "NO"
},
{
"input": "inites inites inites",
"output": "NO"
}
] | 1,695,543,386
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 92
| 102,400
|
s = [i for i in input().split()]
suffix = ["lios", "liala", "etr", "etra", "initis", "inites"]
n = len(s)
f = [0] * n
for i in range(n):
for j in range(6):
if s[i][-len(suffix[j]) :] == suffix[j]:
f[i] = j + 1
if n == 1:
if f[i] != 0:
print("YES")
else:
print("NO")
exit()
sf = gf = True
af = nf = vf = 0
g = f[0] % 2
for i in range(n):
if f[i] == 0:
sf = False
break
if f[i] % 2 != g:
gf = False
break
if af == 0 and f[i] in range(1, 3):
af = 1
if nf == 0 and f[i] in range(3, 5):
nf = 2
elif nf == 2 and f[i] in range(3, 5):
nf = 0
if vf == 0 and f[i] in range(5, 7):
vf = 1
if af == 1 and f[i] not in range(1, 3):
af = 2
if vf == 1 and (f[i] not in range(5, 7) or i == n - 1):
vf = 2
if sf == True and gf == True:
if af == 0 and nf == 2 and vf == 0:
print("YES")
elif af == nf == vf == 2:
print("YES")
else:
print("NO")
else:
print("NO")
|
Title: Grammar Lessons
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of speech: the adjective, the noun, the verb. Each word in his language is an adjective, noun or verb. - There are two genders: masculine and feminine. Each word in his language has gender either masculine or feminine. - Masculine adjectives end with -lios, and feminine adjectives end with -liala. - Masculine nouns end with -etr, and feminime nouns end with -etra. - Masculine verbs end with -initis, and feminime verbs end with -inites. - Thus, each word in the Petya's language has one of the six endings, given above. There are no other endings in Petya's language. - It is accepted that the whole word consists of an ending. That is, words "lios", "liala", "etr" and so on belong to the Petya's language. - There aren't any punctuation marks, grammatical tenses, singular/plural forms or other language complications. - A sentence is either exactly one valid language word or exactly one statement.
Statement is any sequence of the Petya's language, that satisfy both conditions:
- Words in statement follow in the following order (from the left to the right): zero or more adjectives followed by exactly one noun followed by zero or more verbs. - All words in the statement should have the same gender.
After Petya's friend Vasya wrote instant messenger (an instant messaging program) that supported the Petya's language, Petya wanted to add spelling and grammar checking to the program. As Vasya was in the country and Petya didn't feel like waiting, he asked you to help him with this problem. Your task is to define by a given sequence of words, whether it is true that the given text represents exactly one sentence in Petya's language.
Input Specification:
The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible that given words do not belong to the Petya's language.
Output Specification:
If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes).
Demo Input:
['petr\n', 'etis atis animatis etis atis amatis\n', 'nataliala kataliala vetra feinites\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
s = [i for i in input().split()]
suffix = ["lios", "liala", "etr", "etra", "initis", "inites"]
n = len(s)
f = [0] * n
for i in range(n):
for j in range(6):
if s[i][-len(suffix[j]) :] == suffix[j]:
f[i] = j + 1
if n == 1:
if f[i] != 0:
print("YES")
else:
print("NO")
exit()
sf = gf = True
af = nf = vf = 0
g = f[0] % 2
for i in range(n):
if f[i] == 0:
sf = False
break
if f[i] % 2 != g:
gf = False
break
if af == 0 and f[i] in range(1, 3):
af = 1
if nf == 0 and f[i] in range(3, 5):
nf = 2
elif nf == 2 and f[i] in range(3, 5):
nf = 0
if vf == 0 and f[i] in range(5, 7):
vf = 1
if af == 1 and f[i] not in range(1, 3):
af = 2
if vf == 1 and (f[i] not in range(5, 7) or i == n - 1):
vf = 2
if sf == True and gf == True:
if af == 0 and nf == 2 and vf == 0:
print("YES")
elif af == nf == vf == 2:
print("YES")
else:
print("NO")
else:
print("NO")
```
| 0
|
929
|
A
|
Прокат велосипедов
|
PROGRAMMING
| 1,400
|
[
"*special",
"greedy",
"implementation"
] | null | null |
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
|
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
|
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
|
[
"4 4\n3 6 8 10\n",
"2 9\n10 20\n",
"12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n"
] |
[
"2\n",
"-1\n",
"6\n"
] |
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
| 500
|
[
{
"input": "4 4\n3 6 8 10",
"output": "2"
},
{
"input": "2 9\n10 20",
"output": "-1"
},
{
"input": "12 3\n4 6 7 9 10 11 13 15 17 18 20 21",
"output": "6"
},
{
"input": "2 1\n11164 11165",
"output": "1"
},
{
"input": "3 7\n45823 45825 45829",
"output": "1"
},
{
"input": "2 100000\n0 100000",
"output": "1"
},
{
"input": "50 15001\n1269 1580 5431 5916 6642 14145 15760 19922 20589 22062 24138 33454 33505 35916 37012 42577 43140 49457 54720 55188 56205 56639 56870 57997 58379 59088 59297 61805 61861 63005 64013 68848 71182 71497 72715 73008 75616 76042 76614 80690 83089 85033 86408 92392 92763 93833 95085 95815 97152 99379",
"output": "8"
},
{
"input": "5 7\n6219 6222 6229 6231 6236",
"output": "3"
},
{
"input": "10 448\n78449 78573 78599 78742 78748 78759 78853 79091 79298 79324",
"output": "3"
},
{
"input": "20 19191\n11272 12386 14108 24663 24932 26547 29656 44677 45436 45654 48562 62367 71049 71238 78182 84042 88736 92026 96836 99343",
"output": "6"
},
{
"input": "2 100000\n1 100000",
"output": "1"
},
{
"input": "2 99999\n0 100000",
"output": "-1"
},
{
"input": "2 2\n1 3",
"output": "1"
},
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "2 2\n1 4",
"output": "-1"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
}
] | 1,520,079,179
| 74,279
|
Python 3
|
OK
|
TESTS
| 52
| 358
| 5,632,000
|
n, m = map(int, input().split())
a = [int(s) for s in input().split()]
x = 0
step = 0
for i in range(1, n):
if a[i] - a[i-1] > m:
x = -1
break
if x == 0:
while step != n-1:
for i in range(m, 0, -1):
if a[step] + i in a:
step = a.index(a[step] + i)
x += 1
break
print(x)
|
Title: Прокат велосипедов
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
Input Specification:
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
Output Specification:
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
Demo Input:
['4 4\n3 6 8 10\n', '2 9\n10 20\n', '12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
|
```python
n, m = map(int, input().split())
a = [int(s) for s in input().split()]
x = 0
step = 0
for i in range(1, n):
if a[i] - a[i-1] > m:
x = -1
break
if x == 0:
while step != n-1:
for i in range(m, 0, -1):
if a[step] + i in a:
step = a.index(a[step] + i)
x += 1
break
print(x)
```
| 3
|
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,609,139,938
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 218
| 0
|
s = input()
t = input()
n = len(t)
res = 1
index = 0
for i in range(n):
if t[i] == s[index]:
res += 1
index += 1
print(res)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
s = input()
t = input()
n = len(t)
res = 1
index = 0
for i in range(n):
if t[i] == s[index]:
res += 1
index += 1
print(res)
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,618,313,420
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Tue Mar 30 19:32:51 2021
@author: nehas
"""
a=input()
b=input()
if(b==''.join(reversed(a))):
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Tue Mar 30 19:32:51 2021
@author: nehas
"""
a=input()
b=input()
if(b==''.join(reversed(a))):
print("YES")
else:
print("NO")
```
| 3.969
|
787
|
A
|
The Monster
|
PROGRAMMING
| 1,200
|
[
"brute force",
"math",
"number theory"
] | null | null |
A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.
|
The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100).
|
Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time.
|
[
"20 2\n9 19\n",
"2 1\n16 12\n"
] |
[
"82\n",
"-1\n"
] |
In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time.
| 500
|
[
{
"input": "20 2\n9 19",
"output": "82"
},
{
"input": "2 1\n16 12",
"output": "-1"
},
{
"input": "39 52\n88 78",
"output": "1222"
},
{
"input": "59 96\n34 48",
"output": "1748"
},
{
"input": "87 37\n91 29",
"output": "211"
},
{
"input": "11 81\n49 7",
"output": "301"
},
{
"input": "39 21\n95 89",
"output": "3414"
},
{
"input": "59 70\n48 54",
"output": "1014"
},
{
"input": "87 22\n98 32",
"output": "718"
},
{
"input": "15 63\n51 13",
"output": "-1"
},
{
"input": "39 7\n97 91",
"output": "1255"
},
{
"input": "18 18\n71 71",
"output": "1278"
},
{
"input": "46 71\n16 49",
"output": "209"
},
{
"input": "70 11\n74 27",
"output": "2321"
},
{
"input": "94 55\n20 96",
"output": "-1"
},
{
"input": "18 4\n77 78",
"output": "1156"
},
{
"input": "46 44\n23 55",
"output": "-1"
},
{
"input": "74 88\n77 37",
"output": "1346"
},
{
"input": "94 37\n34 7",
"output": "789"
},
{
"input": "22 81\n80 88",
"output": "-1"
},
{
"input": "46 30\n34 62",
"output": "674"
},
{
"input": "40 4\n81 40",
"output": "364"
},
{
"input": "69 48\n39 9",
"output": "48"
},
{
"input": "89 93\n84 87",
"output": "5967"
},
{
"input": "17 45\n42 65",
"output": "317"
},
{
"input": "41 85\n95 46",
"output": "331"
},
{
"input": "69 30\n41 16",
"output": "1410"
},
{
"input": "93 74\n99 93",
"output": "-1"
},
{
"input": "17 19\n44 75",
"output": "427"
},
{
"input": "45 63\n98 53",
"output": "3483"
},
{
"input": "69 11\n48 34",
"output": "-1"
},
{
"input": "55 94\n3 96",
"output": "204"
},
{
"input": "100 100\n100 100",
"output": "100"
},
{
"input": "1 1\n1 1",
"output": "1"
},
{
"input": "1 1\n1 100",
"output": "100"
},
{
"input": "1 100\n100 1",
"output": "101"
},
{
"input": "98 1\n99 100",
"output": "9703"
},
{
"input": "98 1\n99 2",
"output": "9605"
},
{
"input": "97 2\n99 100",
"output": "4852"
},
{
"input": "3 3\n3 1",
"output": "-1"
},
{
"input": "3 2\n7 2",
"output": "2"
},
{
"input": "2 3\n2 5",
"output": "5"
},
{
"input": "2 3\n2 3",
"output": "3"
},
{
"input": "100 3\n100 5",
"output": "-1"
},
{
"input": "6 10\n12 14",
"output": "-1"
},
{
"input": "4 2\n4 4",
"output": "-1"
},
{
"input": "2 3\n2 2",
"output": "-1"
},
{
"input": "2 3\n4 99",
"output": "99"
},
{
"input": "1 5\n1 5",
"output": "5"
},
{
"input": "1 100\n3 1",
"output": "100"
},
{
"input": "2 2\n2 1",
"output": "-1"
},
{
"input": "2 10\n6 20",
"output": "20"
},
{
"input": "2 2\n2 10",
"output": "10"
},
{
"input": "3 7\n3 6",
"output": "-1"
},
{
"input": "1 100\n1 100",
"output": "100"
},
{
"input": "7 25\n39 85",
"output": "319"
},
{
"input": "84 82\n38 6",
"output": "82"
},
{
"input": "7 7\n7 14",
"output": "14"
}
] | 1,535,614,008
| 1,128
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 22
| 109
| 0
|
def gcd(x,y):
if(y==0):
return x
return gcd(y,x%y)
a,b=map(int,input().split())
c,d=map(int,input().split())
if(d>b):
a,b,c,d=c,d,a,b
if(b==d):
print(b)
exit()
if((b-d)%gcd(a,c)):
print(-1)
exit()
while(b!=d):
b+=a
if((b-d)%c==0):
d=b
else:
d+=c*((b-d)//c + 1)
print(b)
|
Title: The Monster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.
Input Specification:
The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100).
Output Specification:
Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time.
Demo Input:
['20 2\n9 19\n', '2 1\n16 12\n']
Demo Output:
['82\n', '-1\n']
Note:
In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time.
|
```python
def gcd(x,y):
if(y==0):
return x
return gcd(y,x%y)
a,b=map(int,input().split())
c,d=map(int,input().split())
if(d>b):
a,b,c,d=c,d,a,b
if(b==d):
print(b)
exit()
if((b-d)%gcd(a,c)):
print(-1)
exit()
while(b!=d):
b+=a
if((b-d)%c==0):
d=b
else:
d+=c*((b-d)//c + 1)
print(b)
```
| 0
|
|
415
|
A
|
Mashmokh and Lights
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to *n*. There are *n* buttons in Mashmokh's room indexed from 1 to *n* as well. If Mashmokh pushes button with index *i*, then each light with index not less than *i* that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed *m* distinct buttons *b*1,<=*b*2,<=...,<=*b**m* (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button *b**i* is actually *b**i*, not *i*.
Please, help Mashmokh, print these indices.
|
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of the factory lights and the pushed buttons respectively. The next line contains *m* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*).
It is guaranteed that all lights will be turned off after pushing all buttons.
|
Output *n* space-separated integers where the *i*-th number is index of the button that turns the *i*-th light off.
|
[
"5 4\n4 3 1 2\n",
"5 5\n5 4 3 2 1\n"
] |
[
"1 1 3 4 4 \n",
"1 2 3 4 5 \n"
] |
In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.
| 500
|
[
{
"input": "5 4\n4 3 1 2",
"output": "1 1 3 4 4 "
},
{
"input": "5 5\n5 4 3 2 1",
"output": "1 2 3 4 5 "
},
{
"input": "16 11\n8 5 12 10 14 2 6 3 15 9 1",
"output": "1 2 2 2 5 5 5 8 8 8 8 8 8 8 8 8 "
},
{
"input": "79 22\n76 32 48 28 33 44 58 59 1 51 77 13 15 64 49 72 74 21 61 12 60 57",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 28 28 28 28 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 76 76 76 76 "
},
{
"input": "25 19\n3 12 21 11 19 6 5 15 4 16 20 8 9 1 22 23 25 18 13",
"output": "1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "48 8\n42 27 40 1 18 3 19 2",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 27 27 27 27 27 27 27 27 27 27 27 27 27 27 27 42 42 42 42 42 42 42 "
},
{
"input": "44 19\n13 20 7 10 9 14 43 17 18 39 21 42 37 1 33 8 35 4 6",
"output": "1 1 1 1 1 1 7 7 7 7 7 7 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 "
},
{
"input": "80 29\n79 51 28 73 65 39 10 1 59 29 7 70 64 3 35 17 24 71 74 2 6 49 66 80 13 18 60 15 12",
"output": "1 1 1 1 1 1 1 1 1 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 79 79 "
},
{
"input": "31 4\n8 18 30 1",
"output": "1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 "
},
{
"input": "62 29\n61 55 35 13 51 56 23 6 8 26 27 40 48 11 18 12 19 50 54 14 24 21 32 17 43 33 1 2 3",
"output": "1 1 1 1 1 6 6 6 6 6 6 6 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 55 55 55 55 55 55 61 61 "
},
{
"input": "5 4\n2 3 4 1",
"output": "1 2 2 2 2 "
},
{
"input": "39 37\n2 5 17 24 19 33 35 16 20 3 1 34 10 36 15 37 14 8 28 21 13 31 30 29 7 25 32 12 6 27 22 4 11 39 18 9 26",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100 100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "18 3\n18 1 11",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 18 "
},
{
"input": "67 20\n66 23 40 49 3 39 60 43 52 47 16 36 22 5 41 10 55 34 64 1",
"output": "1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 66 66 "
},
{
"input": "92 52\n9 85 44 13 27 61 8 1 28 41 6 14 70 67 39 71 56 80 34 21 5 10 40 73 63 38 90 57 37 36 82 86 65 46 7 54 81 12 45 49 83 59 64 26 62 25 60 24 91 47 53 55",
"output": "1 1 1 1 1 1 1 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "66 36\n44 62 32 29 3 15 47 30 50 42 35 2 33 65 10 13 56 12 1 16 7 36 39 11 25 28 20 52 46 38 37 8 61 49 48 14",
"output": "1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 29 29 29 32 32 32 32 32 32 32 32 32 32 32 32 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 "
},
{
"input": "32 8\n27 23 1 13 18 24 17 26",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 23 23 23 23 27 27 27 27 27 27 "
},
{
"input": "26 13\n1 14 13 2 4 24 21 22 16 3 10 12 6",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "31 20\n10 11 20 2 4 26 31 7 13 12 28 1 30 18 21 8 3 16 15 19",
"output": "1 2 2 2 2 2 2 2 2 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "86 25\n22 62 8 23 53 77 9 31 43 1 58 16 72 11 15 35 60 39 79 4 82 64 76 63 59",
"output": "1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8 8 8 8 8 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 "
},
{
"input": "62 54\n2 5 4 47 40 61 37 31 41 16 44 42 48 32 10 6 62 38 52 49 11 20 55 22 3 36 25 21 50 8 28 14 18 39 34 54 53 19 46 27 15 23 12 24 60 17 33 57 58 1 35 29 51 7",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "57 19\n43 45 37 40 42 55 16 33 47 32 34 35 9 41 1 6 8 15 5",
"output": "1 1 1 1 1 1 1 1 9 9 9 9 9 9 9 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 37 37 37 37 37 37 43 43 43 43 43 43 43 43 43 43 43 43 43 43 43 "
},
{
"input": "32 14\n4 7 13 1 25 22 9 27 6 28 30 2 14 21",
"output": "1 1 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 "
},
{
"input": "57 12\n8 53 51 38 1 6 16 33 13 46 28 35",
"output": "1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 "
},
{
"input": "87 9\n57 34 78 1 52 67 56 6 54",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 "
},
{
"input": "88 42\n85 45 52 14 63 53 70 71 16 86 66 47 12 22 10 72 4 31 3 69 11 77 17 25 46 75 23 1 21 84 44 20 18 33 48 88 41 83 67 61 73 34",
"output": "1 1 3 4 4 4 4 4 4 10 10 12 12 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 85 85 85 85 "
},
{
"input": "27 25\n9 21 17 5 16 3 23 7 12 4 14 11 13 1 15 19 27 8 20 10 22 25 6 18 26",
"output": "1 1 3 3 5 5 5 5 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "89 28\n5 22 79 42 16 35 66 48 57 55 1 37 29 31 40 38 45 62 41 87 64 89 81 13 60 44 71 82",
"output": "1 1 1 1 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 "
},
{
"input": "17 4\n4 3 1 2",
"output": "1 1 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 "
}
] | 1,571,502,750
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 124
| 0
|
n,m = map(int,input().split())
q = list(map(int,input().split()))
e = [0] * n
for i in range(0,len(q)):
j = q[i] - 1
while (j<len(e)) and (e[j] == 0):
e[j] = q[i]
j += 1
print(*e)
|
Title: Mashmokh and Lights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to *n*. There are *n* buttons in Mashmokh's room indexed from 1 to *n* as well. If Mashmokh pushes button with index *i*, then each light with index not less than *i* that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed *m* distinct buttons *b*1,<=*b*2,<=...,<=*b**m* (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button *b**i* is actually *b**i*, not *i*.
Please, help Mashmokh, print these indices.
Input Specification:
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of the factory lights and the pushed buttons respectively. The next line contains *m* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*).
It is guaranteed that all lights will be turned off after pushing all buttons.
Output Specification:
Output *n* space-separated integers where the *i*-th number is index of the button that turns the *i*-th light off.
Demo Input:
['5 4\n4 3 1 2\n', '5 5\n5 4 3 2 1\n']
Demo Output:
['1 1 3 4 4 \n', '1 2 3 4 5 \n']
Note:
In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.
|
```python
n,m = map(int,input().split())
q = list(map(int,input().split()))
e = [0] * n
for i in range(0,len(q)):
j = q[i] - 1
while (j<len(e)) and (e[j] == 0):
e[j] = q[i]
j += 1
print(*e)
```
| 3
|
|
218
|
A
|
Mountain Scenery
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=≤<=*i*<=≤<=2*n*) the following condition holds: *y**i*<=-<=1<=<<=*y**i* and *y**i*<=><=*y**i*<=+<=1.
We shall call a vertex of a polyline with an even *x* coordinate a mountain peak.
Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1).
Given Bolek's final picture, restore the initial one.
|
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=≤<=*r**i*<=≤<=100) — the *y* coordinates of the polyline vertices on Bolek's picture.
It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
|
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 — the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
|
[
"3 2\n0 5 3 5 1 5 2\n",
"1 1\n0 2 0\n"
] |
[
"0 5 3 4 1 4 2 \n",
"0 1 0 \n"
] |
none
| 500
|
[
{
"input": "3 2\n0 5 3 5 1 5 2",
"output": "0 5 3 4 1 4 2 "
},
{
"input": "1 1\n0 2 0",
"output": "0 1 0 "
},
{
"input": "1 1\n1 100 0",
"output": "1 99 0 "
},
{
"input": "3 1\n0 1 0 1 0 2 0",
"output": "0 1 0 1 0 1 0 "
},
{
"input": "3 1\n0 1 0 2 0 1 0",
"output": "0 1 0 1 0 1 0 "
},
{
"input": "3 3\n0 100 35 67 40 60 3",
"output": "0 99 35 66 40 59 3 "
},
{
"input": "7 3\n1 2 1 3 1 2 1 2 1 3 1 3 1 2 1",
"output": "1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 "
},
{
"input": "100 100\n1 3 1 3 1 3 0 2 0 3 1 3 1 3 1 3 0 3 1 3 0 2 0 2 0 3 0 2 0 2 0 3 1 3 1 3 1 3 1 3 0 2 0 3 1 3 0 2 0 2 0 2 0 2 0 2 0 3 0 3 0 3 0 3 0 2 0 3 1 3 1 3 1 3 0 3 0 2 0 2 0 2 0 2 0 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 0 2 0 3 1 3 0 3 0 3 0 2 0 2 0 2 0 3 0 3 1 3 1 3 0 3 1 3 1 3 1 3 0 2 0 3 0 2 0 3 1 3 0 3 0 3 1 3 0 2 0 3 0 2 0 2 0 2 0 2 0 3 1 3 0 3 1 3 1",
"output": "1 2 1 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 0 1 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 1 2 0 1 0 2 1 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 0 1 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 2 0 1 0 2 1 2 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 1 0 1 0 2 1 2 0 2 1 2 1 "
},
{
"input": "30 20\n1 3 1 3 0 2 0 4 1 3 0 3 1 3 1 4 2 3 1 2 0 4 2 4 0 4 1 3 0 4 1 4 2 4 2 4 0 3 1 2 1 4 0 3 0 4 1 3 1 4 1 3 0 1 0 4 0 3 2 3 1",
"output": "1 3 1 3 0 2 0 4 1 2 0 2 1 2 1 3 2 3 1 2 0 3 2 3 0 3 1 2 0 3 1 3 2 3 2 3 0 2 1 2 1 3 0 2 0 3 1 2 1 3 1 2 0 1 0 3 0 3 2 3 1 "
},
{
"input": "10 6\n0 5 2 4 1 5 2 5 2 4 2 5 3 5 0 2 0 1 0 1 0",
"output": "0 5 2 4 1 4 2 4 2 3 2 4 3 4 0 1 0 1 0 1 0 "
},
{
"input": "11 6\n3 5 1 4 3 5 0 2 0 2 0 4 0 3 0 4 1 5 2 4 0 4 0",
"output": "3 5 1 4 3 5 0 2 0 2 0 3 0 2 0 3 1 4 2 3 0 3 0 "
},
{
"input": "12 6\n1 2 1 5 0 2 0 4 1 3 1 4 2 4 0 4 0 4 2 4 0 4 0 5 3",
"output": "1 2 1 5 0 2 0 4 1 3 1 4 2 3 0 3 0 3 2 3 0 3 0 4 3 "
},
{
"input": "13 6\n3 5 2 5 0 3 0 1 0 2 0 1 0 1 0 2 1 4 3 5 1 3 1 3 2 3 1",
"output": "3 4 2 4 0 2 0 1 0 1 0 1 0 1 0 2 1 4 3 4 1 2 1 3 2 3 1 "
},
{
"input": "24 7\n3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 4 0 3 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 5 1 3 0 1 0 2 0 3 1 3 1",
"output": "3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 3 0 2 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 4 1 2 0 1 0 1 0 2 1 2 1 "
},
{
"input": "25 8\n3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 5 3 5 0 4 2 3 2 4 1 4 0 4 1 4 0 1 0 4 2",
"output": "3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 4 3 4 0 3 2 3 2 3 1 3 0 3 1 3 0 1 0 3 2 "
},
{
"input": "26 9\n3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 5 1 4 3 5 0 5 2 3 0 3 1 4 1 3 1 4 2 3 1 4 3 4 1 3 2 4 1 3 2 5 1 2 0",
"output": "3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 4 1 4 3 4 0 4 2 3 0 2 1 3 1 2 1 3 2 3 1 4 3 4 1 3 2 3 1 3 2 4 1 2 0 "
},
{
"input": "27 10\n3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 4 2 3 0 4 2 5 3 4 3 4 1 5 3 4 1 2 1 5 0 3 0 5 0 5 3 4 0 1 0 2 0 2 1 4 0 2 1",
"output": "3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 3 2 3 0 3 2 4 3 4 3 4 1 4 3 4 1 2 1 4 0 2 0 4 0 4 3 4 0 1 0 1 0 2 1 3 0 2 1 "
},
{
"input": "40 1\n0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 2 0 1 0 2 1 2 0",
"output": "0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 1 0 1 0 2 1 2 0 "
},
{
"input": "40 2\n0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 3 1 3 0",
"output": "0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 2 1 2 0 "
},
{
"input": "40 3\n1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 2 0 3 1 2 0 3 0",
"output": "1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 1 0 2 1 2 0 2 0 "
},
{
"input": "50 40\n1 4 2 4 1 2 1 4 1 4 2 3 1 2 1 4 1 3 0 2 1 4 0 1 0 3 1 3 1 3 0 4 2 4 2 4 2 4 2 4 2 4 2 4 0 4 1 3 1 3 0 4 1 4 2 3 2 3 0 3 0 3 0 4 1 4 1 3 1 4 1 3 0 4 0 3 0 2 0 2 0 4 1 4 0 2 0 4 1 4 0 3 0 2 1 3 0 2 0 4 0",
"output": "1 4 2 4 1 2 1 3 1 3 2 3 1 2 1 3 1 2 0 2 1 3 0 1 0 2 1 2 1 2 0 3 2 3 2 3 2 3 2 3 2 3 2 3 0 3 1 2 1 2 0 3 1 3 2 3 2 3 0 2 0 2 0 3 1 3 1 2 1 3 1 2 0 3 0 2 0 1 0 1 0 3 1 3 0 1 0 3 1 3 0 2 0 2 1 2 0 1 0 3 0 "
},
{
"input": "100 2\n1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 3 1 2 0 3 1 2 0",
"output": "1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 2 1 2 0 2 1 2 0 "
},
{
"input": "100 3\n0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 3 1 3 1 3 0",
"output": "0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 2 1 2 1 2 0 "
},
{
"input": "100 20\n0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 3 0 2 0 2 1 2 1 3 0 3 1 2 1 3 1 3 1 2 1 2 0 2 1 3 0 2 0 3 0 1 0 3 0 3 0 1 0 4 1 3 0 1 0 1 0 2 1 2 0 2 1 4 1 3 0 2 1 3 1 3 1 3 0 3 0 2 0 1 0 2 1 2 1",
"output": "0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 2 0 1 0 2 1 2 1 2 0 2 1 2 1 2 1 2 1 2 1 2 0 2 1 2 0 1 0 2 0 1 0 2 0 2 0 1 0 3 1 2 0 1 0 1 0 2 1 2 0 2 1 3 1 2 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 2 1 2 1 "
},
{
"input": "100 20\n2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 4 0 9 2 7 6 7 2 8 0 8 2 7 5 10 1 2 0 2 0 4 3 5 4 7 0 10 2 10 3 6 3 7 1 4 0 9 1 4 3 8 1 10 1 10 0 3 2 5 3 9 0 7 4 5 0 1 0",
"output": "2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 3 0 8 2 7 6 7 2 7 0 7 2 6 5 9 1 2 0 1 0 4 3 5 4 6 0 9 2 9 3 5 3 6 1 3 0 8 1 4 3 7 1 9 1 9 0 3 2 4 3 8 0 6 4 5 0 1 0 "
},
{
"input": "98 3\n1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 2 0 2 0",
"output": "1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 1 0 "
},
{
"input": "2 1\n0 2 1 4 1",
"output": "0 2 1 3 1 "
},
{
"input": "2 1\n0 2 1 5 1",
"output": "0 2 1 4 1 "
},
{
"input": "3 3\n1 12 9 11 6 8 1",
"output": "1 11 9 10 6 7 1 "
},
{
"input": "3 2\n0 7 4 7 1 3 2",
"output": "0 6 4 6 1 3 2 "
},
{
"input": "2 1\n1 3 2 4 1",
"output": "1 3 2 3 1 "
},
{
"input": "4 1\n5 6 5 6 5 6 1 3 1",
"output": "5 6 5 6 5 6 1 2 1 "
},
{
"input": "2 1\n0 2 1 3 0",
"output": "0 2 1 2 0 "
},
{
"input": "2 2\n98 100 1 7 2",
"output": "98 99 1 6 2 "
},
{
"input": "3 1\n8 10 9 10 3 5 1",
"output": "8 10 9 10 3 4 1 "
},
{
"input": "3 2\n0 4 3 5 2 5 2",
"output": "0 4 3 4 2 4 2 "
},
{
"input": "2 1\n4 5 2 4 2",
"output": "4 5 2 3 2 "
},
{
"input": "3 1\n0 2 1 2 0 2 0",
"output": "0 2 1 2 0 1 0 "
},
{
"input": "1 1\n5 7 2",
"output": "5 6 2 "
},
{
"input": "2 1\n3 4 1 3 1",
"output": "3 4 1 2 1 "
},
{
"input": "3 1\n0 4 3 5 0 5 0",
"output": "0 4 3 5 0 4 0 "
},
{
"input": "3 1\n1 3 2 3 1 3 1",
"output": "1 3 2 3 1 2 1 "
},
{
"input": "2 1\n0 8 7 100 0",
"output": "0 8 7 99 0 "
},
{
"input": "2 1\n1 3 2 5 1",
"output": "1 3 2 4 1 "
}
] | 1,581,589,652
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 186
| 819,200
|
import string
n = input()
txt = n[0]
x = txt.capitalize()
print(x+n[1:])
|
Title: Mountain Scenery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=≤<=*i*<=≤<=2*n*) the following condition holds: *y**i*<=-<=1<=<<=*y**i* and *y**i*<=><=*y**i*<=+<=1.
We shall call a vertex of a polyline with an even *x* coordinate a mountain peak.
Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1).
Given Bolek's final picture, restore the initial one.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=≤<=*r**i*<=≤<=100) — the *y* coordinates of the polyline vertices on Bolek's picture.
It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
Output Specification:
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 — the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
Demo Input:
['3 2\n0 5 3 5 1 5 2\n', '1 1\n0 2 0\n']
Demo Output:
['0 5 3 4 1 4 2 \n', '0 1 0 \n']
Note:
none
|
```python
import string
n = input()
txt = n[0]
x = txt.capitalize()
print(x+n[1:])
```
| 0
|
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,666,305,349
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
n, m = [int(x) for x in input().split()]
prix = sorted([int(x) for x in input().split()])
s=0
test=[]
for p in prix[:m]:
s+=p
test.append(s)
print(-min(test))
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n, m = [int(x) for x in input().split()]
prix = sorted([int(x) for x in input().split()])
s=0
test=[]
for p in prix[:m]:
s+=p
test.append(s)
print(-min(test))
```
| 0
|
350
|
A
|
TL
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
|
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
|
If there is a valid TL value, print it. Otherwise, print -1.
|
[
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] |
[
"5",
"-1\n"
] |
none
| 500
|
[
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43 42 38 40 22 41 1 49 47 9 8 19 15 29 30 16 18\n91 58 86 51 94 94 73 84 98 69 74 56 52 80 88 61 53 99 88 50 55 95 65 84 87 79 51 52 69 60 74 73 93 61 73 59 64 56 95 78 86 72 79 70 93 78 54 61 71 50",
"output": "49"
},
{
"input": "55 44\n93 17 74 15 34 16 41 80 26 54 94 94 86 93 20 44 63 72 39 43 67 4 37 49 76 94 5 51 64 74 11 47 77 97 57 30 42 72 71 26 8 14 67 64 49 57 30 23 40 4 76 78 87 78 79\n38 55 17 65 26 7 36 65 48 28 49 93 18 98 31 90 26 57 1 26 88 56 48 56 23 13 8 67 80 2 51 3 21 33 20 54 2 45 21 36 3 98 62 2",
"output": "-1"
},
{
"input": "32 100\n30 8 4 35 18 41 18 12 33 39 39 18 39 19 33 46 45 33 34 27 14 39 40 21 38 9 42 35 27 10 14 14\n65 49 89 64 47 78 59 52 73 51 84 82 88 63 91 99 67 87 53 99 75 47 85 82 58 47 80 50 65 91 83 90 77 52 100 88 97 74 98 99 50 93 65 61 65 65 65 96 61 51 84 67 79 90 92 83 100 100 100 95 80 54 77 51 98 64 74 62 60 96 73 74 94 55 89 60 92 65 74 79 66 81 53 47 71 51 54 85 74 97 68 72 88 94 100 85 65 63 65 90",
"output": "46"
},
{
"input": "1 50\n7\n65 52 99 78 71 19 96 72 80 15 50 94 20 35 79 95 44 41 45 53 77 50 74 66 59 96 26 84 27 48 56 84 36 78 89 81 67 34 79 74 99 47 93 92 90 96 72 28 78 66",
"output": "14"
},
{
"input": "1 1\n4\n9",
"output": "8"
},
{
"input": "1 1\n2\n4",
"output": "-1"
},
{
"input": "22 56\n49 20 42 68 15 46 98 78 82 8 7 33 50 30 75 96 36 88 35 99 19 87\n15 18 81 24 35 89 25 32 23 3 48 24 52 69 18 32 23 61 48 98 50 38 5 17 70 20 38 32 49 54 68 11 51 81 46 22 19 59 29 38 45 83 18 13 91 17 84 62 25 60 97 32 23 13 83 58",
"output": "-1"
},
{
"input": "1 1\n50\n100",
"output": "-1"
},
{
"input": "1 1\n49\n100",
"output": "98"
},
{
"input": "1 1\n100\n100",
"output": "-1"
},
{
"input": "1 1\n99\n100",
"output": "-1"
},
{
"input": "8 4\n1 2 49 99 99 95 78 98\n100 100 100 100",
"output": "99"
},
{
"input": "68 85\n43 55 2 4 72 45 19 56 53 81 18 90 11 87 47 8 94 88 24 4 67 9 21 70 25 66 65 27 46 13 8 51 65 99 37 43 71 59 71 79 32 56 49 43 57 85 95 81 40 28 60 36 72 81 60 40 16 78 61 37 29 26 15 95 70 27 50 97\n6 6 48 72 54 31 1 50 29 64 93 9 29 93 66 63 25 90 52 1 66 13 70 30 24 87 32 90 84 72 44 13 25 45 31 16 92 60 87 40 62 7 20 63 86 78 73 88 5 36 74 100 64 34 9 5 62 29 58 48 81 46 84 56 27 1 60 14 54 88 31 93 62 7 9 69 27 48 10 5 33 10 53 66 2",
"output": "-1"
},
{
"input": "5 100\n1 1 1 1 1\n77 53 38 29 97 33 64 17 78 100 27 12 42 44 20 24 44 68 58 57 65 90 8 24 4 6 74 68 61 43 25 69 8 62 36 85 67 48 69 30 35 41 42 12 87 66 50 92 53 76 38 67 85 7 80 78 53 76 94 8 37 50 4 100 4 71 10 48 34 47 83 42 25 81 64 72 25 51 53 75 43 98 53 77 94 38 81 15 89 91 72 76 7 36 27 41 88 18 19 75",
"output": "2"
},
{
"input": "3 3\n2 3 4\n8 9 10",
"output": "4"
},
{
"input": "2 1\n2 3\n15",
"output": "4"
},
{
"input": "2 1\n2 4\n4",
"output": "-1"
},
{
"input": "2 3\n4 5\n10 11 12",
"output": "8"
},
{
"input": "3 1\n2 3 3\n5",
"output": "4"
},
{
"input": "2 1\n9 10\n100",
"output": "18"
},
{
"input": "3 3\n3 12 15\n7 8 9",
"output": "-1"
},
{
"input": "2 2\n3 5\n7 8",
"output": "6"
},
{
"input": "3 3\n4 5 6\n10 11 12",
"output": "8"
},
{
"input": "3 5\n2 3 3\n6 6 6 6 2",
"output": "-1"
},
{
"input": "3 6\n4 5 3\n8 9 7 10 7 11",
"output": "6"
},
{
"input": "3 6\n4 5 2\n8 9 6 10 7 4",
"output": "-1"
},
{
"input": "2 1\n4 6\n10",
"output": "8"
},
{
"input": "1 2\n1\n3 1",
"output": "-1"
},
{
"input": "2 1\n5 6\n20",
"output": "10"
},
{
"input": "2 1\n1 5\n5",
"output": "-1"
},
{
"input": "3 2\n10 20 30\n30 40",
"output": "-1"
},
{
"input": "2 2\n5 6\n7 100",
"output": "-1"
},
{
"input": "2 1\n2 5\n7",
"output": "5"
},
{
"input": "1 1\n5\n20",
"output": "10"
},
{
"input": "2 1\n10 11\n100",
"output": "20"
},
{
"input": "1 1\n1\n10",
"output": "2"
},
{
"input": "1 1\n10\n100",
"output": "20"
}
] | 1,603,107,237
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 280
| 0
|
n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
c=max(a)
d=min(a)
if(2*d<=c):
print(c)
else:
print(-1)
|
Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none
|
```python
n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
c=max(a)
d=min(a)
if(2*d<=c):
print(c)
else:
print(-1)
```
| 0
|
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,571,623,709
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 124
| 0
|
n = int(input())
count = 0
k = 1
while count < n:
count += 5*k
k += 1
n = n - int((k-2)*(k-1)*5/2)
if n <= k-1:
print('Sheldon')
if k-1<n <= 2*k-2:
print('Leonard')
if 2*k-2 < n <= 3*k-3:
print('Penny')
if 3*k-3 < n <= 4*k-4:
print('Rajesh')
if 4*k-4 < n <= 5*k-5:
print('Howard')
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
n = int(input())
count = 0
k = 1
while count < n:
count += 5*k
k += 1
n = n - int((k-2)*(k-1)*5/2)
if n <= k-1:
print('Sheldon')
if k-1<n <= 2*k-2:
print('Leonard')
if 2*k-2 < n <= 3*k-3:
print('Penny')
if 3*k-3 < n <= 4*k-4:
print('Rajesh')
if 4*k-4 < n <= 5*k-5:
print('Howard')
```
| 0
|
743
|
B
|
Chloe and the sequence
|
PROGRAMMING
| 1,200
|
[
"binary search",
"bitmasks",
"constructive algorithms",
"implementation"
] | null | null |
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
|
Print single integer — the integer at the *k*-th position in the obtained sequence.
|
[
"3 2\n",
"4 8\n"
] |
[
"2",
"4"
] |
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
| 1,000
|
[
{
"input": "3 2",
"output": "2"
},
{
"input": "4 8",
"output": "4"
},
{
"input": "5 27",
"output": "1"
},
{
"input": "7 44",
"output": "3"
},
{
"input": "15 18432",
"output": "12"
},
{
"input": "20 259676",
"output": "3"
},
{
"input": "30 671088640",
"output": "28"
},
{
"input": "38 137438953472",
"output": "38"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "4 12",
"output": "3"
},
{
"input": "8 224",
"output": "6"
},
{
"input": "13 1368",
"output": "4"
},
{
"input": "16 49152",
"output": "15"
},
{
"input": "19 1024",
"output": "11"
},
{
"input": "24 15204352",
"output": "20"
},
{
"input": "27 6586544",
"output": "5"
},
{
"input": "31 536870912",
"output": "30"
},
{
"input": "38 94489280512",
"output": "34"
},
{
"input": "42 215268308020",
"output": "3"
},
{
"input": "42 3452074721280",
"output": "21"
},
{
"input": "46 34394312982528",
"output": "15"
},
{
"input": "48 133865540681728",
"output": "39"
},
{
"input": "49 76541041985542",
"output": "2"
},
{
"input": "49 104822971826176",
"output": "34"
},
{
"input": "49 351843720888320",
"output": "47"
},
{
"input": "50 743042492421629",
"output": "1"
},
{
"input": "50 666266740436818",
"output": "2"
},
{
"input": "50 704239287953456",
"output": "5"
},
{
"input": "50 116938486513664",
"output": "17"
},
{
"input": "50 806423059496960",
"output": "37"
},
{
"input": "50 985162418487296",
"output": "48"
},
{
"input": "50 844424930131968",
"output": "49"
},
{
"input": "50 562949953421312",
"output": "50"
},
{
"input": "50 1",
"output": "1"
},
{
"input": "50 1125899906842623",
"output": "1"
},
{
"input": "50 1125899906842620",
"output": "3"
},
{
"input": "39 549755813887",
"output": "1"
},
{
"input": "50 100000000000",
"output": "12"
}
] | 1,683,347,384
| 2,147,483,647
|
Python 3
|
MEMORY_LIMIT_EXCEEDED
|
TESTS
| 6
| 763
| 268,390,400
|
n, k = map(int, input().split())
seq = [1]
for i in range(n - 1):
new_num = min(set(range(1, 2*i+3)) - set(seq))
seq = seq + [new_num] + seq[::-1]
#print(seq)
print(seq[k - 1])
|
Title: Chloe and the sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
Output Specification:
Print single integer — the integer at the *k*-th position in the obtained sequence.
Demo Input:
['3 2\n', '4 8\n']
Demo Output:
['2', '4']
Note:
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
|
```python
n, k = map(int, input().split())
seq = [1]
for i in range(n - 1):
new_num = min(set(range(1, 2*i+3)) - set(seq))
seq = seq + [new_num] + seq[::-1]
#print(seq)
print(seq[k - 1])
```
| 0
|
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,570,027,399
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 312
| 0
|
n,m= map(int, input().split())
l= list(map(int, input().split()))
l.sort()
r= [x for x in l if x<0]
if len(r)>m:
ans= sum(l[:m])
print(abs(ans))
else:
print(abs(sum(r)))
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n,m= map(int, input().split())
l= list(map(int, input().split()))
l.sort()
r= [x for x in l if x<0]
if len(r)>m:
ans= sum(l[:m])
print(abs(ans))
else:
print(abs(sum(r)))
```
| 3.922
|
259
|
B
|
Little Elephant and Magic Square
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
|
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
|
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
|
[
"0 1 1\n1 0 1\n1 1 0\n",
"0 3 6\n5 0 5\n4 7 0\n"
] |
[
"1 1 1\n1 1 1\n1 1 1\n",
"6 3 6\n5 5 5\n4 7 4\n"
] |
none
| 1,000
|
[
{
"input": "0 1 1\n1 0 1\n1 1 0",
"output": "1 1 1\n1 1 1\n1 1 1"
},
{
"input": "0 3 6\n5 0 5\n4 7 0",
"output": "6 3 6\n5 5 5\n4 7 4"
},
{
"input": "0 4 4\n4 0 4\n4 4 0",
"output": "4 4 4\n4 4 4\n4 4 4"
},
{
"input": "0 54 48\n36 0 78\n66 60 0",
"output": "69 54 48\n36 57 78\n66 60 45"
},
{
"input": "0 17 14\n15 0 15\n16 13 0",
"output": "14 17 14\n15 15 15\n16 13 16"
},
{
"input": "0 97 56\n69 0 71\n84 43 0",
"output": "57 97 56\n69 70 71\n84 43 83"
},
{
"input": "0 1099 1002\n1027 0 1049\n1074 977 0",
"output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063"
},
{
"input": "0 98721 99776\n99575 0 99123\n98922 99977 0",
"output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148"
},
{
"input": "0 6361 2304\n1433 0 8103\n7232 3175 0",
"output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897"
},
{
"input": "0 99626 99582\n99766 0 99258\n99442 99398 0",
"output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696"
},
{
"input": "0 99978 99920\n99950 0 99918\n99948 99890 0",
"output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964"
},
{
"input": "0 840 666\n612 0 948\n894 720 0",
"output": "834 840 666\n612 780 948\n894 720 726"
},
{
"input": "0 28 10\n12 0 24\n26 8 0",
"output": "16 28 10\n12 18 24\n26 8 20"
},
{
"input": "0 120 83\n98 0 90\n105 68 0",
"output": "79 120 83\n98 94 90\n105 68 109"
},
{
"input": "0 86900 85807\n85836 0 86842\n86871 85778 0",
"output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368"
},
{
"input": "0 74 78\n78 0 74\n74 78 0",
"output": "76 74 78\n78 76 74\n74 78 76"
},
{
"input": "0 505 681\n605 0 657\n581 757 0",
"output": "707 505 681\n605 631 657\n581 757 555"
},
{
"input": "0 662 918\n822 0 854\n758 1014 0",
"output": "934 662 918\n822 838 854\n758 1014 742"
},
{
"input": "0 93 95\n93 0 97\n95 97 0",
"output": "97 93 95\n93 95 97\n95 97 93"
},
{
"input": "0 709 712\n719 0 695\n702 705 0",
"output": "700 709 712\n719 707 695\n702 705 714"
},
{
"input": "0 7 6\n9 0 1\n4 3 0",
"output": "2 7 6\n9 5 1\n4 3 8"
},
{
"input": "0 9 2\n3 0 7\n8 1 0",
"output": "4 9 2\n3 5 7\n8 1 6"
},
{
"input": "0 1 43\n13 0 61\n31 73 0",
"output": "67 1 43\n13 37 61\n31 73 7"
},
{
"input": "0 100000 100000\n100000 0 100000\n100000 100000 0",
"output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000"
}
] | 1,534,791,557
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 248
| 0
|
L = []
for i in range(3):
L.append(list(map(int, input().split())))
L[0][0] = (L[1][2] + L[2][1])//2
L[1][1] = (L[0][2] + L[2][0])//2
L[2][2] = (L[0][1] + L[1][0])//2
for i in range(3):
print(*L[i])
|
Title: Little Elephant and Magic Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
Input Specification:
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Output Specification:
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
Demo Input:
['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n']
Demo Output:
['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n']
Note:
none
|
```python
L = []
for i in range(3):
L.append(list(map(int, input().split())))
L[0][0] = (L[1][2] + L[2][1])//2
L[1][1] = (L[0][2] + L[2][0])//2
L[2][2] = (L[0][1] + L[1][0])//2
for i in range(3):
print(*L[i])
```
| 3
|
|
149
|
A
|
Business trip
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
|
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
|
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
|
[
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] |
[
"2\n",
"0\n",
"3\n"
] |
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
| 500
|
[
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 17 21 10 11 100 23 10",
"output": "1"
},
{
"input": "52\n1 12 3 11 4 5 10 6 9 7 8 2",
"output": "6"
},
{
"input": "50\n2 2 3 4 5 4 4 5 7 3 2 7",
"output": "-1"
},
{
"input": "0\n55 81 28 48 99 20 67 95 6 19 10 93",
"output": "0"
},
{
"input": "93\n85 40 93 66 92 43 61 3 64 51 90 21",
"output": "1"
},
{
"input": "99\n36 34 22 0 0 0 52 12 0 0 33 47",
"output": "2"
},
{
"input": "99\n28 32 31 0 10 35 11 18 0 0 32 28",
"output": "3"
},
{
"input": "99\n19 17 0 1 18 11 29 9 29 22 0 8",
"output": "4"
},
{
"input": "76\n2 16 11 10 12 0 20 4 4 14 11 14",
"output": "5"
},
{
"input": "41\n2 1 7 7 4 2 4 4 9 3 10 0",
"output": "6"
},
{
"input": "47\n8 2 2 4 3 1 9 4 2 7 7 8",
"output": "7"
},
{
"input": "58\n6 11 7 0 5 6 3 9 4 9 5 1",
"output": "8"
},
{
"input": "32\n5 2 4 1 5 0 5 1 4 3 0 3",
"output": "9"
},
{
"input": "31\n6 1 0 4 4 5 1 0 5 3 2 0",
"output": "9"
},
{
"input": "35\n2 3 0 0 6 3 3 4 3 5 0 6",
"output": "9"
},
{
"input": "41\n3 1 3 4 3 6 6 1 4 4 0 6",
"output": "11"
},
{
"input": "97\n0 5 3 12 10 16 22 8 21 17 21 10",
"output": "5"
},
{
"input": "100\n21 21 0 0 4 13 0 26 0 0 0 15",
"output": "6"
},
{
"input": "100\n0 0 16 5 22 0 5 0 25 0 14 13",
"output": "7"
},
{
"input": "97\n17 0 10 0 0 0 18 0 14 23 15 0",
"output": "6"
},
{
"input": "100\n0 9 0 18 7 0 0 14 33 3 0 16",
"output": "7"
},
{
"input": "95\n5 2 13 0 15 18 17 0 6 11 0 8",
"output": "9"
},
{
"input": "94\n11 13 0 9 15 8 8 16 3 7 1 3",
"output": "11"
},
{
"input": "96\n8 4 12 15 8 0 4 10 6 6 12 11",
"output": "11"
},
{
"input": "100\n5 5 3 8 6 5 0 3 3 8 1 3",
"output": "-1"
},
{
"input": "100\n1 0 0 1 1 0 1 1 1 1 2 1",
"output": "-1"
},
{
"input": "100\n6 3 2 0 4 1 2 2 2 2 1 1",
"output": "-1"
},
{
"input": "0\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "0\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 1 1 1",
"output": "12"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 2",
"output": "12"
},
{
"input": "15\n10 1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
}
] | 1,640,941,224
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
t=int(input())
arr=list(map(int,input().split()))
c=0
f=0
if(t==0):
print(0)
for i,k in enumerate(reversed(sorted(arr))):
c+=k
if(c>=t):
print(i+1)
f=1
break
if(f==0):
print(-1)
|
Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
|
```python
t=int(input())
arr=list(map(int,input().split()))
c=0
f=0
if(t==0):
print(0)
for i,k in enumerate(reversed(sorted(arr))):
c+=k
if(c>=t):
print(i+1)
f=1
break
if(f==0):
print(-1)
```
| 0
|
|
217
|
A
|
Ice Skating
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null |
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
|
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
|
[
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] |
[
"1\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182 184\n806 663\n468 459",
"output": "21"
},
{
"input": "17\n660 646\n440 442\n689 618\n441 415\n922 865\n950 972\n312 366\n203 229\n873 860\n219 199\n344 308\n169 176\n961 992\n153 84\n201 230\n987 938\n834 815",
"output": "16"
},
{
"input": "11\n798 845\n722 911\n374 270\n629 537\n748 856\n831 885\n486 641\n751 829\n609 492\n98 27\n654 663",
"output": "10"
},
{
"input": "1\n321 88",
"output": "0"
},
{
"input": "9\n811 859\n656 676\n76 141\n945 951\n497 455\n18 55\n335 294\n267 275\n656 689",
"output": "7"
},
{
"input": "7\n948 946\n130 130\n761 758\n941 938\n971 971\n387 385\n509 510",
"output": "6"
},
{
"input": "6\n535 699\n217 337\n508 780\n180 292\n393 112\n732 888",
"output": "5"
},
{
"input": "14\n25 23\n499 406\n193 266\n823 751\n219 227\n101 138\n978 992\n43 74\n997 932\n237 189\n634 538\n774 740\n842 767\n742 802",
"output": "13"
},
{
"input": "12\n548 506\n151 198\n370 380\n655 694\n654 690\n407 370\n518 497\n819 827\n765 751\n802 771\n741 752\n653 662",
"output": "11"
},
{
"input": "40\n685 711\n433 403\n703 710\n491 485\n616 619\n288 282\n884 871\n367 352\n500 511\n977 982\n51 31\n576 564\n508 519\n755 762\n22 20\n368 353\n232 225\n953 955\n452 436\n311 330\n967 988\n369 364\n791 803\n150 149\n651 661\n118 93\n398 387\n748 766\n852 852\n230 228\n555 545\n515 519\n667 678\n867 862\n134 146\n859 863\n96 99\n486 469\n303 296\n780 786",
"output": "38"
},
{
"input": "3\n175 201\n907 909\n388 360",
"output": "2"
},
{
"input": "7\n312 298\n86 78\n73 97\n619 594\n403 451\n538 528\n71 86",
"output": "6"
},
{
"input": "19\n802 820\n368 248\n758 794\n455 378\n876 888\n771 814\n245 177\n586 555\n844 842\n364 360\n820 856\n731 624\n982 975\n825 856\n122 121\n862 896\n42 4\n792 841\n828 820",
"output": "16"
},
{
"input": "32\n643 877\n842 614\n387 176\n99 338\n894 798\n652 728\n611 648\n622 694\n579 781\n243 46\n322 305\n198 438\n708 579\n246 325\n536 459\n874 593\n120 277\n989 907\n223 110\n35 130\n761 692\n690 661\n518 766\n226 93\n678 597\n725 617\n661 574\n775 496\n56 416\n14 189\n358 359\n898 901",
"output": "31"
},
{
"input": "32\n325 327\n20 22\n72 74\n935 933\n664 663\n726 729\n785 784\n170 171\n315 314\n577 580\n984 987\n313 317\n434 435\n962 961\n55 54\n46 44\n743 742\n434 433\n617 612\n332 332\n883 886\n940 936\n793 792\n645 644\n611 607\n418 418\n465 465\n219 218\n167 164\n56 54\n403 405\n210 210",
"output": "29"
},
{
"input": "32\n652 712\n260 241\n27 154\n188 16\n521 351\n518 356\n452 540\n790 827\n339 396\n336 551\n897 930\n828 627\n27 168\n180 113\n134 67\n794 671\n812 711\n100 241\n686 813\n138 289\n384 506\n884 932\n913 959\n470 508\n730 734\n373 478\n788 862\n392 426\n148 68\n113 49\n713 852\n924 894",
"output": "29"
},
{
"input": "14\n685 808\n542 677\n712 747\n832 852\n187 410\n399 338\n626 556\n530 635\n267 145\n215 209\n559 684\n944 949\n753 596\n601 823",
"output": "13"
},
{
"input": "5\n175 158\n16 2\n397 381\n668 686\n957 945",
"output": "4"
},
{
"input": "5\n312 284\n490 509\n730 747\n504 497\n782 793",
"output": "4"
},
{
"input": "2\n802 903\n476 348",
"output": "1"
},
{
"input": "4\n325 343\n425 442\n785 798\n275 270",
"output": "3"
},
{
"input": "28\n462 483\n411 401\n118 94\n111 127\n5 6\n70 52\n893 910\n73 63\n818 818\n182 201\n642 633\n900 886\n893 886\n684 700\n157 173\n953 953\n671 660\n224 225\n832 801\n152 157\n601 585\n115 101\n739 722\n611 606\n659 642\n461 469\n702 689\n649 653",
"output": "25"
},
{
"input": "36\n952 981\n885 900\n803 790\n107 129\n670 654\n143 132\n66 58\n813 819\n849 837\n165 198\n247 228\n15 39\n619 618\n105 138\n868 855\n965 957\n293 298\n613 599\n227 212\n745 754\n723 704\n877 858\n503 487\n678 697\n592 595\n155 135\n962 982\n93 89\n660 673\n225 212\n967 987\n690 680\n804 813\n489 518\n240 221\n111 124",
"output": "34"
},
{
"input": "30\n89 3\n167 156\n784 849\n943 937\n144 95\n24 159\n80 120\n657 683\n585 596\n43 147\n909 964\n131 84\n345 389\n333 321\n91 126\n274 325\n859 723\n866 922\n622 595\n690 752\n902 944\n127 170\n426 383\n905 925\n172 284\n793 810\n414 510\n890 884\n123 24\n267 255",
"output": "29"
},
{
"input": "5\n664 666\n951 941\n739 742\n844 842\n2 2",
"output": "4"
},
{
"input": "3\n939 867\n411 427\n757 708",
"output": "2"
},
{
"input": "36\n429 424\n885 972\n442 386\n512 511\n751 759\n4 115\n461 497\n496 408\n8 23\n542 562\n296 331\n448 492\n412 395\n109 166\n622 640\n379 355\n251 262\n564 586\n66 115\n275 291\n666 611\n629 534\n510 567\n635 666\n738 803\n420 369\n92 17\n101 144\n141 92\n258 258\n184 235\n492 456\n311 210\n394 357\n531 512\n634 636",
"output": "34"
},
{
"input": "29\n462 519\n871 825\n127 335\n156 93\n576 612\n885 830\n634 779\n340 105\n744 795\n716 474\n93 139\n563 805\n137 276\n177 101\n333 14\n391 437\n873 588\n817 518\n460 597\n572 670\n140 303\n392 441\n273 120\n862 578\n670 639\n410 161\n544 577\n193 116\n252 195",
"output": "28"
},
{
"input": "23\n952 907\n345 356\n812 807\n344 328\n242 268\n254 280\n1000 990\n80 78\n424 396\n595 608\n755 813\n383 380\n55 56\n598 633\n203 211\n508 476\n600 593\n206 192\n855 882\n517 462\n967 994\n642 657\n493 488",
"output": "22"
},
{
"input": "10\n579 816\n806 590\n830 787\n120 278\n677 800\n16 67\n188 251\n559 560\n87 67\n104 235",
"output": "8"
},
{
"input": "23\n420 424\n280 303\n515 511\n956 948\n799 803\n441 455\n362 369\n299 289\n823 813\n982 967\n876 878\n185 157\n529 551\n964 989\n655 656\n1 21\n114 112\n45 56\n935 937\n1000 997\n934 942\n360 366\n648 621",
"output": "22"
},
{
"input": "23\n102 84\n562 608\n200 127\n952 999\n465 496\n322 367\n728 690\n143 147\n855 867\n861 866\n26 59\n300 273\n255 351\n192 246\n70 111\n365 277\n32 104\n298 319\n330 354\n241 141\n56 125\n315 298\n412 461",
"output": "22"
},
{
"input": "7\n429 506\n346 307\n99 171\n853 916\n322 263\n115 157\n906 924",
"output": "6"
},
{
"input": "3\n1 1\n2 1\n2 2",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "5\n1 1\n1 2\n2 2\n3 1\n3 3",
"output": "0"
},
{
"input": "6\n1 1\n1 2\n2 2\n3 1\n3 2\n3 3",
"output": "0"
},
{
"input": "20\n1 1\n2 2\n3 3\n3 9\n4 4\n5 2\n5 5\n5 7\n5 8\n6 2\n6 6\n6 9\n7 7\n8 8\n9 4\n9 7\n9 9\n10 2\n10 9\n10 10",
"output": "1"
},
{
"input": "21\n1 1\n1 9\n2 1\n2 2\n2 5\n2 6\n2 9\n3 3\n3 8\n4 1\n4 4\n5 5\n5 8\n6 6\n7 7\n8 8\n9 9\n10 4\n10 10\n11 5\n11 11",
"output": "1"
},
{
"input": "22\n1 1\n1 3\n1 4\n1 8\n1 9\n1 11\n2 2\n3 3\n4 4\n4 5\n5 5\n6 6\n6 8\n7 7\n8 3\n8 4\n8 8\n9 9\n10 10\n11 4\n11 9\n11 11",
"output": "3"
},
{
"input": "50\n1 1\n2 2\n2 9\n3 3\n4 4\n4 9\n4 16\n4 24\n5 5\n6 6\n7 7\n8 8\n8 9\n8 20\n9 9\n10 10\n11 11\n12 12\n13 13\n14 7\n14 14\n14 16\n14 25\n15 4\n15 6\n15 15\n15 22\n16 6\n16 16\n17 17\n18 18\n19 6\n19 19\n20 20\n21 21\n22 6\n22 22\n23 23\n24 6\n24 7\n24 8\n24 9\n24 24\n25 1\n25 3\n25 5\n25 7\n25 23\n25 24\n25 25",
"output": "7"
},
{
"input": "55\n1 1\n1 14\n2 2\n2 19\n3 1\n3 3\n3 8\n3 14\n3 23\n4 1\n4 4\n5 5\n5 8\n5 15\n6 2\n6 3\n6 4\n6 6\n7 7\n8 8\n8 21\n9 9\n10 1\n10 10\n11 9\n11 11\n12 12\n13 13\n14 14\n15 15\n15 24\n16 5\n16 16\n17 5\n17 10\n17 17\n17 18\n17 22\n17 27\n18 18\n19 19\n20 20\n21 20\n21 21\n22 22\n23 23\n24 14\n24 24\n25 25\n26 8\n26 11\n26 26\n27 3\n27 27\n28 28",
"output": "5"
},
{
"input": "3\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "6\n4 4\n3 4\n5 4\n4 5\n4 3\n3 1",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n1 2",
"output": "0"
},
{
"input": "8\n1 3\n1 1\n4 1\n2 2\n2 5\n5 9\n5 1\n5 4",
"output": "1"
},
{
"input": "10\n1 1\n1 2\n1 3\n1 4\n5 5\n6 6\n7 7\n8 8\n9 9\n100 100",
"output": "6"
},
{
"input": "7\n1 1\n2 2\n3 3\n4 4\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "6\n1 1\n2 1\n2 2\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "4\n3 1\n2 1\n2 2\n1 2",
"output": "0"
},
{
"input": "6\n1 1\n2 2\n2 1\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "3\n1 2\n1 3\n1 4",
"output": "0"
},
{
"input": "4\n1 1\n2 2\n1 2\n2 1",
"output": "0"
},
{
"input": "4\n1 3\n2 1\n3 2\n3 1",
"output": "1"
},
{
"input": "7\n1 1\n1 2\n2 2\n3 3\n3 4\n4 4\n1 4",
"output": "0"
},
{
"input": "21\n12 12\n13 12\n12 11\n13 13\n10 10\n11 10\n11 11\n501 500\n501 501\n503 502\n500 500\n503 503\n502 501\n502 502\n700 700\n702 702\n703 702\n701 701\n702 701\n703 703\n701 700",
"output": "2"
},
{
"input": "6\n1 11\n6 8\n11 10\n1 10\n11 11\n6 9",
"output": "1"
},
{
"input": "4\n1 1\n2 2\n3 2\n3 1",
"output": "0"
},
{
"input": "3\n1 2\n3 4\n3 2",
"output": "0"
},
{
"input": "3\n1 1\n1 2\n2 2",
"output": "0"
},
{
"input": "4\n5 5\n5 4\n6 3\n6 4",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n2 1",
"output": "0"
}
] | 1,544,849,101
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 186
| 0
|
numberofsnowdrift = int(input())
snowDrifts = [[] for sd in range(numberofsnowdrift)]
snowDriftsX = [[] for i in range(101)]
snowDriftsY = [[] for i in range(101)]
connections = [i for i in range(numberofsnowdrift)]
ranks = [0 for i in range(numberofsnowdrift)]
def findSet(friends , p):
if friends[p] != p:
friends[p] = findSet(friends, friends[p])
return friends[p]
def Union(connections, ranks, p1, p2):
p1F = findSet(connections, p1)
p2F = findSet(connections, p2)
if p1F == p2F:
return
if ranks[p1F] > ranks[p2F]:
connections[p2F] = p1F
elif ranks[p1F] < ranks[p2F]:
connections[p1F] = p2F
else:
connections[p1F] = p2F
ranks[p2F] += 1
for sd in range(numberofsnowdrift):
x, y = map(int,input().split())
snowDrifts[sd] = [x, y]
if len(snowDriftsX[x]) > 0:
Union(connections, ranks, sd, snowDriftsX[x][0])
elif len(snowDriftsY[y]) > 0:
Union(connections, ranks, sd, snowDriftsY[y][0])
snowDriftsX[x].append(sd)
snowDriftsY[y].append(sd)
count = 0
for sd in range(numberofsnowdrift):
if(connections[sd] == sd):
count += 1
print(count - 1)
|
Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
numberofsnowdrift = int(input())
snowDrifts = [[] for sd in range(numberofsnowdrift)]
snowDriftsX = [[] for i in range(101)]
snowDriftsY = [[] for i in range(101)]
connections = [i for i in range(numberofsnowdrift)]
ranks = [0 for i in range(numberofsnowdrift)]
def findSet(friends , p):
if friends[p] != p:
friends[p] = findSet(friends, friends[p])
return friends[p]
def Union(connections, ranks, p1, p2):
p1F = findSet(connections, p1)
p2F = findSet(connections, p2)
if p1F == p2F:
return
if ranks[p1F] > ranks[p2F]:
connections[p2F] = p1F
elif ranks[p1F] < ranks[p2F]:
connections[p1F] = p2F
else:
connections[p1F] = p2F
ranks[p2F] += 1
for sd in range(numberofsnowdrift):
x, y = map(int,input().split())
snowDrifts[sd] = [x, y]
if len(snowDriftsX[x]) > 0:
Union(connections, ranks, sd, snowDriftsX[x][0])
elif len(snowDriftsY[y]) > 0:
Union(connections, ranks, sd, snowDriftsY[y][0])
snowDriftsX[x].append(sd)
snowDriftsY[y].append(sd)
count = 0
for sd in range(numberofsnowdrift):
if(connections[sd] == sd):
count += 1
print(count - 1)
```
| -1
|
|
96
|
A
|
Football
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] |
A. Football
|
2
|
256
|
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
|
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
|
Print "YES" if the situation is dangerous. Otherwise, print "NO".
|
[
"001001\n",
"1000000001\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,694,002,287
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
n = input()
seven = 0
for i in n:
if int(i) == 1:
seven += 1
if seven == 7:
result = "YES"
break
else:
seven = 0
result = "NO"
print(result)
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
n = input()
seven = 0
for i in n:
if int(i) == 1:
seven += 1
if seven == 7:
result = "YES"
break
else:
seven = 0
result = "NO"
print(result)
```
| 0
|
47
|
A
|
Triangular numbers
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Triangular numbers
|
2
|
256
|
A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).
Your task is to find out if a given integer is a triangular number.
|
The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer.
|
If the given integer is a triangular number output YES, otherwise output NO.
|
[
"1\n",
"2\n",
"3\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "YES"
},
{
"input": "2",
"output": "NO"
},
{
"input": "3",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "NO"
},
{
"input": "8",
"output": "NO"
},
{
"input": "12",
"output": "NO"
},
{
"input": "10",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "9",
"output": "NO"
},
{
"input": "14",
"output": "NO"
},
{
"input": "15",
"output": "YES"
},
{
"input": "16",
"output": "NO"
},
{
"input": "20",
"output": "NO"
},
{
"input": "21",
"output": "YES"
},
{
"input": "22",
"output": "NO"
},
{
"input": "121",
"output": "NO"
},
{
"input": "135",
"output": "NO"
},
{
"input": "136",
"output": "YES"
},
{
"input": "137",
"output": "NO"
},
{
"input": "152",
"output": "NO"
},
{
"input": "153",
"output": "YES"
},
{
"input": "154",
"output": "NO"
},
{
"input": "171",
"output": "YES"
},
{
"input": "189",
"output": "NO"
},
{
"input": "190",
"output": "YES"
},
{
"input": "191",
"output": "NO"
},
{
"input": "210",
"output": "YES"
},
{
"input": "211",
"output": "NO"
},
{
"input": "231",
"output": "YES"
},
{
"input": "232",
"output": "NO"
},
{
"input": "252",
"output": "NO"
},
{
"input": "253",
"output": "YES"
},
{
"input": "254",
"output": "NO"
},
{
"input": "275",
"output": "NO"
},
{
"input": "276",
"output": "YES"
},
{
"input": "277",
"output": "NO"
},
{
"input": "299",
"output": "NO"
},
{
"input": "300",
"output": "YES"
},
{
"input": "301",
"output": "NO"
},
{
"input": "324",
"output": "NO"
},
{
"input": "325",
"output": "YES"
},
{
"input": "326",
"output": "NO"
},
{
"input": "350",
"output": "NO"
},
{
"input": "351",
"output": "YES"
},
{
"input": "352",
"output": "NO"
},
{
"input": "377",
"output": "NO"
},
{
"input": "378",
"output": "YES"
},
{
"input": "406",
"output": "YES"
},
{
"input": "434",
"output": "NO"
},
{
"input": "435",
"output": "YES"
},
{
"input": "464",
"output": "NO"
},
{
"input": "465",
"output": "YES"
},
{
"input": "495",
"output": "NO"
},
{
"input": "496",
"output": "YES"
},
{
"input": "497",
"output": "NO"
},
{
"input": "331",
"output": "NO"
},
{
"input": "413",
"output": "NO"
},
{
"input": "177",
"output": "NO"
},
{
"input": "122",
"output": "NO"
},
{
"input": "478",
"output": "NO"
},
{
"input": "188",
"output": "NO"
},
{
"input": "21",
"output": "YES"
},
{
"input": "474",
"output": "NO"
},
{
"input": "21",
"output": "YES"
},
{
"input": "48",
"output": "NO"
},
{
"input": "498",
"output": "NO"
},
{
"input": "499",
"output": "NO"
},
{
"input": "500",
"output": "NO"
}
] | 1,690,828,892
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
from math import isqrt
n = int(input())
n *= 2
for i in range(isqrt(n)):
if i*(i+1) == n:
print("YES")
exit()
print("NO")
|
Title: Triangular numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).
Your task is to find out if a given integer is a triangular number.
Input Specification:
The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer.
Output Specification:
If the given integer is a triangular number output YES, otherwise output NO.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
from math import isqrt
n = int(input())
n *= 2
for i in range(isqrt(n)):
if i*(i+1) == n:
print("YES")
exit()
print("NO")
```
| 0
|
792
|
B
|
Counting-out Rhyme
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
*n* children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to *n*. In the beginning, the first child is considered the leader. The game is played in *k* steps. In the *i*-th step the leader counts out *a**i* people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.
For example, if there are children with numbers [8,<=10,<=13,<=14,<=16] currently in the circle, the leader is child 13 and *a**i*<==<=12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.
You have to write a program which prints the number of the child to be eliminated on every step.
|
The first line contains two integer numbers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1).
The next line contains *k* integer numbers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=109).
|
Print *k* numbers, the *i*-th one corresponds to the number of child to be eliminated at the *i*-th step.
|
[
"7 5\n10 4 11 4 1\n",
"3 2\n2 5\n"
] |
[
"4 2 5 6 1 \n",
"3 2 \n"
] |
Let's consider first example:
- In the first step child 4 is eliminated, child 5 becomes the leader. - In the second step child 2 is eliminated, child 3 becomes the leader. - In the third step child 5 is eliminated, child 6 becomes the leader. - In the fourth step child 6 is eliminated, child 7 becomes the leader. - In the final step child 1 is eliminated, child 3 becomes the leader.
| 0
|
[
{
"input": "7 5\n10 4 11 4 1",
"output": "4 2 5 6 1 "
},
{
"input": "3 2\n2 5",
"output": "3 2 "
},
{
"input": "2 1\n1",
"output": "2 "
},
{
"input": "2 1\n2",
"output": "1 "
},
{
"input": "2 1\n3",
"output": "2 "
},
{
"input": "10 7\n5 10 4 3 8 10 6",
"output": "6 8 3 9 2 4 10 "
},
{
"input": "10 8\n12 6 12 15 20 8 17 12",
"output": "3 10 6 8 2 9 4 5 "
},
{
"input": "12 10\n76 58 82 54 97 46 17 40 36 15",
"output": "5 9 12 1 3 10 8 11 2 4 "
},
{
"input": "12 6\n76 61 94 15 66 26",
"output": "5 12 6 2 7 3 "
},
{
"input": "90 10\n1045 8705 6077 3282 1459 9809 383 6206 2674 7274",
"output": "56 39 45 20 17 55 14 85 51 33 "
},
{
"input": "100 30\n601771 913885 829106 91674 465657 367068 142461 873149 294276 916519 720701 370006 551782 321506 68525 570684 81178 724855 564907 661130 10112 983124 799801 100639 766045 862312 513021 232094 979480 408554",
"output": "72 89 16 26 85 73 29 99 63 30 8 46 70 19 100 93 36 54 65 77 17 79 62 64 21 69 42 82 68 1 "
},
{
"input": "3 2\n20148340 81473314",
"output": "2 3 "
},
{
"input": "3 2\n301633543 643389490",
"output": "2 3 "
},
{
"input": "6 5\n532623340 628883728 583960589 690950241 488468353",
"output": "5 3 6 1 4 "
},
{
"input": "6 2\n458995521 294343587",
"output": "4 1 "
},
{
"input": "68 1\n5",
"output": "6 "
}
] | 1,689,411,082
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 7
| 61
| 0
|
n, k = map(int, input().split())
a = [int(x) for x in input().split()]
ch = list(range(n+1))
#print(ch)
leader = 1
def findnext(pos):
p = pos + 1
if p == n+1:
p = 1
while ch[p]==0:
p = p + 1
if p == n+1:
p = 1
return p
elim = []
for j in range(k):
a[j] %= (n-j) # children left
# print(ch)
# print("a[j]", a[j], "leader", leader)
if a[j]==0:
elim.append(leader)
leader = findnext(leader)
continue
pos = findnext(leader)
for i in range(1, a[j]):
pos = findnext(pos)
elim.append(pos)
ch[pos] = 0
leader = findnext(pos)
print(*elim)
|
Title: Counting-out Rhyme
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to *n*. In the beginning, the first child is considered the leader. The game is played in *k* steps. In the *i*-th step the leader counts out *a**i* people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.
For example, if there are children with numbers [8,<=10,<=13,<=14,<=16] currently in the circle, the leader is child 13 and *a**i*<==<=12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.
You have to write a program which prints the number of the child to be eliminated on every step.
Input Specification:
The first line contains two integer numbers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1).
The next line contains *k* integer numbers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print *k* numbers, the *i*-th one corresponds to the number of child to be eliminated at the *i*-th step.
Demo Input:
['7 5\n10 4 11 4 1\n', '3 2\n2 5\n']
Demo Output:
['4 2 5 6 1 \n', '3 2 \n']
Note:
Let's consider first example:
- In the first step child 4 is eliminated, child 5 becomes the leader. - In the second step child 2 is eliminated, child 3 becomes the leader. - In the third step child 5 is eliminated, child 6 becomes the leader. - In the fourth step child 6 is eliminated, child 7 becomes the leader. - In the final step child 1 is eliminated, child 3 becomes the leader.
|
```python
n, k = map(int, input().split())
a = [int(x) for x in input().split()]
ch = list(range(n+1))
#print(ch)
leader = 1
def findnext(pos):
p = pos + 1
if p == n+1:
p = 1
while ch[p]==0:
p = p + 1
if p == n+1:
p = 1
return p
elim = []
for j in range(k):
a[j] %= (n-j) # children left
# print(ch)
# print("a[j]", a[j], "leader", leader)
if a[j]==0:
elim.append(leader)
leader = findnext(leader)
continue
pos = findnext(leader)
for i in range(1, a[j]):
pos = findnext(pos)
elim.append(pos)
ch[pos] = 0
leader = findnext(pos)
print(*elim)
```
| 0
|
|
268
|
B
|
Buttons
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.
Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.
Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.
|
A single line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of buttons the lock has.
|
In a single line print the number of times Manao has to push a button in the worst-case scenario.
|
[
"2\n",
"3\n"
] |
[
"3\n",
"7\n"
] |
Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.
| 1,000
|
[
{
"input": "2",
"output": "3"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "14"
},
{
"input": "1",
"output": "1"
},
{
"input": "10",
"output": "175"
},
{
"input": "2000",
"output": "1333335000"
},
{
"input": "1747",
"output": "888644743"
},
{
"input": "889",
"output": "117099969"
},
{
"input": "1999",
"output": "1331335999"
},
{
"input": "914",
"output": "127259419"
},
{
"input": "996",
"output": "164675486"
},
{
"input": "17",
"output": "833"
},
{
"input": "50",
"output": "20875"
},
{
"input": "91",
"output": "125671"
},
{
"input": "92",
"output": "129858"
},
{
"input": "256",
"output": "2796416"
},
{
"input": "512",
"output": "22370048"
},
{
"input": "666",
"output": "49235271"
},
{
"input": "667",
"output": "49457383"
},
{
"input": "314",
"output": "5160119"
},
{
"input": "1241",
"output": "318541121"
},
{
"input": "1500",
"output": "562501250"
},
{
"input": "1837",
"output": "1033182073"
},
{
"input": "1000",
"output": "166667500"
}
] | 1,688,488,775
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 60
| 0
|
from math import *
n = int(input())
ans = 0
for i in range(1,n+1):
ans+= comb(n,i)
print(ans)
|
Title: Buttons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.
Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.
Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.
Input Specification:
A single line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of buttons the lock has.
Output Specification:
In a single line print the number of times Manao has to push a button in the worst-case scenario.
Demo Input:
['2\n', '3\n']
Demo Output:
['3\n', '7\n']
Note:
Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.
|
```python
from math import *
n = int(input())
ans = 0
for i in range(1,n+1):
ans+= comb(n,i)
print(ans)
```
| 0
|
|
392
|
C
|
Yet Another Number Sequence
|
PROGRAMMING
| 0
|
[
"combinatorics",
"math",
"matrices"
] | null | null |
Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation:
We'll define a new number sequence *A**i*(*k*) by the formula:
In this problem, your task is to calculate the following sum: *A*1(*k*)<=+<=*A*2(*k*)<=+<=...<=+<=*A**n*(*k*). The answer can be very large, so print it modulo 1000000007 (109<=+<=7).
|
The first line contains two space-separated integers *n*, *k* (1<=≤<=*n*<=≤<=1017; 1<=≤<=*k*<=≤<=40).
|
Print a single integer — the sum of the first *n* elements of the sequence *A**i*(*k*) modulo 1000000007 (109<=+<=7).
|
[
"1 1\n",
"4 1\n",
"5 2\n",
"7 4\n"
] |
[
"1\n",
"34\n",
"316\n",
"73825\n"
] |
none
| 1,500
|
[] | 1,543,070,705
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 1,000
| 614,400
|
# coding: utf-8
# In[14]:
#Codeforces Round #230 (Div. 1)
#Problem C
n , k = input().split()
n = int(n)
k = int(k)
a = 1
b = 1
i = 1
sum = 0
while(n):
#print(b)
c = a + b
a = b
b = c
#print(b,i)
sum = sum + a * (i**k)
i = i +1
n = n - 1
print(sum%1000000007)
|
Title: Yet Another Number Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation:
We'll define a new number sequence *A**i*(*k*) by the formula:
In this problem, your task is to calculate the following sum: *A*1(*k*)<=+<=*A*2(*k*)<=+<=...<=+<=*A**n*(*k*). The answer can be very large, so print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two space-separated integers *n*, *k* (1<=≤<=*n*<=≤<=1017; 1<=≤<=*k*<=≤<=40).
Output Specification:
Print a single integer — the sum of the first *n* elements of the sequence *A**i*(*k*) modulo 1000000007 (109<=+<=7).
Demo Input:
['1 1\n', '4 1\n', '5 2\n', '7 4\n']
Demo Output:
['1\n', '34\n', '316\n', '73825\n']
Note:
none
|
```python
# coding: utf-8
# In[14]:
#Codeforces Round #230 (Div. 1)
#Problem C
n , k = input().split()
n = int(n)
k = int(k)
a = 1
b = 1
i = 1
sum = 0
while(n):
#print(b)
c = a + b
a = b
b = c
#print(b,i)
sum = sum + a * (i**k)
i = i +1
n = n - 1
print(sum%1000000007)
```
| 0
|
|
604
|
B
|
More Cowbell
|
PROGRAMMING
| 1,400
|
[
"binary search",
"greedy"
] | null | null |
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
|
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
|
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
|
[
"2 1\n2 5\n",
"4 3\n2 3 5 9\n",
"3 2\n3 5 7\n"
] |
[
"7\n",
"9\n",
"8\n"
] |
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
| 1,000
|
[
{
"input": "2 1\n2 5",
"output": "7"
},
{
"input": "4 3\n2 3 5 9",
"output": "9"
},
{
"input": "3 2\n3 5 7",
"output": "8"
},
{
"input": "20 11\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 10\n3 15 31 61 63 63 68 94 98 100",
"output": "100"
},
{
"input": "100 97\n340 402 415 466 559 565 649 689 727 771 774 776 789 795 973 1088 1212 1293 1429 1514 1587 1599 1929 1997 2278 2529 2656 2677 2839 2894 2951 3079 3237 3250 3556 3568 3569 3578 3615 3641 3673 3892 4142 4418 4515 4766 4846 4916 5225 5269 5352 5460 5472 5635 5732 5886 5941 5976 5984 6104 6113 6402 6409 6460 6550 6563 6925 7006 7289 7401 7441 7451 7709 7731 7742 7750 7752 7827 8101 8154 8376 8379 8432 8534 8578 8630 8706 8814 8882 8972 9041 9053 9109 9173 9473 9524 9547 9775 9791 9983",
"output": "9983"
},
{
"input": "10 9\n7 29 35 38 41 47 54 56 73 74",
"output": "74"
},
{
"input": "1 2342\n12345",
"output": "12345"
},
{
"input": "10 5\n15 15 20 28 38 44 46 52 69 94",
"output": "109"
},
{
"input": "10 9\n6 10 10 32 36 38 69 80 82 93",
"output": "93"
},
{
"input": "10 10\n4 19 22 24 25 43 49 56 78 88",
"output": "88"
},
{
"input": "100 89\n474 532 759 772 803 965 1043 1325 1342 1401 1411 1452 1531 1707 1906 1928 2034 2222 2335 2606 2757 2968 2978 3211 3513 3734 3772 3778 3842 3948 3976 4038 4055 4113 4182 4267 4390 4408 4478 4595 4668 4792 4919 5133 5184 5255 5312 5341 5476 5628 5683 5738 5767 5806 5973 6051 6134 6254 6266 6279 6314 6342 6599 6676 6747 6777 6827 6842 7057 7097 7259 7340 7378 7405 7510 7520 7698 7796 8148 8351 8507 8601 8805 8814 8826 8978 9116 9140 9174 9338 9394 9403 9407 9423 9429 9519 9764 9784 9838 9946",
"output": "9946"
},
{
"input": "100 74\n10 211 323 458 490 592 979 981 1143 1376 1443 1499 1539 1612 1657 1874 2001 2064 2123 2274 2346 2471 2522 2589 2879 2918 2933 2952 3160 3164 3167 3270 3382 3404 3501 3522 3616 3802 3868 3985 4007 4036 4101 4580 4687 4713 4714 4817 4955 5257 5280 5343 5428 5461 5566 5633 5727 5874 5925 6233 6309 6389 6500 6701 6731 6847 6916 7088 7088 7278 7296 7328 7564 7611 7646 7887 7887 8065 8075 8160 8300 8304 8316 8355 8404 8587 8758 8794 8890 9038 9163 9235 9243 9339 9410 9587 9868 9916 9923 9986",
"output": "9986"
},
{
"input": "100 61\n82 167 233 425 432 456 494 507 562 681 683 921 1218 1323 1395 1531 1586 1591 1675 1766 1802 1842 2116 2625 2697 2735 2739 3337 3349 3395 3406 3596 3610 3721 4059 4078 4305 4330 4357 4379 4558 4648 4651 4784 4819 4920 5049 5312 5361 5418 5440 5463 5547 5594 5821 5951 5972 6141 6193 6230 6797 6842 6853 6854 7017 7026 7145 7322 7391 7460 7599 7697 7756 7768 7872 7889 8094 8215 8408 8440 8462 8714 8756 8760 8881 9063 9111 9184 9281 9373 9406 9417 9430 9511 9563 9634 9660 9788 9883 9927",
"output": "9927"
},
{
"input": "100 84\n53 139 150 233 423 570 786 861 995 1017 1072 1196 1276 1331 1680 1692 1739 1748 1826 2067 2280 2324 2368 2389 2607 2633 2760 2782 2855 2996 3030 3093 3513 3536 3557 3594 3692 3707 3823 3832 4009 4047 4088 4095 4408 4537 4565 4601 4784 4878 4935 5029 5252 5322 5389 5407 5511 5567 5857 6182 6186 6198 6280 6290 6353 6454 6458 6567 6843 7166 7216 7257 7261 7375 7378 7539 7542 7762 7771 7797 7980 8363 8606 8612 8663 8801 8808 8823 8918 8975 8997 9240 9245 9259 9356 9755 9759 9760 9927 9970",
"output": "9970"
},
{
"input": "100 50\n130 248 312 312 334 589 702 916 921 1034 1047 1346 1445 1500 1585 1744 1951 2123 2273 2362 2400 2455 2496 2530 2532 2944 3074 3093 3094 3134 3698 3967 4047 4102 4109 4260 4355 4466 4617 4701 4852 4892 4915 4917 4936 4981 4999 5106 5152 5203 5214 5282 5412 5486 5525 5648 5897 5933 5969 6251 6400 6421 6422 6558 6805 6832 6908 6924 6943 6980 7092 7206 7374 7417 7479 7546 7672 7756 7973 8020 8028 8079 8084 8085 8137 8153 8178 8239 8639 8667 8829 9263 9333 9370 9420 9579 9723 9784 9841 9993",
"output": "11103"
},
{
"input": "100 50\n156 182 208 409 496 515 659 761 772 794 827 912 1003 1236 1305 1388 1412 1422 1428 1465 1613 2160 2411 2440 2495 2684 2724 2925 3033 3035 3155 3260 3378 3442 3483 3921 4031 4037 4091 4113 4119 4254 4257 4442 4559 4614 4687 4839 4896 5054 5246 5316 5346 5859 5928 5981 6148 6250 6422 6433 6448 6471 6473 6485 6503 6779 6812 7050 7064 7074 7141 7378 7424 7511 7574 7651 7808 7858 8286 8291 8446 8536 8599 8628 8636 8768 8900 8981 9042 9055 9114 9146 9186 9411 9480 9590 9681 9749 9757 9983",
"output": "10676"
},
{
"input": "100 50\n145 195 228 411 577 606 629 775 1040 1040 1058 1187 1307 1514 1784 1867 1891 2042 2042 2236 2549 2555 2560 2617 2766 2807 2829 2917 3070 3072 3078 3095 3138 3147 3149 3196 3285 3287 3309 3435 3531 3560 3563 3769 3830 3967 4081 4158 4315 4387 4590 4632 4897 4914 5128 5190 5224 5302 5402 5416 5420 5467 5517 5653 5820 5862 5941 6053 6082 6275 6292 6316 6490 6530 6619 6632 6895 7071 7234 7323 7334 7412 7626 7743 8098 8098 8136 8158 8264 8616 8701 8718 8770 8803 8809 8983 9422 9530 9811 9866",
"output": "10011"
},
{
"input": "100 50\n56 298 387 456 518 532 589 792 870 1041 1055 1122 1141 1166 1310 1329 1523 1548 1626 1730 1780 1833 1850 1911 2006 2157 2303 2377 2403 2442 2450 2522 2573 2822 2994 3200 3238 3252 3280 3311 3345 3422 3429 3506 3526 3617 3686 3791 4134 4467 4525 4614 4633 4792 5017 5220 5243 5338 5445 5536 5639 5675 5763 5875 6129 6220 6228 6287 6385 6616 6789 6822 6940 6959 6985 7297 7304 7391 7443 7580 7824 7884 7981 8055 8063 8223 8280 8322 8346 8473 8688 8986 9087 9323 9477 9489 9518 9621 9699 9804",
"output": "10009"
},
{
"input": "100 49990\n221 257 543 613 818 855 884 1195 1233 1293 1327 1474 1567 1568 1728 1789 1809 1957 1994 2198 2217 2289 2298 2360 2441 2454 2517 2519 2597 2652 2909 2958 2970 3271 3310 3409 3450 3646 3950 3965 4016 4116 4134 4159 4462 4624 4631 4783 4804 5031 5276 5538 5729 5845 5880 5894 5959 5961 6028 6105 6184 6335 6370 6523 6532 6677 6762 6868 6998 7066 7245 7360 7490 7492 7645 7677 7786 7804 8038 8131 8207 8400 8507 8556 8575 8599 8673 8676 8712 8816 9123 9158 9216 9369 9628 9707 9768 9800 9876 9934",
"output": "9934"
},
{
"input": "1 50000\n2",
"output": "2"
},
{
"input": "1 40000\n23456",
"output": "23456"
},
{
"input": "1 2\n10",
"output": "10"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n2 3",
"output": "3"
},
{
"input": "4 2\n1 2 8 9",
"output": "10"
}
] | 1,620,482,909
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 8
| 155
| 30,310,400
|
import math
n, k = map(int, input().split())
arr = [int(z) for z in input().split()]
pairs = math.ceil(k/n)
singles = n - 2*pairs
pairedup = arr[:n-singles]
print(max(arr[-1], pairedup[0] + pairedup[-1]))
|
Title: More Cowbell
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
Input Specification:
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
Output Specification:
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
Demo Input:
['2 1\n2 5\n', '4 3\n2 3 5 9\n', '3 2\n3 5 7\n']
Demo Output:
['7\n', '9\n', '8\n']
Note:
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
|
```python
import math
n, k = map(int, input().split())
arr = [int(z) for z in input().split()]
pairs = math.ceil(k/n)
singles = n - 2*pairs
pairedup = arr[:n-singles]
print(max(arr[-1], pairedup[0] + pairedup[-1]))
```
| 0
|
|
607
|
A
|
Chain Reaction
|
PROGRAMMING
| 1,600
|
[
"binary search",
"dp"
] | null | null |
There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.
Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons.
The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*.
|
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.
|
[
"4\n1 9\n3 1\n6 1\n7 4\n",
"7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n"
] |
[
"1\n",
"3\n"
] |
For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2.
For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.
| 500
|
[
{
"input": "4\n1 9\n3 1\n6 1\n7 4",
"output": "1"
},
{
"input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1",
"output": "3"
},
{
"input": "1\n0 1",
"output": "0"
},
{
"input": "1\n0 1000000",
"output": "0"
},
{
"input": "1\n1000000 1000000",
"output": "0"
},
{
"input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 6\n7 7",
"output": "4"
},
{
"input": "5\n1 1\n3 1\n5 1\n7 10\n8 10",
"output": "2"
},
{
"input": "11\n110 90\n100 70\n90 10\n80 10\n70 1\n60 1\n50 10\n40 1\n30 1\n10 1\n20 1",
"output": "4"
}
] | 1,628,234,717
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 1,326
| 89,600,000
|
from collections import defaultdict
n=int(input())
pos=defaultdict(lambda:False)
pw=defaultdict(lambda:0)
dp=defaultdict(lambda:0)
ans=0;
maxdist=0
for _ in range(n):
a,b=map(int,input().split())
pos[a]=True;
pw[a]=b;
maxdist=max(maxdist,a);
for x in range(maxdist+1):
if(pos[x]):
dp[x]=1+dp[x-pw[x]-1]
else:
dp[x]=dp[x-1]
ans=0
for x in range(maxdist+1):
ans=max(ans,dp[x])
#print(pos)
#print(pw)
#print(dp)
ans=n-ans
print(ans)
|
Title: Chain Reaction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.
Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons.
The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*.
Output Specification:
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.
Demo Input:
['4\n1 9\n3 1\n6 1\n7 4\n', '7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n']
Demo Output:
['1\n', '3\n']
Note:
For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2.
For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.
|
```python
from collections import defaultdict
n=int(input())
pos=defaultdict(lambda:False)
pw=defaultdict(lambda:0)
dp=defaultdict(lambda:0)
ans=0;
maxdist=0
for _ in range(n):
a,b=map(int,input().split())
pos[a]=True;
pw[a]=b;
maxdist=max(maxdist,a);
for x in range(maxdist+1):
if(pos[x]):
dp[x]=1+dp[x-pw[x]-1]
else:
dp[x]=dp[x-1]
ans=0
for x in range(maxdist+1):
ans=max(ans,dp[x])
#print(pos)
#print(pw)
#print(dp)
ans=n-ans
print(ans)
```
| 3
|
|
900
|
C
|
Remove Extra One
|
PROGRAMMING
| 1,700
|
[
"brute force",
"data structures",
"math"
] | null | null |
You are given a permutation *p* of length *n*. Remove one element from permutation to make the number of records the maximum possible.
We remind that in a sequence of numbers *a*1,<=*a*2,<=...,<=*a**k* the element *a**i* is a record if for every integer *j* (1<=≤<=*j*<=<<=*i*) the following holds: *a**j*<=<<=*a**i*.
|
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=105) — the length of the permutation.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the permutation. All the integers are distinct.
|
Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.
|
[
"1\n1\n",
"5\n5 1 2 3 4\n"
] |
[
"1\n",
"5\n"
] |
In the first example the only element can be removed.
| 1,500
|
[
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n5 1 2 3 4",
"output": "5"
},
{
"input": "5\n4 3 5 1 2",
"output": "1"
},
{
"input": "9\n9 5 8 6 3 2 4 1 7",
"output": "9"
},
{
"input": "3\n3 2 1",
"output": "1"
},
{
"input": "7\n1 6 7 4 2 5 3",
"output": "2"
},
{
"input": "48\n38 6 31 19 45 28 27 43 11 35 36 20 9 16 42 48 14 22 39 18 12 10 34 25 13 26 40 29 17 8 33 46 24 30 37 44 1 15 2 21 3 5 4 47 32 23 41 7",
"output": "38"
},
{
"input": "26\n23 14 15 19 9 22 20 12 5 4 21 1 16 8 6 11 3 17 2 10 24 26 13 18 25 7",
"output": "23"
},
{
"input": "46\n32 25 11 1 3 10 8 12 18 42 28 16 35 30 41 38 43 4 13 23 6 17 36 34 39 22 26 14 45 20 33 44 21 7 15 5 40 46 2 29 37 9 31 19 27 24",
"output": "42"
},
{
"input": "24\n20 3 22 10 2 14 7 18 6 23 17 12 5 11 15 13 19 24 16 1 21 4 8 9",
"output": "1"
},
{
"input": "57\n40 11 43 39 13 29 18 57 54 48 17 4 22 5 38 15 36 53 33 3 51 41 30 9 26 10 55 27 35 56 23 20 1 8 12 46 21 28 6 19 34 2 45 31 49 42 50 16 44 7 25 52 14 32 47 37 24",
"output": "57"
},
{
"input": "85\n82 72 24 38 81 18 49 62 37 28 41 57 10 55 83 67 56 2 73 44 26 85 78 14 27 40 51 61 54 29 16 25 5 31 71 42 21 30 3 74 6 63 76 33 39 68 66 23 53 20 22 43 45 52 80 60 1 59 50 58 12 77 65 36 15 19 46 17 79 9 47 8 70 75 34 7 69 32 4 84 64 35 11 13 48",
"output": "82"
},
{
"input": "5\n2 3 4 1 5",
"output": "1"
},
{
"input": "87\n66 53 79 35 24 61 22 70 29 43 6 21 75 4 85 2 37 18 65 49 40 82 58 73 33 87 71 19 34 83 84 25 56 48 9 63 38 20 67 32 74 42 51 39 11 1 78 86 44 64 81 17 62 72 47 54 52 23 7 5 41 46 3 28 77 57 13 15 59 68 14 36 50 27 80 31 26 10 55 60 69 76 16 12 8 45 30",
"output": "79"
},
{
"input": "92\n42 64 33 89 57 9 24 44 87 67 92 84 39 88 26 27 85 62 22 83 23 71 14 13 73 79 15 49 2 12 76 53 81 40 31 3 72 58 1 61 7 82 20 54 46 77 11 16 28 48 6 45 36 43 60 38 18 4 32 74 10 91 19 86 75 51 50 52 78 25 65 8 55 30 90 69 59 63 56 80 29 68 70 17 35 41 37 47 66 34 5 21",
"output": "1"
},
{
"input": "5\n1 2 3 4 5",
"output": "1"
}
] | 1,514,918,399
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n=int(input())
A=[int(i) for i in input().split(" ")]
if n=1 for n==2:
print(min(A))
else:
rem=[n]*n
|
Title: Remove Extra One
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a permutation *p* of length *n*. Remove one element from permutation to make the number of records the maximum possible.
We remind that in a sequence of numbers *a*1,<=*a*2,<=...,<=*a**k* the element *a**i* is a record if for every integer *j* (1<=≤<=*j*<=<<=*i*) the following holds: *a**j*<=<<=*a**i*.
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=105) — the length of the permutation.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the permutation. All the integers are distinct.
Output Specification:
Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.
Demo Input:
['1\n1\n', '5\n5 1 2 3 4\n']
Demo Output:
['1\n', '5\n']
Note:
In the first example the only element can be removed.
|
```python
n=int(input())
A=[int(i) for i in input().split(" ")]
if n=1 for n==2:
print(min(A))
else:
rem=[n]*n
```
| -1
|
|
629
|
A
|
Far Relative’s Birthday Cake
|
PROGRAMMING
| 800
|
[
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null |
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
|
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
|
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
|
[
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] |
[
"4\n",
"9\n"
] |
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)
| 500
|
[
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},
{
"input": "8\n..C....C\nC.CCC.CC\n.C..C.CC\nCC......\nC..C..CC\nC.C...C.\nC.C..C..\nC...C.C.",
"output": "80"
},
{
"input": "9\n.C...CCCC\nC.CCCC...\n....C..CC\n.CC.CCC..\n.C.C..CC.\nC...C.CCC\nCCC.C...C\nCCCC....C\n..C..C..C",
"output": "144"
},
{
"input": "10\n..C..C.C..\n..CC..C.CC\n.C.C...C.C\n..C.CC..CC\n....C..C.C\n...C..C..C\nCC.CC....C\n..CCCC.C.C\n..CC.CCC..\nCCCC..C.CC",
"output": "190"
},
{
"input": "11\nC.CC...C.CC\nCC.C....C.C\n.....C..CCC\n....C.CC.CC\nC..C..CC...\nC...C...C..\nCC..CCC.C.C\n..C.CC.C..C\nC...C.C..CC\n.C.C..CC..C\n.C.C.CC.C..",
"output": "228"
},
{
"input": "21\n...CCC.....CC..C..C.C\n..CCC...CC...CC.CCC.C\n....C.C.C..CCC..C.C.C\n....CCC..C..C.CC.CCC.\n...CCC.C..C.C.....CCC\n.CCC.....CCC..C...C.C\nCCCC.C...CCC.C...C.CC\nC..C...C.CCC..CC..C..\nC...CC..C.C.CC..C.CC.\nCC..CCCCCCCCC..C....C\n.C..CCCC.CCCC.CCC...C\nCCC...CCC...CCC.C..C.\n.CCCCCCCC.CCCC.CC.C..\n.C.C..C....C.CCCCCC.C\n...C...C.CCC.C.CC..C.\nCCC...CC..CC...C..C.C\n.CCCCC...C.C..C.CC.C.\n..CCC.C.C..CCC.CCC...\n..C..C.C.C.....CC.C..\n.CC.C...C.CCC.C....CC\n...C..CCCC.CCC....C..",
"output": "2103"
},
{
"input": "20\nC.C.CCC.C....C.CCCCC\nC.CC.C..CCC....CCCC.\n.CCC.CC...CC.CCCCCC.\n.C...CCCC..C....CCC.\n.C..CCCCCCC.C.C.....\nC....C.C..CCC.C..CCC\n...C.C.CC..CC..CC...\nC...CC.C.CCCCC....CC\n.CC.C.CCC....C.CCC.C\nCC...CC...CC..CC...C\nC.C..CC.C.CCCC.C.CC.\n..CCCCC.C.CCC..CCCC.\n....C..C..C.CC...C.C\nC..CCC..CC..C.CC..CC\n...CC......C.C..C.C.\nCC.CCCCC.CC.CC...C.C\n.C.CC..CC..CCC.C.CCC\nC..C.CC....C....C...\n..CCC..CCC...CC..C.C\n.C.CCC.CCCCCCCCC..CC",
"output": "2071"
},
{
"input": "17\nCCC..C.C....C.C.C\n.C.CC.CC...CC..C.\n.CCCC.CC.C..CCC.C\n...CCC.CC.CCC.C.C\nCCCCCCCC..C.CC.CC\n...C..C....C.CC.C\nCC....CCC...C.CC.\n.CC.C.CC..C......\n.CCCCC.C.CC.CCCCC\n..CCCC...C..CC..C\nC.CC.C.CC..C.C.C.\nC..C..C..CCC.C...\n.C..CCCC..C......\n.CC.C...C..CC.CC.\nC..C....CC...CC..\nC.CC.CC..C.C..C..\nCCCC...C.C..CCCC.",
"output": "1160"
},
{
"input": "15\nCCCC.C..CCC....\nCCCCCC.CC.....C\n...C.CC.C.C.CC.\nCCCCCCC..C..C..\nC..CCC..C.CCCC.\n.CC..C.C.C.CC.C\n.C.C..C..C.C..C\n...C...C..CCCC.\n.....C.C..CC...\nCC.C.C..CC.C..C\n..CCCCC..CCC...\nCC.CC.C..CC.CCC\n..CCC...CC.C..C\nCC..C.C..CCC..C\n.C.C....CCC...C",
"output": "789"
},
{
"input": "1\n.",
"output": "0"
},
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "13\nC.C...C.C.C..\nCC.CCCC.CC..C\n.C.CCCCC.CC..\nCCCC..C...C..\n...CC.C.C...C\n.CC.CCC...CC.\nCC.CCCCCC....\n.C...C..CC..C\nCCCC.CC...C..\n.C.CCC..C.CC.\n..C...CC..C.C\n..C.CCC..CC.C\n.C...CCC.CC.C",
"output": "529"
},
{
"input": "16\n.C.C.C.C.C...C.C\n..C..C.CCCCCC...\n..C.C.C.C..C..C.\n.CC....C.CCC..C.\n.C.CCC..C....CCC\nCC..C.CC..C.C.CC\n...C..C..CC..CC.\n.CCC..C.CC.C.C..\n.CC.C..........C\nC...C....CC..C..\nC.CCC.C..C..C...\n.CCCCCCCCCCCC..C\n..C.C.CC.CC.CCC.\nCC..C.C....C..CC\nC.CCC..C..C.C.CC\n.C.CCC.CC..CCC.C",
"output": "874"
},
{
"input": "2\nCC\nCC",
"output": "4"
},
{
"input": "3\nC..\nC..\nC..",
"output": "3"
}
] | 1,455,986,893
| 793
|
Python 3
|
OK
|
TESTS
| 48
| 62
| 0
|
n = int(input())
grid = [input() for i in range(n)]
ans = 0
for i in range(n):
c = grid[i].count('C')
ans += c * (c - 1) // 2
c = 0
for j in range(n):
if grid[j][i] == 'C':
c += 1
ans += c * (c - 1) // 2
print(ans)
|
Title: Far Relative’s Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)
|
```python
n = int(input())
grid = [input() for i in range(n)]
ans = 0
for i in range(n):
c = grid[i].count('C')
ans += c * (c - 1) // 2
c = 0
for j in range(n):
if grid[j][i] == 'C':
c += 1
ans += c * (c - 1) // 2
print(ans)
```
| 3
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,688,981,484
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
s = set(input())
l = s.copy()
t =0
sum = 0
for x in s:
l.discard(x)
l.add(x)
l.discard(',')
print(len(l))
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
s = set(input())
l = s.copy()
t =0
sum = 0
for x in s:
l.discard(x)
l.add(x)
l.discard(',')
print(len(l))
```
| 0
|
|
214
|
B
|
Hometask
|
PROGRAMMING
| 1,600
|
[
"brute force",
"constructive algorithms",
"greedy",
"math"
] | null | null |
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a set of digits, your task is to find the maximum integer that you can make from these digits. The made number must be divisible by 2, 3, 5 without a residue. It is permitted to use not all digits from the set, it is forbidden to use leading zeroes.
Each digit is allowed to occur in the number the same number of times it occurs in the set.
|
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of digits in the set. The second line contains *n* digits, the digits are separated by a single space.
|
On a single line print the answer to the problem. If such number does not exist, then you should print -1.
|
[
"1\n0\n",
"11\n3 4 5 4 5 3 5 3 4 4 0\n",
"8\n3 2 5 1 5 2 2 3\n"
] |
[
"0\n",
"5554443330\n",
"-1\n"
] |
In the first sample there is only one number you can make — 0. In the second sample the sought number is 5554443330. In the third sample it is impossible to make the required number.
| 1,000
|
[
{
"input": "1\n0",
"output": "0"
},
{
"input": "11\n3 4 5 4 5 3 5 3 4 4 0",
"output": "5554443330"
},
{
"input": "8\n3 2 5 1 5 2 2 3",
"output": "-1"
},
{
"input": "12\n5 3 3 3 2 5 5 1 2 1 4 1",
"output": "-1"
},
{
"input": "8\n5 5 4 1 5 5 5 3",
"output": "-1"
},
{
"input": "12\n3 1 2 3 2 0 2 2 2 0 2 3",
"output": "33322222200"
},
{
"input": "12\n5 1 4 4 2 1 7 7 4 2 5 1",
"output": "-1"
},
{
"input": "5\n3 6 1 6 2",
"output": "-1"
},
{
"input": "11\n3 9 9 6 4 3 6 4 9 6 0",
"output": "999666330"
},
{
"input": "5\n9 6 6 6 1",
"output": "-1"
},
{
"input": "10\n2 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "10\n1 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "5\n1 1 0 0 0",
"output": "0"
},
{
"input": "5\n0 0 2 2 0",
"output": "0"
},
{
"input": "6\n3 3 2 2 2 0",
"output": "332220"
},
{
"input": "7\n3 3 2 2 2 2 0",
"output": "332220"
},
{
"input": "6\n0 3 3 1 1 1",
"output": "331110"
},
{
"input": "7\n0 3 3 1 1 1 1",
"output": "331110"
},
{
"input": "7\n0 3 3 4 4 4 4",
"output": "444330"
},
{
"input": "7\n0 3 3 2 2 4 4",
"output": "4433220"
},
{
"input": "7\n4 2 3 3 0 0 0",
"output": "4332000"
},
{
"input": "4\n1 1 0 3",
"output": "30"
},
{
"input": "4\n3 0 2 2",
"output": "30"
},
{
"input": "8\n3 3 3 5 5 0 0 0",
"output": "333000"
},
{
"input": "8\n3 3 6 3 0 7 7 9",
"output": "963330"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "-1"
},
{
"input": "9\n9 9 9 9 9 9 9 9 9",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n9 0",
"output": "90"
},
{
"input": "10\n3 0 2 2 2 2 2 2 2 2",
"output": "32222220"
},
{
"input": "10\n3 0 1 1 1 1 1 1 1 1",
"output": "31111110"
},
{
"input": "10\n3 0 4 4 4 4 4 4 4 4",
"output": "44444430"
},
{
"input": "10\n2 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "10\n2 2 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "4\n5 5 0 0",
"output": "0"
},
{
"input": "3\n1 4 0",
"output": "0"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "4\n0 1 4 3",
"output": "30"
},
{
"input": "3\n2 0 0",
"output": "0"
},
{
"input": "4\n0 1 2 3",
"output": "3210"
},
{
"input": "4\n1 0 0 0",
"output": "0"
},
{
"input": "5\n8 0 0 0 0",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 8 5 6",
"output": "600"
},
{
"input": "4\n5 8 3 0",
"output": "30"
},
{
"input": "4\n1 4 0 0",
"output": "0"
},
{
"input": "3\n0 0 1",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "3\n1 0 0",
"output": "0"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "3\n0 0 4",
"output": "0"
},
{
"input": "2\n0 1",
"output": "0"
},
{
"input": "4\n1 1 0 0",
"output": "0"
},
{
"input": "6\n2 2 0 0 0 0",
"output": "0"
},
{
"input": "5\n3 2 5 0 0",
"output": "300"
},
{
"input": "4\n5 3 2 0",
"output": "30"
},
{
"input": "5\n0 0 0 2 2",
"output": "0"
},
{
"input": "5\n0 0 0 0 1",
"output": "0"
},
{
"input": "4\n0 3 5 8",
"output": "30"
}
] | 1,695,923,836
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = 1e9 + 7;
void solve()
{
int n;
cin >> n;
vector<int> arr(n);
int zero = 0;
vector<int> hsh(10, 0);
int sum = 0;
for (int i = 0; i < n; i++)
{
cin >> arr[i];
sum += arr[i];
if (arr[i] == 0)
zero++;
hsh[arr[i]]++;
}
if (zero == 0)
cout << -1;
else if (zero == n)
cout << 0;
else
{
bool flag = false;
if (sum % 3 == 1)
{
for (int i = 1; i <= 10; i += 3)
{
if (hsh[i] != 0)
{
flag = true;
hsh[i]--;
break;
}
}
if (!flag)
for (int i = 2; i <= 10; i++)
{
if (hsh[i] != 0 && hsh[i] > 1)
{
flag = true;
hsh[i] = hsh[i] - 2;
break;
}
}
}
if (sum % 3 == 2)
{
for (int i = 2; i <= 10; i += 3)
{
if (hsh[i] != 0)
{
flag = true;
hsh[i]--;
break;
}
}
if (!flag)
for (int i = 1; i <= 10; i++)
{
if (hsh[i] != 0 && hsh[i] > 1)
{
flag = true;
hsh[i] = hsh[i] - 2;
break;
}
}
}
else
flag = true;
if (flag)
{
int temp = 0;
for (int i = 0; i < 10; i++)
{
temp += hsh[i];
}
hsh[0] = zero;
if (temp == 0)
cout << 0;
else
{
for (int i = 9; i >= 0; i--)
{
for (int j = 0; j < hsh[i]; j++)
cout << i;
}
}
}
else
cout << 0;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t = 1;
while (t--)
{
solve();
}
return 0;
}
|
Title: Hometask
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a set of digits, your task is to find the maximum integer that you can make from these digits. The made number must be divisible by 2, 3, 5 without a residue. It is permitted to use not all digits from the set, it is forbidden to use leading zeroes.
Each digit is allowed to occur in the number the same number of times it occurs in the set.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of digits in the set. The second line contains *n* digits, the digits are separated by a single space.
Output Specification:
On a single line print the answer to the problem. If such number does not exist, then you should print -1.
Demo Input:
['1\n0\n', '11\n3 4 5 4 5 3 5 3 4 4 0\n', '8\n3 2 5 1 5 2 2 3\n']
Demo Output:
['0\n', '5554443330\n', '-1\n']
Note:
In the first sample there is only one number you can make — 0. In the second sample the sought number is 5554443330. In the third sample it is impossible to make the required number.
|
```python
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = 1e9 + 7;
void solve()
{
int n;
cin >> n;
vector<int> arr(n);
int zero = 0;
vector<int> hsh(10, 0);
int sum = 0;
for (int i = 0; i < n; i++)
{
cin >> arr[i];
sum += arr[i];
if (arr[i] == 0)
zero++;
hsh[arr[i]]++;
}
if (zero == 0)
cout << -1;
else if (zero == n)
cout << 0;
else
{
bool flag = false;
if (sum % 3 == 1)
{
for (int i = 1; i <= 10; i += 3)
{
if (hsh[i] != 0)
{
flag = true;
hsh[i]--;
break;
}
}
if (!flag)
for (int i = 2; i <= 10; i++)
{
if (hsh[i] != 0 && hsh[i] > 1)
{
flag = true;
hsh[i] = hsh[i] - 2;
break;
}
}
}
if (sum % 3 == 2)
{
for (int i = 2; i <= 10; i += 3)
{
if (hsh[i] != 0)
{
flag = true;
hsh[i]--;
break;
}
}
if (!flag)
for (int i = 1; i <= 10; i++)
{
if (hsh[i] != 0 && hsh[i] > 1)
{
flag = true;
hsh[i] = hsh[i] - 2;
break;
}
}
}
else
flag = true;
if (flag)
{
int temp = 0;
for (int i = 0; i < 10; i++)
{
temp += hsh[i];
}
hsh[0] = zero;
if (temp == 0)
cout << 0;
else
{
for (int i = 9; i >= 0; i--)
{
for (int j = 0; j < hsh[i]; j++)
cout << i;
}
}
}
else
cout << 0;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t = 1;
while (t--)
{
solve();
}
return 0;
}
```
| -1
|
|
294
|
A
|
Shaass and Oskols
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
|
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
|
On the *i*-th line of the output print the number of birds on the *i*-th wire.
|
[
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] |
[
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] |
none
| 500
|
[
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,675,227,886
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 31
| 122
| 0
|
import sys
def sol():
read = sys.stdin.readline
wires: int = int(read())
birds: list[int] = [int(i) for i in read().split()]
instructions: int = int(read())
for _ in range(instructions):
wire_idx, bird_idx = (int(i) for i in read().split())
left_birds = bird_idx - 1
if left_birds > 0 and wire_idx > 1:
birds[wire_idx - 2] += left_birds
right_birds = birds[wire_idx - 1] - bird_idx
if right_birds > 0 and wire_idx < len(birds):
birds[wire_idx] += right_birds
# Clear the current
birds[wire_idx - 1] = 0
for birt in birds:
print(birt)
if __name__ == "__main__":
sol()
|
Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none
|
```python
import sys
def sol():
read = sys.stdin.readline
wires: int = int(read())
birds: list[int] = [int(i) for i in read().split()]
instructions: int = int(read())
for _ in range(instructions):
wire_idx, bird_idx = (int(i) for i in read().split())
left_birds = bird_idx - 1
if left_birds > 0 and wire_idx > 1:
birds[wire_idx - 2] += left_birds
right_birds = birds[wire_idx - 1] - bird_idx
if right_birds > 0 and wire_idx < len(birds):
birds[wire_idx] += right_birds
# Clear the current
birds[wire_idx - 1] = 0
for birt in birds:
print(birt)
if __name__ == "__main__":
sol()
```
| 3
|
|
916
|
A
|
Jamie and Alarm Snooze
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.
A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky.
Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*.
Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'.
Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
|
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60).
The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
|
Print the minimum number of times he needs to press the button.
|
[
"3\n11 23\n",
"5\n01 07\n"
] |
[
"2\n",
"0\n"
] |
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.
In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
| 500
|
[
{
"input": "3\n11 23",
"output": "2"
},
{
"input": "5\n01 07",
"output": "0"
},
{
"input": "34\n09 24",
"output": "3"
},
{
"input": "2\n14 37",
"output": "0"
},
{
"input": "14\n19 54",
"output": "9"
},
{
"input": "42\n15 44",
"output": "12"
},
{
"input": "46\n02 43",
"output": "1"
},
{
"input": "14\n06 41",
"output": "1"
},
{
"input": "26\n04 58",
"output": "26"
},
{
"input": "54\n16 47",
"output": "0"
},
{
"input": "38\n20 01",
"output": "3"
},
{
"input": "11\n02 05",
"output": "8"
},
{
"input": "55\n22 10",
"output": "5"
},
{
"input": "23\n10 08",
"output": "6"
},
{
"input": "23\n23 14",
"output": "9"
},
{
"input": "51\n03 27",
"output": "0"
},
{
"input": "35\n15 25",
"output": "13"
},
{
"input": "3\n12 15",
"output": "6"
},
{
"input": "47\n00 28",
"output": "3"
},
{
"input": "31\n13 34",
"output": "7"
},
{
"input": "59\n17 32",
"output": "0"
},
{
"input": "25\n11 03",
"output": "8"
},
{
"input": "9\n16 53",
"output": "4"
},
{
"input": "53\n04 06",
"output": "3"
},
{
"input": "37\n00 12",
"output": "5"
},
{
"input": "5\n13 10",
"output": "63"
},
{
"input": "50\n01 59",
"output": "10"
},
{
"input": "34\n06 13",
"output": "4"
},
{
"input": "2\n18 19",
"output": "1"
},
{
"input": "46\n06 16",
"output": "17"
},
{
"input": "14\n03 30",
"output": "41"
},
{
"input": "40\n13 37",
"output": "0"
},
{
"input": "24\n17 51",
"output": "0"
},
{
"input": "8\n14 57",
"output": "0"
},
{
"input": "52\n18 54",
"output": "2"
},
{
"input": "20\n15 52",
"output": "24"
},
{
"input": "20\n03 58",
"output": "30"
},
{
"input": "48\n07 11",
"output": "0"
},
{
"input": "32\n04 01",
"output": "2"
},
{
"input": "60\n08 15",
"output": "1"
},
{
"input": "44\n20 20",
"output": "4"
},
{
"input": "55\n15 35",
"output": "9"
},
{
"input": "55\n03 49",
"output": "11"
},
{
"input": "23\n16 39",
"output": "4"
},
{
"input": "7\n20 36",
"output": "7"
},
{
"input": "35\n16 42",
"output": "1"
},
{
"input": "35\n05 56",
"output": "21"
},
{
"input": "3\n17 45",
"output": "0"
},
{
"input": "47\n05 59",
"output": "6"
},
{
"input": "15\n10 13",
"output": "9"
},
{
"input": "59\n06 18",
"output": "9"
},
{
"input": "34\n17 18",
"output": "0"
},
{
"input": "18\n05 23",
"output": "2"
},
{
"input": "46\n17 21",
"output": "0"
},
{
"input": "30\n06 27",
"output": "0"
},
{
"input": "14\n18 40",
"output": "3"
},
{
"input": "58\n22 54",
"output": "6"
},
{
"input": "26\n19 44",
"output": "5"
},
{
"input": "10\n15 57",
"output": "0"
},
{
"input": "54\n20 47",
"output": "0"
},
{
"input": "22\n08 45",
"output": "3"
},
{
"input": "48\n18 08",
"output": "1"
},
{
"input": "32\n07 06",
"output": "0"
},
{
"input": "60\n19 19",
"output": "2"
},
{
"input": "45\n07 25",
"output": "0"
},
{
"input": "29\n12 39",
"output": "8"
},
{
"input": "13\n08 28",
"output": "3"
},
{
"input": "41\n21 42",
"output": "5"
},
{
"input": "41\n09 32",
"output": "3"
},
{
"input": "9\n21 45",
"output": "2"
},
{
"input": "37\n10 43",
"output": "5"
},
{
"input": "3\n20 50",
"output": "1"
},
{
"input": "47\n00 04",
"output": "1"
},
{
"input": "15\n13 10",
"output": "21"
},
{
"input": "15\n17 23",
"output": "0"
},
{
"input": "43\n22 13",
"output": "2"
},
{
"input": "27\n10 26",
"output": "6"
},
{
"input": "55\n22 24",
"output": "5"
},
{
"input": "55\n03 30",
"output": "11"
},
{
"input": "24\n23 27",
"output": "0"
},
{
"input": "52\n11 33",
"output": "3"
},
{
"input": "18\n22 48",
"output": "17"
},
{
"input": "1\n12 55",
"output": "8"
},
{
"input": "1\n04 27",
"output": "0"
},
{
"input": "1\n12 52",
"output": "5"
},
{
"input": "1\n20 16",
"output": "9"
},
{
"input": "1\n04 41",
"output": "4"
},
{
"input": "1\n20 21",
"output": "4"
},
{
"input": "1\n04 45",
"output": "8"
},
{
"input": "1\n12 18",
"output": "1"
},
{
"input": "1\n04 42",
"output": "5"
},
{
"input": "1\n02 59",
"output": "2"
},
{
"input": "1\n18 24",
"output": "7"
},
{
"input": "1\n02 04",
"output": "7"
},
{
"input": "1\n18 28",
"output": "1"
},
{
"input": "1\n18 01",
"output": "2"
},
{
"input": "1\n10 25",
"output": "8"
},
{
"input": "1\n02 49",
"output": "2"
},
{
"input": "1\n02 30",
"output": "3"
},
{
"input": "1\n18 54",
"output": "7"
},
{
"input": "1\n02 19",
"output": "2"
},
{
"input": "1\n05 25",
"output": "8"
},
{
"input": "60\n23 55",
"output": "6"
},
{
"input": "60\n08 19",
"output": "1"
},
{
"input": "60\n00 00",
"output": "7"
},
{
"input": "60\n08 24",
"output": "1"
},
{
"input": "60\n16 13",
"output": "9"
},
{
"input": "60\n08 21",
"output": "1"
},
{
"input": "60\n16 45",
"output": "9"
},
{
"input": "60\n08 26",
"output": "1"
},
{
"input": "60\n08 50",
"output": "1"
},
{
"input": "60\n05 21",
"output": "12"
},
{
"input": "60\n13 29",
"output": "6"
},
{
"input": "60\n05 18",
"output": "12"
},
{
"input": "60\n13 42",
"output": "6"
},
{
"input": "60\n05 07",
"output": "0"
},
{
"input": "60\n05 47",
"output": "0"
},
{
"input": "60\n21 55",
"output": "4"
},
{
"input": "60\n05 36",
"output": "12"
},
{
"input": "60\n21 08",
"output": "4"
},
{
"input": "60\n21 32",
"output": "4"
},
{
"input": "60\n16 31",
"output": "9"
},
{
"input": "5\n00 00",
"output": "73"
},
{
"input": "2\n06 58",
"output": "390"
},
{
"input": "60\n00 00",
"output": "7"
},
{
"input": "2\n00 00",
"output": "181"
},
{
"input": "10\n00 00",
"output": "37"
},
{
"input": "60\n01 00",
"output": "8"
},
{
"input": "12\n00 06",
"output": "31"
},
{
"input": "1\n00 01",
"output": "4"
},
{
"input": "5\n00 05",
"output": "74"
},
{
"input": "60\n01 01",
"output": "8"
},
{
"input": "11\n18 11",
"output": "2"
},
{
"input": "60\n01 15",
"output": "8"
},
{
"input": "10\n00 16",
"output": "38"
},
{
"input": "60\n00 59",
"output": "7"
},
{
"input": "30\n00 00",
"output": "13"
},
{
"input": "60\n01 05",
"output": "8"
},
{
"input": "4\n00 03",
"output": "4"
},
{
"input": "4\n00 00",
"output": "91"
},
{
"input": "60\n00 01",
"output": "7"
},
{
"input": "6\n00 03",
"output": "1"
},
{
"input": "13\n00 00",
"output": "1"
},
{
"input": "1\n18 01",
"output": "2"
},
{
"input": "5\n06 00",
"output": "145"
},
{
"input": "60\n04 08",
"output": "11"
},
{
"input": "5\n01 55",
"output": "96"
},
{
"input": "8\n00 08",
"output": "47"
},
{
"input": "23\n18 23",
"output": "2"
},
{
"input": "6\n00 06",
"output": "62"
},
{
"input": "59\n18 59",
"output": "2"
},
{
"input": "11\n00 10",
"output": "3"
},
{
"input": "10\n00 01",
"output": "37"
},
{
"input": "59\n00 00",
"output": "7"
},
{
"input": "10\n18 10",
"output": "2"
},
{
"input": "5\n00 01",
"output": "73"
},
{
"input": "1\n00 00",
"output": "3"
},
{
"input": "8\n00 14",
"output": "47"
},
{
"input": "60\n03 00",
"output": "10"
},
{
"input": "60\n00 10",
"output": "7"
},
{
"input": "5\n01 13",
"output": "87"
},
{
"input": "30\n02 43",
"output": "18"
},
{
"input": "17\n00 08",
"output": "3"
},
{
"input": "3\n00 00",
"output": "1"
},
{
"input": "60\n00 05",
"output": "7"
},
{
"input": "5\n18 05",
"output": "2"
},
{
"input": "30\n00 30",
"output": "14"
},
{
"input": "1\n00 06",
"output": "9"
},
{
"input": "55\n00 00",
"output": "7"
},
{
"input": "8\n02 08",
"output": "62"
},
{
"input": "7\n00 00",
"output": "9"
},
{
"input": "6\n08 06",
"output": "2"
},
{
"input": "48\n06 24",
"output": "16"
},
{
"input": "8\n06 58",
"output": "98"
},
{
"input": "3\n12 00",
"output": "1"
},
{
"input": "5\n01 06",
"output": "86"
},
{
"input": "2\n00 08",
"output": "185"
},
{
"input": "3\n18 03",
"output": "2"
},
{
"input": "1\n17 00",
"output": "0"
},
{
"input": "59\n00 48",
"output": "7"
},
{
"input": "5\n12 01",
"output": "49"
},
{
"input": "55\n01 25",
"output": "9"
},
{
"input": "2\n07 23",
"output": "0"
},
{
"input": "10\n01 10",
"output": "44"
},
{
"input": "2\n00 01",
"output": "2"
},
{
"input": "59\n00 01",
"output": "6"
},
{
"input": "5\n00 02",
"output": "1"
},
{
"input": "4\n01 02",
"output": "106"
},
{
"input": "5\n00 06",
"output": "74"
},
{
"input": "42\n00 08",
"output": "9"
},
{
"input": "60\n01 20",
"output": "8"
},
{
"input": "3\n06 00",
"output": "1"
},
{
"input": "4\n00 01",
"output": "1"
},
{
"input": "2\n00 06",
"output": "184"
},
{
"input": "1\n00 57",
"output": "0"
},
{
"input": "6\n00 00",
"output": "61"
},
{
"input": "5\n08 40",
"output": "9"
},
{
"input": "58\n00 55",
"output": "1"
},
{
"input": "2\n00 02",
"output": "182"
},
{
"input": "1\n08 01",
"output": "2"
},
{
"input": "10\n10 10",
"output": "14"
},
{
"input": "60\n01 11",
"output": "8"
},
{
"input": "2\n07 00",
"output": "0"
},
{
"input": "15\n00 03",
"output": "25"
},
{
"input": "6\n04 34",
"output": "106"
},
{
"input": "16\n00 16",
"output": "24"
},
{
"input": "2\n00 59",
"output": "1"
},
{
"input": "59\n00 08",
"output": "7"
},
{
"input": "10\n03 10",
"output": "56"
},
{
"input": "3\n08 03",
"output": "2"
},
{
"input": "20\n06 11",
"output": "37"
},
{
"input": "4\n01 00",
"output": "106"
},
{
"input": "38\n01 08",
"output": "12"
},
{
"input": "60\n00 06",
"output": "7"
},
{
"input": "5\n12 00",
"output": "49"
},
{
"input": "6\n01 42",
"output": "78"
},
{
"input": "4\n00 04",
"output": "92"
},
{
"input": "60\n04 05",
"output": "11"
},
{
"input": "1\n00 53",
"output": "6"
},
{
"input": "5\n08 05",
"output": "2"
},
{
"input": "60\n18 45",
"output": "1"
},
{
"input": "60\n06 23",
"output": "13"
},
{
"input": "6\n00 15",
"output": "3"
},
{
"input": "58\n00 06",
"output": "7"
},
{
"input": "2\n06 44",
"output": "383"
},
{
"input": "1\n08 00",
"output": "1"
},
{
"input": "10\n06 58",
"output": "78"
},
{
"input": "59\n00 58",
"output": "8"
},
{
"input": "1\n18 00",
"output": "1"
},
{
"input": "50\n00 42",
"output": "9"
},
{
"input": "30\n18 30",
"output": "2"
},
{
"input": "60\n21 59",
"output": "4"
},
{
"input": "2\n10 52",
"output": "87"
},
{
"input": "56\n00 00",
"output": "7"
},
{
"input": "16\n18 16",
"output": "2"
},
{
"input": "5\n01 05",
"output": "86"
},
{
"input": "5\n05 00",
"output": "133"
},
{
"input": "5\n23 59",
"output": "72"
},
{
"input": "7\n17 13",
"output": "0"
},
{
"input": "58\n00 00",
"output": "7"
},
{
"input": "15\n00 07",
"output": "0"
},
{
"input": "59\n08 00",
"output": "1"
},
{
"input": "46\n00 00",
"output": "8"
},
{
"input": "59\n01 05",
"output": "2"
},
{
"input": "2\n01 00",
"output": "211"
},
{
"input": "60\n00 24",
"output": "7"
},
{
"input": "10\n00 08",
"output": "37"
},
{
"input": "10\n00 06",
"output": "37"
},
{
"input": "60\n01 24",
"output": "8"
},
{
"input": "50\n00 10",
"output": "8"
},
{
"input": "2\n03 00",
"output": "271"
},
{
"input": "4\n19 04",
"output": "17"
},
{
"input": "25\n00 23",
"output": "16"
},
{
"input": "10\n01 01",
"output": "43"
}
] | 1,516,890,588
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 254
| 77
| 5,632,000
|
x = int(input())
h, m = map(int, input().split(" "))
c = 0
while (h%10!=7 and m%10!=7):
c+= 1
m-=x
if (m < 0):
h -= 1
m = 60 + m
if (h < 0):
h = 23
print(c)
|
Title: Jamie and Alarm Snooze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.
A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky.
Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*.
Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'.
Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60).
The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
Output Specification:
Print the minimum number of times he needs to press the button.
Demo Input:
['3\n11 23\n', '5\n01 07\n']
Demo Output:
['2\n', '0\n']
Note:
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.
In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
|
```python
x = int(input())
h, m = map(int, input().split(" "))
c = 0
while (h%10!=7 and m%10!=7):
c+= 1
m-=x
if (m < 0):
h -= 1
m = 60 + m
if (h < 0):
h = 23
print(c)
```
| 3
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,698,397,830
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
n = int(input())
count = 0
sol = [int(x) for x in input().strip().split()]
for i in range(n):
if(sol.count(1) >= 2):
count += 1
print(count)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
count = 0
sol = [int(x) for x in input().strip().split()]
for i in range(n):
if(sol.count(1) >= 2):
count += 1
print(count)
```
| 0
|
|
839
|
C
|
Journey
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"dp",
"graphs",
"probabilities",
"trees"
] | null | null |
There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities.
Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road.
It is guaranteed that one can reach any city from any other by the roads.
|
Print a number — the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"4\n1 2\n1 3\n2 4\n",
"5\n1 2\n1 3\n3 4\n2 5\n"
] |
[
"1.500000000000000\n",
"2.000000000000000\n"
] |
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
| 1,500
|
[
{
"input": "4\n1 2\n1 3\n2 4",
"output": "1.500000000000000"
},
{
"input": "5\n1 2\n1 3\n3 4\n2 5",
"output": "2.000000000000000"
},
{
"input": "70\n1 25\n57 1\n18 1\n65 1\n38 1\n1 41\n1 5\n1 69\n1 3\n31 1\n1 8\n1 9\n53 1\n70 1\n45 1\n1 24\n1 42\n1 30\n1 12\n1 37\n64 1\n1 28\n1 58\n1 22\n11 1\n1 4\n1 27\n1 16\n1 21\n54 1\n1 51\n1 43\n29 1\n56 1\n1 39\n32 1\n1 15\n1 17\n1 19\n1 40\n36 1\n48 1\n63 1\n1 7\n1 47\n1 13\n1 46\n60 1\n1 6\n23 1\n20 1\n1 52\n2 1\n26 1\n1 59\n1 66\n10 1\n1 62\n1 68\n1 55\n50 1\n33 1\n44 1\n1 34\n1 35\n1 61\n14 1\n67 1\n49 1",
"output": "1.000000000000000"
},
{
"input": "10\n8 6\n9 10\n8 7\n1 4\n1 8\n9 5\n9 8\n2 5\n3 1",
"output": "1.500000000000000"
},
{
"input": "1",
"output": "0.000000000000000"
}
] | 1,642,965,231
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 920
| 39,526,400
|
def dfs(start):
visited = [0, True] + [False]*(n-1)
stack = [(x, 1, 1/len(tree[start])) for x in tree[start]]
while stack:
v, l, p = stack.pop()
if leafs[v]:
L.append(l*p)
if not visited[v]:
visited[v] = True
for u in tree[v]:
if not visited[u]:
stack.append((u, l+1, p/(len(tree[v])-1)))
n = int(input())
tree = {x: [] for x in range(1, n+1)}
L = []
for _ in range(n-1):
u, v = map(int, input().split())
tree[u].append(v)
tree[v].append(u)
leafs = [False, True] + [len(tree[x]) == 1 for x in range(2, n+1)]
dfs(1)
print(sum(L))
|
Title: Journey
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities.
Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output Specification:
Print a number — the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['4\n1 2\n1 3\n2 4\n', '5\n1 2\n1 3\n3 4\n2 5\n']
Demo Output:
['1.500000000000000\n', '2.000000000000000\n']
Note:
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
|
```python
def dfs(start):
visited = [0, True] + [False]*(n-1)
stack = [(x, 1, 1/len(tree[start])) for x in tree[start]]
while stack:
v, l, p = stack.pop()
if leafs[v]:
L.append(l*p)
if not visited[v]:
visited[v] = True
for u in tree[v]:
if not visited[u]:
stack.append((u, l+1, p/(len(tree[v])-1)))
n = int(input())
tree = {x: [] for x in range(1, n+1)}
L = []
for _ in range(n-1):
u, v = map(int, input().split())
tree[u].append(v)
tree[v].append(u)
leafs = [False, True] + [len(tree[x]) == 1 for x in range(2, n+1)]
dfs(1)
print(sum(L))
```
| 3
|
|
886
|
A
|
ACM ICPC
|
PROGRAMMING
| 1,000
|
[
"brute force"
] | null | null |
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
|
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
|
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
|
[
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
| 500
|
[
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,689,873,405
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
team = list(map(int,input().split()))
if sum(team) %2 == 0:
s1 = sum(team[0:2])
s2 = sum(team[2:])
if abs(s1-s2) in team:print("YES")
else:print("NO")
else:print("NO")
|
Title: ACM ICPC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
|
```python
team = list(map(int,input().split()))
if sum(team) %2 == 0:
s1 = sum(team[0:2])
s2 = sum(team[2:])
if abs(s1-s2) in team:print("YES")
else:print("NO")
else:print("NO")
```
| 0
|
|
476
|
A
|
Dreamoon and Stairs
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
|
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
|
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
|
[
"10 2\n",
"3 5\n"
] |
[
"6\n",
"-1\n"
] |
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
| 500
|
[
{
"input": "10 2",
"output": "6"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "29 7",
"output": "21"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "10000 2",
"output": "5000"
},
{
"input": "10000 3",
"output": "5001"
},
{
"input": "10000 10",
"output": "5000"
},
{
"input": "9999 3",
"output": "5001"
},
{
"input": "9999 2",
"output": "5000"
},
{
"input": "9999 10",
"output": "5000"
},
{
"input": "9999 9",
"output": "5004"
},
{
"input": "18 10",
"output": "10"
},
{
"input": "19 10",
"output": "10"
},
{
"input": "20 10",
"output": "10"
},
{
"input": "21 10",
"output": "20"
},
{
"input": "7688 5",
"output": "3845"
},
{
"input": "4608 5",
"output": "2305"
},
{
"input": "3979 2",
"output": "1990"
},
{
"input": "9985 6",
"output": "4998"
},
{
"input": "3230 8",
"output": "1616"
},
{
"input": "24 9",
"output": "18"
},
{
"input": "3275 8",
"output": "1640"
},
{
"input": "6240 7",
"output": "3122"
},
{
"input": "5227 4",
"output": "2616"
},
{
"input": "3832 6",
"output": "1920"
},
{
"input": "4 2",
"output": "2"
},
{
"input": "6 3",
"output": "3"
},
{
"input": "10 5",
"output": "5"
},
{
"input": "3 2",
"output": "2"
},
{
"input": "6 4",
"output": "4"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "9 9",
"output": "9"
},
{
"input": "4 5",
"output": "-1"
}
] | 1,668,881,098
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 46
| 0
|
n, m = map(int, input().split())
t = True
sm = n / 2
if sm != int(sm):
sm += 0.5
while t:
number = n / m
if sm > n * 2 or n < m:
print(-1)
break
if sm % m == 0:
t = False
print(int(sm))
sm += 1
|
Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
Output Specification:
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
Demo Input:
['10 2\n', '3 5\n']
Demo Output:
['6\n', '-1\n']
Note:
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
|
```python
n, m = map(int, input().split())
t = True
sm = n / 2
if sm != int(sm):
sm += 0.5
while t:
number = n / m
if sm > n * 2 or n < m:
print(-1)
break
if sm % m == 0:
t = False
print(int(sm))
sm += 1
```
| 3
|
|
347
|
A
|
Difference Row
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"implementation",
"sortings"
] | null | null |
You want to arrange *n* integers *a*1,<=*a*2,<=...,<=*a**n* in some order in a row. Let's define the value of an arrangement as the sum of differences between all pairs of adjacent integers.
More formally, let's denote some arrangement as a sequence of integers *x*1,<=*x*2,<=...,<=*x**n*, where sequence *x* is a permutation of sequence *a*. The value of such an arrangement is (*x*1<=-<=*x*2)<=+<=(*x*2<=-<=*x*3)<=+<=...<=+<=(*x**n*<=-<=1<=-<=*x**n*).
Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence *x* that corresponds to an arrangement of the largest possible value.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *a*1, *a*2, ..., *a**n* (|*a**i*|<=≤<=1000).
|
Print the required sequence *x*1,<=*x*2,<=...,<=*x**n*. Sequence *x* should be the lexicographically smallest permutation of *a* that corresponds to an arrangement of the largest possible value.
|
[
"5\n100 -100 50 0 -50\n"
] |
[
"100 -50 0 50 -100 \n"
] |
In the sample test case, the value of the output arrangement is (100 - ( - 50)) + (( - 50) - 0) + (0 - 50) + (50 - ( - 100)) = 200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.
Sequence *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*p*</sub> is lexicographically smaller than sequence *y*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">2</sub>, ... , *y*<sub class="lower-index">*p*</sub> if there exists an integer *r* (0 ≤ *r* < *p*) such that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> < *y*<sub class="lower-index">*r* + 1</sub>.
| 500
|
[
{
"input": "5\n100 -100 50 0 -50",
"output": "100 -50 0 50 -100 "
},
{
"input": "10\n764 -367 0 963 -939 -795 -26 -49 948 -282",
"output": "963 -795 -367 -282 -49 -26 0 764 948 -939 "
},
{
"input": "20\n262 -689 -593 161 -678 -555 -633 -697 369 258 673 50 833 737 -650 198 -651 -621 -396 939",
"output": "939 -689 -678 -651 -650 -633 -621 -593 -555 -396 50 161 198 258 262 369 673 737 833 -697 "
},
{
"input": "50\n-262 -377 -261 903 547 759 -800 -53 670 92 758 109 547 877 152 -901 -318 -527 -388 24 139 -227 413 -135 811 -886 -22 -526 -643 -431 284 609 -745 -62 323 -441 743 -800 86 862 587 -513 -468 -651 -760 197 141 -414 -909 438",
"output": "903 -901 -886 -800 -800 -760 -745 -651 -643 -527 -526 -513 -468 -441 -431 -414 -388 -377 -318 -262 -261 -227 -135 -62 -53 -22 24 86 92 109 139 141 152 197 284 323 413 438 547 547 587 609 670 743 758 759 811 862 877 -909 "
},
{
"input": "100\n144 -534 -780 -1 -259 -945 -992 -967 -679 -239 -22 387 130 -908 140 -270 16 646 398 599 -631 -231 687 -505 89 77 584 162 124 132 33 271 212 734 350 -678 969 43 487 -689 -432 -225 -603 801 -828 -684 349 318 109 723 33 -247 719 368 -286 217 260 77 -618 955 408 994 -313 -341 578 609 60 900 222 -779 -507 464 -147 -789 -477 -235 -407 -432 35 300 -53 -896 -476 927 -293 -869 -852 -566 -759 95 506 -914 -405 -621 319 -622 -49 -334 328 -104",
"output": "994 -967 -945 -914 -908 -896 -869 -852 -828 -789 -780 -779 -759 -689 -684 -679 -678 -631 -622 -621 -618 -603 -566 -534 -507 -505 -477 -476 -432 -432 -407 -405 -341 -334 -313 -293 -286 -270 -259 -247 -239 -235 -231 -225 -147 -104 -53 -49 -22 -1 16 33 33 35 43 60 77 77 89 95 109 124 130 132 140 144 162 212 217 222 260 271 300 318 319 328 349 350 368 387 398 408 464 487 506 578 584 599 609 646 687 719 723 734 801 900 927 955 969 -992 "
},
{
"input": "100\n-790 341 910 905 -779 279 696 -375 525 -21 -2 751 -887 764 520 -844 850 -537 -882 -183 139 -397 561 -420 -991 691 587 -93 -701 -957 -89 227 233 545 934 309 -26 454 -336 -994 -135 -840 -320 -387 -943 650 628 -583 701 -708 -881 287 -932 -265 -312 -757 695 985 -165 -329 -4 -462 -627 798 -124 -539 843 -492 -967 -782 879 -184 -351 -385 -713 699 -477 828 219 961 -170 -542 877 -718 417 152 -905 181 301 920 685 -502 518 -115 257 998 -112 -234 -223 -396",
"output": "998 -991 -967 -957 -943 -932 -905 -887 -882 -881 -844 -840 -790 -782 -779 -757 -718 -713 -708 -701 -627 -583 -542 -539 -537 -502 -492 -477 -462 -420 -397 -396 -387 -385 -375 -351 -336 -329 -320 -312 -265 -234 -223 -184 -183 -170 -165 -135 -124 -115 -112 -93 -89 -26 -21 -4 -2 139 152 181 219 227 233 257 279 287 301 309 341 417 454 518 520 525 545 561 587 628 650 685 691 695 696 699 701 751 764 798 828 843 850 877 879 905 910 920 934 961 985 -994 "
},
{
"input": "100\n720 331 -146 -935 399 248 525 -669 614 -245 320 229 842 -894 -73 584 -458 -975 -604 -78 607 -120 -377 409 -743 862 -969 980 105 841 -795 996 696 -759 -482 624 -578 421 -717 -553 -652 -268 405 426 642 870 -650 -812 178 -882 -237 -737 -724 358 407 714 759 779 -899 -726 398 -663 -56 -736 -825 313 -746 117 -457 330 -925 497 332 -794 -506 -811 -990 -799 -343 -380 598 926 671 967 -573 -687 741 484 -641 -698 -251 -391 23 692 337 -639 126 8 -915 -386",
"output": "996 -975 -969 -935 -925 -915 -899 -894 -882 -825 -812 -811 -799 -795 -794 -759 -746 -743 -737 -736 -726 -724 -717 -698 -687 -669 -663 -652 -650 -641 -639 -604 -578 -573 -553 -506 -482 -458 -457 -391 -386 -380 -377 -343 -268 -251 -245 -237 -146 -120 -78 -73 -56 8 23 105 117 126 178 229 248 313 320 330 331 332 337 358 398 399 405 407 409 421 426 484 497 525 584 598 607 614 624 642 671 692 696 714 720 741 759 779 841 842 862 870 926 967 980 -990 "
},
{
"input": "100\n-657 320 -457 -472 -423 -227 -902 -520 702 -27 -103 149 268 -922 307 -292 377 730 117 1000 935 459 -502 796 -494 892 -523 866 166 -248 57 -606 -96 -948 988 194 -687 832 -425 28 -356 -884 688 353 225 204 -68 960 -929 -312 -479 381 512 -274 -505 -260 -506 572 226 -822 -13 325 -370 403 -714 494 339 283 356 327 159 -151 -13 -760 -159 -991 498 19 -159 583 178 -50 -421 -679 -978 334 688 -99 117 -988 371 693 946 -58 -699 -133 62 693 535 -375",
"output": "1000 -988 -978 -948 -929 -922 -902 -884 -822 -760 -714 -699 -687 -679 -657 -606 -523 -520 -506 -505 -502 -494 -479 -472 -457 -425 -423 -421 -375 -370 -356 -312 -292 -274 -260 -248 -227 -159 -159 -151 -133 -103 -99 -96 -68 -58 -50 -27 -13 -13 19 28 57 62 117 117 149 159 166 178 194 204 225 226 268 283 307 320 325 327 334 339 353 356 371 377 381 403 459 494 498 512 535 572 583 688 688 693 693 702 730 796 832 866 892 935 946 960 988 -991 "
},
{
"input": "100\n853 752 931 -453 -943 -118 -772 -814 791 191 -83 -373 -748 -136 -286 250 627 292 -48 -896 -296 736 -628 -376 -246 -495 366 610 228 664 -951 -952 811 192 -730 -377 319 799 753 166 827 501 157 -834 -776 424 655 -827 549 -487 608 -643 419 349 -88 95 231 -520 -508 -105 -727 568 -241 286 586 -956 -880 892 866 22 658 832 -216 -54 491 -500 -687 393 24 129 946 303 931 563 -269 -203 -251 647 -824 -163 248 -896 -133 749 -619 -212 -2 491 287 219",
"output": "946 -952 -951 -943 -896 -896 -880 -834 -827 -824 -814 -776 -772 -748 -730 -727 -687 -643 -628 -619 -520 -508 -500 -495 -487 -453 -377 -376 -373 -296 -286 -269 -251 -246 -241 -216 -212 -203 -163 -136 -133 -118 -105 -88 -83 -54 -48 -2 22 24 95 129 157 166 191 192 219 228 231 248 250 286 287 292 303 319 349 366 393 419 424 491 491 501 549 563 568 586 608 610 627 647 655 658 664 736 749 752 753 791 799 811 827 832 853 866 892 931 931 -956 "
},
{
"input": "100\n9 857 227 -593 -983 -439 17 -523 -354 -189 780 -267 771 -981 943 620 -832 79 761 -943 218 -966 75 131 -596 534 51 796 -612 -381 -690 -353 -170 648 804 -256 257 -16 964 -728 310 50 453 737 -228 -625 618 841 -102 974 -850 -641 -788 231 -982 -84 -917 942 -913 -768 -83 298 388 447 -490 271 -949 976 -820 -876 -822 -188 -306 877 219 854 561 -307 -920 916 -925 -591 -149 -166 -572 860 -217 -831 -552 822 355 -150 203 -710 530 910 889 964 -125 -597",
"output": "976 -982 -981 -966 -949 -943 -925 -920 -917 -913 -876 -850 -832 -831 -822 -820 -788 -768 -728 -710 -690 -641 -625 -612 -597 -596 -593 -591 -572 -552 -523 -490 -439 -381 -354 -353 -307 -306 -267 -256 -228 -217 -189 -188 -170 -166 -150 -149 -125 -102 -84 -83 -16 9 17 50 51 75 79 131 203 218 219 227 231 257 271 298 310 355 388 447 453 530 534 561 618 620 648 737 761 771 780 796 804 822 841 854 857 860 877 889 910 916 942 943 964 964 974 -983 "
},
{
"input": "2\n-1000 1000",
"output": "1000 -1000 "
},
{
"input": "2\n1000 -1000",
"output": "1000 -1000 "
},
{
"input": "2\n0 0",
"output": "0 0 "
},
{
"input": "5\n1 2 3 4 5",
"output": "5 2 3 4 1 "
},
{
"input": "6\n1 1 1 2 2 2",
"output": "2 1 1 2 2 1 "
},
{
"input": "3\n-1 -1 -1",
"output": "-1 -1 -1 "
}
] | 1,405,736,127
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
n=int(input())
A=list(map(int,input().split()))
B=[]
C=[]
_0=False
for a in A:
if a>0:
B.append(a)
elif a<0:
C.append(a)
else:
_0=True
B.sort(reverse=False)
C.sort(reverse=True)
A=[]
if B and C:
A.append(B[-1])
if C!=C[0]:
A=A+C[1:]
if _0:
A.append(0)
if B[0:-1]!=B[-1]:
A=A+B[0:-1]
A.append(C[0])
if B and not C:
B.sort(reverse=True)
if _0:
A.append(0)
if C and not B:
C.sort(reverse=False)
if _0:
A.append(0)
print(" ".join(list(map(str,A))))
|
Title: Difference Row
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You want to arrange *n* integers *a*1,<=*a*2,<=...,<=*a**n* in some order in a row. Let's define the value of an arrangement as the sum of differences between all pairs of adjacent integers.
More formally, let's denote some arrangement as a sequence of integers *x*1,<=*x*2,<=...,<=*x**n*, where sequence *x* is a permutation of sequence *a*. The value of such an arrangement is (*x*1<=-<=*x*2)<=+<=(*x*2<=-<=*x*3)<=+<=...<=+<=(*x**n*<=-<=1<=-<=*x**n*).
Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence *x* that corresponds to an arrangement of the largest possible value.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *a*1, *a*2, ..., *a**n* (|*a**i*|<=≤<=1000).
Output Specification:
Print the required sequence *x*1,<=*x*2,<=...,<=*x**n*. Sequence *x* should be the lexicographically smallest permutation of *a* that corresponds to an arrangement of the largest possible value.
Demo Input:
['5\n100 -100 50 0 -50\n']
Demo Output:
['100 -50 0 50 -100 \n']
Note:
In the sample test case, the value of the output arrangement is (100 - ( - 50)) + (( - 50) - 0) + (0 - 50) + (50 - ( - 100)) = 200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.
Sequence *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*p*</sub> is lexicographically smaller than sequence *y*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">2</sub>, ... , *y*<sub class="lower-index">*p*</sub> if there exists an integer *r* (0 ≤ *r* < *p*) such that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> < *y*<sub class="lower-index">*r* + 1</sub>.
|
```python
n=int(input())
A=list(map(int,input().split()))
B=[]
C=[]
_0=False
for a in A:
if a>0:
B.append(a)
elif a<0:
C.append(a)
else:
_0=True
B.sort(reverse=False)
C.sort(reverse=True)
A=[]
if B and C:
A.append(B[-1])
if C!=C[0]:
A=A+C[1:]
if _0:
A.append(0)
if B[0:-1]!=B[-1]:
A=A+B[0:-1]
A.append(C[0])
if B and not C:
B.sort(reverse=True)
if _0:
A.append(0)
if C and not B:
C.sort(reverse=False)
if _0:
A.append(0)
print(" ".join(list(map(str,A))))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
The employees of the F company have lots of ways to entertain themselves. Today they invited a famous magician who shows a trick with plastic cups and a marble.
The point is to trick the spectator's attention. Initially, the spectator stands in front of a line of *n* plastic cups. Then the magician places a small marble under one cup and shuffles the cups. Then the spectator should guess which cup hides the marble.
But the head coder of the F company isn't easy to trick. When he saw the performance, he noticed several important facts:
- each cup contains a mark — a number from 1 to *n*; all marks on the cups are distinct; - the magician shuffles the cups in *m* operations, each operation looks like that: take a cup marked *x**i*, sitting at position *y**i* in the row of cups (the positions are numbered from left to right, starting from 1) and shift it to the very beginning of the cup row (on the first position).
When the head coder came home after work he wanted to re-do the trick. Unfortunately, he didn't remember the starting or the final position of the cups. He only remembered which operations the magician performed. Help the coder: given the operations in the order they were made find at least one initial permutation of the cups that can go through the described operations in the given order. Otherwise, state that such permutation doesn't exist.
|
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=106). Each of the next *m* lines contains a couple of integers. The *i*-th line contains integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the description of the *i*-th operation of the magician. Note that the operations are given in the order in which the magician made them and the coder wants to make them in the same order.
|
If the described permutation doesn't exist (the programmer remembered wrong operations), print -1. Otherwise, print *n* distinct integers, each from 1 to *n*: the *i*-th number should represent the mark on the cup that initially is in the row in position *i*.
If there are multiple correct answers, you should print the lexicographically minimum one.
|
[
"2 1\n2 1\n",
"3 2\n1 2\n1 1\n",
"3 3\n1 3\n2 3\n1 3\n"
] |
[
"2 1 \n",
"2 1 3 \n",
"-1\n"
] |
none
| 0
|
[] | 1,398,175,114
| 6,214
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 4
| 62
| 0
|
__author__ = 'Lipen'
def main():
n, m = map(int, input().split())
data = [0]*n
offset = 0
for i in range(m):
x, y = map(int, input().split())
if x in data and data.index(x)!=y-1 + offset: #???
print(-1)
return
if x not in data:
data[y-1 + offset] = x
if y!=1:
offset += 1
k = data.count(0)
if k>0:
j = 0
for i in range(1, n+1):
if i not in data:
data[data.index(0)] = i
j+=1
if j>k:
break
print(' '.join(str(item) for item in data))
try:
main()
except:
print(-1)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The employees of the F company have lots of ways to entertain themselves. Today they invited a famous magician who shows a trick with plastic cups and a marble.
The point is to trick the spectator's attention. Initially, the spectator stands in front of a line of *n* plastic cups. Then the magician places a small marble under one cup and shuffles the cups. Then the spectator should guess which cup hides the marble.
But the head coder of the F company isn't easy to trick. When he saw the performance, he noticed several important facts:
- each cup contains a mark — a number from 1 to *n*; all marks on the cups are distinct; - the magician shuffles the cups in *m* operations, each operation looks like that: take a cup marked *x**i*, sitting at position *y**i* in the row of cups (the positions are numbered from left to right, starting from 1) and shift it to the very beginning of the cup row (on the first position).
When the head coder came home after work he wanted to re-do the trick. Unfortunately, he didn't remember the starting or the final position of the cups. He only remembered which operations the magician performed. Help the coder: given the operations in the order they were made find at least one initial permutation of the cups that can go through the described operations in the given order. Otherwise, state that such permutation doesn't exist.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=106). Each of the next *m* lines contains a couple of integers. The *i*-th line contains integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the description of the *i*-th operation of the magician. Note that the operations are given in the order in which the magician made them and the coder wants to make them in the same order.
Output Specification:
If the described permutation doesn't exist (the programmer remembered wrong operations), print -1. Otherwise, print *n* distinct integers, each from 1 to *n*: the *i*-th number should represent the mark on the cup that initially is in the row in position *i*.
If there are multiple correct answers, you should print the lexicographically minimum one.
Demo Input:
['2 1\n2 1\n', '3 2\n1 2\n1 1\n', '3 3\n1 3\n2 3\n1 3\n']
Demo Output:
['2 1 \n', '2 1 3 \n', '-1\n']
Note:
none
|
```python
__author__ = 'Lipen'
def main():
n, m = map(int, input().split())
data = [0]*n
offset = 0
for i in range(m):
x, y = map(int, input().split())
if x in data and data.index(x)!=y-1 + offset: #???
print(-1)
return
if x not in data:
data[y-1 + offset] = x
if y!=1:
offset += 1
k = data.count(0)
if k>0:
j = 0
for i in range(1, n+1):
if i not in data:
data[data.index(0)] = i
j+=1
if j>k:
break
print(' '.join(str(item) for item in data))
try:
main()
except:
print(-1)
```
| 0
|
|
848
|
B
|
Rooter's Song
|
PROGRAMMING
| 1,900
|
[
"constructive algorithms",
"data structures",
"geometry",
"implementation",
"sortings",
"two pointers"
] | null | null |
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size *w*<=×<=*h*, represented by a rectangle with corners (0,<=0), (*w*,<=0), (*w*,<=*h*) and (0,<=*h*). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand *n* dancers. The *i*-th of them falls into one of the following groups:
- Vertical: stands at (*x**i*,<=0), moves in positive *y* direction (upwards); - Horizontal: stands at (0,<=*y**i*), moves in positive *x* direction (rightwards).
According to choreography, the *i*-th dancer should stand still for the first *t**i* milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
|
The first line of input contains three space-separated positive integers *n*, *w* and *h* (1<=≤<=*n*<=≤<=100<=000, 2<=≤<=*w*,<=*h*<=≤<=100<=000) — the number of dancers and the width and height of the stage, respectively.
The following *n* lines each describes a dancer: the *i*-th among them contains three space-separated integers *g**i*, *p**i*, and *t**i* (1<=≤<=*g**i*<=≤<=2, 1<=≤<=*p**i*<=≤<=99<=999, 0<=≤<=*t**i*<=≤<=100<=000), describing a dancer's group *g**i* (*g**i*<==<=1 — vertical, *g**i*<==<=2 — horizontal), position, and waiting time. If *g**i*<==<=1 then *p**i*<==<=*x**i*; otherwise *p**i*<==<=*y**i*. It's guaranteed that 1<=≤<=*x**i*<=≤<=*w*<=-<=1 and 1<=≤<=*y**i*<=≤<=*h*<=-<=1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
|
Output *n* lines, the *i*-th of which contains two space-separated integers (*x**i*,<=*y**i*) — the stopping position of the *i*-th dancer in the input.
|
[
"8 10 8\n1 1 10\n1 4 13\n1 7 1\n1 8 2\n2 2 0\n2 5 14\n2 6 0\n2 6 1\n",
"3 2 3\n1 1 2\n2 1 1\n1 1 5\n"
] |
[
"4 8\n10 5\n8 8\n10 6\n10 2\n1 8\n7 8\n10 6\n",
"1 3\n2 1\n1 3\n"
] |
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
In the second example, no dancers collide.
| 1,000
|
[
{
"input": "8 10 8\n1 1 10\n1 4 13\n1 7 1\n1 8 2\n2 2 0\n2 5 14\n2 6 0\n2 6 1",
"output": "4 8\n10 5\n8 8\n10 6\n10 2\n1 8\n7 8\n10 6"
},
{
"input": "3 2 3\n1 1 2\n2 1 1\n1 1 5",
"output": "1 3\n2 1\n1 3"
},
{
"input": "1 10 10\n1 8 1",
"output": "8 10"
},
{
"input": "3 4 5\n1 3 9\n2 1 9\n1 2 8",
"output": "3 5\n4 1\n2 5"
},
{
"input": "10 500 500\n2 88 59\n2 470 441\n1 340 500\n2 326 297\n1 74 45\n1 302 273\n1 132 103\n2 388 359\n1 97 68\n2 494 465",
"output": "500 494\n97 500\n340 500\n302 500\n500 470\n500 88\n500 326\n132 500\n500 388\n74 500"
},
{
"input": "20 50000 50000\n2 45955 55488\n1 19804 29337\n2 3767 90811\n2 24025 33558\n1 46985 56518\n2 21094 30627\n2 5787 15320\n1 4262 91306\n2 37231 46764\n1 18125 27658\n1 36532 12317\n1 31330 40863\n1 18992 28525\n1 29387 38920\n1 44654 54187\n2 45485 55018\n2 36850 46383\n1 44649 54182\n1 40922 50455\n2 12781 99825",
"output": "18125 50000\n50000 45955\n50000 12781\n31330 50000\n50000 5787\n40922 50000\n44649 50000\n50000 3767\n19804 50000\n44654 50000\n36532 50000\n50000 37231\n46985 50000\n50000 45485\n50000 21094\n18992 50000\n29387 50000\n50000 24025\n50000 36850\n4262 50000"
},
{
"input": "20 15 15\n2 7 100000\n1 2 100000\n2 1 100000\n1 9 100000\n2 4 100000\n2 3 100000\n2 14 100000\n1 6 100000\n1 10 100000\n2 5 100000\n2 13 100000\n1 8 100000\n1 13 100000\n1 14 100000\n2 10 100000\n1 5 100000\n1 11 100000\n1 12 100000\n1 1 100000\n2 2 100000",
"output": "15 7\n15 2\n1 15\n9 15\n15 4\n15 3\n14 15\n6 15\n15 10\n5 15\n13 15\n8 15\n15 13\n15 14\n10 15\n15 5\n11 15\n12 15\n15 1\n2 15"
},
{
"input": "5 20 20\n1 15 3\n2 15 3\n2 3 1\n2 1 0\n1 16 4",
"output": "16 20\n15 20\n20 3\n20 1\n20 15"
},
{
"input": "15 80 80\n2 36 4\n2 65 5\n1 31 2\n2 3 1\n2 62 0\n2 37 5\n1 16 4\n2 47 2\n1 17 5\n1 9 5\n2 2 0\n2 62 5\n2 34 2\n1 33 1\n2 69 3",
"output": "80 37\n80 65\n31 80\n80 3\n80 62\n33 80\n16 80\n80 47\n17 80\n9 80\n80 2\n80 62\n80 36\n80 34\n80 69"
},
{
"input": "15 15 15\n1 10 1\n2 11 0\n2 6 4\n1 1 0\n1 7 5\n1 14 3\n1 3 1\n1 4 2\n1 9 0\n2 10 1\n1 12 1\n2 2 0\n1 5 3\n2 3 0\n2 4 2",
"output": "15 10\n12 15\n3 15\n1 15\n15 2\n15 11\n7 15\n15 6\n10 15\n9 15\n14 15\n5 15\n15 4\n15 3\n4 15"
},
{
"input": "5 5 5\n1 1 0\n2 1 0\n2 2 1\n1 2 1\n2 4 3",
"output": "5 2\n5 4\n2 5\n5 1\n1 5"
}
] | 1,586,359,429
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 23
| 1,200
| 39,628,800
|
# https://codeforces.com/problemset/problem/848/B
def push(d, val, type_, arr):
# pos index
type_ %= 2
if val not in d:
d[val] = [[],[]]
d[val][type_].append(arr)
d = {}
n, w, h = map(int, input().split())
for index in range(n):
g, p, t = map(int, input().split())
push(d, p-t, g, [p ,index])
for k, v in d.items():
v[0]=sorted(v[0], key = lambda x: x[0], reverse=True)
v[1]=sorted(v[1], key = lambda x: x[0], reverse=False)
ans = [0] * n
for v in d.values():
cur=0
bound = len(v[1])
step = len(v[0])
merge = v[0]+v[1]
n_ = len(merge)
for pos, index in merge:
if cur<bound:
ans[index]=str(merge[(step+cur)%n_][0])+' '+str(h)
else:
ans[index]=str(w)+' '+str(merge[(step+cur)%n_][0])
cur+=1
print('\n'.join(ans))
|
Title: Rooter's Song
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size *w*<=×<=*h*, represented by a rectangle with corners (0,<=0), (*w*,<=0), (*w*,<=*h*) and (0,<=*h*). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand *n* dancers. The *i*-th of them falls into one of the following groups:
- Vertical: stands at (*x**i*,<=0), moves in positive *y* direction (upwards); - Horizontal: stands at (0,<=*y**i*), moves in positive *x* direction (rightwards).
According to choreography, the *i*-th dancer should stand still for the first *t**i* milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input Specification:
The first line of input contains three space-separated positive integers *n*, *w* and *h* (1<=≤<=*n*<=≤<=100<=000, 2<=≤<=*w*,<=*h*<=≤<=100<=000) — the number of dancers and the width and height of the stage, respectively.
The following *n* lines each describes a dancer: the *i*-th among them contains three space-separated integers *g**i*, *p**i*, and *t**i* (1<=≤<=*g**i*<=≤<=2, 1<=≤<=*p**i*<=≤<=99<=999, 0<=≤<=*t**i*<=≤<=100<=000), describing a dancer's group *g**i* (*g**i*<==<=1 — vertical, *g**i*<==<=2 — horizontal), position, and waiting time. If *g**i*<==<=1 then *p**i*<==<=*x**i*; otherwise *p**i*<==<=*y**i*. It's guaranteed that 1<=≤<=*x**i*<=≤<=*w*<=-<=1 and 1<=≤<=*y**i*<=≤<=*h*<=-<=1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output Specification:
Output *n* lines, the *i*-th of which contains two space-separated integers (*x**i*,<=*y**i*) — the stopping position of the *i*-th dancer in the input.
Demo Input:
['8 10 8\n1 1 10\n1 4 13\n1 7 1\n1 8 2\n2 2 0\n2 5 14\n2 6 0\n2 6 1\n', '3 2 3\n1 1 2\n2 1 1\n1 1 5\n']
Demo Output:
['4 8\n10 5\n8 8\n10 6\n10 2\n1 8\n7 8\n10 6\n', '1 3\n2 1\n1 3\n']
Note:
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
In the second example, no dancers collide.
|
```python
# https://codeforces.com/problemset/problem/848/B
def push(d, val, type_, arr):
# pos index
type_ %= 2
if val not in d:
d[val] = [[],[]]
d[val][type_].append(arr)
d = {}
n, w, h = map(int, input().split())
for index in range(n):
g, p, t = map(int, input().split())
push(d, p-t, g, [p ,index])
for k, v in d.items():
v[0]=sorted(v[0], key = lambda x: x[0], reverse=True)
v[1]=sorted(v[1], key = lambda x: x[0], reverse=False)
ans = [0] * n
for v in d.values():
cur=0
bound = len(v[1])
step = len(v[0])
merge = v[0]+v[1]
n_ = len(merge)
for pos, index in merge:
if cur<bound:
ans[index]=str(merge[(step+cur)%n_][0])+' '+str(h)
else:
ans[index]=str(w)+' '+str(merge[(step+cur)%n_][0])
cur+=1
print('\n'.join(ans))
```
| 3
|
|
617
|
A
|
Elephant
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
|
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
|
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
|
[
"5\n",
"12\n"
] |
[
"1\n",
"3\n"
] |
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
| 500
|
[
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "534204",
"output": "106841"
},
{
"input": "469569",
"output": "93914"
},
{
"input": "502877",
"output": "100576"
},
{
"input": "942212",
"output": "188443"
},
{
"input": "97",
"output": "20"
},
{
"input": "53",
"output": "11"
},
{
"input": "89",
"output": "18"
},
{
"input": "574",
"output": "115"
},
{
"input": "716",
"output": "144"
},
{
"input": "729",
"output": "146"
},
{
"input": "8901",
"output": "1781"
},
{
"input": "3645",
"output": "729"
},
{
"input": "4426",
"output": "886"
},
{
"input": "46573",
"output": "9315"
},
{
"input": "86380",
"output": "17276"
},
{
"input": "94190",
"output": "18838"
},
{
"input": "999990",
"output": "199998"
},
{
"input": "999991",
"output": "199999"
},
{
"input": "999992",
"output": "199999"
},
{
"input": "999993",
"output": "199999"
},
{
"input": "999994",
"output": "199999"
},
{
"input": "999995",
"output": "199999"
},
{
"input": "999996",
"output": "200000"
},
{
"input": "999997",
"output": "200000"
},
{
"input": "999998",
"output": "200000"
}
] | 1,699,939,285
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
import math
x = int(input())
min_steps = math.ceil(x / 5)
# Print the result
print(min_steps)
|
Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
|
```python
import math
x = int(input())
min_steps = math.ceil(x / 5)
# Print the result
print(min_steps)
```
| 3
|
|
992
|
C
|
Nastya and a Wardrobe
|
PROGRAMMING
| 1,600
|
[
"math"
] | null | null |
Nastya received a gift on New Year — a magic wardrobe. It is magic because in the end of each month the number of dresses in it doubles (i.e. the number of dresses becomes twice as large as it is in the beginning of the month).
Unfortunately, right after the doubling the wardrobe eats one of the dresses (if any) with the 50% probability. It happens every month except the last one in the year.
Nastya owns *x* dresses now, so she became interested in the [expected number](https://en.wikipedia.org/wiki/Expected_value) of dresses she will have in one year. Nastya lives in Byteland, so the year lasts for *k*<=+<=1 months.
Nastya is really busy, so she wants you to solve this problem. You are the programmer, after all. Also, you should find the answer modulo 109<=+<=7, because it is easy to see that it is always integer.
|
The only line contains two integers *x* and *k* (0<=≤<=*x*,<=*k*<=≤<=1018), where *x* is the initial number of dresses and *k*<=+<=1 is the number of months in a year in Byteland.
|
In the only line print a single integer — the expected number of dresses Nastya will own one year later modulo 109<=+<=7.
|
[
"2 0\n",
"2 1\n",
"3 2\n"
] |
[
"4\n",
"7\n",
"21\n"
] |
In the first example a year consists on only one month, so the wardrobe does not eat dresses at all.
In the second example after the first month there are 3 dresses with 50% probability and 4 dresses with 50% probability. Thus, in the end of the year there are 6 dresses with 50% probability and 8 dresses with 50% probability. This way the answer for this test is (6 + 8) / 2 = 7.
| 1,500
|
[
{
"input": "2 0",
"output": "4"
},
{
"input": "2 1",
"output": "7"
},
{
"input": "3 2",
"output": "21"
},
{
"input": "1 411",
"output": "485514976"
},
{
"input": "1 692",
"output": "860080936"
},
{
"input": "16 8",
"output": "7937"
},
{
"input": "18 12",
"output": "143361"
},
{
"input": "1 1000000000000000000",
"output": "719476261"
},
{
"input": "0 24",
"output": "0"
},
{
"input": "24 0",
"output": "48"
},
{
"input": "1000000000000000000 1",
"output": "195"
},
{
"input": "348612312017571993 87570063840727716",
"output": "551271547"
},
{
"input": "314647997243943415 107188213956410843",
"output": "109575135"
},
{
"input": "375000003 2",
"output": "0"
},
{
"input": "451 938",
"output": "598946958"
},
{
"input": "4 1669",
"output": "185365669"
},
{
"input": "24 347",
"output": "860029201"
},
{
"input": "1619 1813",
"output": "481568710"
},
{
"input": "280 472",
"output": "632090765"
},
{
"input": "1271 237",
"output": "27878991"
},
{
"input": "626 560",
"output": "399405853"
},
{
"input": "167 887",
"output": "983959273"
},
{
"input": "1769 422",
"output": "698926874"
},
{
"input": "160 929",
"output": "752935252"
},
{
"input": "1075 274",
"output": "476211777"
},
{
"input": "1332 332",
"output": "47520583"
},
{
"input": "103872254428948073 97291596742897547",
"output": "283633261"
},
{
"input": "157600018563121064 54027847222622605",
"output": "166795759"
},
{
"input": "514028642164226185 95344332761644668",
"output": "718282571"
},
{
"input": "91859547444219924 75483775868568438",
"output": "462306789"
},
{
"input": "295961633522750187 84483303945499729",
"output": "11464805"
},
{
"input": "8814960236468055 86463151557693391",
"output": "430718856"
},
{
"input": "672751296745170589 13026894786355983",
"output": "260355651"
},
{
"input": "909771081413191574 18862935031728197",
"output": "800873185"
},
{
"input": "883717267463724670 29585639347346605",
"output": "188389362"
},
{
"input": "431620727626880523 47616788361847228",
"output": "311078131"
},
{
"input": "816689044159694273 6475970360049048",
"output": "211796030"
},
{
"input": "313779810374175108 13838123840048842",
"output": "438854949"
},
{
"input": "860936792402722414 59551033597232946",
"output": "359730003"
},
{
"input": "332382902893992163 15483141652464187",
"output": "719128379"
},
{
"input": "225761360057436129 49203610094504526",
"output": "54291755"
},
{
"input": "216006901533424028 8313457244750219",
"output": "362896012"
},
{
"input": "568001660010321225 97167523790774710",
"output": "907490480"
},
{
"input": "904089164817530426 53747406876903279",
"output": "702270335"
},
{
"input": "647858974461637674 18385058205826214",
"output": "375141527"
},
{
"input": "720433754707338458 94180351080265292",
"output": "273505123"
},
{
"input": "268086842387268316 76502855388264782",
"output": "288717798"
},
{
"input": "488603693655520686 79239542983498430",
"output": "316399174"
},
{
"input": "152455635055802121 50394545488662355",
"output": "697051907"
},
{
"input": "585664029992038779 34972826534657555",
"output": "699566354"
},
{
"input": "349532090641396787 12248820623854158",
"output": "233938854"
},
{
"input": "353579407209009179 74469254935824590",
"output": "771349161"
},
{
"input": "491414900908765740 49509676303815755",
"output": "237095803"
},
{
"input": "91142854626119420 900651524977956",
"output": "211575546"
},
{
"input": "73543340229981083 66918326344192076",
"output": "710215652"
},
{
"input": "463958371369193376 89203995753927042",
"output": "41857490"
},
{
"input": "911873413622533246 54684577459651780",
"output": "926432198"
},
{
"input": "316313018463929883 78259904441946885",
"output": "36284201"
},
{
"input": "889560480100219043 54181377424922141",
"output": "281123162"
},
{
"input": "0 3259862395629356",
"output": "0"
},
{
"input": "1 3",
"output": "9"
},
{
"input": "3 1",
"output": "11"
},
{
"input": "1000000007 1",
"output": "1000000006"
},
{
"input": "1000000007 2",
"output": "1000000004"
},
{
"input": "1000000007 0",
"output": "0"
},
{
"input": "1000000007 12",
"output": "999995912"
},
{
"input": "1000000007 70",
"output": "729983755"
},
{
"input": "250000002 1",
"output": "0"
},
{
"input": "1000000007 3",
"output": "1000000000"
},
{
"input": "999999999 0",
"output": "999999991"
},
{
"input": "1000000007 5",
"output": "999999976"
},
{
"input": "1000000007 1000000007",
"output": "1000000006"
},
{
"input": "10000000000000000 0",
"output": "860000007"
},
{
"input": "1000000000000 0",
"output": "999986007"
},
{
"input": "99999999999999999 0",
"output": "600000012"
},
{
"input": "1000000000000000 0",
"output": "986000007"
}
] | 1,536,148,213
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 7
| 108
| 307,200
|
# cook your dish here
def power(x, y, p) :
return (1<<y)%p
d,n=map(int,input().split())
ans =power(2,n+1,1000000007);
ans1=power(2,n,1000000007);
print(((ans*(d%1000000007))%1000000007 - ans1 +1)%1000000007)
|
Title: Nastya and a Wardrobe
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya received a gift on New Year — a magic wardrobe. It is magic because in the end of each month the number of dresses in it doubles (i.e. the number of dresses becomes twice as large as it is in the beginning of the month).
Unfortunately, right after the doubling the wardrobe eats one of the dresses (if any) with the 50% probability. It happens every month except the last one in the year.
Nastya owns *x* dresses now, so she became interested in the [expected number](https://en.wikipedia.org/wiki/Expected_value) of dresses she will have in one year. Nastya lives in Byteland, so the year lasts for *k*<=+<=1 months.
Nastya is really busy, so she wants you to solve this problem. You are the programmer, after all. Also, you should find the answer modulo 109<=+<=7, because it is easy to see that it is always integer.
Input Specification:
The only line contains two integers *x* and *k* (0<=≤<=*x*,<=*k*<=≤<=1018), where *x* is the initial number of dresses and *k*<=+<=1 is the number of months in a year in Byteland.
Output Specification:
In the only line print a single integer — the expected number of dresses Nastya will own one year later modulo 109<=+<=7.
Demo Input:
['2 0\n', '2 1\n', '3 2\n']
Demo Output:
['4\n', '7\n', '21\n']
Note:
In the first example a year consists on only one month, so the wardrobe does not eat dresses at all.
In the second example after the first month there are 3 dresses with 50% probability and 4 dresses with 50% probability. Thus, in the end of the year there are 6 dresses with 50% probability and 8 dresses with 50% probability. This way the answer for this test is (6 + 8) / 2 = 7.
|
```python
# cook your dish here
def power(x, y, p) :
return (1<<y)%p
d,n=map(int,input().split())
ans =power(2,n+1,1000000007);
ans1=power(2,n,1000000007);
print(((ans*(d%1000000007))%1000000007 - ans1 +1)%1000000007)
```
| -1
|
|
519
|
C
|
A and B and Team Training
|
PROGRAMMING
| 1,300
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are *n* experienced members and *m* newbies on the training session. Can you calculate what maximum number of teams can be formed?
|
The first line contains two integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=5·105) — the number of experienced participants and newbies that are present at the training session.
|
Print the maximum number of teams that can be formed.
|
[
"2 6\n",
"4 5\n"
] |
[
"2\n",
"3\n"
] |
Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB).
| 1,500
|
[
{
"input": "2 6",
"output": "2"
},
{
"input": "4 5",
"output": "3"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "500000 500000",
"output": "333333"
},
{
"input": "70 100",
"output": "56"
},
{
"input": "5 12525",
"output": "5"
},
{
"input": "10 5",
"output": "5"
},
{
"input": "5 10",
"output": "5"
},
{
"input": "0 0",
"output": "0"
},
{
"input": "0 1",
"output": "0"
},
{
"input": "1 0",
"output": "0"
},
{
"input": "0 21233",
"output": "0"
},
{
"input": "12523 0",
"output": "0"
},
{
"input": "1231 1253",
"output": "828"
},
{
"input": "500000 0",
"output": "0"
},
{
"input": "1 500000",
"output": "1"
},
{
"input": "250000 500000",
"output": "250000"
},
{
"input": "500000 250000",
"output": "250000"
},
{
"input": "33333 77777",
"output": "33333"
},
{
"input": "30900 174529",
"output": "30900"
},
{
"input": "89979 57154",
"output": "49044"
},
{
"input": "231646 398487",
"output": "210044"
},
{
"input": "332019 281112",
"output": "204377"
},
{
"input": "473686 122443",
"output": "122443"
},
{
"input": "481245 86879",
"output": "86879"
},
{
"input": "39935 123534",
"output": "39935"
},
{
"input": "10000 20000",
"output": "10000"
},
{
"input": "10000 20001",
"output": "10000"
},
{
"input": "10000 20002",
"output": "10000"
},
{
"input": "10000 20003",
"output": "10000"
},
{
"input": "10000 20004",
"output": "10000"
},
{
"input": "10001 20000",
"output": "10000"
},
{
"input": "10001 20001",
"output": "10000"
},
{
"input": "10001 20002",
"output": "10001"
},
{
"input": "10001 20003",
"output": "10001"
},
{
"input": "10001 20004",
"output": "10001"
},
{
"input": "20000 10000",
"output": "10000"
},
{
"input": "20001 10000",
"output": "10000"
},
{
"input": "20002 10000",
"output": "10000"
},
{
"input": "20003 10000",
"output": "10000"
},
{
"input": "20004 10000",
"output": "10000"
},
{
"input": "20000 10001",
"output": "10000"
},
{
"input": "20001 10001",
"output": "10000"
},
{
"input": "20002 10001",
"output": "10001"
},
{
"input": "20003 10001",
"output": "10001"
},
{
"input": "20004 10001",
"output": "10001"
},
{
"input": "10 0",
"output": "0"
},
{
"input": "0 6",
"output": "0"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "0 0",
"output": "0"
}
] | 1,663,597,473
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 52
| 186
| 4,915,200
|
from collections import deque,Counter
from math import *
import sys
import random
from bisect import *
from functools import reduce
from sys import stdin
from heapq import *
import copy
input = lambda: sys.stdin.readline().rstrip()
a,b = map(int,input().split())
ans = 0
while a>=1 or b>=1:
if a<b and a>=1 and b>=2:
a-=1
b-=2
elif a>=2 and b>=1:
a-=2
b-=1
else:
break
ans+=1
print(ans)
|
Title: A and B and Team Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are *n* experienced members and *m* newbies on the training session. Can you calculate what maximum number of teams can be formed?
Input Specification:
The first line contains two integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=5·105) — the number of experienced participants and newbies that are present at the training session.
Output Specification:
Print the maximum number of teams that can be formed.
Demo Input:
['2 6\n', '4 5\n']
Demo Output:
['2\n', '3\n']
Note:
Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB).
|
```python
from collections import deque,Counter
from math import *
import sys
import random
from bisect import *
from functools import reduce
from sys import stdin
from heapq import *
import copy
input = lambda: sys.stdin.readline().rstrip()
a,b = map(int,input().split())
ans = 0
while a>=1 or b>=1:
if a<b and a>=1 and b>=2:
a-=1
b-=2
elif a>=2 and b>=1:
a-=2
b-=1
else:
break
ans+=1
print(ans)
```
| 3
|
|
253
|
A
|
Boys and Girls
|
PROGRAMMING
| 1,100
|
[
"greedy"
] | null | null |
There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
|
The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space.
|
Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them.
|
[
"3 3\n",
"4 2\n"
] |
[
"GBGBGB\n",
"BGBGBB\n"
] |
In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal.
| 500
|
[
{
"input": "3 3",
"output": "GBGBGB"
},
{
"input": "4 2",
"output": "BGBGBB"
},
{
"input": "5 5",
"output": "GBGBGBGBGB"
},
{
"input": "6 4",
"output": "BGBGBGBGBB"
},
{
"input": "100 1",
"output": "BGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "76 48",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "100 90",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBB"
},
{
"input": "90 100",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGG"
},
{
"input": "1 98",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 100",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "56 98",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "89 89",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGB"
},
{
"input": "18 94",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "84 27",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "1 1",
"output": "GB"
},
{
"input": "1 2",
"output": "GBG"
},
{
"input": "2 1",
"output": "BGB"
},
{
"input": "1 34",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "46 2",
"output": "BGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "99 3",
"output": "BGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 100",
"output": "GBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "100 100",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGB"
},
{
"input": "1 4",
"output": "GBGGG"
}
] | 1,613,552,531
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 92
| 0
|
boys,girls=list(map(int,input().split(" ")))
if boys>girls:
res="BG"
else:
res="GB"
for x in range(min(boys,girls)):
print(res,end="")
for x in range(abs(boys-girls)):
print(res[0])
|
Title: Boys and Girls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
Input Specification:
The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them.
Demo Input:
['3 3\n', '4 2\n']
Demo Output:
['GBGBGB\n', 'BGBGBB\n']
Note:
In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal.
|
```python
boys,girls=list(map(int,input().split(" ")))
if boys>girls:
res="BG"
else:
res="GB"
for x in range(min(boys,girls)):
print(res,end="")
for x in range(abs(boys-girls)):
print(res[0])
```
| -1
|
|
330
|
A
|
Cakeminator
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
|
The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
|
Output the maximum number of cake cells that the cakeminator can eat.
|
[
"3 4\nS...\n....\n..S.\n"
] |
[
"8\n"
] |
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).
| 500
|
[
{
"input": "3 4\nS...\n....\n..S.",
"output": "8"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "2 2\nSS\nSS",
"output": "0"
},
{
"input": "7 3\nS..\nS..\nS..\nS..\nS..\nS..\nS..",
"output": "14"
},
{
"input": "3 5\n..S..\nSSSSS\n..S..",
"output": "0"
},
{
"input": "10 10\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS",
"output": "0"
},
{
"input": "10 10\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS",
"output": "30"
},
{
"input": "10 10\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..",
"output": "80"
},
{
"input": "9 5\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS",
"output": "0"
},
{
"input": "9 9\n...S.....\nS.S.....S\n.S....S..\n.S.....SS\n.........\n..S.S..S.\n.SS......\n....S....\n..S...S..",
"output": "17"
},
{
"input": "5 6\nSSSSSS\nSSSSSS\nSSSSSS\nSS.S..\nS.S.SS",
"output": "0"
},
{
"input": "9 8\n........\n.......S\n........\nS.......\n........\n........\nS.......\n........\n.......S",
"output": "64"
},
{
"input": "9 7\n......S\n......S\nS.S.S..\n.......\n.......\n.S.....\n.S....S\n..S....\n.S....S",
"output": "28"
},
{
"input": "10 10\n.....S....\n....SS..S.\n.S...S....\n........SS\n.S.......S\nSS..S.....\n.SS.....SS\nS..S......\n.......SSS\nSSSSS....S",
"output": "10"
},
{
"input": "6 7\n..S.SS.\n......S\n....S.S\nSS..S..\nS..SS.S\n.....S.",
"output": "0"
},
{
"input": "10 6\n.SSSSS\nSSS.SS\nSSSSSS\nS.SSSS\nSSSSS.\nS.SSSS\nSS.SSS\n.SSS.S\n.SSS..\nSS..SS",
"output": "0"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "4 3\n.S.\nS.S\n.S.\nS.S",
"output": "0"
},
{
"input": "2 3\n...\nSSS",
"output": "3"
},
{
"input": "2 4\nS.SS\nS.SS",
"output": "2"
},
{
"input": "2 2\n..\n.S",
"output": "3"
},
{
"input": "3 2\n.S\n.S\nSS",
"output": "0"
},
{
"input": "2 4\nSS.S\n..S.",
"output": "0"
},
{
"input": "2 3\n...\nS..",
"output": "5"
}
] | 1,647,876,796
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 60
| 0
|
r,c = [int(i) for i in input().split()]
mat = []
for i in range(r) :
x = [i for i in input()]
mat.append(x)
a,b = [] , []
for i in range(r) :
if 'S' in mat[i] :
a.append(i)
b.append(mat[i].index('S'))
cords = []
for i in a:
for j in b:
if (str(i) + '-' + str(j)) not in cords:
cords.append((str(i) + '-' + str(j)))
print(r*c - len(cords))
|
Title: Cakeminator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
Input Specification:
The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
Output Specification:
Output the maximum number of cake cells that the cakeminator can eat.
Demo Input:
['3 4\nS...\n....\n..S.\n']
Demo Output:
['8\n']
Note:
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).
|
```python
r,c = [int(i) for i in input().split()]
mat = []
for i in range(r) :
x = [i for i in input()]
mat.append(x)
a,b = [] , []
for i in range(r) :
if 'S' in mat[i] :
a.append(i)
b.append(mat[i].index('S'))
cords = []
for i in a:
for j in b:
if (str(i) + '-' + str(j)) not in cords:
cords.append((str(i) + '-' + str(j)))
print(r*c - len(cords))
```
| 0
|
|
579
|
A
|
Raising Bacteria
|
PROGRAMMING
| 1,000
|
[
"bitmasks"
] | null | null |
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
|
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
|
The only line containing one integer: the answer.
|
[
"5\n",
"8\n"
] |
[
"2\n",
"1\n"
] |
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
| 250
|
[
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697681824",
"output": "14"
},
{
"input": "41313494",
"output": "14"
},
{
"input": "673935585",
"output": "13"
},
{
"input": "117422204",
"output": "19"
},
{
"input": "954746654",
"output": "15"
},
{
"input": "536838144",
"output": "14"
},
{
"input": "1000000000",
"output": "13"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "6",
"output": "2"
},
{
"input": "7",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "999999999",
"output": "21"
},
{
"input": "999999998",
"output": "20"
},
{
"input": "999999997",
"output": "20"
},
{
"input": "999999996",
"output": "19"
},
{
"input": "999999995",
"output": "20"
},
{
"input": "999999994",
"output": "19"
},
{
"input": "999999993",
"output": "19"
},
{
"input": "999999992",
"output": "18"
},
{
"input": "999999991",
"output": "20"
},
{
"input": "999999990",
"output": "19"
},
{
"input": "536870910",
"output": "28"
},
{
"input": "536870912",
"output": "1"
},
{
"input": "536870913",
"output": "2"
},
{
"input": "536870914",
"output": "2"
},
{
"input": "22",
"output": "3"
},
{
"input": "14",
"output": "3"
}
] | 1,663,250,550
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 46
| 0
|
n = int(input())
k = 30
s=0
for i in range(30,-1,-1):
if n//(2**i)>0:
s=s+(n//2**i)
n = n%(2**i)
print(s)
|
Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
|
```python
n = int(input())
k = 30
s=0
for i in range(30,-1,-1):
if n//(2**i)>0:
s=s+(n//2**i)
n = n%(2**i)
print(s)
```
| 3
|
|
902
|
B
|
Coloring a Tree
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"dsu",
"greedy"
] | null | null |
You are given a rooted tree with *n* vertices. The vertices are numbered from 1 to *n*, the root is the vertex number 1.
Each vertex has a color, let's denote the color of vertex *v* by *c**v*. Initially *c**v*<==<=0.
You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex *v* and a color *x*, and then color all vectices in the subtree of *v* (including *v* itself) in color *x*. In other words, for every vertex *u*, such that the path from root to *u* passes through *v*, set *c**u*<==<=*x*.
It is guaranteed that you have to color each vertex in a color different from 0.
You can learn what a rooted tree is using the link: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)).
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=104) — the number of vertices in the tree.
The second line contains *n*<=-<=1 integers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=<<=*i*), where *p**i* means that there is an edge between vertices *i* and *p**i*.
The third line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*n*), where *c**i* is the color you should color the *i*-th vertex into.
It is guaranteed that the given graph is a tree.
|
Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.
|
[
"6\n1 2 2 1 5\n2 1 1 1 1 1\n",
"7\n1 1 2 3 1 4\n3 3 1 1 1 2 3\n"
] |
[
"3\n",
"5\n"
] |
The tree from the first sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/10324ccdc37f95343acc4f3c6050d8c334334ffa.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/1c7bb267e2c1a006132248a43121400189309e2f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On seond step we color all vertices in the subtree of vertex 5 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/2201a6d49b89ba850ff0d0bdcbb3f8e9dd3871a8.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 2 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/6fa977fcdebdde94c47695151e0427b33d0102c5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The tree from the second sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/d70f9ae72a2ed429dd6531cac757e375dd3c953d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/7289e8895d0dd56c47b6b17969b9cf77b36786b5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On second step we color all vertices in the subtree of vertex 3 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/819001df7229138db3a407713744d1e3be88b64e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 6 into color 2:
<img class="tex-graphics" src="https://espresso.codeforces.com/80ebbd870a0a339636a21b9acdaf9de046458b43.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fourth step we color all vertices in the subtree of vertex 4 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/ed836aa723ac0176abde4e32988e3ac205014e93.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fith step we color all vertices in the subtree of vertex 7 into color 3:
<img class="tex-graphics" src="https://espresso.codeforces.com/8132909e11b41c27b8df2f0b0c10bc841f35e58a.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "6\n1 2 2 1 5\n2 1 1 1 1 1",
"output": "3"
},
{
"input": "7\n1 1 2 3 1 4\n3 3 1 1 1 2 3",
"output": "5"
},
{
"input": "2\n1\n2 2",
"output": "1"
},
{
"input": "3\n1 1\n2 2 2",
"output": "1"
},
{
"input": "4\n1 2 1\n1 2 3 4",
"output": "4"
},
{
"input": "4\n1 2 3\n4 1 2 4",
"output": "4"
},
{
"input": "5\n1 2 1 4\n1 1 1 2 2",
"output": "2"
},
{
"input": "3\n1 2\n2 1 1",
"output": "2"
},
{
"input": "4\n1 1 1\n3 1 3 1",
"output": "3"
},
{
"input": "4\n1 1 2\n4 1 4 1",
"output": "2"
},
{
"input": "4\n1 2 2\n3 1 2 3",
"output": "4"
},
{
"input": "3\n1 1\n1 2 2",
"output": "3"
}
] | 1,568,022,599
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 50
| 171
| 4,403,200
|
ii=lambda:int(input())
kk=lambda:map(int,input().split())
ll=lambda:list(kk())
n=ii()
parents = [-1]*n
edges = {i:[] for i in range(n)}
ls = kk()
for i in range(1, n):
o = ls.__next__()-1
edges[i].append(o)
edges[o].append(i)
parents[0]=0
rcol = ll()
lowest = [0]
cnt=1
while lowest:
l2 = []
for l in lowest:
if rcol[parents[l]]!=rcol[l]: cnt+=1
for child in edges[l]:
if child == parents[l]: continue
parents[child] = l
l2.append(child)
lowest=l2
print(cnt)
|
Title: Coloring a Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rooted tree with *n* vertices. The vertices are numbered from 1 to *n*, the root is the vertex number 1.
Each vertex has a color, let's denote the color of vertex *v* by *c**v*. Initially *c**v*<==<=0.
You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex *v* and a color *x*, and then color all vectices in the subtree of *v* (including *v* itself) in color *x*. In other words, for every vertex *u*, such that the path from root to *u* passes through *v*, set *c**u*<==<=*x*.
It is guaranteed that you have to color each vertex in a color different from 0.
You can learn what a rooted tree is using the link: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)).
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=104) — the number of vertices in the tree.
The second line contains *n*<=-<=1 integers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=<<=*i*), where *p**i* means that there is an edge between vertices *i* and *p**i*.
The third line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*n*), where *c**i* is the color you should color the *i*-th vertex into.
It is guaranteed that the given graph is a tree.
Output Specification:
Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.
Demo Input:
['6\n1 2 2 1 5\n2 1 1 1 1 1\n', '7\n1 1 2 3 1 4\n3 3 1 1 1 2 3\n']
Demo Output:
['3\n', '5\n']
Note:
The tree from the first sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/10324ccdc37f95343acc4f3c6050d8c334334ffa.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/1c7bb267e2c1a006132248a43121400189309e2f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On seond step we color all vertices in the subtree of vertex 5 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/2201a6d49b89ba850ff0d0bdcbb3f8e9dd3871a8.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 2 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/6fa977fcdebdde94c47695151e0427b33d0102c5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The tree from the second sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/d70f9ae72a2ed429dd6531cac757e375dd3c953d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/7289e8895d0dd56c47b6b17969b9cf77b36786b5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On second step we color all vertices in the subtree of vertex 3 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/819001df7229138db3a407713744d1e3be88b64e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 6 into color 2:
<img class="tex-graphics" src="https://espresso.codeforces.com/80ebbd870a0a339636a21b9acdaf9de046458b43.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fourth step we color all vertices in the subtree of vertex 4 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/ed836aa723ac0176abde4e32988e3ac205014e93.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fith step we color all vertices in the subtree of vertex 7 into color 3:
<img class="tex-graphics" src="https://espresso.codeforces.com/8132909e11b41c27b8df2f0b0c10bc841f35e58a.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
ii=lambda:int(input())
kk=lambda:map(int,input().split())
ll=lambda:list(kk())
n=ii()
parents = [-1]*n
edges = {i:[] for i in range(n)}
ls = kk()
for i in range(1, n):
o = ls.__next__()-1
edges[i].append(o)
edges[o].append(i)
parents[0]=0
rcol = ll()
lowest = [0]
cnt=1
while lowest:
l2 = []
for l in lowest:
if rcol[parents[l]]!=rcol[l]: cnt+=1
for child in edges[l]:
if child == parents[l]: continue
parents[child] = l
l2.append(child)
lowest=l2
print(cnt)
```
| 3
|
|
911
|
B
|
Two Cakes
|
PROGRAMMING
| 1,200
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into *a* pieces, and the second one — into *b* pieces.
Ivan knows that there will be *n* people at the celebration (including himself), so Ivan has set *n* plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:
1. Each piece of each cake is put on some plate; 1. Each plate contains at least one piece of cake; 1. No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number *x* such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least *x* pieces of cake.
Help Ivan to calculate this number *x*!
|
The first line contains three integers *n*, *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100, 2<=≤<=*n*<=≤<=*a*<=+<=*b*) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.
|
Print the maximum possible number *x* such that Ivan can distribute the cake in such a way that each plate will contain at least *x* pieces of cake.
|
[
"5 2 3\n",
"4 7 10\n"
] |
[
"1\n",
"3\n"
] |
In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.
In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.
| 0
|
[
{
"input": "5 2 3",
"output": "1"
},
{
"input": "4 7 10",
"output": "3"
},
{
"input": "100 100 100",
"output": "2"
},
{
"input": "10 100 3",
"output": "3"
},
{
"input": "2 9 29",
"output": "9"
},
{
"input": "4 6 10",
"output": "3"
},
{
"input": "3 70 58",
"output": "35"
},
{
"input": "5 7 10",
"output": "3"
},
{
"input": "5 30 22",
"output": "10"
},
{
"input": "5 5 6",
"output": "2"
},
{
"input": "2 4 3",
"output": "3"
},
{
"input": "10 10 31",
"output": "3"
},
{
"input": "2 1 1",
"output": "1"
},
{
"input": "10 98 99",
"output": "19"
},
{
"input": "4 10 16",
"output": "5"
},
{
"input": "11 4 8",
"output": "1"
},
{
"input": "5 10 14",
"output": "4"
},
{
"input": "6 7 35",
"output": "7"
},
{
"input": "5 6 7",
"output": "2"
},
{
"input": "4 15 3",
"output": "3"
},
{
"input": "7 48 77",
"output": "16"
},
{
"input": "4 4 10",
"output": "3"
},
{
"input": "4 7 20",
"output": "6"
},
{
"input": "5 2 8",
"output": "2"
},
{
"input": "3 2 3",
"output": "1"
},
{
"input": "14 95 1",
"output": "1"
},
{
"input": "99 82 53",
"output": "1"
},
{
"input": "10 71 27",
"output": "9"
},
{
"input": "5 7 8",
"output": "2"
},
{
"input": "11 77 77",
"output": "12"
},
{
"input": "10 5 28",
"output": "3"
},
{
"input": "7 3 12",
"output": "2"
},
{
"input": "10 15 17",
"output": "3"
},
{
"input": "7 7 7",
"output": "1"
},
{
"input": "4 11 18",
"output": "6"
},
{
"input": "3 3 4",
"output": "2"
},
{
"input": "9 2 10",
"output": "1"
},
{
"input": "100 90 20",
"output": "1"
},
{
"input": "3 2 2",
"output": "1"
},
{
"input": "12 45 60",
"output": "8"
},
{
"input": "3 94 79",
"output": "47"
},
{
"input": "41 67 34",
"output": "2"
},
{
"input": "9 3 23",
"output": "2"
},
{
"input": "10 20 57",
"output": "7"
},
{
"input": "55 27 30",
"output": "1"
},
{
"input": "100 100 10",
"output": "1"
},
{
"input": "20 8 70",
"output": "3"
},
{
"input": "3 3 3",
"output": "1"
},
{
"input": "4 9 15",
"output": "5"
},
{
"input": "3 1 3",
"output": "1"
},
{
"input": "2 94 94",
"output": "94"
},
{
"input": "5 3 11",
"output": "2"
},
{
"input": "4 3 2",
"output": "1"
},
{
"input": "12 12 100",
"output": "9"
},
{
"input": "6 75 91",
"output": "25"
},
{
"input": "3 4 3",
"output": "2"
},
{
"input": "3 2 5",
"output": "2"
},
{
"input": "6 5 15",
"output": "3"
},
{
"input": "4 3 6",
"output": "2"
},
{
"input": "3 9 9",
"output": "4"
},
{
"input": "26 93 76",
"output": "6"
},
{
"input": "41 34 67",
"output": "2"
},
{
"input": "6 12 6",
"output": "3"
},
{
"input": "5 20 8",
"output": "5"
},
{
"input": "2 1 3",
"output": "1"
},
{
"input": "35 66 99",
"output": "4"
},
{
"input": "30 7 91",
"output": "3"
},
{
"input": "5 22 30",
"output": "10"
},
{
"input": "8 19 71",
"output": "10"
},
{
"input": "3 5 6",
"output": "3"
},
{
"input": "5 3 8",
"output": "2"
},
{
"input": "2 4 2",
"output": "2"
},
{
"input": "4 3 7",
"output": "2"
},
{
"input": "5 20 10",
"output": "5"
},
{
"input": "5 100 50",
"output": "25"
},
{
"input": "6 3 10",
"output": "2"
},
{
"input": "2 90 95",
"output": "90"
},
{
"input": "4 8 6",
"output": "3"
},
{
"input": "6 10 3",
"output": "2"
},
{
"input": "3 3 5",
"output": "2"
},
{
"input": "5 33 33",
"output": "11"
},
{
"input": "5 5 8",
"output": "2"
},
{
"input": "19 24 34",
"output": "3"
},
{
"input": "5 5 12",
"output": "3"
},
{
"input": "8 7 10",
"output": "2"
},
{
"input": "5 56 35",
"output": "17"
},
{
"input": "4 3 5",
"output": "1"
},
{
"input": "18 100 50",
"output": "8"
},
{
"input": "5 6 8",
"output": "2"
},
{
"input": "5 98 100",
"output": "33"
},
{
"input": "6 5 8",
"output": "2"
},
{
"input": "3 40 80",
"output": "40"
},
{
"input": "4 8 11",
"output": "4"
},
{
"input": "66 100 99",
"output": "3"
},
{
"input": "17 100 79",
"output": "10"
},
{
"input": "3 2 10",
"output": "2"
},
{
"input": "99 100 99",
"output": "2"
},
{
"input": "21 100 5",
"output": "5"
},
{
"input": "3 10 2",
"output": "2"
},
{
"input": "4 100 63",
"output": "33"
},
{
"input": "2 2 10",
"output": "2"
},
{
"input": "5 94 79",
"output": "31"
},
{
"input": "4 12 5",
"output": "4"
},
{
"input": "5 5 40",
"output": "5"
},
{
"input": "99 99 99",
"output": "1"
},
{
"input": "8 97 44",
"output": "16"
},
{
"input": "11 4 10",
"output": "1"
},
{
"input": "6 3 3",
"output": "1"
},
{
"input": "7 3 4",
"output": "1"
},
{
"input": "8 4 4",
"output": "1"
},
{
"input": "9 4 5",
"output": "1"
},
{
"input": "12 6 6",
"output": "1"
},
{
"input": "4 48 89",
"output": "29"
},
{
"input": "8 3 6",
"output": "1"
},
{
"input": "4 6 3",
"output": "2"
},
{
"input": "5 5 1",
"output": "1"
},
{
"input": "11 6 5",
"output": "1"
},
{
"input": "4 5 4",
"output": "2"
},
{
"input": "6 6 4",
"output": "1"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "4 1 3",
"output": "1"
},
{
"input": "3 3 1",
"output": "1"
},
{
"input": "9 4 6",
"output": "1"
},
{
"input": "6 5 6",
"output": "1"
},
{
"input": "2 2 3",
"output": "2"
},
{
"input": "4 5 1",
"output": "1"
},
{
"input": "13 6 7",
"output": "1"
},
{
"input": "14 7 7",
"output": "1"
},
{
"input": "12 97 13",
"output": "8"
},
{
"input": "4 2 9",
"output": "2"
},
{
"input": "10 20 59",
"output": "7"
},
{
"input": "12 34 56",
"output": "7"
},
{
"input": "4 5 9",
"output": "3"
},
{
"input": "2 2 2",
"output": "2"
},
{
"input": "4 66 41",
"output": "22"
}
] | 1,515,004,879
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 3
| 61
| 5,632,000
|
l = input().split(' ')
n = int(l[0]); a = int(l[1]); b = int(l[2])
p = (a+b)//n
m = n - a//p - b//p
c = a%p; d = b%p
if m == 1 :
print(max(a//(a//p+1), b//(b//p+1)))
else :
print(min(p + c//(a//p),p + d//(b//p)))
|
Title: Two Cakes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into *a* pieces, and the second one — into *b* pieces.
Ivan knows that there will be *n* people at the celebration (including himself), so Ivan has set *n* plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:
1. Each piece of each cake is put on some plate; 1. Each plate contains at least one piece of cake; 1. No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number *x* such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least *x* pieces of cake.
Help Ivan to calculate this number *x*!
Input Specification:
The first line contains three integers *n*, *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100, 2<=≤<=*n*<=≤<=*a*<=+<=*b*) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.
Output Specification:
Print the maximum possible number *x* such that Ivan can distribute the cake in such a way that each plate will contain at least *x* pieces of cake.
Demo Input:
['5 2 3\n', '4 7 10\n']
Demo Output:
['1\n', '3\n']
Note:
In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.
In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.
|
```python
l = input().split(' ')
n = int(l[0]); a = int(l[1]); b = int(l[2])
p = (a+b)//n
m = n - a//p - b//p
c = a%p; d = b%p
if m == 1 :
print(max(a//(a//p+1), b//(b//p+1)))
else :
print(min(p + c//(a//p),p + d//(b//p)))
```
| -1
|
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,615,122,810
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 124
| 0
|
n=int(input())
array=list(map(int,input().split()))
idx=0
while n>0:
if array[idx]>=n:
print(idx+1)
break
n-=array[idx]
idx+=1
if idx==7:
idx=0
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
n=int(input())
array=list(map(int,input().split()))
idx=0
while n>0:
if array[idx]>=n:
print(idx+1)
break
n-=array[idx]
idx+=1
if idx==7:
idx=0
```
| 3
|
|
313
|
A
|
Ilya and Bank Account
|
PROGRAMMING
| 900
|
[
"implementation",
"number theory"
] | null | null |
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
|
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
|
In a single line print an integer — the maximum state of the bank account that Ilya can get.
|
[
"2230\n",
"-10\n",
"-100003\n"
] |
[
"2230\n",
"0\n",
"-10000\n"
] |
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
| 500
|
[
{
"input": "2230",
"output": "2230"
},
{
"input": "-10",
"output": "0"
},
{
"input": "-100003",
"output": "-10000"
},
{
"input": "544883178",
"output": "544883178"
},
{
"input": "-847251738",
"output": "-84725173"
},
{
"input": "423654797",
"output": "423654797"
},
{
"input": "-623563697",
"output": "-62356367"
},
{
"input": "645894116",
"output": "645894116"
},
{
"input": "-384381709",
"output": "-38438170"
},
{
"input": "437587210",
"output": "437587210"
},
{
"input": "-297534606",
"output": "-29753460"
},
{
"input": "891773002",
"output": "891773002"
},
{
"input": "-56712976",
"output": "-5671296"
},
{
"input": "963662765",
"output": "963662765"
},
{
"input": "-272656295",
"output": "-27265625"
},
{
"input": "383441522",
"output": "383441522"
},
{
"input": "-477665112",
"output": "-47766511"
},
{
"input": "791725034",
"output": "791725034"
},
{
"input": "-812168727",
"output": "-81216872"
},
{
"input": "528894922",
"output": "528894922"
},
{
"input": "-479977172",
"output": "-47997712"
},
{
"input": "568044564",
"output": "568044564"
},
{
"input": "-392784794",
"output": "-39278474"
},
{
"input": "925596634",
"output": "925596634"
},
{
"input": "-836078769",
"output": "-83607876"
},
{
"input": "71036059",
"output": "71036059"
},
{
"input": "-337396162",
"output": "-33739612"
},
{
"input": "87129297",
"output": "87129297"
},
{
"input": "-648171877",
"output": "-64817187"
},
{
"input": "20218400",
"output": "20218400"
},
{
"input": "10",
"output": "10"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "-1000000000",
"output": "-100000000"
},
{
"input": "-102",
"output": "-10"
},
{
"input": "-120",
"output": "-10"
},
{
"input": "-20",
"output": "0"
},
{
"input": "-15",
"output": "-1"
},
{
"input": "-5575533",
"output": "-557553"
},
{
"input": "-50",
"output": "0"
},
{
"input": "-31",
"output": "-1"
},
{
"input": "-55",
"output": "-5"
},
{
"input": "-81",
"output": "-1"
},
{
"input": "-99",
"output": "-9"
},
{
"input": "-23",
"output": "-2"
},
{
"input": "-41",
"output": "-1"
},
{
"input": "-24",
"output": "-2"
},
{
"input": "46",
"output": "46"
}
] | 1,696,967,670
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
def bank_account():
status = "-100023"
status = list(status)
print(status[-2])
if int(status[-2]) > 0:
status.remove(status[-2])
else:
status.remove(status[-1])
print("".join(status))
bank_account()
|
Title: Ilya and Bank Account
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
Input Specification:
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
Output Specification:
In a single line print an integer — the maximum state of the bank account that Ilya can get.
Demo Input:
['2230\n', '-10\n', '-100003\n']
Demo Output:
['2230\n', '0\n', '-10000\n']
Note:
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
|
```python
def bank_account():
status = "-100023"
status = list(status)
print(status[-2])
if int(status[-2]) > 0:
status.remove(status[-2])
else:
status.remove(status[-1])
print("".join(status))
bank_account()
```
| 0
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,548,929,882
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 0
|
n = int(input())
a = list(map(int,input().split()))
a = sorted(a)
for i in range(n-1,-1,-1):
if(sum(a[0:i]) < sum(a[i:n])):
break
print(n-i)
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
a = list(map(int,input().split()))
a = sorted(a)
for i in range(n-1,-1,-1):
if(sum(a[0:i]) < sum(a[i:n])):
break
print(n-i)
```
| 0
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,634,572,685
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 154
| 0
|
c = max(list(map(int,input().split())))
k = list(range(c,7))
l = len(k)
if max(k)==1:
print('1/1')
elif c==6:
print('0/1')
else:
if l==2:
print('1/3')
elif l==3:
print('1/2')
elif l==4:
print('2/3')
else:
print('{}/6'.format(l))
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
c = max(list(map(int,input().split())))
k = list(range(c,7))
l = len(k)
if max(k)==1:
print('1/1')
elif c==6:
print('0/1')
else:
if l==2:
print('1/3')
elif l==3:
print('1/2')
elif l==4:
print('2/3')
else:
print('{}/6'.format(l))
```
| 0
|
7
|
D
|
Palindrome Degree
|
PROGRAMMING
| 2,200
|
[
"hashing",
"strings"
] |
D. Palindrome Degree
|
1
|
256
|
String *s* of length *n* is called *k*-palindrome, if it is a palindrome itself, and its prefix and suffix of length are (*k*<=-<=1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string *s* such a maximum number *k*, for which *s* is *k*-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
|
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
|
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
|
[
"a2A\n",
"abacaba\n"
] |
[
"1",
"6"
] |
none
| 0
|
[
{
"input": "a2A",
"output": "1"
},
{
"input": "abacaba",
"output": "6"
},
{
"input": "CCeCeCCCee",
"output": "4"
},
{
"input": "opooppppopppopoppopoooppopopooopopppooopppoppoppoppppoooppooooooopppoopoopooooppooooppppppppooopooop",
"output": "3"
},
{
"input": "odribmizzsgholprdsth",
"output": "1"
},
{
"input": "z",
"output": "1"
}
] | 1,463,048,621
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 7
| 1,000
| 5,632,000
|
s = input()
n = len(s)
calculated = [1]*n
total = 0
for i in range(1,n):
for j in range(i+1):
if s[j] != s[i-j]:
calculated[i] = 0
break
else:
calculated[i] = calculated[((i+1)//2)-1]+1
total += calculated[i]
print(total+1)
|
Title: Palindrome Degree
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
String *s* of length *n* is called *k*-palindrome, if it is a palindrome itself, and its prefix and suffix of length are (*k*<=-<=1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string *s* such a maximum number *k*, for which *s* is *k*-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
Input Specification:
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
Output Specification:
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
Demo Input:
['a2A\n', 'abacaba\n']
Demo Output:
['1', '6']
Note:
none
|
```python
s = input()
n = len(s)
calculated = [1]*n
total = 0
for i in range(1,n):
for j in range(i+1):
if s[j] != s[i-j]:
calculated[i] = 0
break
else:
calculated[i] = calculated[((i+1)//2)-1]+1
total += calculated[i]
print(total+1)
```
| 0
|
495
|
A
|
Digital Counter
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Malek lives in an apartment block with 100 floors numbered from 0 to 99. The apartment has an elevator with a digital counter showing the floor that the elevator is currently on. The elevator shows each digit of a number with 7 light sticks by turning them on or off. The picture below shows how the elevator shows each digit.
One day when Malek wanted to go from floor 88 to floor 0 using the elevator he noticed that the counter shows number 89 instead of 88. Then when the elevator started moving the number on the counter changed to 87. After a little thinking Malek came to the conclusion that there is only one explanation for this: One of the sticks of the counter was broken. Later that day Malek was thinking about the broken stick and suddenly he came up with the following problem.
Suppose the digital counter is showing number *n*. Malek calls an integer *x* (0<=≤<=*x*<=≤<=99) good if it's possible that the digital counter was supposed to show *x* but because of some(possibly none) broken sticks it's showing *n* instead. Malek wants to know number of good integers for a specific *n*. So you must write a program that calculates this number. Please note that the counter always shows two digits.
|
The only line of input contains exactly two digits representing number *n* (0<=≤<=*n*<=≤<=99). Note that *n* may have a leading zero.
|
In the only line of the output print the number of good integers.
|
[
"89\n",
"00\n",
"73\n"
] |
[
"2\n",
"4\n",
"15\n"
] |
In the first sample the counter may be supposed to show 88 or 89.
In the second sample the good integers are 00, 08, 80 and 88.
In the third sample the good integers are 03, 08, 09, 33, 38, 39, 73, 78, 79, 83, 88, 89, 93, 98, 99.
| 500
|
[
{
"input": "89",
"output": "2"
},
{
"input": "00",
"output": "4"
},
{
"input": "73",
"output": "15"
},
{
"input": "08",
"output": "2"
},
{
"input": "26",
"output": "4"
},
{
"input": "49",
"output": "6"
},
{
"input": "88",
"output": "1"
},
{
"input": "04",
"output": "6"
},
{
"input": "60",
"output": "4"
},
{
"input": "11",
"output": "49"
},
{
"input": "22",
"output": "4"
},
{
"input": "33",
"output": "9"
},
{
"input": "44",
"output": "9"
},
{
"input": "55",
"output": "16"
},
{
"input": "66",
"output": "4"
},
{
"input": "77",
"output": "25"
},
{
"input": "88",
"output": "1"
},
{
"input": "99",
"output": "4"
},
{
"input": "80",
"output": "2"
},
{
"input": "78",
"output": "5"
},
{
"input": "67",
"output": "10"
},
{
"input": "89",
"output": "2"
},
{
"input": "46",
"output": "6"
},
{
"input": "90",
"output": "4"
},
{
"input": "92",
"output": "4"
},
{
"input": "35",
"output": "12"
},
{
"input": "05",
"output": "8"
},
{
"input": "57",
"output": "20"
},
{
"input": "20",
"output": "4"
}
] | 1,522,257,741
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 93
| 7,065,600
|
n=list(map(int,input()))
a=[2,7,2,3,2,2,2,5,1,2]
print(a[n[0]]*a[n[1]])
|
Title: Digital Counter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Malek lives in an apartment block with 100 floors numbered from 0 to 99. The apartment has an elevator with a digital counter showing the floor that the elevator is currently on. The elevator shows each digit of a number with 7 light sticks by turning them on or off. The picture below shows how the elevator shows each digit.
One day when Malek wanted to go from floor 88 to floor 0 using the elevator he noticed that the counter shows number 89 instead of 88. Then when the elevator started moving the number on the counter changed to 87. After a little thinking Malek came to the conclusion that there is only one explanation for this: One of the sticks of the counter was broken. Later that day Malek was thinking about the broken stick and suddenly he came up with the following problem.
Suppose the digital counter is showing number *n*. Malek calls an integer *x* (0<=≤<=*x*<=≤<=99) good if it's possible that the digital counter was supposed to show *x* but because of some(possibly none) broken sticks it's showing *n* instead. Malek wants to know number of good integers for a specific *n*. So you must write a program that calculates this number. Please note that the counter always shows two digits.
Input Specification:
The only line of input contains exactly two digits representing number *n* (0<=≤<=*n*<=≤<=99). Note that *n* may have a leading zero.
Output Specification:
In the only line of the output print the number of good integers.
Demo Input:
['89\n', '00\n', '73\n']
Demo Output:
['2\n', '4\n', '15\n']
Note:
In the first sample the counter may be supposed to show 88 or 89.
In the second sample the good integers are 00, 08, 80 and 88.
In the third sample the good integers are 03, 08, 09, 33, 38, 39, 73, 78, 79, 83, 88, 89, 93, 98, 99.
|
```python
n=list(map(int,input()))
a=[2,7,2,3,2,2,2,5,1,2]
print(a[n[0]]*a[n[1]])
```
| 0
|
|
322
|
B
|
Ciel and Flowers
|
PROGRAMMING
| 1,600
|
[
"combinatorics",
"math"
] | null | null |
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
|
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
|
Print the maximal number of bouquets Fox Ciel can make.
|
[
"3 6 9\n",
"4 4 4\n",
"0 0 0\n"
] |
[
"6\n",
"4\n",
"0\n"
] |
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
| 1,000
|
[
{
"input": "3 6 9",
"output": "6"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 3 6",
"output": "3"
},
{
"input": "7 8 9",
"output": "7"
},
{
"input": "8 8 9",
"output": "8"
},
{
"input": "15 3 999",
"output": "339"
},
{
"input": "32 62 92",
"output": "62"
},
{
"input": "123456789 123456789 123456789",
"output": "123456789"
},
{
"input": "3 5 5",
"output": "4"
},
{
"input": "666806767 385540591 357848286",
"output": "470065214"
},
{
"input": "80010646 727118126 817880463",
"output": "541669744"
},
{
"input": "829651016 732259171 572879931",
"output": "711596705"
},
{
"input": "242854896 442432924 180395753",
"output": "288561190"
},
{
"input": "139978911 5123031 935395222",
"output": "360165721"
},
{
"input": "553182792 10264076 395427398",
"output": "319624755"
},
{
"input": "597790453 720437830 855459575",
"output": "724562619"
},
{
"input": "494914467 356982656 757942689",
"output": "536613270"
},
{
"input": "908118348 67156409 217974865",
"output": "397749873"
},
{
"input": "952726009 629846517 972974334",
"output": "851848953"
},
{
"input": "775140200 616574841 630329230",
"output": "674014756"
},
{
"input": "524780569 326748594 90361407",
"output": "313963523"
},
{
"input": "937984449 184405994 992844522",
"output": "705078321"
},
{
"input": "835108464 525983528 452876698",
"output": "604656229"
},
{
"input": "879716125 531124573 207876166",
"output": "539572288"
},
{
"input": "292920005 241298326 667908343",
"output": "400708891"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 999999999 999999998",
"output": "999999998"
},
{
"input": "999999998 999999998 999999999",
"output": "999999998"
},
{
"input": "0 1 1",
"output": "0"
},
{
"input": "0 1000000000 0",
"output": "333333333"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "1 0 0",
"output": "0"
},
{
"input": "0 2 2",
"output": "0"
},
{
"input": "3 3 5",
"output": "3"
},
{
"input": "2 2 0",
"output": "0"
},
{
"input": "0 5 5",
"output": "2"
},
{
"input": "2 0 11",
"output": "3"
},
{
"input": "9 9 7",
"output": "8"
},
{
"input": "65 30 74",
"output": "56"
}
] | 1,573,472,682
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 186
| 0
|
r,g,b = map(int, input().split())
k = min(r,g,b)
r-=k
g-=k
b-=k
k += r//3 + g//3 + b//3
print(k)
|
Title: Ciel and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Output Specification:
Print the maximal number of bouquets Fox Ciel can make.
Demo Input:
['3 6 9\n', '4 4 4\n', '0 0 0\n']
Demo Output:
['6\n', '4\n', '0\n']
Note:
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
|
```python
r,g,b = map(int, input().split())
k = min(r,g,b)
r-=k
g-=k
b-=k
k += r//3 + g//3 + b//3
print(k)
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,511,535,248
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 77
| 5,632,000
|
# your code goes here
a = input()
b = input()
c = int(a, 2)^int(b, 2)
c = (bin(c).split('0b'))[1]
dif = len(a) - len(c)
while dif:
c = '0' + c
dif -= 1
print(c)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
# your code goes here
a = input()
b = input()
c = int(a, 2)^int(b, 2)
c = (bin(c).split('0b'))[1]
dif = len(a) - len(c)
while dif:
c = '0' + c
dif -= 1
print(c)
```
| 3.97026
|
279
|
B
|
Books
|
PROGRAMMING
| 1,400
|
[
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null |
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
|
Print a single integer — the maximum number of books Valera can read.
|
[
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] |
[
"3\n",
"1\n"
] |
none
| 1,000
|
[
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4"
},
{
"input": "7 13\n6 8 14 9 4 11 10",
"output": "2"
},
{
"input": "10 15\n10 9 1 1 5 10 5 3 7 2",
"output": "3"
},
{
"input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4",
"output": "6"
},
{
"input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19",
"output": "4"
},
{
"input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52",
"output": "3"
},
{
"input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708",
"output": "0"
},
{
"input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550",
"output": "1"
}
] | 1,694,983,685
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 8
| 154
| 13,107,200
|
a , b = map(int,input().split())
li = list(map(int,input().split()))
l , r , c = 0 , 0 , 0
ans = 0
while r < a:
if c < b:
c+=li[r]
r+=1
if c <= b : ans = r - l
else: c -= li[l] ; l +=1
print(ans)
|
Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
a , b = map(int,input().split())
li = list(map(int,input().split()))
l , r , c = 0 , 0 , 0
ans = 0
while r < a:
if c < b:
c+=li[r]
r+=1
if c <= b : ans = r - l
else: c -= li[l] ; l +=1
print(ans)
```
| 0
|
|
439
|
B
|
Devu, the Dumb Guy
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] | null | null |
Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him *n* subjects, the *i**th* subject has *c**i* chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is *x* hours. In other words he can learn a chapter of a particular subject in *x* hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the *n* subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type.
|
The first line will contain two space separated integers *n*, *x* (1<=≤<=*n*,<=*x*<=≤<=105). The next line will contain *n* space separated integers: *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105).
|
Output a single integer representing the answer to the problem.
|
[
"2 3\n4 1\n",
"4 2\n5 1 2 1\n",
"3 3\n1 1 1\n"
] |
[
"11\n",
"10\n",
"6\n"
] |
Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours.
| 1,000
|
[
{
"input": "2 3\n4 1",
"output": "11"
},
{
"input": "4 2\n5 1 2 1",
"output": "10"
},
{
"input": "3 3\n1 1 1",
"output": "6"
},
{
"input": "20 4\n1 1 3 5 5 1 3 4 2 5 2 4 3 1 3 3 3 3 4 3",
"output": "65"
},
{
"input": "20 10\n6 6 1 2 6 4 5 3 6 5 4 5 6 5 4 6 6 2 3 3",
"output": "196"
},
{
"input": "1 1\n9273",
"output": "9273"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 2\n1",
"output": "2"
},
{
"input": "1 2\n2",
"output": "4"
},
{
"input": "2 1\n1 2",
"output": "3"
}
] | 1,664,606,172
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 171
| 7,270,400
|
n, x = map(int,input().split())
c = list(map(int,input().split()))
c.sort()
ans = 0
i = 0
n = len(c)
while i<n and x>1:
ans += c[i]*x
x-=1
i+=1
if i<n:
while i<n:
ans += c[i]
i+=1
print(ans)
|
Title: Devu, the Dumb Guy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him *n* subjects, the *i**th* subject has *c**i* chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is *x* hours. In other words he can learn a chapter of a particular subject in *x* hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the *n* subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type.
Input Specification:
The first line will contain two space separated integers *n*, *x* (1<=≤<=*n*,<=*x*<=≤<=105). The next line will contain *n* space separated integers: *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105).
Output Specification:
Output a single integer representing the answer to the problem.
Demo Input:
['2 3\n4 1\n', '4 2\n5 1 2 1\n', '3 3\n1 1 1\n']
Demo Output:
['11\n', '10\n', '6\n']
Note:
Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours.
|
```python
n, x = map(int,input().split())
c = list(map(int,input().split()))
c.sort()
ans = 0
i = 0
n = len(c)
while i<n and x>1:
ans += c[i]*x
x-=1
i+=1
if i<n:
while i<n:
ans += c[i]
i+=1
print(ans)
```
| 3
|
|
924
|
B
|
Three-level Laser
|
PROGRAMMING
| 1,600
|
[
"binary search",
"greedy",
"two pointers"
] | null | null |
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
|
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
|
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
|
[
"4 4\n1 3 5 7\n",
"10 8\n10 13 15 16 17 19 20 22 24 25\n",
"3 1\n2 5 10\n"
] |
[
"0.5\n",
"0.875\n",
"-1\n"
] |
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 1,000
|
[
{
"input": "4 4\n1 3 5 7",
"output": "0.5"
},
{
"input": "10 8\n10 13 15 16 17 19 20 22 24 25",
"output": "0.875"
},
{
"input": "3 1\n2 5 10",
"output": "-1"
},
{
"input": "5 3\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 128\n110 121 140 158 174 188 251 271 272 277",
"output": "0.86554621848739499157"
},
{
"input": "20 17\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.92857142857142860315"
},
{
"input": "30 23\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.95652173913043481157"
},
{
"input": "50 64\n257 258 350 375 1014 1017 1051 1097 1169 1177 1223 1836 1942 1983 2111 2131 2341 2418 2593 2902 2948 3157 3243 3523 3566 4079 4499 4754 5060 5624 6279 6976 7011 7071 7278 7366 7408 7466 7526 7837 7934 8532 8577 8680 9221 9271 9327 9411 9590 9794",
"output": "0.91891891891891896993"
},
{
"input": "5 2\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 2\n110 121 140 158 174 188 251 271 272 277",
"output": "-1"
},
{
"input": "30 5\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.80000000000000004441"
},
{
"input": "10 6\n110 121 140 158 174 188 251 271 272 277",
"output": "0.83333333333333337034"
},
{
"input": "20 4\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.25"
},
{
"input": "3 1000000000\n1 2 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "3 1\n1 2 3",
"output": "-1"
},
{
"input": "5 1000000000\n1 2 3 999999999 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "10 199\n1 3 190 191 193 195 196 197 199 200",
"output": "0.98994974874371854945"
},
{
"input": "10 300\n80 100 103 140 146 159 392 393 396 398",
"output": "0.98993288590604022747"
},
{
"input": "10 92\n44 119 252 281 303 323 351 363 377 392",
"output": "0.77528089887640450062"
},
{
"input": "4 2\n1 3 5 7",
"output": "-1"
},
{
"input": "8 2\n1 3 7 9 15 17 23 25",
"output": "-1"
},
{
"input": "3 5\n1 2 10",
"output": "-1"
},
{
"input": "4 7\n1 5 8 9",
"output": "0.42857142857142854764"
}
] | 1,676,443,838
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 76
| 93
| 14,643,200
|
def read_ints():
return map(int, input().split())
def read_int():
return int(input())
n, max_delta = read_ints()
a = list(read_ints())
ans = -1
right = 0
for left in range(n):
while right + 1 < n and a[right + 1] - a[left] <= max_delta:
right += 1
if right > left + 1:
res = (a[right] - a[left + 1]) / (a[right] - a[left])
ans = max(ans, res)
print(ans)
|
Title: Three-level Laser
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
Input Specification:
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
Output Specification:
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['4 4\n1 3 5 7\n', '10 8\n10 13 15 16 17 19 20 22 24 25\n', '3 1\n2 5 10\n']
Demo Output:
['0.5\n', '0.875\n', '-1\n']
Note:
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
def read_ints():
return map(int, input().split())
def read_int():
return int(input())
n, max_delta = read_ints()
a = list(read_ints())
ans = -1
right = 0
for left in range(n):
while right + 1 < n and a[right + 1] - a[left] <= max_delta:
right += 1
if right > left + 1:
res = (a[right] - a[left + 1]) / (a[right] - a[left])
ans = max(ans, res)
print(ans)
```
| 3
|
|
165
|
A
|
Supercentral Point
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
|
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
|
Print the only number — the number of supercentral points of the given set.
|
[
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
| 500
|
[
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,676,133,897
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 154
| 0
|
n = int(input())
pairs = []
for i in range(n):
pairs.append(tuple(map(int, input().split(" "))))
pairs.sort(key=lambda x: 1000 * x[0] + x[1])
pairsY = sorted(pairs, key=lambda y: 1000 * y[1] + y[0])
validX = set()
validY = set()
for i in range(1, n-1):
if(pairs[i-1][0] == pairs[i][0] and pairs[i+1][0] == pairs[i][0]):
validX.add(pairs[i])
if(pairsY[i-1][1] == pairsY[i][1] and pairsY[i+1][1] == pairsY[i][1]):
validY.add(pairsY[i])
print(len(validX.intersection(validY)))
|
Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
|
```python
n = int(input())
pairs = []
for i in range(n):
pairs.append(tuple(map(int, input().split(" "))))
pairs.sort(key=lambda x: 1000 * x[0] + x[1])
pairsY = sorted(pairs, key=lambda y: 1000 * y[1] + y[0])
validX = set()
validY = set()
for i in range(1, n-1):
if(pairs[i-1][0] == pairs[i][0] and pairs[i+1][0] == pairs[i][0]):
validX.add(pairs[i])
if(pairsY[i-1][1] == pairsY[i][1] and pairsY[i+1][1] == pairsY[i][1]):
validY.add(pairsY[i])
print(len(validX.intersection(validY)))
```
| 0
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,548,063,682
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 0
|
def sayHello(word):
hello = "hello"
count = 0
j = 0
for i in word:
if i == hello[j]:
count += 1
if count == len("hello"):
return "YES"
j += 1
return "NO"
word = input()
print(sayHello(word))
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
def sayHello(word):
hello = "hello"
count = 0
j = 0
for i in word:
if i == hello[j]:
count += 1
if count == len("hello"):
return "YES"
j += 1
return "NO"
word = input()
print(sayHello(word))
```
| 3.938
|
839
|
B
|
Game of the Rows
|
PROGRAMMING
| 1,900
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
Daenerys Targaryen has an army consisting of *k* groups of soldiers, the *i*-th group contains *a**i* soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has *n* rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1,<=2}, {3,<=4}, {4,<=5}, {5,<=6} or {7,<=8}.
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10000, 1<=≤<=*k*<=≤<=100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains *k* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=10000), where *a**i* denotes the number of soldiers in the *i*-th group.
It is guaranteed that *a*1<=+<=*a*2<=+<=...<=+<=*a**k*<=≤<=8·*n*.
|
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
|
[
"2 2\n5 8\n",
"1 2\n7 1\n",
"1 2\n4 4\n",
"1 4\n2 2 1 2\n"
] |
[
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] |
In the first sample, Daenerys can place the soldiers like in the figure below:
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
| 1,000
|
[
{
"input": "2 2\n5 8",
"output": "YES"
},
{
"input": "1 2\n7 1",
"output": "NO"
},
{
"input": "1 2\n4 4",
"output": "YES"
},
{
"input": "1 4\n2 2 1 2",
"output": "YES"
},
{
"input": "10000 100\n749 2244 949 2439 2703 44 2394 124 285 3694 3609 717 1413 155 974 1778 1448 1327 1487 3458 319 1395 3783 2184 2062 43 826 38 3276 807 1837 4635 171 1386 1768 1128 2020 2536 800 782 3058 174 455 83 647 595 658 109 33 23 70 39 38 1 6 35 94 9 22 12 6 1 2 2 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 9938",
"output": "YES"
},
{
"input": "100 15\n165 26 83 64 235 48 36 51 3 18 5 10 9 6 5",
"output": "YES"
},
{
"input": "1 4\n2 2 2 2",
"output": "NO"
},
{
"input": "5691 91\n6573 1666 2158 2591 4636 886 263 4217 389 29 1513 1172 617 2012 1855 798 1588 979 152 37 890 375 1091 839 385 382 1 255 117 289 119 224 182 69 19 71 115 13 4 22 35 2 60 12 6 12 19 9 3 2 2 6 5 1 7 7 3 1 5 1 7 1 4 1 1 3 2 1 2 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5631",
"output": "NO"
},
{
"input": "2000 50\n203 89 1359 3105 898 1381 248 365 108 766 961 630 265 819 838 125 1751 289 177 81 131 564 102 95 49 74 92 101 19 17 156 5 5 4 20 9 25 16 16 2 8 5 4 2 1 3 4 1 3 2",
"output": "NO"
},
{
"input": "10000 100\n800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800 800",
"output": "YES"
},
{
"input": "10000 100\n749 2244 949 2439 2703 44 2394 124 285 3694 3609 717 1413 155 974 1778 1448 1327 1487 3458 319 1395 3783 2184 2062 43 826 38 3276 807 1837 4635 171 1386 1768 1128 2020 2536 2050 1074 605 979 1724 1608 672 88 1243 129 718 544 3590 37 187 600 738 34 64 316 58 6 84 252 75 68 40 68 4 29 29 8 13 11 5 1 5 1 3 2 1 1 1 2 3 4 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 3",
"output": "NO"
},
{
"input": "8459 91\n778 338 725 1297 115 540 1452 2708 193 1806 1496 1326 2648 176 199 93 342 3901 2393 2718 800 3434 657 4037 291 690 1957 3280 73 6011 2791 1987 440 455 444 155 261 234 829 1309 1164 616 34 627 107 213 52 110 323 81 98 8 7 73 20 12 56 3 40 12 8 7 69 1 14 3 6 2 6 8 3 5 4 4 3 1 1 4 2 1 1 1 8 2 2 2 1 1 1 2 8421",
"output": "NO"
},
{
"input": "1 3\n2 3 2",
"output": "YES"
},
{
"input": "10000 91\n2351 1402 1137 2629 4718 1138 1839 1339 2184 2387 165 370 918 1476 2717 879 1152 5367 3940 608 941 766 1256 656 2768 916 4176 489 1989 1633 2725 2329 2795 1970 667 340 1275 120 870 488 225 59 64 255 207 3 37 127 19 224 34 283 144 50 132 60 57 29 18 6 7 4 4 15 3 5 1 10 5 2 3 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 9948",
"output": "YES"
},
{
"input": "10000 83\n64 612 2940 2274 1481 1713 860 1264 104 5616 2574 5292 4039 292 1416 854 3854 1140 4344 3904 1720 1968 442 884 2032 875 291 677 2780 3074 3043 2997 407 727 344 511 156 321 134 51 382 336 591 52 134 39 104 10 20 15 24 2 70 39 14 16 16 25 1 6 2 2 1 1 1 2 4 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 9968",
"output": "YES"
},
{
"input": "4000 71\n940 1807 57 715 532 212 3842 2180 2283 744 1453 800 1945 380 2903 293 633 391 2866 256 102 46 228 1099 434 210 244 14 27 4 63 53 3 9 36 25 1 12 2 14 12 28 2 28 8 5 11 8 2 3 6 4 1 1 1 3 2 1 1 1 2 2 1 1 1 1 1 2 1 1 3966",
"output": "YES"
},
{
"input": "3403 59\n1269 1612 453 795 1216 941 19 44 1796 324 2019 1397 651 382 841 2003 3013 638 1007 1001 351 95 394 149 125 13 116 183 20 78 208 19 152 10 151 177 16 23 17 22 8 1 3 2 6 1 5 3 13 1 8 4 3 4 4 4 2 2 3378",
"output": "YES"
},
{
"input": "2393 33\n1381 2210 492 3394 912 2927 1189 269 66 102 104 969 395 385 369 354 251 28 203 334 20 10 156 29 61 13 30 4 1 32 2 2 2436",
"output": "YES"
},
{
"input": "10000 100\n749 2244 949 2439 2703 44 2394 124 285 3694 3609 717 1413 155 974 1778 1448 1327 1487 3458 319 1395 3783 2184 2062 43 826 38 3276 807 1837 4635 171 1386 1768 1128 2020 2536 800 782 3058 174 455 83 647 595 658 109 33 23 70 39 38 1 6 35 94 9 22 12 6 1 2 2 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 9939",
"output": "NO"
},
{
"input": "10000 89\n1001 1531 2489 457 1415 617 2057 2658 3030 789 2500 3420 1550 376 720 78 506 293 1978 383 3195 2036 891 1741 1817 486 2650 360 2250 2531 3250 1612 2759 603 5321 1319 791 1507 265 174 877 1861 572 172 580 536 777 165 169 11 125 31 186 113 78 27 25 37 8 21 48 24 4 33 35 13 15 1 3 2 2 8 3 5 1 1 6 1 1 2 1 1 2 2 1 1 2 1 9953",
"output": "NO"
},
{
"input": "4 16\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "NO"
},
{
"input": "10000 71\n110 14 2362 260 423 881 1296 3904 1664 849 57 631 1922 917 4832 1339 3398 4578 59 2663 2223 698 4002 3013 747 699 1230 2750 239 1409 6291 2133 1172 5824 181 797 26 281 574 557 19 82 624 387 278 53 64 163 22 617 15 35 42 48 14 140 171 36 28 22 5 49 17 5 10 14 13 1 3 3 9979",
"output": "NO"
},
{
"input": "3495 83\n2775 2523 1178 512 3171 1159 1382 2146 2192 1823 799 231 502 16 99 309 656 665 222 285 11 106 244 137 241 45 41 29 485 6 62 38 94 5 7 93 48 5 10 13 2 1 2 1 4 8 5 9 4 6 1 1 1 3 4 3 7 1 2 3 1 1 7 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 3443",
"output": "NO"
},
{
"input": "1000 40\n1701 1203 67 464 1884 761 11 559 29 115 405 133 174 63 147 93 41 19 1 15 41 8 33 4 4 1 4 1 1 2 1 2 1 1 2 1 1 2 1 4",
"output": "NO"
},
{
"input": "347 20\n55 390 555 426 140 360 29 115 23 113 58 30 33 1 23 3 35 5 7 363",
"output": "NO"
},
{
"input": "10000 100\n749 2244 949 2439 2703 44 2394 124 285 3694 3609 717 1413 155 974 1778 1448 1327 1487 3458 319 1395 3783 2184 2062 43 826 38 3276 807 1837 4635 171 1386 1768 1128 2020 2536 800 782 3058 174 455 83 647 595 658 109 33 23 70 39 38 1 6 35 94 9 22 12 6 1 2 2 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 9940",
"output": "NO"
},
{
"input": "10000 93\n1388 119 711 23 4960 4002 2707 188 813 1831 334 543 338 3402 1808 3368 1428 971 985 220 1521 457 457 140 332 1503 1539 2095 1891 269 5223 226 1528 190 428 5061 410 1587 1149 1934 2275 1337 1828 275 181 85 499 29 585 808 751 401 635 461 181 164 274 36 401 255 38 60 76 16 6 35 79 46 1 39 11 2 8 2 4 14 3 1 1 1 1 1 2 1 3 1 1 1 1 2 1 1 9948",
"output": "NO"
},
{
"input": "4981 51\n5364 2166 223 742 350 1309 15 229 4100 3988 227 1719 9 125 787 427 141 842 171 2519 32 2554 2253 721 775 88 720 9 397 513 100 291 111 32 238 42 152 108 5 58 96 53 7 19 11 2 5 5 6 2 4966",
"output": "NO"
},
{
"input": "541 31\n607 204 308 298 398 213 1182 58 162 46 64 12 38 91 29 2 4 12 19 3 7 9 3 6 1 1 2 1 3 1 529",
"output": "YES"
},
{
"input": "100 100\n6 129 61 6 87 104 45 28 3 35 2 14 1 37 2 4 24 4 3 1 6 4 2 1 1 3 1 2 2 9 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 22",
"output": "NO"
},
{
"input": "1 4\n2 2 2 1",
"output": "YES"
},
{
"input": "1 3\n2 2 2",
"output": "YES"
},
{
"input": "2 5\n8 2 2 2 2",
"output": "YES"
},
{
"input": "1 4\n1 1 2 2",
"output": "YES"
},
{
"input": "1 3\n2 2 3",
"output": "YES"
},
{
"input": "1 3\n4 2 2",
"output": "YES"
},
{
"input": "1 4\n2 1 2 2",
"output": "YES"
},
{
"input": "1 3\n3 2 2",
"output": "YES"
},
{
"input": "2 8\n2 2 2 2 2 2 1 1",
"output": "YES"
},
{
"input": "2 6\n2 2 2 2 2 2",
"output": "YES"
},
{
"input": "1 4\n1 2 2 2",
"output": "YES"
},
{
"input": "1 4\n1 1 1 1",
"output": "YES"
},
{
"input": "2 7\n2 2 2 2 2 2 2",
"output": "YES"
},
{
"input": "2 8\n1 1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "3 7\n12 2 2 2 2 2 2",
"output": "YES"
},
{
"input": "2 6\n4 1 3 1 1 3",
"output": "NO"
},
{
"input": "1 3\n2 2 4",
"output": "YES"
},
{
"input": "5 15\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "YES"
},
{
"input": "2 8\n2 2 2 2 1 1 1 1",
"output": "YES"
},
{
"input": "1 2\n6 2",
"output": "YES"
},
{
"input": "4 13\n2 2 2 2 2 2 2 2 2 2 2 2 4",
"output": "YES"
},
{
"input": "2 7\n1 1 1 4 2 2 2",
"output": "YES"
},
{
"input": "3 8\n8 2 2 2 2 2 2 2",
"output": "YES"
},
{
"input": "2 8\n1 1 1 1 2 2 2 2",
"output": "YES"
},
{
"input": "2 8\n2 2 2 2 1 1 2 2",
"output": "YES"
},
{
"input": "1 4\n2 2 1 1",
"output": "YES"
},
{
"input": "3 9\n2 2 2 2 2 2 2 2 2",
"output": "YES"
},
{
"input": "2 6\n2 2 2 2 2 5",
"output": "YES"
},
{
"input": "1 1\n6",
"output": "YES"
},
{
"input": "2 1\n16",
"output": "YES"
},
{
"input": "1 1\n2",
"output": "YES"
},
{
"input": "2 8\n2 2 2 2 2 2 2 1",
"output": "NO"
},
{
"input": "4 16\n1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "YES"
},
{
"input": "2 7\n4 1 1 1 1 2 2",
"output": "YES"
},
{
"input": "2 6\n2 2 2 5 2 2",
"output": "YES"
},
{
"input": "3 1\n22",
"output": "YES"
},
{
"input": "2 8\n2 2 2 2 1 1 1 3",
"output": "NO"
},
{
"input": "3 12\n2 1 2 2 2 1 2 2 2 1 2 2",
"output": "YES"
},
{
"input": "1 4\n2 2 3 1",
"output": "NO"
},
{
"input": "2 6\n5 2 2 2 2 2",
"output": "YES"
},
{
"input": "20 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "1 3\n2 2 1",
"output": "YES"
},
{
"input": "1 2\n3 3",
"output": "YES"
},
{
"input": "2 6\n2 3 2 2 3 2",
"output": "YES"
},
{
"input": "2 8\n2 2 1 1 2 2 2 2",
"output": "YES"
},
{
"input": "2 6\n3 3 2 2 2 2",
"output": "YES"
},
{
"input": "3 12\n2 2 2 2 2 2 2 2 2 1 1 1",
"output": "YES"
},
{
"input": "3 10\n2 2 2 2 2 2 2 2 2 3",
"output": "YES"
}
] | 1,601,323,556
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 139
| 0
|
n,k=map(int,input().split())
s=list(map(int,input().split()))
q=n
double=2*n
for i in range(k):
p=min(double,s[i]//2)
double-=p
s[i]-=(p*2)
for i in range(k):
p=min(q,s[i]//4)
q-=p
s[i]-=(p*4)
if q!=0:
for i in s:
if i==0:
continue
if i==1 and q!=0:
q-=0.5
elif i==2 :
if q%1==0.5 :
q-=0.5
elif q>0:
q-=1
else:
print('NO')
exit()
elif i>=3 and q>=1 :
q-=1
else:
print('NO')
exit()
print('YES')
exit()
elif q==0:
for i in s:
if i!=0:
print('NO')
exit()
print('YES')
|
Title: Game of the Rows
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Daenerys Targaryen has an army consisting of *k* groups of soldiers, the *i*-th group contains *a**i* soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has *n* rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1,<=2}, {3,<=4}, {4,<=5}, {5,<=6} or {7,<=8}.
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10000, 1<=≤<=*k*<=≤<=100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains *k* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=10000), where *a**i* denotes the number of soldiers in the *i*-th group.
It is guaranteed that *a*1<=+<=*a*2<=+<=...<=+<=*a**k*<=≤<=8·*n*.
Output Specification:
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['2 2\n5 8\n', '1 2\n7 1\n', '1 2\n4 4\n', '1 4\n2 2 1 2\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample, Daenerys can place the soldiers like in the figure below:
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
|
```python
n,k=map(int,input().split())
s=list(map(int,input().split()))
q=n
double=2*n
for i in range(k):
p=min(double,s[i]//2)
double-=p
s[i]-=(p*2)
for i in range(k):
p=min(q,s[i]//4)
q-=p
s[i]-=(p*4)
if q!=0:
for i in s:
if i==0:
continue
if i==1 and q!=0:
q-=0.5
elif i==2 :
if q%1==0.5 :
q-=0.5
elif q>0:
q-=1
else:
print('NO')
exit()
elif i>=3 and q>=1 :
q-=1
else:
print('NO')
exit()
print('YES')
exit()
elif q==0:
for i in s:
if i!=0:
print('NO')
exit()
print('YES')
```
| 0
|
|
958
|
D1
|
Hyperspace Jump (easy)
|
PROGRAMMING
| 1,400
|
[
"expression parsing",
"math"
] | null | null |
The Rebel fleet is on the run. It consists of *m* ships currently gathered around a single planet. Just a few seconds ago, the vastly more powerful Empire fleet has appeared in the same solar system, and the Rebels will need to escape into hyperspace. In order to spread the fleet, the captain of each ship has independently come up with the coordinate to which that ship will jump. In the obsolete navigation system used by the Rebels, this coordinate is given as the value of an arithmetic expression of the form .
To plan the future of the resistance movement, Princess Heidi needs to know, for each ship, how many ships are going to end up at the same coordinate after the jump. You are her only hope!
|
The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) – the number of ships. The next *m* lines describe one jump coordinate each, given as an arithmetic expression. An expression has the form (a+b)/c. Namely, it consists of: an opening parenthesis (, a positive integer *a* of up to two decimal digits, a plus sign +, a positive integer *b* of up to two decimal digits, a closing parenthesis ), a slash /, and a positive integer *c* of up to two decimal digits.
|
Print a single line consisting of *m* space-separated integers. The *i*-th integer should be equal to the number of ships whose coordinate is equal to that of the *i*-th ship (including the *i*-th ship itself).
|
[
"4\n(99+98)/97\n(26+4)/10\n(12+33)/15\n(5+1)/7\n"
] |
[
"1 2 2 1 "
] |
In the sample testcase, the second and the third ship will both end up at the coordinate 3.
Note that this problem has only two versions – easy and hard.
| 0
|
[
{
"input": "4\n(99+98)/97\n(26+4)/10\n(12+33)/15\n(5+1)/7",
"output": "1 2 2 1 "
},
{
"input": "10\n(44+98)/19\n(36+58)/47\n(62+74)/68\n(69+95)/82\n(26+32)/29\n(32+46)/39\n(32+24)/28\n(47+61)/54\n(39+13)/26\n(98+98)/98",
"output": "1 9 9 9 9 9 9 9 9 9 "
},
{
"input": "30\n(89+76)/87\n(81+78)/18\n(60+97)/32\n(41+14)/48\n(55+65)/27\n(29+15)/95\n(64+13)/96\n(78+30)/75\n(43+6)/60\n(69+34)/48\n(62+2)/97\n(85+42)/3\n(4+97)/42\n(1+18)/39\n(46+55)/76\n(22+59)/24\n(62+81)/98\n(64+8)/51\n(9+59)/48\n(47+2)/80\n(33+74)/76\n(61+83)/44\n(86+4)/51\n(65+41)/49\n(53+36)/45\n(6+19)/15\n(51+21)/68\n(98+36)/86\n(92+65)/86\n(27+58)/78",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
}
] | 1,523,691,926
| 2,426
|
PyPy 3
|
OK
|
TESTS
| 16
| 2,370
| 48,537,600
|
import re
from collections import defaultdict
P = re.compile(r'\((\d+)\+(\d+)\)/(\d+)')
m = int(input())
s = defaultdict(int)
xs = []
for _ in range(m):
a, b, c = map(int, P.match(input()).groups())
x = (a + b) / c
s[x] += 1
xs.append(x)
print(' '.join(str(s[x]) for x in xs))
|
Title: Hyperspace Jump (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Rebel fleet is on the run. It consists of *m* ships currently gathered around a single planet. Just a few seconds ago, the vastly more powerful Empire fleet has appeared in the same solar system, and the Rebels will need to escape into hyperspace. In order to spread the fleet, the captain of each ship has independently come up with the coordinate to which that ship will jump. In the obsolete navigation system used by the Rebels, this coordinate is given as the value of an arithmetic expression of the form .
To plan the future of the resistance movement, Princess Heidi needs to know, for each ship, how many ships are going to end up at the same coordinate after the jump. You are her only hope!
Input Specification:
The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) – the number of ships. The next *m* lines describe one jump coordinate each, given as an arithmetic expression. An expression has the form (a+b)/c. Namely, it consists of: an opening parenthesis (, a positive integer *a* of up to two decimal digits, a plus sign +, a positive integer *b* of up to two decimal digits, a closing parenthesis ), a slash /, and a positive integer *c* of up to two decimal digits.
Output Specification:
Print a single line consisting of *m* space-separated integers. The *i*-th integer should be equal to the number of ships whose coordinate is equal to that of the *i*-th ship (including the *i*-th ship itself).
Demo Input:
['4\n(99+98)/97\n(26+4)/10\n(12+33)/15\n(5+1)/7\n']
Demo Output:
['1 2 2 1 ']
Note:
In the sample testcase, the second and the third ship will both end up at the coordinate 3.
Note that this problem has only two versions – easy and hard.
|
```python
import re
from collections import defaultdict
P = re.compile(r'\((\d+)\+(\d+)\)/(\d+)')
m = int(input())
s = defaultdict(int)
xs = []
for _ in range(m):
a, b, c = map(int, P.match(input()).groups())
x = (a + b) / c
s[x] += 1
xs.append(x)
print(' '.join(str(s[x]) for x in xs))
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,658,777,545
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
word1 = list(input())
word2 = list(input())
if len(word1) != len(word2):
correct = True
for i in range(len(word1)):
if word1[i] != word2[-i]:
correct = False
break
if correct:
print("YES")
else:
print("NO")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
word1 = list(input())
word2 = list(input())
if len(word1) != len(word2):
correct = True
for i in range(len(word1)):
if word1[i] != word2[-i]:
correct = False
break
if correct:
print("YES")
else:
print("NO")
else:
print("NO")
```
| 0
|
366
|
B
|
Dima and To-do List
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.
Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete *k*<=-<=1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.
Overall, Dima has *n* tasks to do, each task has a unique number from 1 to *n*. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.
Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.
|
The first line of the input contains two integers *n*,<=*k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103), where *a**i* is the power Inna tells Dima off with if she is present in the room while he is doing the *i*-th task.
It is guaranteed that *n* is divisible by *k*.
|
In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.
|
[
"6 2\n3 2 1 6 5 4\n",
"10 5\n1 3 5 7 9 9 4 1 8 5\n"
] |
[
"1\n",
"3\n"
] |
Explanation of the first example.
If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as *k* = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.
Explanation of the second example.
In the second example *k* = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.
| 1,000
|
[
{
"input": "6 2\n3 2 1 6 5 4",
"output": "1"
},
{
"input": "10 5\n1 3 5 7 9 9 4 1 8 5",
"output": "3"
},
{
"input": "20 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 10\n8 4 5 7 6 9 2 2 3 5",
"output": "7"
},
{
"input": "50 10\n1 2 3 4 5 6 7 8 9 10 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "4 2\n2 1 1 3",
"output": "1"
},
{
"input": "15 5\n5 5 5 5 5 1 2 3 4 5 1 2 3 4 5",
"output": "1"
},
{
"input": "20 10\n3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 6 4",
"output": "1"
}
] | 1,455,470,709
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 139
| 6,963,200
|
def main():
n, k = map(int, input().split())
aa = list(map(int, input().split()))
res = [sum(aa[i::k]) for i in range(k)]
print(min(range(k), key=res.__getitem__) + 1)
if __name__ == '__main__':
main()
|
Title: Dima and To-do List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.
Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete *k*<=-<=1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.
Overall, Dima has *n* tasks to do, each task has a unique number from 1 to *n*. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.
Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.
Input Specification:
The first line of the input contains two integers *n*,<=*k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103), where *a**i* is the power Inna tells Dima off with if she is present in the room while he is doing the *i*-th task.
It is guaranteed that *n* is divisible by *k*.
Output Specification:
In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.
Demo Input:
['6 2\n3 2 1 6 5 4\n', '10 5\n1 3 5 7 9 9 4 1 8 5\n']
Demo Output:
['1\n', '3\n']
Note:
Explanation of the first example.
If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as *k* = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.
Explanation of the second example.
In the second example *k* = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.
|
```python
def main():
n, k = map(int, input().split())
aa = list(map(int, input().split()))
res = [sum(aa[i::k]) for i in range(k)]
print(min(range(k), key=res.__getitem__) + 1)
if __name__ == '__main__':
main()
```
| 3
|
|
257
|
A
|
Sockets
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Vasya has got many devices that work on electricity. He's got *n* supply-line filters to plug the devices, the *i*-th supply-line filter has *a**i* sockets.
Overall Vasya has got *m* devices and *k* electrical sockets in his flat, he can plug the devices or supply-line filters directly. Of course, he can plug the supply-line filter to any other supply-line filter. The device (or the supply-line filter) is considered plugged to electricity if it is either plugged to one of *k* electrical sockets, or if it is plugged to some supply-line filter that is in turn plugged to electricity.
What minimum number of supply-line filters from the given set will Vasya need to plug all the devices he has to electricity? Note that all devices and supply-line filters take one socket for plugging and that he can use one socket to plug either one device or one supply-line filter.
|
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=50) — the number of supply-line filters, the number of devices and the number of sockets that he can plug to directly, correspondingly. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=50) — number *a**i* stands for the number of sockets on the *i*-th supply-line filter.
|
Print a single number — the minimum number of supply-line filters that is needed to plug all the devices to electricity. If it is impossible to plug all the devices even using all the supply-line filters, print -1.
|
[
"3 5 3\n3 1 2\n",
"4 7 2\n3 3 2 4\n",
"5 5 1\n1 3 1 2 1\n"
] |
[
"1\n",
"2\n",
"-1\n"
] |
In the first test case he can plug the first supply-line filter directly to electricity. After he plug it, he get 5 (3 on the supply-line filter and 2 remaining sockets for direct plugging) available sockets to plug. Thus, one filter is enough to plug 5 devices.
One of the optimal ways in the second test sample is to plug the second supply-line filter directly and plug the fourth supply-line filter to one of the sockets in the second supply-line filter. Thus, he gets exactly 7 sockets, available to plug: one to plug to the electricity directly, 2 on the second supply-line filter, 4 on the fourth supply-line filter. There's no way he can plug 7 devices if he use one supply-line filter.
| 500
|
[
{
"input": "3 5 3\n3 1 2",
"output": "1"
},
{
"input": "4 7 2\n3 3 2 4",
"output": "2"
},
{
"input": "5 5 1\n1 3 1 2 1",
"output": "-1"
},
{
"input": "4 5 8\n3 2 4 3",
"output": "0"
},
{
"input": "5 10 1\n4 3 4 2 4",
"output": "3"
},
{
"input": "7 13 2\n5 3 4 1 2 1 2",
"output": "5"
},
{
"input": "7 17 5\n1 6 2 1 1 4 3",
"output": "-1"
},
{
"input": "10 25 7\n5 7 4 8 3 3 5 4 5 5",
"output": "4"
},
{
"input": "10 8 4\n1 1 2 1 3 1 3 1 4 2",
"output": "2"
},
{
"input": "13 20 9\n2 9 2 2 5 11 10 10 13 4 6 11 14",
"output": "1"
},
{
"input": "9 30 8\n3 6 10 8 1 5 3 9 3",
"output": "3"
},
{
"input": "15 26 4\n3 6 7 1 5 2 4 4 7 3 8 7 2 4 8",
"output": "4"
},
{
"input": "20 20 3\n6 6 5 1 7 8 8 6 10 7 8 5 6 8 1 7 10 6 2 7",
"output": "2"
},
{
"input": "10 30 5\n4 5 3 3 4 4 4 3 5 1",
"output": "9"
},
{
"input": "20 30 1\n12 19 16 2 11 19 1 15 13 13 3 10 1 18 7 5 6 8 9 1",
"output": "2"
},
{
"input": "50 50 2\n2 2 4 5 2 1 5 4 5 4 5 2 1 2 3 3 5 1 2 2 1 3 4 5 5 4 3 2 2 1 3 2 3 2 4 4 1 3 5 4 3 2 4 3 4 4 4 4 3 4",
"output": "14"
},
{
"input": "5 50 6\n2 1 3 1 3",
"output": "-1"
},
{
"input": "20 50 10\n5 4 3 6 3 7 2 3 7 8 6 3 8 3 3 5 1 9 6 2",
"output": "7"
},
{
"input": "40 40 3\n2 1 4 2 4 2 3 3 3 3 1 2 3 2 2 3 4 2 3 1 2 4 1 4 1 4 3 3 1 1 3 1 3 4 4 3 1 1 2 4",
"output": "14"
},
{
"input": "33 49 16\n40 16 48 49 30 28 8 6 48 39 48 6 24 28 30 35 12 23 49 29 31 8 40 18 16 34 43 15 12 33 14 24 13",
"output": "1"
},
{
"input": "10 49 11\n5 18 1 19 11 11 16 5 6 6",
"output": "3"
},
{
"input": "50 30 1\n2 1 2 1 2 3 3 1 2 2 3 2 1 3 1 3 1 2 2 3 2 1 3 1 1 2 3 2 2 1 1 3 3 2 2 2 3 2 3 3 3 3 1 1 3 1 1 3 1 3",
"output": "15"
},
{
"input": "50 50 2\n1 2 3 2 1 2 4 2 3 4 3 1 3 2 2 3 1 4 2 1 4 4 2 2 2 3 2 3 1 1 4 4 1 1 2 3 4 2 2 3 4 3 4 3 3 3 2 3 1 1",
"output": "19"
},
{
"input": "49 49 3\n8 8 8 7 5 6 6 8 1 3 1 8 8 3 2 1 2 2 5 4 4 7 8 7 6 4 2 5 7 3 4 2 3 2 3 4 5 7 3 3 4 5 5 8 2 5 1 1 7",
"output": "7"
},
{
"input": "4 50 1\n3 11 5 5",
"output": "-1"
},
{
"input": "50 5 1\n1 1 1 1 1 1 1 2 1 1 2 2 2 2 1 2 1 2 1 2 2 1 1 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 1 2 1 1 2 1 1 2 1 2",
"output": "4"
},
{
"input": "50 23 2\n2 1 2 2 3 2 1 2 3 3 3 1 3 1 3 2 1 3 2 2 1 1 3 2 2 1 1 3 2 1 2 3 2 2 2 2 1 3 1 2 2 3 3 1 3 3 3 1 2 3",
"output": "11"
},
{
"input": "49 19 2\n3 2 3 2 3 3 2 3 3 1 2 1 2 3 2 3 1 3 1 3 1 3 2 2 2 2 3 3 1 1 2 3 2 3 1 2 3 3 1 1 3 1 1 3 1 1 1 2 3",
"output": "9"
},
{
"input": "10 50 5\n23 18 15 23 26 23 4 29 15 25",
"output": "2"
},
{
"input": "15 38 3\n3 5 5 5 4 1 3 1 5 4 2 4 3 1 1",
"output": "-1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "3 6 3\n2 2 2",
"output": "3"
},
{
"input": "1 7 4\n1",
"output": "-1"
},
{
"input": "47 7 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
}
] | 1,695,563,569
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 0
|
s=input()
t=input()
i=0
j=0
k=1
while i<len(t):
if s[i]==t[j]:
j=j+1
i=i+1
k=k+1
else:
i=i+1
print(k)
|
Title: Sockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got many devices that work on electricity. He's got *n* supply-line filters to plug the devices, the *i*-th supply-line filter has *a**i* sockets.
Overall Vasya has got *m* devices and *k* electrical sockets in his flat, he can plug the devices or supply-line filters directly. Of course, he can plug the supply-line filter to any other supply-line filter. The device (or the supply-line filter) is considered plugged to electricity if it is either plugged to one of *k* electrical sockets, or if it is plugged to some supply-line filter that is in turn plugged to electricity.
What minimum number of supply-line filters from the given set will Vasya need to plug all the devices he has to electricity? Note that all devices and supply-line filters take one socket for plugging and that he can use one socket to plug either one device or one supply-line filter.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=50) — the number of supply-line filters, the number of devices and the number of sockets that he can plug to directly, correspondingly. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=50) — number *a**i* stands for the number of sockets on the *i*-th supply-line filter.
Output Specification:
Print a single number — the minimum number of supply-line filters that is needed to plug all the devices to electricity. If it is impossible to plug all the devices even using all the supply-line filters, print -1.
Demo Input:
['3 5 3\n3 1 2\n', '4 7 2\n3 3 2 4\n', '5 5 1\n1 3 1 2 1\n']
Demo Output:
['1\n', '2\n', '-1\n']
Note:
In the first test case he can plug the first supply-line filter directly to electricity. After he plug it, he get 5 (3 on the supply-line filter and 2 remaining sockets for direct plugging) available sockets to plug. Thus, one filter is enough to plug 5 devices.
One of the optimal ways in the second test sample is to plug the second supply-line filter directly and plug the fourth supply-line filter to one of the sockets in the second supply-line filter. Thus, he gets exactly 7 sockets, available to plug: one to plug to the electricity directly, 2 on the second supply-line filter, 4 on the fourth supply-line filter. There's no way he can plug 7 devices if he use one supply-line filter.
|
```python
s=input()
t=input()
i=0
j=0
k=1
while i<len(t):
if s[i]==t[j]:
j=j+1
i=i+1
k=k+1
else:
i=i+1
print(k)
```
| 0
|
|
380
|
C
|
Sereja and Brackets
|
PROGRAMMING
| 2,000
|
[
"data structures",
"schedules"
] | null | null |
Sereja has a bracket sequence *s*1,<=*s*2,<=...,<=*s**n*, or, in other words, a string *s* of length *n*, consisting of characters "(" and ")".
Sereja needs to answer *m* queries, each of them is described by two integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). The answer to the *i*-th query is the length of the maximum correct bracket subsequence of sequence *s**l**i*,<=*s**l**i*<=+<=1,<=...,<=*s**r**i*. Help Sereja answer all queries.
You can find the definitions for a subsequence and a correct bracket sequence in the notes.
|
The first line contains a sequence of characters *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*n*<=≤<=106) without any spaces. Each character is either a "(" or a ")". The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains a pair of integers. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) — the description of the *i*-th query.
|
Print the answer to each question on a single line. Print the answers in the order they go in the input.
|
[
"())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10\n"
] |
[
"0\n0\n2\n10\n4\n6\n6\n"
] |
A subsequence of length |*x*| of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is string *x* = *s*<sub class="lower-index">*k*<sub class="lower-index">1</sub></sub>*s*<sub class="lower-index">*k*<sub class="lower-index">2</sub></sub>... *s*<sub class="lower-index">*k*<sub class="lower-index">|*x*|</sub></sub> (1 ≤ *k*<sub class="lower-index">1</sub> < *k*<sub class="lower-index">2</sub> < ... < *k*<sub class="lower-index">|*x*|</sub> ≤ |*s*|).
A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
For the third query required sequence will be «()».
For the fourth query required sequence will be «()(())(())».
| 1,500
|
[
{
"input": "())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10",
"output": "0\n0\n2\n10\n4\n6\n6"
},
{
"input": "(((((()((((((((((()((()(((((\n1\n8 15",
"output": "0"
},
{
"input": "((()((())(((((((((()(()(()(((((((((((((((()(()((((((((((((((()(((((((((((((((((((()(((\n39\n28 56\n39 46\n57 63\n29 48\n51 75\n14 72\n5 70\n51 73\n10 64\n31 56\n50 54\n15 78\n78 82\n1 11\n1 70\n1 19\n10 22\n13 36\n3 10\n34 40\n51 76\n64 71\n36 75\n24 71\n1 63\n5 14\n46 67\n32 56\n39 43\n43 56\n61 82\n2 78\n1 21\n10 72\n49 79\n12 14\n53 79\n15 31\n7 47",
"output": "4\n4\n2\n4\n2\n12\n16\n2\n12\n4\n0\n12\n0\n6\n18\n6\n2\n6\n6\n0\n2\n0\n6\n8\n18\n4\n2\n4\n2\n2\n2\n18\n8\n12\n2\n0\n2\n6\n12"
},
{
"input": "))(()))))())())))))())((()()))))()))))))))))))\n9\n26 42\n21 22\n6 22\n7 26\n43 46\n25 27\n32 39\n22 40\n2 45",
"output": "4\n0\n6\n8\n0\n2\n2\n10\n20"
},
{
"input": "(()((((()(())((((((((()((((((()((((\n71\n15 29\n17 18\n5 26\n7 10\n16 31\n26 35\n2 30\n16 24\n2 24\n7 12\n15 18\n12 13\n25 30\n1 30\n12 13\n16 20\n6 35\n20 28\n18 23\n9 31\n12 35\n14 17\n8 16\n3 10\n12 33\n7 19\n2 33\n7 17\n21 27\n10 30\n29 32\n9 28\n18 32\n28 31\n31 33\n4 26\n15 27\n10 17\n8 14\n11 28\n8 23\n17 33\n4 14\n3 6\n6 34\n19 23\n4 21\n16 27\n14 27\n6 19\n31 32\n29 32\n9 17\n1 21\n2 31\n18 29\n16 26\n15 18\n4 5\n13 20\n9 28\n18 30\n1 32\n2 9\n16 24\n1 20\n4 15\n16 23\n19 34\n5 22\n5 23",
"output": "2\n0\n8\n2\n4\n2\n10\n2\n10\n4\n0\n0\n0\n10\n0\n0\n10\n2\n2\n8\n4\n0\n6\n2\n4\n6\n12\n6\n2\n6\n2\n6\n4\n2\n0\n8\n2\n4\n6\n4\n8\n4\n6\n0\n10\n2\n6\n2\n2\n6\n0\n2\n4\n8\n12\n2\n2\n0\n0\n0\n6\n2\n12\n4\n2\n8\n6\n2\n4\n6\n8"
},
{
"input": "(((())((((()()((((((()((()(((((((((((()((\n6\n20 37\n28 32\n12 18\n7 25\n21 33\n4 5",
"output": "4\n0\n2\n6\n4\n2"
},
{
"input": "(((()((((()()()(()))((((()(((()))()((((()))()((())\n24\n37 41\n13 38\n31 34\n14 16\n29 29\n12 46\n1 26\n15 34\n8 47\n11 23\n6 32\n2 22\n9 27\n17 40\n6 15\n4 49\n12 33\n3 48\n22 47\n19 48\n10 27\n23 25\n4 44\n27 48",
"output": "2\n16\n0\n2\n0\n26\n16\n12\n30\n8\n18\n14\n14\n12\n6\n34\n16\n32\n18\n18\n12\n0\n30\n16"
},
{
"input": ")()((((((((((((((((()(((()()(()((((((()(((((((()()))((((())(((((((((()(((((((((\n51\n29 53\n31 69\n54 59\n3 52\n26 46\n14 62\n6 54\n39 56\n17 27\n46 74\n60 72\n18 26\n38 46\n4 27\n22 52\n44 49\n42 77\n2 20\n39 57\n61 70\n33 54\n10 30\n67 70\n46 66\n17 77\n5 52\n33 77\n26 32\n1 72\n40 78\n38 68\n19 47\n30 53\n19 29\n52 71\n1 11\n22 53\n17 42\n2 51\n4 12\n24 76\n22 34\n21 69\n11 69\n36 52\n17 31\n57 58\n54 62\n23 71\n5 46\n51 53",
"output": "12\n14\n4\n18\n6\n22\n18\n8\n4\n12\n2\n4\n2\n4\n16\n2\n14\n2\n8\n2\n10\n6\n2\n10\n24\n18\n16\n4\n26\n14\n14\n10\n12\n6\n6\n2\n16\n10\n18\n0\n22\n6\n20\n22\n10\n8\n2\n4\n22\n10\n0"
},
{
"input": "(\n1\n1 1",
"output": "0"
},
{
"input": ")\n1\n1 1",
"output": "0"
},
{
"input": "()\n1\n1 2",
"output": "2"
},
{
"input": ")(\n1\n1 2",
"output": "0"
}
] | 1,584,775,506
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 108
| 307,200
|
b = input()
k = int(input())
open_b = 0
final = [0 for _ in range(len(b))]
initial = [0 for _ in range(len(b))]
aux1 = 0
aux2 = 0
for i in range(len(b)):
if b[i] == '(':
open_b += 1
aux2 += 2
elif b[i] == ')':
if open_b > 0:
open_b -= 1
aux1 += 2
final[i] = aux1
initial[i] = aux2
for _ in range(k):
l, r = list(map(int, input().split(' ')))
x, y = final[r-1], final[l-1]
c1, c2 = initial[r-1], initial[l-1]
print((((x + c1)//2)//2)*2 - (((y + c2)//2)//2)*2)
|
Title: Sereja and Brackets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has a bracket sequence *s*1,<=*s*2,<=...,<=*s**n*, or, in other words, a string *s* of length *n*, consisting of characters "(" and ")".
Sereja needs to answer *m* queries, each of them is described by two integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). The answer to the *i*-th query is the length of the maximum correct bracket subsequence of sequence *s**l**i*,<=*s**l**i*<=+<=1,<=...,<=*s**r**i*. Help Sereja answer all queries.
You can find the definitions for a subsequence and a correct bracket sequence in the notes.
Input Specification:
The first line contains a sequence of characters *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*n*<=≤<=106) without any spaces. Each character is either a "(" or a ")". The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains a pair of integers. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) — the description of the *i*-th query.
Output Specification:
Print the answer to each question on a single line. Print the answers in the order they go in the input.
Demo Input:
['())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10\n']
Demo Output:
['0\n0\n2\n10\n4\n6\n6\n']
Note:
A subsequence of length |*x*| of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is string *x* = *s*<sub class="lower-index">*k*<sub class="lower-index">1</sub></sub>*s*<sub class="lower-index">*k*<sub class="lower-index">2</sub></sub>... *s*<sub class="lower-index">*k*<sub class="lower-index">|*x*|</sub></sub> (1 ≤ *k*<sub class="lower-index">1</sub> < *k*<sub class="lower-index">2</sub> < ... < *k*<sub class="lower-index">|*x*|</sub> ≤ |*s*|).
A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
For the third query required sequence will be «()».
For the fourth query required sequence will be «()(())(())».
|
```python
b = input()
k = int(input())
open_b = 0
final = [0 for _ in range(len(b))]
initial = [0 for _ in range(len(b))]
aux1 = 0
aux2 = 0
for i in range(len(b)):
if b[i] == '(':
open_b += 1
aux2 += 2
elif b[i] == ')':
if open_b > 0:
open_b -= 1
aux1 += 2
final[i] = aux1
initial[i] = aux2
for _ in range(k):
l, r = list(map(int, input().split(' ')))
x, y = final[r-1], final[l-1]
c1, c2 = initial[r-1], initial[l-1]
print((((x + c1)//2)//2)*2 - (((y + c2)//2)//2)*2)
```
| 0
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,695,024,259
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 46
| 0
|
a,b=map(int,input().split())
i=1
while 0!=i*a%10!=b:i+=1
print(i)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
a,b=map(int,input().split())
i=1
while 0!=i*a%10!=b:i+=1
print(i)
```
| 3
|
|
650
|
A
|
Watchmen
|
PROGRAMMING
| 1,400
|
[
"data structures",
"geometry",
"math"
] | null | null |
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
|
The first line of the input contains the single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109).
Some positions may coincide.
|
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
|
[
"3\n1 1\n7 5\n1 5\n",
"6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n"
] |
[
"2\n",
"11\n"
] |
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
| 500
|
[
{
"input": "3\n1 1\n7 5\n1 5",
"output": "2"
},
{
"input": "6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1",
"output": "11"
},
{
"input": "10\n46 -55\n46 45\n46 45\n83 -55\n46 45\n83 -55\n46 45\n83 45\n83 45\n46 -55",
"output": "33"
},
{
"input": "1\n-5 -90",
"output": "0"
},
{
"input": "2\n315 845\n-669 -762",
"output": "0"
},
{
"input": "3\n8911 7861\n-6888 7861\n8911 7861",
"output": "3"
},
{
"input": "2\n-1 1000000000\n0 -1",
"output": "0"
},
{
"input": "2\n1000000000 0\n-7 1",
"output": "0"
},
{
"input": "2\n1 4\n2 1",
"output": "0"
},
{
"input": "2\n1 0\n0 2333333",
"output": "0"
},
{
"input": "2\n2 1\n1 2",
"output": "0"
},
{
"input": "2\n1 1000000000\n2 -1000000000",
"output": "0"
},
{
"input": "2\n0 1000000000\n1 -7",
"output": "0"
},
{
"input": "2\n1 0\n0 19990213",
"output": "0"
}
] | 1,588,079,092
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 3,000
| 1,331,200
|
n = int(input())
a,b,c =dict(),dict(),[]
e,f,g = 0,0,0
for i in range(n):
v1,v2 = map(int,input().split())
if v1 not in a:
a[v1]=1
else:
e+=a[v1]
a[v1]+=1
if v2 not in b:
b[v2]=1
else:
e+=b[v2]
b[v2]+=1
f+=c.count([v1,v2])
c.append([v1,v2])
print(e-f)
|
Title: Watchmen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
Input Specification:
The first line of the input contains the single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109).
Some positions may coincide.
Output Specification:
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
Demo Input:
['3\n1 1\n7 5\n1 5\n', '6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n']
Demo Output:
['2\n', '11\n']
Note:
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
|
```python
n = int(input())
a,b,c =dict(),dict(),[]
e,f,g = 0,0,0
for i in range(n):
v1,v2 = map(int,input().split())
if v1 not in a:
a[v1]=1
else:
e+=a[v1]
a[v1]+=1
if v2 not in b:
b[v2]=1
else:
e+=b[v2]
b[v2]+=1
f+=c.count([v1,v2])
c.append([v1,v2])
print(e-f)
```
| 0
|
|
1,000
|
B
|
Light It Up
|
PROGRAMMING
| 1,500
|
[
"greedy"
] | null | null |
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
|
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$.
|
Print the only integer — maximum possible total time when the lamp is lit.
|
[
"3 10\n4 6 7\n",
"2 12\n1 10\n",
"2 7\n3 4\n"
] |
[
"8\n",
"9\n",
"6\n"
] |
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
| 0
|
[
{
"input": "3 10\n4 6 7",
"output": "8"
},
{
"input": "2 12\n1 10",
"output": "9"
},
{
"input": "2 7\n3 4",
"output": "6"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "5 10\n1 3 5 6 8",
"output": "6"
},
{
"input": "7 1000000000\n1 10001 10011 20011 20021 40021 40031",
"output": "999999969"
},
{
"input": "7 1000000000\n3 10001 10011 20011 20021 40021 40031",
"output": "999999969"
},
{
"input": "1 10\n1",
"output": "9"
},
{
"input": "1 10000000\n1",
"output": "9999999"
},
{
"input": "1 8\n1",
"output": "7"
},
{
"input": "7 17\n1 5 9 10 11 14 16",
"output": "9"
},
{
"input": "4 17\n1 5 9 10",
"output": "12"
},
{
"input": "5 12\n1 2 3 4 5",
"output": "9"
},
{
"input": "2 1000000000\n100 111",
"output": "999999989"
},
{
"input": "3 90591\n90579 90580 90581",
"output": "90589"
},
{
"input": "5 16\n1 2 3 4 5",
"output": "13"
},
{
"input": "6 981231233\n1 6 15 30 130 1875",
"output": "981229468"
},
{
"input": "1 3\n1",
"output": "2"
},
{
"input": "3 12\n4 7 11",
"output": "8"
},
{
"input": "1 100000\n1",
"output": "99999"
},
{
"input": "2 5\n1 3",
"output": "3"
},
{
"input": "1 1000000000\n1",
"output": "999999999"
},
{
"input": "1 1000000000\n999999999",
"output": "999999999"
},
{
"input": "3 4\n1 2 3",
"output": "2"
},
{
"input": "3 5\n1 2 3",
"output": "3"
},
{
"input": "1 4\n2",
"output": "3"
},
{
"input": "3 5\n1 3 4",
"output": "3"
},
{
"input": "9 20\n5 9 11 12 14 15 16 17 19",
"output": "12"
},
{
"input": "2 4\n1 2",
"output": "3"
},
{
"input": "1 10\n2",
"output": "9"
},
{
"input": "2 6\n2 4",
"output": "4"
},
{
"input": "2 4\n1 3",
"output": "2"
},
{
"input": "3 6\n1 2 4",
"output": "4"
},
{
"input": "7 9863\n65 96 97 98 101 112 1115",
"output": "9819"
}
] | 1,666,177,345
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 19 11:48:45 2022
@author: 帝落梦魂
"""
n,m = map(int,input().split())
a = [0]+[int(i) for i in input().split()]+[m]
light = 0
if n%2 == 1:
deltL = [m-a[-2]]
else:
deltL = [0]
ml = 1
for i in range(3+n%2,n+2,2):
deltL = [deltL[0]+2*a[-i+1]-a[-i+2]-a[-i]] + deltL
light = light + a[-i+2]-a[-i+1]
if deltL[0] > ml:
if a[-i+1]-a[-i]>1 or a[-i+2]-a[-i+1]>1:
ml = deltL[0]
light = light + a[1] + ml - 1
print(light)
|
Title: Light It Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
Input Specification:
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$.
Output Specification:
Print the only integer — maximum possible total time when the lamp is lit.
Demo Input:
['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n']
Demo Output:
['8\n', '9\n', '6\n']
Note:
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
|
```python
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 19 11:48:45 2022
@author: 帝落梦魂
"""
n,m = map(int,input().split())
a = [0]+[int(i) for i in input().split()]+[m]
light = 0
if n%2 == 1:
deltL = [m-a[-2]]
else:
deltL = [0]
ml = 1
for i in range(3+n%2,n+2,2):
deltL = [deltL[0]+2*a[-i+1]-a[-i+2]-a[-i]] + deltL
light = light + a[-i+2]-a[-i+1]
if deltL[0] > ml:
if a[-i+1]-a[-i]>1 or a[-i+2]-a[-i+1]>1:
ml = deltL[0]
light = light + a[1] + ml - 1
print(light)
```
| 0
|
|
877
|
A
|
Alex and broken contest
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
|
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.
|
Print "YES", if problem is from this contest, and "NO" otherwise.
|
[
"Alex_and_broken_contest\n",
"NikitaAndString\n",
"Danil_and_Olya\n"
] |
[
"NO",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "Alex_and_broken_contest",
"output": "NO"
},
{
"input": "NikitaAndString",
"output": "YES"
},
{
"input": "Danil_and_Olya",
"output": "NO"
},
{
"input": "Slava____and_the_game",
"output": "YES"
},
{
"input": "Olya_and_energy_drinks",
"output": "YES"
},
{
"input": "Danil_and_part_time_job",
"output": "YES"
},
{
"input": "Ann_and_books",
"output": "YES"
},
{
"input": "Olya",
"output": "YES"
},
{
"input": "Nikita",
"output": "YES"
},
{
"input": "Slava",
"output": "YES"
},
{
"input": "Vanya",
"output": "NO"
},
{
"input": "I_dont_know_what_to_write_here",
"output": "NO"
},
{
"input": "danil_and_work",
"output": "NO"
},
{
"input": "Ann",
"output": "YES"
},
{
"input": "Batman_Nananananananan_Batman",
"output": "NO"
},
{
"input": "Olya_Nikita_Ann_Slava_Danil",
"output": "NO"
},
{
"input": "its_me_Mario",
"output": "NO"
},
{
"input": "A",
"output": "NO"
},
{
"input": "Wake_up_Neo",
"output": "NO"
},
{
"input": "Hardest_problem_ever",
"output": "NO"
},
{
"input": "Nikita_Nikita",
"output": "NO"
},
{
"input": "____________________________________________________________________________________________________",
"output": "NO"
},
{
"input": "Nikitb",
"output": "NO"
},
{
"input": "Unn",
"output": "NO"
},
{
"input": "oLya_adn_smth",
"output": "NO"
},
{
"input": "FloorISLava",
"output": "NO"
},
{
"input": "ann",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "AAnnnnn",
"output": "YES"
},
{
"input": "AnnAnn",
"output": "NO"
},
{
"input": "Annn",
"output": "YES"
},
{
"input": "Dilzhan",
"output": "NO"
},
{
"input": "Danilaaa",
"output": "YES"
},
{
"input": "AndAnn",
"output": "YES"
},
{
"input": "OlyaAnnAnn",
"output": "NO"
},
{
"input": "DanilDanilOlya",
"output": "NO"
},
{
"input": "DDanil",
"output": "YES"
},
{
"input": "AnnAnnDanil",
"output": "NO"
},
{
"input": "And_Danil",
"output": "YES"
},
{
"input": "abcddddDanil",
"output": "YES"
},
{
"input": "DanilOlyaOlya",
"output": "NO"
},
{
"input": "Nikitaaa",
"output": "YES"
},
{
"input": "aaabbba",
"output": "NO"
},
{
"input": "Ann_Ann_Danil",
"output": "NO"
},
{
"input": "Danil_Danil_Nikita",
"output": "NO"
},
{
"input": "AlexaaaaaaBBBBBOlyaDDDDD",
"output": "YES"
},
{
"input": "IloveDaniland",
"output": "YES"
},
{
"input": "AnAnn",
"output": "YES"
},
{
"input": "Danil_Danil_Olya",
"output": "NO"
},
{
"input": "DanilDanilSlava",
"output": "NO"
},
{
"input": "DanilDanil",
"output": "NO"
},
{
"input": "OlyOlya",
"output": "YES"
},
{
"input": "NikitaNikitb",
"output": "YES"
},
{
"input": "ababaca",
"output": "NO"
},
{
"input": "AnnNikitaNikitaNikitaNikita__good_luck",
"output": "NO"
}
] | 1,515,913,174
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 5,632,000
|
a=input()
k=0
for i in range(len(a)-4):
if a[i]=='D' and a[i+1]=='a' and a[i+2]=='n' and a[i+3]=='i' and a[i+4]=='l':
k+=1
for i in range(len(a)-3):
if a[i]=='O' and a[i+1]=='l' and a[i+2]=='y' and a[i+3]=='a':
k+=1
for i in range(len(a)-4):
if a[i]=='S' and a[i+1]=='l' and a[i+2]=='a' and [i+3]=='v' and a[i+4]=='a':
k+=1
for i in range(len(a)-2):
if a[i]=='A' and a[i+1]=='n' and a[i+2]=='n':
k+=1
for i in range(len(a)-5):
if a[i]=='N' and a[i+1]=='i' and a[i+2]=='k' and a[i+3]=='i' and a[i+4]=='t' and a[i+5]=='a':
k+=1
if k==1:
print('YES')
else:
print('NO')
|
Title: Alex and broken contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input Specification:
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.
Output Specification:
Print "YES", if problem is from this contest, and "NO" otherwise.
Demo Input:
['Alex_and_broken_contest\n', 'NikitaAndString\n', 'Danil_and_Olya\n']
Demo Output:
['NO', 'YES', 'NO']
Note:
none
|
```python
a=input()
k=0
for i in range(len(a)-4):
if a[i]=='D' and a[i+1]=='a' and a[i+2]=='n' and a[i+3]=='i' and a[i+4]=='l':
k+=1
for i in range(len(a)-3):
if a[i]=='O' and a[i+1]=='l' and a[i+2]=='y' and a[i+3]=='a':
k+=1
for i in range(len(a)-4):
if a[i]=='S' and a[i+1]=='l' and a[i+2]=='a' and [i+3]=='v' and a[i+4]=='a':
k+=1
for i in range(len(a)-2):
if a[i]=='A' and a[i+1]=='n' and a[i+2]=='n':
k+=1
for i in range(len(a)-5):
if a[i]=='N' and a[i+1]=='i' and a[i+2]=='k' and a[i+3]=='i' and a[i+4]=='t' and a[i+5]=='a':
k+=1
if k==1:
print('YES')
else:
print('NO')
```
| 0
|
|
862
|
A
|
Mahmoud and Ehab and the MEX
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
|
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
|
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
|
[
"5 3\n0 4 5 6 7\n",
"1 0\n0\n",
"5 0\n1 2 3 4 5\n"
] |
[
"2\n",
"1\n",
"0\n"
] |
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
| 500
|
[
{
"input": "5 3\n0 4 5 6 7",
"output": "2"
},
{
"input": "1 0\n0",
"output": "1"
},
{
"input": "5 0\n1 2 3 4 5",
"output": "0"
},
{
"input": "10 5\n57 1 47 9 93 37 76 70 78 15",
"output": "4"
},
{
"input": "10 5\n99 98 93 97 95 100 92 94 91 96",
"output": "5"
},
{
"input": "10 5\n1 2 3 4 59 45 0 58 51 91",
"output": "0"
},
{
"input": "100 100\n79 13 21 11 3 87 28 40 29 4 96 34 8 78 61 46 33 45 99 30 92 67 22 97 39 86 73 31 74 44 62 55 57 2 54 63 80 69 25 48 77 98 17 93 15 16 89 12 43 23 37 95 14 38 83 90 49 56 72 10 20 0 50 71 70 88 19 1 76 81 52 41 82 68 85 47 6 7 35 60 18 64 75 84 27 9 65 91 94 42 53 24 66 26 59 36 51 32 5 58",
"output": "0"
},
{
"input": "100 50\n95 78 46 92 80 18 79 58 30 72 19 89 39 29 44 65 15 100 59 8 96 9 62 67 41 42 82 14 57 32 71 77 40 5 7 51 28 53 85 23 16 35 3 91 6 11 75 61 17 66 13 47 36 56 10 22 83 60 48 24 26 97 4 33 76 86 70 0 34 64 52 43 21 49 55 74 1 73 81 25 54 63 94 84 20 68 87 12 31 88 38 93 37 90 98 69 99 45 27 2",
"output": "0"
},
{
"input": "100 33\n28 11 79 92 88 62 77 72 7 41 96 97 67 84 44 8 81 35 38 1 64 68 46 17 98 83 31 12 74 21 2 22 47 6 36 75 65 61 37 26 25 45 59 48 100 51 93 76 78 49 3 57 16 4 87 29 55 82 70 39 53 0 60 15 24 71 58 20 66 89 95 42 13 43 63 90 85 52 50 30 54 40 56 23 27 34 32 18 10 19 69 9 99 73 91 14 5 80 94 86",
"output": "0"
},
{
"input": "99 33\n25 76 41 95 55 20 47 59 58 84 87 92 16 27 35 65 72 63 93 54 36 96 15 86 5 69 24 46 67 73 48 60 40 6 61 74 97 10 100 8 52 26 77 18 7 62 37 2 14 66 11 56 68 91 0 64 75 99 30 21 53 1 89 81 3 98 12 88 39 38 29 83 22 90 9 28 45 43 78 44 32 57 4 50 70 17 13 51 80 85 71 94 82 19 34 42 23 79 49",
"output": "1"
},
{
"input": "100 100\n65 56 84 46 44 33 99 74 62 72 93 67 43 92 75 88 38 34 66 12 55 76 58 90 78 8 14 45 97 59 48 32 64 18 39 89 31 51 54 81 29 36 70 77 40 22 49 27 3 1 73 13 98 42 87 37 2 57 4 6 50 25 23 79 28 86 68 61 80 17 19 10 15 63 52 11 35 60 21 16 24 85 30 91 7 5 69 20 71 82 53 94 41 95 96 9 26 83 0 47",
"output": "0"
},
{
"input": "100 100\n58 88 12 71 22 1 40 19 73 20 67 48 57 17 69 36 100 35 33 37 72 55 52 8 89 85 47 42 78 70 81 86 11 9 68 99 6 16 21 61 53 98 23 62 32 59 51 0 87 24 50 30 65 10 80 95 7 92 25 74 60 79 91 5 13 31 75 38 90 94 46 66 93 34 14 41 28 2 76 84 43 96 3 56 49 82 27 77 64 63 4 45 18 29 54 39 15 26 83 44",
"output": "2"
},
{
"input": "89 100\n58 96 17 41 86 34 28 84 18 40 8 77 87 89 68 79 33 35 53 49 0 6 22 12 72 90 48 55 21 50 56 62 75 2 37 95 69 74 14 20 44 46 27 32 31 59 63 60 10 85 71 70 38 52 94 30 61 51 80 26 36 23 39 47 76 45 100 57 15 78 97 66 54 13 99 16 93 73 24 4 83 5 98 81 92 25 29 88 65",
"output": "13"
},
{
"input": "100 50\n7 95 24 76 81 78 60 69 83 84 100 1 65 31 48 92 73 39 18 89 38 97 10 42 8 55 98 51 21 90 62 77 16 91 0 94 4 37 19 17 67 35 45 41 56 20 15 85 75 28 59 27 12 54 61 68 36 5 79 93 66 11 70 49 50 34 30 25 96 46 64 14 32 22 47 40 58 23 43 9 87 82 26 53 80 52 3 86 13 99 33 71 6 88 57 74 2 44 72 63",
"output": "2"
},
{
"input": "77 0\n27 8 20 92 21 41 53 98 17 65 67 35 81 11 55 49 61 44 2 66 51 89 40 28 52 62 86 91 64 24 18 5 94 82 96 99 71 6 39 83 26 29 16 30 45 97 80 90 69 12 13 33 76 73 46 19 78 56 88 38 42 34 57 77 47 4 59 58 7 100 95 72 9 74 15 43 54",
"output": "0"
},
{
"input": "100 50\n55 36 0 32 81 6 17 43 24 13 30 19 8 59 71 45 15 74 3 41 99 42 86 47 2 94 35 1 66 95 38 49 4 27 96 89 34 44 92 25 51 39 54 28 80 77 20 14 48 40 68 56 31 63 33 78 69 37 18 26 83 70 23 82 91 65 67 52 61 53 7 22 60 21 12 73 72 87 75 100 90 29 64 79 98 85 5 62 93 84 50 46 97 58 57 16 9 10 76 11",
"output": "1"
},
{
"input": "77 0\n12 8 19 87 9 54 55 86 97 7 27 85 25 48 94 73 26 1 13 57 72 69 76 39 38 91 75 40 42 28 93 21 70 84 65 11 60 90 20 95 66 89 59 47 34 99 6 61 52 100 50 3 77 81 82 53 15 24 0 45 44 14 68 96 58 5 18 35 10 98 29 74 92 49 83 71 17",
"output": "1"
},
{
"input": "100 70\n25 94 66 65 10 99 89 6 70 31 7 40 20 92 64 27 21 72 77 98 17 43 47 44 48 81 38 56 100 39 90 22 88 76 3 83 86 29 33 55 82 79 49 11 2 16 12 78 85 69 32 97 26 15 53 24 23 91 51 67 34 35 52 5 62 50 95 18 71 13 75 8 30 42 93 36 45 60 63 46 57 41 87 0 84 54 74 37 4 58 28 19 96 61 80 9 1 14 73 68",
"output": "2"
},
{
"input": "89 19\n14 77 85 81 79 38 91 45 55 51 50 11 62 67 73 76 2 27 16 23 3 29 65 98 78 17 4 58 22 20 34 66 64 31 72 5 32 44 12 75 80 47 18 25 99 0 61 56 71 84 48 88 10 7 86 8 49 24 43 21 37 28 33 54 46 57 40 89 36 97 6 96 39 95 26 74 1 69 9 100 52 30 83 87 68 60 92 90 35",
"output": "2"
},
{
"input": "89 100\n69 61 56 45 11 41 42 32 28 29 0 76 7 65 13 35 36 82 10 39 26 34 38 40 92 12 17 54 24 46 88 70 66 27 100 52 85 62 22 48 86 68 21 49 53 94 67 20 1 90 77 84 31 87 58 47 95 33 4 72 93 83 8 51 91 80 99 43 71 19 44 59 98 97 64 9 81 16 79 63 25 37 3 75 2 55 50 6 18",
"output": "13"
},
{
"input": "77 0\n38 76 24 74 42 88 29 75 96 46 90 32 59 97 98 60 41 57 80 37 100 49 25 63 95 31 61 68 53 78 27 66 84 48 94 83 30 26 36 99 71 62 45 47 70 28 35 54 34 85 79 43 91 72 86 33 67 92 77 65 69 52 82 55 87 64 56 40 50 44 51 73 89 81 58 93 39",
"output": "0"
},
{
"input": "89 100\n38 90 80 64 35 44 56 11 15 89 23 12 49 70 72 60 63 85 92 10 45 83 8 88 41 33 16 6 61 76 62 71 87 13 25 77 74 0 1 37 96 93 7 94 21 82 34 78 4 73 65 20 81 95 50 32 48 17 69 55 68 5 51 27 53 43 91 67 59 46 86 84 99 24 22 3 97 98 40 36 26 58 57 9 42 30 52 2 47",
"output": "11"
},
{
"input": "77 0\n55 71 78 86 68 35 53 10 59 32 81 19 74 97 62 61 93 87 96 44 25 18 43 82 84 16 34 48 92 39 64 36 49 91 45 76 95 31 57 29 75 79 13 2 14 24 52 23 33 20 47 99 63 15 5 80 58 67 12 3 85 6 1 27 73 90 4 42 37 70 8 11 89 77 9 22 94",
"output": "0"
},
{
"input": "77 0\n12 75 31 71 44 8 3 82 21 77 50 29 57 74 40 10 15 42 84 2 100 9 28 72 92 0 49 11 90 55 17 36 19 54 68 52 4 69 97 91 5 39 59 45 89 62 53 83 16 94 76 60 95 47 30 51 7 48 20 70 67 32 58 78 63 34 56 93 99 88 24 1 66 22 25 14 13",
"output": "1"
},
{
"input": "100 70\n91 82 8 85 26 25 95 97 40 87 81 93 7 73 38 94 64 96 74 18 90 19 65 68 72 61 23 43 36 41 60 88 30 33 71 24 52 39 15 3 16 89 86 79 55 4 9 58 67 44 46 29 6 48 84 69 27 21 78 54 51 57 80 53 76 50 47 77 45 12 34 10 100 0 17 31 56 99 98 11 92 5 2 42 32 59 66 62 37 63 28 75 35 1 22 13 83 49 20 14",
"output": "0"
},
{
"input": "77 0\n51 5 81 62 30 22 11 0 83 16 79 85 52 70 69 10 8 47 58 3 24 34 44 14 82 66 99 17 28 31 64 67 23 49 94 45 4 12 27 15 21 6 43 72 87 2 63 92 35 39 59 9 90 78 93 20 65 36 60 89 50 41 61 84 77 86 76 100 38 68 53 97 96 95 7 19 88",
"output": "1"
},
{
"input": "1 100\n0",
"output": "99"
},
{
"input": "1 0\n100",
"output": "0"
},
{
"input": "1 100\n100",
"output": "101"
},
{
"input": "2 100\n0 100",
"output": "100"
},
{
"input": "5 5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 3\n0 3 4 5 6",
"output": "3"
},
{
"input": "7 10\n0 1 2 3 4 5 10",
"output": "5"
},
{
"input": "2 2\n0 2",
"output": "2"
},
{
"input": "2 1\n1 2",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 2\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 6\n0 1 2 3 4",
"output": "1"
},
{
"input": "3 2\n3 4 5",
"output": "2"
}
] | 1,624,177,947
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 62
| 0
|
n, x = map(int,input().split())
s = set(map(int,input().split()))
a = len(set(range(0, x)) - s)
print(a+1 if x in s else a)
|
Title: Mahmoud and Ehab and the MEX
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
Input Specification:
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
Output Specification:
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
Demo Input:
['5 3\n0 4 5 6 7\n', '1 0\n0\n', '5 0\n1 2 3 4 5\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
|
```python
n, x = map(int,input().split())
s = set(map(int,input().split()))
a = len(set(range(0, x)) - s)
print(a+1 if x in s else a)
```
| 3
|
|
1
|
C
|
Ancient Berland Circus
|
PROGRAMMING
| 2,100
|
[
"geometry",
"math"
] |
C. Ancient Berland Circus
|
2
|
64
|
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
|
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
|
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
|
[
"0.000000 0.000000\n1.000000 1.000000\n0.000000 1.000000\n"
] |
[
"1.00000000\n"
] | 0
|
[
{
"input": "0.000000 0.000000\n1.000000 1.000000\n0.000000 1.000000",
"output": "1.00000000"
},
{
"input": "71.756151 7.532275\n-48.634784 100.159986\n91.778633 158.107739",
"output": "9991.27897663"
},
{
"input": "18.716839 40.852752\n66.147248 -4.083161\n111.083161 43.347248",
"output": "4268.87997505"
},
{
"input": "-13.242302 -45.014124\n-33.825369 51.083964\n84.512928 -55.134407",
"output": "16617.24002771"
},
{
"input": "115.715093 141.583620\n136.158119 -23.780834\n173.673212 64.802787",
"output": "24043.74046813"
},
{
"input": "17.288379 68.223317\n48.776683 71.688379\n23.170559 106.572762",
"output": "1505.27997374"
},
{
"input": "76.820252 66.709341\n61.392328 82.684207\n44.267775 -2.378694",
"output": "6503.44762335"
},
{
"input": "-46.482632 -31.161247\n19.689679 -70.646972\n-17.902656 -58.455808",
"output": "23949.55226823"
},
{
"input": "34.236058 108.163949\n28.639345 104.566515\n25.610069 86.002927",
"output": "780.93431702"
},
{
"input": "25.428124 39.407248\n17.868098 39.785933\n11.028461 43.028890",
"output": "1152.21351717"
},
{
"input": "36.856072 121.845502\n46.453956 109.898647\n-30.047767 77.590282",
"output": "5339.35578947"
},
{
"input": "-18.643272 56.008305\n9.107608 -22.094058\n-6.456146 70.308320",
"output": "9009.25177521"
},
{
"input": "88.653021 18.024220\n51.942488 -2.527850\n76.164701 24.553012",
"output": "1452.52866331"
},
{
"input": "80.181999 -38.076894\n23.381778 122.535736\n47.118815 140.734014",
"output": "28242.17663744"
},
{
"input": "1.514204 81.400629\n32.168797 100.161401\n7.778734 46.010993",
"output": "3149.43107333"
},
{
"input": "84.409605 38.496141\n77.788313 39.553807\n75.248391 59.413884",
"output": "438.85760782"
},
{
"input": "12.272903 101.825792\n-51.240438 -12.708472\n-29.729299 77.882032",
"output": "24908.67540438"
},
{
"input": "35.661751 27.283571\n96.513550 51.518022\n97.605986 131.258287",
"output": "13324.78113326"
},
{
"input": "-20.003518 -4.671086\n93.588632 6.362759\n-24.748109 24.792124",
"output": "11191.04493104"
},
{
"input": "93.583067 132.858352\n63.834975 19.353720\n33.677824 102.529376",
"output": "10866.49390021"
},
{
"input": "-7.347450 36.971423\n84.498728 89.423536\n75.469963 98.022482",
"output": "8977.83404724"
},
{
"input": "51.679280 56.072393\n-35.819256 73.390532\n-10.661374 129.756454",
"output": "7441.86549199"
},
{
"input": "97.326813 61.492460\n100.982131 57.717635\n68.385216 22.538372",
"output": "1840.59945324"
},
{
"input": "-16.356805 109.310423\n124.529388 25.066276\n-37.892043 80.604904",
"output": "22719.36404168"
},
{
"input": "103.967164 63.475916\n86.466163 59.341930\n69.260229 73.258917",
"output": "1621.96700296"
},
{
"input": "122.381894 -48.763263\n163.634346 -22.427845\n26.099674 73.681862",
"output": "22182.51901824"
},
{
"input": "119.209229 133.905087\n132.001535 22.179509\n96.096673 0.539763",
"output": "16459.52899209"
},
{
"input": "77.145533 85.041789\n67.452820 52.513188\n80.503843 85.000149",
"output": "1034.70898496"
},
{
"input": "28.718442 36.116251\n36.734593 35.617015\n76.193973 99.136077",
"output": "6271.48941610"
},
{
"input": "0.376916 17.054676\n100.187614 85.602831\n1.425829 132.750915",
"output": "13947.47744984"
},
{
"input": "46.172435 -22.819705\n17.485134 -1.663888\n101.027565 111.619705",
"output": "16483.23337238"
},
{
"input": "55.957968 -72.765994\n39.787413 -75.942282\n24.837014 128.144762",
"output": "32799.66697178"
},
{
"input": "40.562163 -47.610606\n10.073051 -54.490068\n54.625875 -40.685797",
"output": "31224.34817875"
},
{
"input": "20.965151 74.716562\n167.264364 81.864800\n5.931644 48.813212",
"output": "30115.26346791"
},
{
"input": "105.530943 80.920069\n40.206723 125.323331\n40.502256 -85.455877",
"output": "36574.64621711"
},
{
"input": "104.636703 49.583778\n85.940583 95.426299\n69.375168 93.234795",
"output": "2632.68754075"
},
{
"input": "72.873708 -59.083734\n110.911118 -6.206576\n-44.292395 13.106202",
"output": "19244.42781859"
},
{
"input": "49.320630 48.119616\n65.888396 93.514980\n27.342377 97.600590",
"output": "2437.50897386"
},
{
"input": "6.949504 69.606390\n26.139268 72.136945\n24.032442 57.407195",
"output": "372.09309018"
},
{
"input": "-21.925928 -24.623076\n-33.673619 -11.677794\n4.692348 52.266292",
"output": "5669.99444283"
},
{
"input": "109.515505 37.575315\n5.377080 101.729711\n17.501630 103.324931",
"output": "25142.85604936"
},
{
"input": "-56.880888 172.997993\n81.126977 42.144034\n-51.413417 17.057807",
"output": "29051.57171313"
},
{
"input": "80.895061 94.491414\n42.361631 65.191687\n77.556800 76.694829",
"output": "2386.01792476"
},
{
"input": "165.094169 94.574129\n46.867578 147.178855\n174.685774 62.705213",
"output": "32087.47120554"
},
{
"input": "146.604506 -3.502359\n24.935572 44.589981\n106.160918 -51.162271",
"output": "13799.61044048"
},
{
"input": "139.847022 19.153937\n104.096879 75.379874\n49.164271 46.404632",
"output": "7083.26303902"
},
{
"input": "31.312532 151.532355\n182.646053 56.534075\n15.953947 127.065925",
"output": "25712.80766033"
},
{
"input": "42.147045 64.165917\n70.260284 4.962470\n10.532991 76.277713",
"output": "14261.92257159"
},
{
"input": "129.400249 -44.695226\n122.278798 -53.696996\n44.828427 -83.507917",
"output": "26227.47891833"
},
{
"input": "28.420253 0.619862\n10.966628 21.724132\n14.618862 10.754642",
"output": "1760.14006648"
}
] | 1,678,768,530
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 1
| 186
| 9,420,800
|
# Submitted using https://github.com/Nirlep5252/codeforces-cli
# >.<
import math
from typing import *
def distance(p1: Tuple[float], p2: Tuple[float]) -> float:
return ((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) ** 0.5
def mid_point(p1: Tuple[float], p2: Tuple[float]) -> Tuple[float]:
return (p1[0] + p2[0]) / 2, (p1[1] + p2[1]) / 2
def area_triangle(a: float, b: float, c: float) -> float:
s = (a + b + c) / 2
return (s * (s - a) * (s - b) * (s - c)) ** 0.5
def get_angle(p1: Tuple[float], p2: Tuple[float], p3: Tuple[float]) -> float:
p12 = distance(p1, p2)
p23 = distance(p2, p3)
p13 = distance(p1, p3)
return math.acos((p12 ** 2 + p13 ** 2 - p23 ** 2) / (2 * p12 * p13))
coords = tuple([tuple(map(float, input().split())) for _ in range(3)])
lines = [distance(*coords[:2]), distance(*coords[1:]), distance(coords[0], coords[-1])]
if len(set(lines)) == 1:
print(area_triangle(*lines))
else:
m = max(lines)
points = []
if m == lines[0]:
p1 = mid_point(*coords[:2])
points.append(coords[:2])
elif m == lines[1]:
p1 = mid_point(*coords[1:])
points.append(coords[1:])
else:
p1 = mid_point(coords[0], coords[-1])
points.append(coords[0], coords[-1])
m = min(lines)
if m == lines[0]:
p2, p3 = coords[:2]
elif m == lines[1]:
p2, p3 = coords[1:]
else:
p2, p3 = coords[0], coords[-1]
if p2 in points:
common_point = p2
other_point = p3
else:
common_point = p3
other_point = p2
# print(p1, common_point, other_point)
# print(math.pi / get_angle(p1, common_point, other_point))
min_angle = min([
get_angle(p1, p2, p3),
get_angle(p1, p3, p2),
get_angle(p2, p1, p3),
get_angle(p2, p3, p1),
get_angle(p3, p1, p2),
get_angle(p3, p2, p1)
])
# print(math.pi / min_angle)
sides = [distance(p1, p2), distance(p2, p3), distance(p1, p3)]
area_smol_triangle = area_triangle(*sides)
print(area_smol_triangle * round(math.pi / min_angle))
|
Title: Ancient Berland Circus
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input Specification:
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output Specification:
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Demo Input:
['0.000000 0.000000\n1.000000 1.000000\n0.000000 1.000000\n']
Demo Output:
['1.00000000\n']
|
```python
# Submitted using https://github.com/Nirlep5252/codeforces-cli
# >.<
import math
from typing import *
def distance(p1: Tuple[float], p2: Tuple[float]) -> float:
return ((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) ** 0.5
def mid_point(p1: Tuple[float], p2: Tuple[float]) -> Tuple[float]:
return (p1[0] + p2[0]) / 2, (p1[1] + p2[1]) / 2
def area_triangle(a: float, b: float, c: float) -> float:
s = (a + b + c) / 2
return (s * (s - a) * (s - b) * (s - c)) ** 0.5
def get_angle(p1: Tuple[float], p2: Tuple[float], p3: Tuple[float]) -> float:
p12 = distance(p1, p2)
p23 = distance(p2, p3)
p13 = distance(p1, p3)
return math.acos((p12 ** 2 + p13 ** 2 - p23 ** 2) / (2 * p12 * p13))
coords = tuple([tuple(map(float, input().split())) for _ in range(3)])
lines = [distance(*coords[:2]), distance(*coords[1:]), distance(coords[0], coords[-1])]
if len(set(lines)) == 1:
print(area_triangle(*lines))
else:
m = max(lines)
points = []
if m == lines[0]:
p1 = mid_point(*coords[:2])
points.append(coords[:2])
elif m == lines[1]:
p1 = mid_point(*coords[1:])
points.append(coords[1:])
else:
p1 = mid_point(coords[0], coords[-1])
points.append(coords[0], coords[-1])
m = min(lines)
if m == lines[0]:
p2, p3 = coords[:2]
elif m == lines[1]:
p2, p3 = coords[1:]
else:
p2, p3 = coords[0], coords[-1]
if p2 in points:
common_point = p2
other_point = p3
else:
common_point = p3
other_point = p2
# print(p1, common_point, other_point)
# print(math.pi / get_angle(p1, common_point, other_point))
min_angle = min([
get_angle(p1, p2, p3),
get_angle(p1, p3, p2),
get_angle(p2, p1, p3),
get_angle(p2, p3, p1),
get_angle(p3, p1, p2),
get_angle(p3, p2, p1)
])
# print(math.pi / min_angle)
sides = [distance(p1, p2), distance(p2, p3), distance(p1, p3)]
area_smol_triangle = area_triangle(*sides)
print(area_smol_triangle * round(math.pi / min_angle))
```
| -1
|
|
554
|
B
|
Ohana Cleans Up
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"strings"
] | null | null |
Ohana Matsumae is trying to clean a room, which is divided up into an *n* by *n* grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean.
|
The first line of input will be a single integer *n* (1<=≤<=*n*<=≤<=100).
The next *n* lines will describe the state of the room. The *i*-th line will contain a binary string with *n* characters denoting the state of the *i*-th row of the room. The *j*-th character on this line is '1' if the *j*-th square in the *i*-th row is clean, and '0' if it is dirty.
|
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
|
[
"4\n0101\n1000\n1111\n0101\n",
"3\n111\n111\n111\n"
] |
[
"2\n",
"3\n"
] |
In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything.
| 500
|
[
{
"input": "4\n0101\n1000\n1111\n0101",
"output": "2"
},
{
"input": "3\n111\n111\n111",
"output": "3"
},
{
"input": "10\n0100000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000",
"output": "9"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n0111010011\n0111010011\n1010010001\n0111010011\n0000110000\n0111010011\n0111010011\n0000110000\n1010010001\n0111010011",
"output": "6"
},
{
"input": "20\n10101011101000011010\n11111010001100110101\n01011100010000001111\n10110100000101010011\n11010001110111101101\n00100110011011101010\n01000110101011001100\n01101100111101101101\n10111010010100111100\n00010010110001101110\n10111110010000101010\n10010111110100100100\n11010111001111110100\n11110111101100000001\n00011010100111011000\n11110001011000011010\n10001101010000011011\n01010101110010000111\n11100110111101101111\n11011111110010001111",
"output": "1"
},
{
"input": "10\n1001000000\n0111101111\n1111001011\n1000010100\n0111101111\n0101100110\n1001000000\n1000010100\n0111101111\n1001000000",
"output": "3"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000",
"output": "10"
},
{
"input": "10\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111",
"output": "10"
},
{
"input": "10\n1000000000\n0100000000\n0010000000\n0001000000\n0000100000\n0000010000\n0000001000\n0000000100\n0000000010\n0000000001",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "4\n0000\n0000\n1111\n1111",
"output": "2"
},
{
"input": "11\n10000000001\n10000000001\n10000000001\n10000000001\n10001000001\n10001000000\n10001000001\n10001000001\n10001000000\n10001000000\n10001000100",
"output": "4"
}
] | 1,440,096,656
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
n=int(input())
l=[]
for i in range(n):
s=input()
l.append(s)
m=0
for i in range(n-1):
if(l[i]!=-1):
k=1
for j in range(i+1,n):
if(l[i]==l[j]):
k+=1
l[j]=-1
if(k>m):
m=k
print(m)
|
Title: Ohana Cleans Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ohana Matsumae is trying to clean a room, which is divided up into an *n* by *n* grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean.
Input Specification:
The first line of input will be a single integer *n* (1<=≤<=*n*<=≤<=100).
The next *n* lines will describe the state of the room. The *i*-th line will contain a binary string with *n* characters denoting the state of the *i*-th row of the room. The *j*-th character on this line is '1' if the *j*-th square in the *i*-th row is clean, and '0' if it is dirty.
Output Specification:
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
Demo Input:
['4\n0101\n1000\n1111\n0101\n', '3\n111\n111\n111\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything.
|
```python
n=int(input())
l=[]
for i in range(n):
s=input()
l.append(s)
m=0
for i in range(n-1):
if(l[i]!=-1):
k=1
for j in range(i+1,n):
if(l[i]==l[j]):
k+=1
l[j]=-1
if(k>m):
m=k
print(m)
```
| 0
|
|
80
|
A
|
Panoramix's Prediction
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Panoramix's Prediction
|
2
|
256
|
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
|
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
|
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
|
[
"3 5\n",
"7 11\n",
"7 9\n"
] |
[
"YES",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,603,367,874
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 154
| 0
|
from sympy import *
n = int(input())
m = int(input())
a = nextprime(n)
if m==a:
print("YES")
else:
print("NO")
|
Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none
|
```python
from sympy import *
n = int(input())
m = int(input())
a = nextprime(n)
if m==a:
print("YES")
else:
print("NO")
```
| -1
|
910
|
A
|
The Way to Home
|
PROGRAMMING
| 800
|
[
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null |
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
|
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
|
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
|
[
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] |
[
"2\n",
"-1\n",
"3\n",
"4\n"
] |
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
| 500
|
[
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
"output": "1"
},
{
"input": "10 7\n1101111011",
"output": "2"
},
{
"input": "10 9\n1110000101",
"output": "1"
},
{
"input": "10 9\n1100000001",
"output": "1"
},
{
"input": "20 5\n11111111110111101001",
"output": "4"
},
{
"input": "20 11\n11100000111000011011",
"output": "2"
},
{
"input": "20 19\n10100000000000000001",
"output": "1"
},
{
"input": "50 13\n10011010100010100111010000010000000000010100000101",
"output": "5"
},
{
"input": "50 8\n11010100000011001100001100010001110000101100110011",
"output": "8"
},
{
"input": "99 4\n111111111111111111111111111111111111111111111111111111111011111111111111111111111111111111111111111",
"output": "25"
},
{
"input": "99 98\n100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 5\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "20"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111",
"output": "25"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111101111111011111111111111111111111111111",
"output": "25"
},
{
"input": "100 3\n1111110111111111111111111111111111111111101111111111111111111111111101111111111111111111111111111111",
"output": "34"
},
{
"input": "100 8\n1111111111101110111111111111111111111111111111111111111111111111111111110011111111111111011111111111",
"output": "13"
},
{
"input": "100 7\n1011111111111111111011101111111011111101111111111101111011110111111111111111111111110111111011111111",
"output": "15"
},
{
"input": "100 9\n1101111110111110101111111111111111011001110111011101011111111111010101111111100011011111111010111111",
"output": "12"
},
{
"input": "100 6\n1011111011111111111011010110011001010101111110111111000111011011111110101101110110101111110000100111",
"output": "18"
},
{
"input": "100 7\n1110001111101001110011111111111101111101101001010001101000101100000101101101011111111101101000100001",
"output": "16"
},
{
"input": "100 11\n1000010100011100011011100000010011001111011110100100001011010100011011111001101101110110010110001101",
"output": "10"
},
{
"input": "100 9\n1001001110000011100100000001000110111101101010101001000101001010011001101100110011011110110011011111",
"output": "13"
},
{
"input": "100 7\n1010100001110101111011000111000001110100100110110001110110011010100001100100001110111100110000101001",
"output": "18"
},
{
"input": "100 10\n1110110000000110000000101110100000111000001011100000100110010001110111001010101000011000000001011011",
"output": "12"
},
{
"input": "100 13\n1000000100000000100011000010010000101010011110000000001000011000110100001000010001100000011001011001",
"output": "9"
},
{
"input": "100 11\n1000000000100000010000100001000100000000010000100100000000100100001000000001011000110001000000000101",
"output": "12"
},
{
"input": "100 22\n1000100000001010000000000000000001000000100000000000000000010000000000001000000000000000000100000001",
"output": "7"
},
{
"input": "100 48\n1000000000000000011000000000000000000000000000000001100000000000000000000000000000000000000000000001",
"output": "3"
},
{
"input": "100 48\n1000000000000000000000100000000000000000000000000000000000000000000001000000000000000000100000000001",
"output": "3"
},
{
"input": "100 75\n1000000100000000000000000000000000000000000000000000000000000000000000000000000001000000000000000001",
"output": "3"
},
{
"input": "100 73\n1000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 99\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "99"
},
{
"input": "100 2\n1111111111111111111111111111111110111111111111111111111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "100 1\n1111111111111111011111111111111111111111111111111111111111111111111101111111111111111111111111111111",
"output": "-1"
},
{
"input": "100 3\n1111111111111111111111111101111111111111111111111011111111111111111111111111111011111111111111111111",
"output": "33"
},
{
"input": "100 1\n1101111111111111111111101111111111111111111111111111111111111011111111101111101111111111111111111111",
"output": "-1"
},
{
"input": "100 6\n1111111111111111111111101111111101011110001111111111111111110111111111111111111111111110010111111111",
"output": "17"
},
{
"input": "100 2\n1111111101111010110111011011110111101111111011111101010101011111011111111111111011111001101111101111",
"output": "-1"
},
{
"input": "100 8\n1100110101111001101001111000111100110100011110111011001011111110000110101000001110111011100111011011",
"output": "14"
},
{
"input": "100 10\n1000111110100000001001101100000010011100010101001100010011111001001101111110110111101111001010001101",
"output": "11"
},
{
"input": "100 7\n1110000011010001110101011010000011110001000000011101110111010110001000011101111010010001101111110001",
"output": "-1"
},
{
"input": "100 3\n1111010001000001011011000011001111000100101000101101000010111101111000010000011110110011001101010111",
"output": "-1"
},
{
"input": "100 9\n1101010101101100010111011000010100001010000101010011001001100010110110000000010000101000000001101101",
"output": "13"
},
{
"input": "100 14\n1010100000000000010101000010001100000000000011100010000001000001011010001110001010100000100001101101",
"output": "9"
},
{
"input": "100 13\n1000000001101001110000010000011001000000000000001010000000100001001010000000000000000100010000000001",
"output": "-1"
},
{
"input": "100 18\n1000000000000000110000000000000000010000000001000001000001000000000100000000000010000000000000000001",
"output": "-1"
},
{
"input": "100 32\n1000000000000000000000000001000000000000000000000101000000000000000000000000000000000001000000000001",
"output": "-1"
},
{
"input": "100 79\n1000000001000000000101000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 41\n1000000000000000000000000000000000010000000000000000000000000000000000000000100000000000000000000001",
"output": "3"
},
{
"input": "100 82\n1000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 96\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "-1"
},
{
"input": "43 30\n1001000001111111010100100100110101011101101",
"output": "2"
},
{
"input": "7 1\n1111111",
"output": "6"
},
{
"input": "9 3\n101000001",
"output": "-1"
},
{
"input": "10 3\n1100000001",
"output": "-1"
},
{
"input": "8 2\n10000101",
"output": "-1"
},
{
"input": "2 1\n11",
"output": "1"
}
] | 1,618,497,591
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 93
| 512,000
|
import re
import sys
exit=sys.exit
from bisect import bisect_left as bsl,bisect_right as bsr
from collections import Counter,defaultdict as ddict,deque
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rln=sys.stdin.readline
rl=lambda:rln().rstrip('\n')
rfs=lambda:rln().split()
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
########################################################################
def solve(n,d,s):
ans=1
jump=maxjump=d
for i in range(1,n-1):
if s[i]=='1':
maxjump=max(maxjump,i+d)
if jump==i:
if maxjump==i:
return -1
ans+=1
jump=maxjump
return ans
n,d=ris()
s=rl()
print(solve(n,d,s))
|
Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
|
```python
import re
import sys
exit=sys.exit
from bisect import bisect_left as bsl,bisect_right as bsr
from collections import Counter,defaultdict as ddict,deque
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rln=sys.stdin.readline
rl=lambda:rln().rstrip('\n')
rfs=lambda:rln().split()
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
########################################################################
def solve(n,d,s):
ans=1
jump=maxjump=d
for i in range(1,n-1):
if s[i]=='1':
maxjump=max(maxjump,i+d)
if jump==i:
if maxjump==i:
return -1
ans+=1
jump=maxjump
return ans
n,d=ris()
s=rl()
print(solve(n,d,s))
```
| 3
|
|
449
|
B
|
Jzzhu and Cities
|
PROGRAMMING
| 2,000
|
[
"graphs",
"greedy",
"shortest paths"
] | null | null |
Jzzhu is the president of country A. There are *n* cities numbered from 1 to *n* in his country. City 1 is the capital of A. Also there are *m* roads connecting the cities. One can go from city *u**i* to *v**i* (and vise versa) using the *i*-th road, the length of this road is *x**i*. Finally, there are *k* train routes in the country. One can use the *i*-th train route to go from capital of the country to city *s**i* (and vise versa), the length of this route is *y**i*.
Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change.
|
The first line contains three integers *n*,<=*m*,<=*k* (2<=≤<=*n*<=≤<=105; 1<=≤<=*m*<=≤<=3·105; 1<=≤<=*k*<=≤<=105).
Each of the next *m* lines contains three integers *u**i*,<=*v**i*,<=*x**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*; 1<=≤<=*x**i*<=≤<=109).
Each of the next *k* lines contains two integers *s**i* and *y**i* (2<=≤<=*s**i*<=≤<=*n*; 1<=≤<=*y**i*<=≤<=109).
It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital.
|
Output a single integer representing the maximum number of the train routes which can be closed.
|
[
"5 5 3\n1 2 1\n2 3 2\n1 3 3\n3 4 4\n1 5 5\n3 5\n4 5\n5 5\n",
"2 2 3\n1 2 2\n2 1 3\n2 1\n2 2\n2 3\n"
] |
[
"2\n",
"2\n"
] |
none
| 1,000
|
[
{
"input": "5 5 3\n1 2 1\n2 3 2\n1 3 3\n3 4 4\n1 5 5\n3 5\n4 5\n5 5",
"output": "2"
},
{
"input": "2 2 3\n1 2 2\n2 1 3\n2 1\n2 2\n2 3",
"output": "2"
},
{
"input": "5 4 3\n1 2 999999999\n2 3 1000000000\n3 4 529529529\n5 1 524524524\n5 524444444\n5 529999999\n2 1000000000",
"output": "2"
},
{
"input": "3 2 5\n1 2 2\n2 3 4\n3 5\n3 5\n3 5\n3 6\n3 7",
"output": "4"
},
{
"input": "5 5 3\n1 2 999999999\n2 3 1000000000\n3 4 529529529\n5 1 524524524\n5 3 1000000000\n5 524444444\n5 529999999\n2 1000000000",
"output": "2"
},
{
"input": "2 1 5\n1 2 4\n2 3\n2 5\n2 4\n2 4\n2 5",
"output": "4"
},
{
"input": "3 3 6\n1 2 499999999\n2 3 500000000\n1 3 999999999\n2 499999999\n2 500000000\n2 499999999\n3 999999999\n3 1000000000\n3 1000000000",
"output": "6"
},
{
"input": "2 1 1\n1 2 1\n2 1000000000",
"output": "1"
},
{
"input": "3 2 2\n1 2 4\n2 3 4\n2 2\n3 6",
"output": "1"
},
{
"input": "5 5 2\n1 2 100\n2 3 100\n3 4 100\n4 5 20\n2 5 5\n5 50\n4 1",
"output": "1"
},
{
"input": "3 2 2\n1 2 100\n2 3 1\n2 1\n3 3",
"output": "1"
}
] | 1,696,497,460
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 307,200
|
import heapq
from collections import *
def dijkstra(graph, start):
distances = {node: float('inf') for node in range(1,n+1)}
distances[start] = 0
visited = set()
priority_queue = [(0, start)]
while priority_queue:
current_distance, current_node = heapq.heappop(priority_queue)
if current_node in visited:
continue
visited.add(current_node)
for neighbor, weight in graph[current_node]:
distance = current_distance + weight
if distance < distances[neighbor]:
distances[neighbor] = distance
heapq.heappush(priority_queue, (distance, neighbor))
return distances
n,m,k = list(map(int,input().split()))
graph = defaultdict(list)
for i in range(m):
start,end,weight = list(map(int,input().split()))
graph[start].append([end,weight])
ans = 0
distances = dijkstra(graph,1)
a = []
for i in range(k):
t,y = list(map(int,input().split()))
a.append([t,y])
if distances[t] <= y:
ans += 1
print(ans)
|
Title: Jzzhu and Cities
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jzzhu is the president of country A. There are *n* cities numbered from 1 to *n* in his country. City 1 is the capital of A. Also there are *m* roads connecting the cities. One can go from city *u**i* to *v**i* (and vise versa) using the *i*-th road, the length of this road is *x**i*. Finally, there are *k* train routes in the country. One can use the *i*-th train route to go from capital of the country to city *s**i* (and vise versa), the length of this route is *y**i*.
Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change.
Input Specification:
The first line contains three integers *n*,<=*m*,<=*k* (2<=≤<=*n*<=≤<=105; 1<=≤<=*m*<=≤<=3·105; 1<=≤<=*k*<=≤<=105).
Each of the next *m* lines contains three integers *u**i*,<=*v**i*,<=*x**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*; 1<=≤<=*x**i*<=≤<=109).
Each of the next *k* lines contains two integers *s**i* and *y**i* (2<=≤<=*s**i*<=≤<=*n*; 1<=≤<=*y**i*<=≤<=109).
It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital.
Output Specification:
Output a single integer representing the maximum number of the train routes which can be closed.
Demo Input:
['5 5 3\n1 2 1\n2 3 2\n1 3 3\n3 4 4\n1 5 5\n3 5\n4 5\n5 5\n', '2 2 3\n1 2 2\n2 1 3\n2 1\n2 2\n2 3\n']
Demo Output:
['2\n', '2\n']
Note:
none
|
```python
import heapq
from collections import *
def dijkstra(graph, start):
distances = {node: float('inf') for node in range(1,n+1)}
distances[start] = 0
visited = set()
priority_queue = [(0, start)]
while priority_queue:
current_distance, current_node = heapq.heappop(priority_queue)
if current_node in visited:
continue
visited.add(current_node)
for neighbor, weight in graph[current_node]:
distance = current_distance + weight
if distance < distances[neighbor]:
distances[neighbor] = distance
heapq.heappush(priority_queue, (distance, neighbor))
return distances
n,m,k = list(map(int,input().split()))
graph = defaultdict(list)
for i in range(m):
start,end,weight = list(map(int,input().split()))
graph[start].append([end,weight])
ans = 0
distances = dijkstra(graph,1)
a = []
for i in range(k):
t,y = list(map(int,input().split()))
a.append([t,y])
if distances[t] <= y:
ans += 1
print(ans)
```
| 0
|
|
533
|
C
|
Board Game
|
PROGRAMMING
| 1,700
|
[
"games",
"greedy",
"implementation",
"math"
] | null | null |
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (*x*,<=*y*) to (*x*<=-<=1,<=*y*) or (*x*,<=*y*<=-<=1). Vasiliy can move his pawn from (*x*,<=*y*) to one of cells: (*x*<=-<=1,<=*y*),<=(*x*<=-<=1,<=*y*<=-<=1) and (*x*,<=*y*<=-<=1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative *x*-coordinate or *y*-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0,<=0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
|
The first line contains four integers: *x**p*,<=*y**p*,<=*x**v*,<=*y**v* (0<=≤<=*x**p*,<=*y**p*,<=*x**v*,<=*y**v*<=≤<=105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0,<=0).
|
Output the name of the winner: "Polycarp" or "Vasiliy".
|
[
"2 1 2 2\n",
"4 7 7 4\n"
] |
[
"Polycarp\n",
"Vasiliy\n"
] |
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
| 250
|
[
{
"input": "2 1 2 2",
"output": "Polycarp"
},
{
"input": "4 7 7 4",
"output": "Vasiliy"
},
{
"input": "20 0 7 22",
"output": "Polycarp"
},
{
"input": "80 100 83 97",
"output": "Vasiliy"
},
{
"input": "80 100 77 103",
"output": "Vasiliy"
},
{
"input": "55000 60000 55003 60100",
"output": "Polycarp"
},
{
"input": "100000 100000 100000 99999",
"output": "Vasiliy"
},
{
"input": "100000 99999 100000 100000",
"output": "Polycarp"
},
{
"input": "0 100000 100000 99999",
"output": "Polycarp"
},
{
"input": "0 100000 99999 100000",
"output": "Polycarp"
},
{
"input": "0 90000 89999 89999",
"output": "Vasiliy"
},
{
"input": "0 1 0 2",
"output": "Polycarp"
},
{
"input": "0 1 1 0",
"output": "Polycarp"
},
{
"input": "0 1 1 1",
"output": "Polycarp"
},
{
"input": "0 1 1 2",
"output": "Polycarp"
},
{
"input": "0 1 2 0",
"output": "Polycarp"
},
{
"input": "0 1 2 1",
"output": "Polycarp"
},
{
"input": "0 1 2 2",
"output": "Polycarp"
},
{
"input": "0 2 0 1",
"output": "Vasiliy"
},
{
"input": "0 2 1 0",
"output": "Vasiliy"
},
{
"input": "0 2 1 1",
"output": "Vasiliy"
},
{
"input": "0 2 1 2",
"output": "Polycarp"
},
{
"input": "0 2 2 0",
"output": "Polycarp"
},
{
"input": "0 2 2 1",
"output": "Polycarp"
},
{
"input": "0 2 2 2",
"output": "Polycarp"
},
{
"input": "1 0 0 1",
"output": "Polycarp"
},
{
"input": "1 0 0 2",
"output": "Polycarp"
},
{
"input": "1 0 1 1",
"output": "Polycarp"
},
{
"input": "1 0 1 2",
"output": "Polycarp"
},
{
"input": "1 0 2 0",
"output": "Polycarp"
},
{
"input": "1 0 2 1",
"output": "Polycarp"
},
{
"input": "1 0 2 2",
"output": "Polycarp"
},
{
"input": "1 1 0 1",
"output": "Vasiliy"
},
{
"input": "1 1 0 2",
"output": "Polycarp"
},
{
"input": "1 1 1 0",
"output": "Vasiliy"
},
{
"input": "1 1 1 2",
"output": "Polycarp"
},
{
"input": "1 1 2 0",
"output": "Polycarp"
},
{
"input": "1 1 2 1",
"output": "Polycarp"
},
{
"input": "1 1 2 2",
"output": "Polycarp"
},
{
"input": "1 2 0 1",
"output": "Vasiliy"
},
{
"input": "1 2 0 2",
"output": "Vasiliy"
},
{
"input": "1 2 1 0",
"output": "Vasiliy"
},
{
"input": "1 2 1 1",
"output": "Vasiliy"
},
{
"input": "1 2 2 0",
"output": "Vasiliy"
},
{
"input": "1 2 2 1",
"output": "Vasiliy"
},
{
"input": "1 2 2 2",
"output": "Polycarp"
},
{
"input": "2 0 0 1",
"output": "Vasiliy"
},
{
"input": "2 0 0 2",
"output": "Polycarp"
},
{
"input": "2 0 1 0",
"output": "Vasiliy"
},
{
"input": "2 0 1 1",
"output": "Vasiliy"
},
{
"input": "2 0 1 2",
"output": "Polycarp"
},
{
"input": "2 0 2 1",
"output": "Polycarp"
},
{
"input": "2 0 2 2",
"output": "Polycarp"
},
{
"input": "2 1 0 1",
"output": "Vasiliy"
},
{
"input": "2 1 0 2",
"output": "Vasiliy"
},
{
"input": "2 1 1 0",
"output": "Vasiliy"
},
{
"input": "2 1 1 1",
"output": "Vasiliy"
},
{
"input": "2 1 1 2",
"output": "Vasiliy"
},
{
"input": "2 1 2 0",
"output": "Vasiliy"
},
{
"input": "2 1 2 2",
"output": "Polycarp"
},
{
"input": "2 2 0 1",
"output": "Vasiliy"
},
{
"input": "2 2 0 2",
"output": "Vasiliy"
},
{
"input": "2 2 1 0",
"output": "Vasiliy"
},
{
"input": "2 2 1 1",
"output": "Vasiliy"
},
{
"input": "2 2 1 2",
"output": "Vasiliy"
},
{
"input": "2 2 2 0",
"output": "Vasiliy"
},
{
"input": "2 2 2 1",
"output": "Vasiliy"
},
{
"input": "13118 79593 32785 22736",
"output": "Vasiliy"
},
{
"input": "23039 21508 54113 76824",
"output": "Polycarp"
},
{
"input": "32959 49970 75441 55257",
"output": "Polycarp"
},
{
"input": "91573 91885 61527 58038",
"output": "Vasiliy"
},
{
"input": "70620 15283 74892 15283",
"output": "Polycarp"
},
{
"input": "43308 1372 53325 1370",
"output": "Polycarp"
},
{
"input": "74005 7316 74004 7412",
"output": "Vasiliy"
},
{
"input": "53208 42123 95332 85846",
"output": "Polycarp"
},
{
"input": "14969 66451 81419 29039",
"output": "Vasiliy"
},
{
"input": "50042 34493 84536 17892",
"output": "Polycarp"
},
{
"input": "67949 70623 71979 70623",
"output": "Polycarp"
},
{
"input": "67603 35151 67603 39519",
"output": "Polycarp"
},
{
"input": "27149 26539 53690 17953",
"output": "Polycarp"
},
{
"input": "36711 38307 75018 72040",
"output": "Polycarp"
},
{
"input": "4650 67347 71998 50474",
"output": "Polycarp"
},
{
"input": "4075 33738 4561 33738",
"output": "Polycarp"
},
{
"input": "35868 55066 47754 55066",
"output": "Polycarp"
},
{
"input": "41150 1761 41152 1841",
"output": "Polycarp"
},
{
"input": "63557 16718 38133 80275",
"output": "Polycarp"
},
{
"input": "8956 24932 30356 33887",
"output": "Polycarp"
},
{
"input": "27338 8401 27337 12321",
"output": "Vasiliy"
},
{
"input": "56613 48665 66408 48665",
"output": "Polycarp"
},
{
"input": "34750 34886 34751 44842",
"output": "Polycarp"
},
{
"input": "7591 24141 31732 23276",
"output": "Polycarp"
},
{
"input": "2333 91141 93473 66469",
"output": "Vasiliy"
},
{
"input": "9 0 8 0",
"output": "Vasiliy"
},
{
"input": "0 1000 100 99",
"output": "Vasiliy"
},
{
"input": "4 4 2 2",
"output": "Vasiliy"
},
{
"input": "0 4 4 3",
"output": "Polycarp"
},
{
"input": "100 1 1 100",
"output": "Vasiliy"
},
{
"input": "9 17 14 16",
"output": "Vasiliy"
},
{
"input": "0 3 3 1",
"output": "Polycarp"
},
{
"input": "10 0 0 10",
"output": "Polycarp"
},
{
"input": "5 0 0 4",
"output": "Vasiliy"
},
{
"input": "2 1 1 3",
"output": "Polycarp"
},
{
"input": "4 5 5 5",
"output": "Polycarp"
},
{
"input": "0 3 2 2",
"output": "Vasiliy"
},
{
"input": "3 0 0 10",
"output": "Polycarp"
}
] | 1,698,771,030
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 105
| 124
| 1,638,400
|
# LUOGU_RID: 132709855
# pypy3
from collections import *
from itertools import *
from functools import *
from bisect import *
from heapq import *
import sys
from math import gcd
IN = lambda: sys.stdin.readline().rstrip("\r\n")
PN = lambda x: sys.stdout.write(x)
I = lambda: int(IN())
S = lambda: IN().split()
M = lambda: map(int, IN().split())
L = lambda: list(map(int, IN().split()))
G = lambda: map(lambda x: int(x) - 1, IN().split())
x0, y0, x1, y1 = M()
if x0 <= x1 and y0 <= y1:
print("Polycarp")
elif x0 + y0 <= max(x1, y1):
print("Polycarp")
else:
print("Vasiliy")
|
Title: Board Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (*x*,<=*y*) to (*x*<=-<=1,<=*y*) or (*x*,<=*y*<=-<=1). Vasiliy can move his pawn from (*x*,<=*y*) to one of cells: (*x*<=-<=1,<=*y*),<=(*x*<=-<=1,<=*y*<=-<=1) and (*x*,<=*y*<=-<=1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative *x*-coordinate or *y*-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0,<=0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input Specification:
The first line contains four integers: *x**p*,<=*y**p*,<=*x**v*,<=*y**v* (0<=≤<=*x**p*,<=*y**p*,<=*x**v*,<=*y**v*<=≤<=105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0,<=0).
Output Specification:
Output the name of the winner: "Polycarp" or "Vasiliy".
Demo Input:
['2 1 2 2\n', '4 7 7 4\n']
Demo Output:
['Polycarp\n', 'Vasiliy\n']
Note:
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
|
```python
# LUOGU_RID: 132709855
# pypy3
from collections import *
from itertools import *
from functools import *
from bisect import *
from heapq import *
import sys
from math import gcd
IN = lambda: sys.stdin.readline().rstrip("\r\n")
PN = lambda x: sys.stdout.write(x)
I = lambda: int(IN())
S = lambda: IN().split()
M = lambda: map(int, IN().split())
L = lambda: list(map(int, IN().split()))
G = lambda: map(lambda x: int(x) - 1, IN().split())
x0, y0, x1, y1 = M()
if x0 <= x1 and y0 <= y1:
print("Polycarp")
elif x0 + y0 <= max(x1, y1):
print("Polycarp")
else:
print("Vasiliy")
```
| 3
|
|
349
|
A
|
Cinema Line
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
|
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
|
[
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"input": "10\n50 50 50 50 50 50 50 50 50 50",
"output": "NO"
},
{
"input": "10\n100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "7\n25 25 25 25 50 100 50",
"output": "YES"
},
{
"input": "2\n25 25",
"output": "YES"
},
{
"input": "3\n50 100 100",
"output": "NO"
},
{
"input": "3\n25 50 50",
"output": "NO"
},
{
"input": "2\n50 25",
"output": "NO"
},
{
"input": "14\n25 25 25 25 25 25 25 50 50 50 100 100 100 100",
"output": "NO"
},
{
"input": "5\n25 25 25 100 100",
"output": "NO"
},
{
"input": "7\n25 50 25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 25 50 50 50",
"output": "NO"
},
{
"input": "5\n25 25 50 100 50",
"output": "NO"
},
{
"input": "5\n25 25 50 50 100",
"output": "NO"
},
{
"input": "1\n50",
"output": "NO"
}
] | 1,626,841,173
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 12
| 248
| 12,185,600
|
t=int(input())
x=list(map(int,input().split()))
t=0
f=0
h=0
if x[0]==25:
for i in range(len(x)):
if x[i]==25:
t+=x[i]
if x[i]==50:
if t:
t-=25
f+=x[i]
else:
print("NO")
break
if x[i]==100:
if t and f:
t-=25
f-=50
h+=x[i]
else:
print("NO")
break
else:
print("YES")
else:
print("NO")
|
Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
t=int(input())
x=list(map(int,input().split()))
t=0
f=0
h=0
if x[0]==25:
for i in range(len(x)):
if x[i]==25:
t+=x[i]
if x[i]==50:
if t:
t-=25
f+=x[i]
else:
print("NO")
break
if x[i]==100:
if t and f:
t-=25
f-=50
h+=x[i]
else:
print("NO")
break
else:
print("YES")
else:
print("NO")
```
| 0
|
|
500
|
A
|
New Year Transportation
|
PROGRAMMING
| 1,000
|
[
"dfs and similar",
"graphs",
"implementation"
] | null | null |
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
|
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
|
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
|
[
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
| 500
|
[
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1",
"output": "YES"
},
{
"input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1",
"output": "NO"
},
{
"input": "10 3\n8 3 5 4 2 3 2 2 1",
"output": "NO"
},
{
"input": "10 9\n8 3 5 4 2 3 2 2 1",
"output": "YES"
},
{
"input": "3 2\n1 1",
"output": "YES"
},
{
"input": "3 2\n2 1",
"output": "NO"
},
{
"input": "4 2\n2 1 1",
"output": "NO"
},
{
"input": "4 4\n2 2 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 1 1",
"output": "YES"
},
{
"input": "3 3\n2 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1",
"output": "YES"
},
{
"input": "8 8\n1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5 5\n1 1 1 1",
"output": "YES"
}
] | 1,695,785,911
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 9,420,800
|
import sys
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
# sys.stderr = open("error.txt", "w")
'''
baabbbeeee
i=0 b
i =1 a
'''
n, t = map(int,input().split())
arr = list(map(int,input().split()))
f = "NO"
for i , j in enumerate(arr) :
if i+arr[i] == t-1:
f="YES"
break
print(f)
|
Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
|
```python
import sys
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
# sys.stderr = open("error.txt", "w")
'''
baabbbeeee
i=0 b
i =1 a
'''
n, t = map(int,input().split())
arr = list(map(int,input().split()))
f = "NO"
for i , j in enumerate(arr) :
if i+arr[i] == t-1:
f="YES"
break
print(f)
```
| 0
|
|
453
|
A
|
Little Pony and Expected Maximum
|
PROGRAMMING
| 1,600
|
[
"probabilities"
] | null | null |
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
|
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
|
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
|
[
"6 1\n",
"6 3\n",
"2 2\n"
] |
[
"3.500000000000\n",
"4.958333333333\n",
"1.750000000000\n"
] |
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
| 500
|
[
{
"input": "6 1",
"output": "3.500000000000"
},
{
"input": "6 3",
"output": "4.958333333333"
},
{
"input": "2 2",
"output": "1.750000000000"
},
{
"input": "5 4",
"output": "4.433600000000"
},
{
"input": "5 8",
"output": "4.814773760000"
},
{
"input": "3 10",
"output": "2.982641534996"
},
{
"input": "3 6",
"output": "2.910836762689"
},
{
"input": "1 8",
"output": "1.000000000000"
},
{
"input": "24438 9",
"output": "21994.699969310015"
},
{
"input": "94444 9",
"output": "85000.099992058866"
},
{
"input": "8 66716",
"output": "8.000000000000"
},
{
"input": "4 25132",
"output": "4.000000000000"
},
{
"input": "51520 73331",
"output": "51519.682650242677"
},
{
"input": "54230 31747",
"output": "54228.743352775018"
},
{
"input": "24236 90163",
"output": "24235.975171545670"
},
{
"input": "26946 99523",
"output": "26945.974480086279"
},
{
"input": "50323 7",
"output": "44033.124988408454"
},
{
"input": "53033 3",
"output": "39775.249995286234"
},
{
"input": "55743 5",
"output": "46452.999992525307"
},
{
"input": "59964 79",
"output": "59214.949890211828"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "3 1",
"output": "2.000000000000"
},
{
"input": "1 2",
"output": "1.000000000000"
},
{
"input": "53513 34040",
"output": "53511.875329020870"
},
{
"input": "100000 100000",
"output": "99999.418033254507"
},
{
"input": "1 100000",
"output": "1.000000000000"
},
{
"input": "100000 1",
"output": "50000.499999999935"
},
{
"input": "2 100000",
"output": "2.000000000000"
},
{
"input": "100000 2",
"output": "66667.166665000332"
},
{
"input": "50000 100000",
"output": "49999.843487110789"
},
{
"input": "99999 1111",
"output": "99909.571915885972"
},
{
"input": "99999 99999",
"output": "99998.418033254609"
},
{
"input": "1000 1000",
"output": "999.419018443269"
},
{
"input": "50000 50000",
"output": "49999.418043215679"
},
{
"input": "88888 88888",
"output": "88887.418034499773"
},
{
"input": "99999 100000",
"output": "99998.418042461126"
},
{
"input": "100000 555",
"output": "99820.643422392372"
},
{
"input": "10000 10000",
"output": "9999.418122897887"
},
{
"input": "100000 5001",
"output": "99980.503829474910"
},
{
"input": "100000 1000",
"output": "99900.599066768002"
}
] | 1,607,090,170
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 1,000
| 1,536,000
|
n, m = eval(', '.join(input().split()))
def pow(a, b):
res = 1
while b:
if b&1:
res *= a
a *= a
b>>=1
return res
s = 0
d = pow(n, m)
for i in range(1, n+1):
s += i*(pow(i, m)-pow(i-1, m))
print(s/d)
|
Title: Little Pony and Expected Maximum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
Input Specification:
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output Specification:
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
Demo Input:
['6 1\n', '6 3\n', '2 2\n']
Demo Output:
['3.500000000000\n', '4.958333333333\n', '1.750000000000\n']
Note:
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
|
```python
n, m = eval(', '.join(input().split()))
def pow(a, b):
res = 1
while b:
if b&1:
res *= a
a *= a
b>>=1
return res
s = 0
d = pow(n, m)
for i in range(1, n+1):
s += i*(pow(i, m)-pow(i-1, m))
print(s/d)
```
| 0
|
|
767
|
A
|
Snacktower
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
|
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
|
[
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] |
[
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] |
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
| 500
|
[
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,629,961,891
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 6,963,200
|
n=int(input())
inputs= input().split()
has=[0]*n
# print(has)
store=[]
for a in inputs:
if int(a)!=n:
store.append(int(a))
str1=""
else:
str1 = str(a)
store.sort(reverse=True)
for j in store:
str1=str1+" "+str(j)
store=[]
print(str1)
has[int(a)-1]=1
while n!=0 and has[n-1]!=0:
n=n-1
|
Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
|
```python
n=int(input())
inputs= input().split()
has=[0]*n
# print(has)
store=[]
for a in inputs:
if int(a)!=n:
store.append(int(a))
str1=""
else:
str1 = str(a)
store.sort(reverse=True)
for j in store:
str1=str1+" "+str(j)
store=[]
print(str1)
has[int(a)-1]=1
while n!=0 and has[n-1]!=0:
n=n-1
```
| 0
|
|
979
|
B
|
Treasure Hunt
|
PROGRAMMING
| 1,800
|
[
"greedy"
] | null | null |
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of $7$ because its subribbon a appears $7$ times, and the ribbon abcdabc has the beauty of $2$ because its subribbon abc appears twice.
The rules are simple. The game will have $n$ turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after $n$ turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
|
The first line contains an integer $n$ ($0 \leq n \leq 10^{9}$) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than $10^{5}$ uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
|
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
|
[
"3\nKuroo\nShiro\nKatie\n",
"7\ntreasurehunt\nthreefriends\nhiCodeforces\n",
"1\nabcabc\ncbabac\nababca\n",
"15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE\n"
] |
[
"Kuro\n",
"Shiro\n",
"Katie\n",
"Draw\n"
] |
In the first example, after $3$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $5$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $4$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is $9$ and the number of turn is $15$, everyone can change their ribbons in some way to reach the maximal beauty of $9$ by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
| 1,000
|
[
{
"input": "3\nKuroo\nShiro\nKatie",
"output": "Kuro"
},
{
"input": "7\ntreasurehunt\nthreefriends\nhiCodeforces",
"output": "Shiro"
},
{
"input": "1\nabcabc\ncbabac\nababca",
"output": "Katie"
},
{
"input": "15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE",
"output": "Draw"
},
{
"input": "1\naaaaaaaaaa\nAAAAAAcAAA\nbbbbbbzzbb",
"output": "Shiro"
},
{
"input": "60\nddcZYXYbZbcXYcZdYbddaddYaZYZdaZdZZdXaaYdaZZZaXZXXaaZbb\ndcdXcYbcaXYaXYcacYabYcbZYdacaYbYdXaccYXZZZdYbbYdcZZZbY\nXaZXbbdcXaadcYdYYcbZdcaXaYZabbXZZYbYbcXbaXabcXbXadbZYZ",
"output": "Draw"
},
{
"input": "9174\nbzbbbzzzbbzzccczzccczzbzbzcbzbbzccbzcccbccczzbbcbbzbzzzcbczbzbzzbbbczbbcbzzzbcbzczbcczb\ndbzzzccdcdczzzzzcdczbbzcdzbcdbzzdczbzddcddbdbzzzczcczzbdcbbzccbzzzdzbzddcbzbdzdcczccbdb\nzdczddzcdddddczdczdczdcdzczddzczdzddczdcdcdzczczzdzccdccczczdzczczdzcdddzddzccddcczczzd",
"output": "Draw"
},
{
"input": "727\nbaabbabbbababbbbaaaabaabbaabababaaababaaababbbbababbbbbbbbbbaaabaabbbbbbbbaaaabaabbaaabaabbabaa\nddcdcccccccdccdcdccdddcddcddcddddcdddcdcdccddcdddddccddcccdcdddcdcccdccccccdcdcdccccccdccccccdc\nfffeefeffeefeeeeffefffeeefffeefffefeefefeeeffefefefefefefffffffeeeeeffffeefeeeeffffeeeeeefeffef",
"output": "Draw"
},
{
"input": "61\nbzqiqprzfwddqwctcrhnkqcsnbmcmfmrgaljwieajfouvuiunmfbrehxchupmsdpwilwu\njyxxujvxkwilikqeegzxlyiugflxqqbwbujzedqnlzucdnuipacatdhcozuvgktwvirhs\ntqiahohijwfcetyyjlkfhfvkhdgllxmhyyhhtlhltcdspusyhwpwqzyagtsbaswaobwub",
"output": "Katie"
},
{
"input": "30\njAjcdwkvcTYSYBBLniJIIIiubKWnqeDtUiaXSIPfhDTOrCWBQetm\nPQPOTgqfBWzQvPNeEaUaPQGdUgldmOZsBtsIqZGGyXozntMpOsyY\nNPfvGxMqIULNWOmUrHJfsqORUHkzKQfecXsTzgFCmUtFmIBudCJr",
"output": "Draw"
},
{
"input": "3\nabcabcabcabcdddabc\nzxytzytxxtytxyzxyt\nfgffghfghffgghghhh",
"output": "Katie"
},
{
"input": "3\naaaaa\naaaaa\naaaab",
"output": "Draw"
},
{
"input": "3\naaaaaaa\naaaabcd\nabcdefg",
"output": "Draw"
},
{
"input": "3\naaaaaaa\naaabcde\nabcdefg",
"output": "Kuro"
},
{
"input": "3\naaaaaaa\naaaabbb\nabcdefg",
"output": "Draw"
},
{
"input": "3\naaa\nbbb\nabc",
"output": "Draw"
},
{
"input": "3\naaaaa\nabcde\nabcde",
"output": "Kuro"
},
{
"input": "3\naaaaa\nqwert\nlkjhg",
"output": "Kuro"
},
{
"input": "3\naaaaa\nbbbbb\naabcd",
"output": "Draw"
},
{
"input": "3\nabcde\nfghij\nkkkkk",
"output": "Katie"
},
{
"input": "4\naaaabcd\naaaabcd\naaaaaaa",
"output": "Draw"
},
{
"input": "3\naaaabb\naabcde\nabcdef",
"output": "Kuro"
},
{
"input": "2\naaab\nabcd\naaaa",
"output": "Draw"
},
{
"input": "3\naaaaaa\naaaaaa\nabcdef",
"output": "Draw"
},
{
"input": "1\nAAAAA\nBBBBB\nABCDE",
"output": "Draw"
},
{
"input": "1\nabcde\naaaaa\naaaaa",
"output": "Draw"
},
{
"input": "4\naaabbb\nabfcde\nabfcde",
"output": "Kuro"
},
{
"input": "0\naaa\naab\nccd",
"output": "Kuro"
},
{
"input": "3\naaaaa\naaaaa\naabbb",
"output": "Draw"
},
{
"input": "3\nxxxxxx\nxxxooo\nabcdef",
"output": "Draw"
},
{
"input": "2\noooo\naaac\nabcd",
"output": "Draw"
},
{
"input": "1\naaaaaaa\naaabcde\nabcdefg",
"output": "Kuro"
},
{
"input": "3\nooooo\naaabb\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaa\nqwert\nqwery",
"output": "Kuro"
},
{
"input": "2\naaaaaa\nbbbbbb\naaaaab",
"output": "Draw"
},
{
"input": "3\naabb\naabb\naabc",
"output": "Draw"
},
{
"input": "2\naaa\naab\naab",
"output": "Draw"
},
{
"input": "3\nbbbbcc\nbbbbbb\nsadfgh",
"output": "Draw"
},
{
"input": "3\naaaaaacc\nxxxxkkkk\nxxxxkkkk",
"output": "Kuro"
},
{
"input": "2\naaaac\nbbbbc\nccccc",
"output": "Draw"
},
{
"input": "3\naaaaaaaaa\naaabbbbbb\nabcdewert",
"output": "Draw"
},
{
"input": "3\naaabc\naaaab\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaaaaa\naaaaaaab\naaaabbbb",
"output": "Draw"
},
{
"input": "2\nabcdefg\nabccccc\nacccccc",
"output": "Draw"
},
{
"input": "3\naaaaa\naabcd\nabcde",
"output": "Draw"
},
{
"input": "4\naaabbb\nabcdef\nabcdef",
"output": "Kuro"
},
{
"input": "4\naaabbb\naabdef\nabcdef",
"output": "Draw"
},
{
"input": "3\nabba\nbbbb\naaaa",
"output": "Draw"
},
{
"input": "3\naaaaa\nbbaaa\nabcde",
"output": "Draw"
},
{
"input": "2\naaa\naaa\nabc",
"output": "Draw"
},
{
"input": "3\naaaaa\nabcda\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaa\nabcde\nbcdef",
"output": "Kuro"
},
{
"input": "3\naaabb\naabbc\nqwert",
"output": "Draw"
},
{
"input": "3\naaaaaa\naabbcc\naabbcc",
"output": "Kuro"
},
{
"input": "3\nAAAAAA\nAAAAAB\nABCDEF",
"output": "Draw"
},
{
"input": "3\nabc\naac\nbbb",
"output": "Draw"
},
{
"input": "2\naaaab\naabbc\naabbc",
"output": "Kuro"
},
{
"input": "2\naaaaaab\naaaaabb\nabcdefg",
"output": "Draw"
},
{
"input": "3\naaaaaaaaaaa\nbbbbbbbbaaa\nqwertyuiasd",
"output": "Draw"
},
{
"input": "3\naaaa\nbbbb\naabb",
"output": "Draw"
},
{
"input": "3\naaaabb\naaabcd\nabcdef",
"output": "Draw"
},
{
"input": "3\naaa\nabc\nbbb",
"output": "Draw"
},
{
"input": "1\naa\nab\nbb",
"output": "Shiro"
},
{
"input": "1\naacb\nabcd\naaaa",
"output": "Draw"
},
{
"input": "3\naaaabb\naaabbb\nabcdef",
"output": "Draw"
},
{
"input": "3\naaaa\naaaa\nabcd",
"output": "Draw"
},
{
"input": "2\nabcd\nabcd\naaad",
"output": "Katie"
},
{
"input": "3\naaa\nbbb\naab",
"output": "Draw"
},
{
"input": "3\naaaaaa\naaaaab\naaaaaa",
"output": "Draw"
},
{
"input": "2\naaab\nabcd\nabcd",
"output": "Kuro"
},
{
"input": "3\nooooo\nShiro\nKatie",
"output": "Kuro"
},
{
"input": "3\naaabb\naabcd\nabcde",
"output": "Draw"
},
{
"input": "4\nabcd\nabcd\naaaa",
"output": "Draw"
},
{
"input": "4\naaa\nbbb\naab",
"output": "Draw"
},
{
"input": "2\nxxxx\nyyyx\nabcd",
"output": "Draw"
},
{
"input": "3\nAAAAA\nAAAAB\nABCDE",
"output": "Draw"
},
{
"input": "3\naaaacdc\naaaaabc\naaaaabc",
"output": "Draw"
},
{
"input": "3\naaaaaa\naabcde\naabcde",
"output": "Kuro"
},
{
"input": "3\naaabb\naaabb\naaaaa",
"output": "Draw"
},
{
"input": "5\nabbbbb\ncbbbbb\nabcdef",
"output": "Draw"
},
{
"input": "3\naaaaaaaaa\naaaaabbbb\naaaaabbbb",
"output": "Kuro"
},
{
"input": "4\naaaaaab\naaabbbb\naaabbbb",
"output": "Draw"
},
{
"input": "3\naaaabb\naaaabb\naaabbb",
"output": "Draw"
},
{
"input": "2\naaaabb\naaaaab\nabcdef",
"output": "Draw"
},
{
"input": "2\naaaaa\naaaae\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaaa\nbbbcde\nabcdef",
"output": "Draw"
},
{
"input": "4\naaaabbb\naabcdef\naabcdef",
"output": "Kuro"
},
{
"input": "2\naaaaa\naaaab\nabcde",
"output": "Draw"
},
{
"input": "3\naabbbbb\naaabbbb\nabcdefg",
"output": "Draw"
},
{
"input": "3\nabcde\naabcd\naaaaa",
"output": "Draw"
},
{
"input": "5\naaabbcc\nabcdefg\nabcdefg",
"output": "Kuro"
},
{
"input": "3\naabbb\nabcde\nabcde",
"output": "Kuro"
},
{
"input": "0\nbbb\nabb\nqer",
"output": "Kuro"
},
{
"input": "5\naabbbbb\naaaaaaa\nabcdefg",
"output": "Draw"
},
{
"input": "2\naaaab\naaaab\naaabb",
"output": "Draw"
},
{
"input": "2\naaaaaab\naaaabbb\naaaaccc",
"output": "Kuro"
},
{
"input": "3\naaaaaaaaaaaa\naaaaaaaaaaab\naaaaaabbbbbb",
"output": "Draw"
},
{
"input": "3\naaabb\nabcde\naaaaa",
"output": "Draw"
},
{
"input": "3\naaaaaac\naaaaebc\naaaaaac",
"output": "Draw"
},
{
"input": "3\naaaaaa\naaabbb\nqwerty",
"output": "Draw"
},
{
"input": "3\ncccca\nabcde\nabcde",
"output": "Kuro"
},
{
"input": "100005\nAA\nBC\nCC",
"output": "Draw"
},
{
"input": "3\naaaa\nbbbb\nccca",
"output": "Draw"
},
{
"input": "3\naaaaa\nbcdef\nbcdef",
"output": "Kuro"
},
{
"input": "2\naaab\naabb\nqwer",
"output": "Draw"
},
{
"input": "3\nabcddd\nabcdef\nbbaaaa",
"output": "Draw"
},
{
"input": "2\naaaa\naaaa\naabc",
"output": "Draw"
},
{
"input": "3\naaaa\naaaa\naaab",
"output": "Draw"
},
{
"input": "3\nabcddd\nabcdef\naaaaaa",
"output": "Draw"
},
{
"input": "1\naaaa\nabcd\naaab",
"output": "Katie"
}
] | 1,589,023,483
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 184
| 139
| 8,704,000
|
n = int(input())
s1, s2, s3 = input(), input(), input()
l = len(s1)
flag = 0
if n>=l:
print("Draw")
flag = 1
if flag == 0:
ans = []
m1 = m2 = m3 = 0
s01 = s02 = s03 = ''
li = [chr(x) for x in range(ord('a'), ord('z') + 1)]
li += [chr(x) for x in range(ord('A'), ord('Z') + 1)]
s_temp = set([char for char in s1])
for item in li:
t1 = s1.count(item)
t2 = s2.count(item)
t3 = s3.count(item)
if t1>m1:
m1 = t1
s01 = item
if t2>m2:
m2 = t2
s02 = item
if t3>m3:
m3 = t3
s03 = item
if (n==1):
if (m1==l):
m1-=1
else:
m1+=1
if (m2==l):
m2-=1
else:
m2+=1
if (m3==l):
m3-=1
else:
m3+=1
else:
if m1+n>l:
m1 = l
else:
m1 += n
if m2+n>l:
m2 = l
else:
m2 += n
if m3+n>l:
m3 = l
else:
m3 += n
if (m1>m2 and m1>m3):
print("Kuro")
elif (m2>m1 and m2>m3):
print("Shiro")
elif (m3>m1 and m3>m2):
print("Katie")
else:
print("Draw")
|
Title: Treasure Hunt
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of $7$ because its subribbon a appears $7$ times, and the ribbon abcdabc has the beauty of $2$ because its subribbon abc appears twice.
The rules are simple. The game will have $n$ turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after $n$ turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input Specification:
The first line contains an integer $n$ ($0 \leq n \leq 10^{9}$) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than $10^{5}$ uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output Specification:
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Demo Input:
['3\nKuroo\nShiro\nKatie\n', '7\ntreasurehunt\nthreefriends\nhiCodeforces\n', '1\nabcabc\ncbabac\nababca\n', '15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE\n']
Demo Output:
['Kuro\n', 'Shiro\n', 'Katie\n', 'Draw\n']
Note:
In the first example, after $3$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $5$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $4$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is $9$ and the number of turn is $15$, everyone can change their ribbons in some way to reach the maximal beauty of $9$ by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
|
```python
n = int(input())
s1, s2, s3 = input(), input(), input()
l = len(s1)
flag = 0
if n>=l:
print("Draw")
flag = 1
if flag == 0:
ans = []
m1 = m2 = m3 = 0
s01 = s02 = s03 = ''
li = [chr(x) for x in range(ord('a'), ord('z') + 1)]
li += [chr(x) for x in range(ord('A'), ord('Z') + 1)]
s_temp = set([char for char in s1])
for item in li:
t1 = s1.count(item)
t2 = s2.count(item)
t3 = s3.count(item)
if t1>m1:
m1 = t1
s01 = item
if t2>m2:
m2 = t2
s02 = item
if t3>m3:
m3 = t3
s03 = item
if (n==1):
if (m1==l):
m1-=1
else:
m1+=1
if (m2==l):
m2-=1
else:
m2+=1
if (m3==l):
m3-=1
else:
m3+=1
else:
if m1+n>l:
m1 = l
else:
m1 += n
if m2+n>l:
m2 = l
else:
m2 += n
if m3+n>l:
m3 = l
else:
m3 += n
if (m1>m2 and m1>m3):
print("Kuro")
elif (m2>m1 and m2>m3):
print("Shiro")
elif (m3>m1 and m3>m2):
print("Katie")
else:
print("Draw")
```
| 3
|
|
433
|
B
|
Kuriyama Mirai's Stones
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation",
"sortings"
] | null | null |
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
|
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
|
[
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] |
[
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] |
Please note that the answers to the questions may overflow 32-bit integer type.
| 1,500
|
[
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1 1 3\n1 1 3\n2 1 4\n1 1 2",
"output": "5\n7\n5\n5\n2\n7\n7\n13\n4"
},
{
"input": "18\n26 46 56 18 78 88 86 93 13 77 21 84 59 61 5 74 72 52\n25\n1 10 10\n1 9 13\n2 13 17\n1 8 14\n2 2 6\n1 12 16\n2 15 17\n2 3 6\n1 3 13\n2 8 9\n2 17 17\n1 17 17\n2 5 10\n2 1 18\n1 4 16\n1 1 13\n1 1 8\n2 7 11\n2 6 12\n1 5 9\n1 4 5\n2 7 15\n1 8 8\n1 8 14\n1 3 7",
"output": "77\n254\n413\n408\n124\n283\n258\n111\n673\n115\n88\n72\n300\n1009\n757\n745\n491\n300\n420\n358\n96\n613\n93\n408\n326"
},
{
"input": "56\n43 100 44 66 65 11 26 75 96 77 5 15 75 96 11 44 11 97 75 53 33 26 32 33 90 26 68 72 5 44 53 26 33 88 68 25 84 21 25 92 1 84 21 66 94 35 76 51 11 95 67 4 61 3 34 18\n27\n1 20 38\n1 11 46\n2 42 53\n1 8 11\n2 11 42\n2 35 39\n2 37 41\n1 48 51\n1 32 51\n1 36 40\n1 31 56\n1 18 38\n2 9 51\n1 7 48\n1 15 52\n1 27 31\n2 5 19\n2 35 50\n1 31 34\n1 2 7\n2 15 33\n2 46 47\n1 26 28\n2 3 29\n1 23 45\n2 29 55\n1 14 29",
"output": "880\n1727\n1026\n253\n1429\n335\n350\n224\n1063\n247\n1236\n1052\n2215\n2128\n1840\n242\n278\n1223\n200\n312\n722\n168\n166\n662\n1151\n2028\n772"
},
{
"input": "18\n38 93 48 14 69 85 26 47 71 11 57 9 38 65 72 78 52 47\n38\n2 10 12\n1 6 18\n2 2 2\n1 3 15\n2 1 16\n2 5 13\n1 9 17\n1 2 15\n2 5 17\n1 15 15\n2 4 11\n2 3 4\n2 2 5\n2 1 17\n2 6 16\n2 8 16\n2 8 14\n1 9 12\n2 8 13\n2 1 14\n2 5 13\n1 2 3\n1 9 14\n2 12 15\n2 3 3\n2 9 13\n2 4 12\n2 11 14\n2 6 16\n1 8 14\n1 12 15\n2 3 4\n1 3 5\n2 4 14\n1 6 6\n2 7 14\n2 7 18\n1 8 12",
"output": "174\n658\n11\n612\n742\n461\n453\n705\n767\n72\n353\n40\n89\n827\n644\n559\n409\n148\n338\n592\n461\n141\n251\n277\n14\n291\n418\n262\n644\n298\n184\n40\n131\n558\n85\n456\n784\n195"
},
{
"input": "1\n2\n10\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1",
"output": "2\n2\n2\n2\n2\n2\n2\n2\n2\n2"
},
{
"input": "2\n1 5\n8\n2 1 2\n1 1 1\n1 1 2\n1 1 1\n2 2 2\n2 1 2\n1 1 1\n1 2 2",
"output": "6\n1\n6\n1\n5\n6\n1\n5"
},
{
"input": "8\n2 6 4 6 8 4 7 7\n9\n2 6 8\n1 2 3\n2 3 4\n2 1 2\n2 4 5\n1 3 8\n2 4 4\n2 2 4\n1 2 4",
"output": "22\n10\n10\n6\n12\n36\n6\n14\n16"
},
{
"input": "4\n1 1 2 8\n9\n1 2 4\n1 2 4\n1 1 2\n2 3 4\n2 1 1\n1 2 3\n2 3 4\n1 1 4\n1 1 4",
"output": "11\n11\n2\n10\n1\n3\n10\n12\n12"
},
{
"input": "5\n8 8 1 10 7\n5\n1 2 2\n1 1 4\n1 3 3\n1 2 2\n1 2 5",
"output": "8\n27\n1\n8\n26"
},
{
"input": "10\n10 3 3 2 1 2 7 1 1 1\n4\n2 10 10\n2 6 6\n1 5 10\n1 6 8",
"output": "10\n2\n13\n10"
},
{
"input": "6\n6 6 3 8 5 4\n3\n2 2 3\n2 1 5\n1 1 5",
"output": "9\n24\n28"
},
{
"input": "1\n1\n2\n1 1 1\n2 1 1",
"output": "1\n1"
}
] | 1,656,209,791
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 31,232,000
|
#begin of codeforces template
# (don't delete):
#
# from collections import *
# from heapq import *
# import bisect
#
#t = int(input()) #input number of test cases
#for _ in range(t): #iter for test cases
# n = int(input()) #input int
# n,m = map(int,input().split()) #input tuple
# L = list(map(int,input().split())) #input list
# s = input() #input string
# ans = solve(s,L) #solve
# print(ans)
#
#end of codeforces template
from collections import *
from itertools import *
n = int(input()) #input number of test cases
A = list(map(int,input().split())) #input list
B = sorted(A)
pre1 = list(accumulate(A,initial=0))
pre2 = list(accumulate(B,initial=0))
q = int(input()) #input number of test cases
for _ in range(q):
t,l,r = map(int,input().split()) #input tuple
if t == 1:
print(pre1[r]-pre1[l-1])
else:
print(pre2[r]-pre2[l-1])
|
Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type.
|
```python
#begin of codeforces template
# (don't delete):
#
# from collections import *
# from heapq import *
# import bisect
#
#t = int(input()) #input number of test cases
#for _ in range(t): #iter for test cases
# n = int(input()) #input int
# n,m = map(int,input().split()) #input tuple
# L = list(map(int,input().split())) #input list
# s = input() #input string
# ans = solve(s,L) #solve
# print(ans)
#
#end of codeforces template
from collections import *
from itertools import *
n = int(input()) #input number of test cases
A = list(map(int,input().split())) #input list
B = sorted(A)
pre1 = list(accumulate(A,initial=0))
pre2 = list(accumulate(B,initial=0))
q = int(input()) #input number of test cases
for _ in range(q):
t,l,r = map(int,input().split()) #input tuple
if t == 1:
print(pre1[r]-pre1[l-1])
else:
print(pre2[r]-pre2[l-1])
```
| -1
|
|
757
|
A
|
Gotta Catch Em' All!
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
|
Input contains a single line containing a string *s* (1<=<=≤<=<=|*s*|<=<=≤<=<=105) — the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
|
Output a single integer, the answer to the problem.
|
[
"Bulbbasaur\n",
"F\n",
"aBddulbasaurrgndgbualdBdsagaurrgndbb\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".
| 500
|
[
{
"input": "Bulbbasaur",
"output": "1"
},
{
"input": "F",
"output": "0"
},
{
"input": "aBddulbasaurrgndgbualdBdsagaurrgndbb",
"output": "2"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbuuuuuuuuuullllllllllssssssssssaaaaaaaaaarrrrrrrrrr",
"output": "5"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuussssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "BBBBBBBBBBssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrr",
"output": "0"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuullllllllllllllllllllssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrr",
"output": "10"
},
{
"input": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuullllllllllllllllllllssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "20"
},
{
"input": "CeSlSwec",
"output": "0"
},
{
"input": "PnMrWPBGzVcmRcO",
"output": "0"
},
{
"input": "hHPWBQeEmCuhdCnzrqYtuFtwxokGhdGkFtsFICVqYfJeUrSBtSxEbzMCblOgqOvjXURhSKivPcseqgiNuUgIboEYMvVeRBbpzCGCfVydDvZNFGSFidwUtNbmPSfSYdMNmHgchIsiVswzFsGQewlMVEzicOagpWMdCWrCdPmexfnM",
"output": "0"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbuuuuuuuuuuuullllllllllllssssssssssssaaaaaaaaaaaarrrrrrrrrrrrZBphUC",
"output": "6"
},
{
"input": "bulsar",
"output": "0"
},
{
"input": "Bblsar",
"output": "0"
},
{
"input": "Bbusar",
"output": "0"
},
{
"input": "Bbular",
"output": "0"
},
{
"input": "Bbulsr",
"output": "0"
},
{
"input": "Bbulsa",
"output": "0"
},
{
"input": "Bbulsar",
"output": "0"
},
{
"input": "Bbulsar",
"output": "0"
},
{
"input": "CaQprCjTiQACZjUJjSmMHVTDorSUugvTtksEjptVzNLhClWaVVWszIixBlqFkvjDmbRjarQoUWhXHoCgYNNjvEgRTgKpbdEMFsmqcTyvJzupKgYiYMtrZWXIAGVhmDURtddbBZIMgIgXqQUmXpssLSaVCDGZDHimNthwiAWabjtcraAQugMCpBPQZbBGZyqUZmzDVSvJZmDWfZEUHGJVtiJANAIbvjTxtvvTbjWRpNQZlxAqpLCLRVwYWqLaHOTvzgeNGdxiBwsAVKKsewXMTwZUUfxYwrwsiaRBwEdvDDoPsQUtinvajBoRzLBUuQekhjsfDAOQzIABSVPitRuhvvqeAahsSELTGbCPh",
"output": "2"
},
{
"input": "Bulbasaur",
"output": "1"
},
{
"input": "BulbasaurBulbasaur",
"output": "2"
},
{
"input": "Bulbbasar",
"output": "0"
},
{
"input": "Bulbasur",
"output": "0"
},
{
"input": "Bulbsaur",
"output": "0"
},
{
"input": "BulbsurBulbsurBulbsurBulbsur",
"output": "0"
},
{
"input": "Blbbasar",
"output": "0"
},
{
"input": "Bulbasar",
"output": "0"
},
{
"input": "BBullllbbaassaauurr",
"output": "1"
},
{
"input": "BulbasaurBulbasar",
"output": "1"
},
{
"input": "BulbasaurBulbsaur",
"output": "1"
},
{
"input": "Bubasaur",
"output": "0"
},
{
"input": "ulbasaurulbasaur",
"output": "0"
},
{
"input": "Bulbasr",
"output": "0"
},
{
"input": "BBBuuulllbbbaaasssaaauuurrr",
"output": "3"
},
{
"input": "BBuuuullbbaaaassrr",
"output": "2"
},
{
"input": "BBBBBBBuuuuuuuullllllllllllbbbbaaaaaassssssssssssssssaaaaauuuuuuuuuuuuurrrrrrrrrrrrrrrr",
"output": "4"
},
{
"input": "BBuullbbaassaarr",
"output": "1"
},
{
"input": "Bulbasau",
"output": "0"
},
{
"input": "BBuullbbaassaauurr",
"output": "2"
},
{
"input": "BulbasauBulbasauBulbasauBulbasauBulbasauBulbasauBulbasauBulbasau",
"output": "0"
},
{
"input": "Blbasaur",
"output": "0"
},
{
"input": "BulbasaurBulbasaurd",
"output": "2"
},
{
"input": "ulbasaur",
"output": "0"
},
{
"input": "Bulbaaur",
"output": "0"
},
{
"input": "BBuuuullbbbbbbbbbbbbbbbaassrr",
"output": "1"
},
{
"input": "Bulbasua",
"output": "0"
},
{
"input": "Bubbasaur",
"output": "0"
},
{
"input": "BulbasauBulbasauBulbasauBulbasauBulbasauBulbasaurrr",
"output": "3"
},
{
"input": "BulbasaurBubasaur",
"output": "1"
},
{
"input": "Baab",
"output": "0"
},
{
"input": "BulbasaurBulbasau",
"output": "1"
},
{
"input": "Bulbasauu",
"output": "0"
},
{
"input": "BulbasauBulbasau",
"output": "0"
},
{
"input": "BBBBBBBBBBB",
"output": "0"
},
{
"input": "Bulbbasau",
"output": "0"
},
{
"input": "BulbbasaurBulbbasar",
"output": "1"
},
{
"input": "Bulaaaasaur",
"output": "0"
},
{
"input": "BulbasaurBulbasauBulbasauBulbasau",
"output": "1"
}
] | 1,484,849,359
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 12
| 124
| 23,654,400
|
# your code goes here
# Bulbbasaur
import sys
def main():
news = input()
# print(news)
dic = make_dict(news)
count = catch_bulb(dic)
print(count)
def make_dict(news):
dic = {}
for s in news:
if s in dic:
dic[s] += 1
else:
dic[s] = 1
return dic
def catch_bulb(dic):
# print(dic)
count = 0
while ('B' in dic and dic['B'] >= 1) and ('u' in dic and dic['u'] > 1) and ('l' in dic and dic['l'] >= 1) and ('b' in dic and dic['b'] > 1) and ('a' in dic and dic['a'] > 1) and ('s' in dic and dic['s'] >= 1) and ('r' in dic and dic['r'] >= 1):
count += 1
dic['B'] -= 1
dic['u'] -= 2
dic['l'] -= 1
dic['b'] -= 2
dic['a'] -= 2
dic['s'] -= 1
dic['r'] -= 1
return count
if __name__ == '__main__':
main()
|
Title: Gotta Catch Em' All!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input Specification:
Input contains a single line containing a string *s* (1<=<=≤<=<=|*s*|<=<=≤<=<=105) — the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
Output Specification:
Output a single integer, the answer to the problem.
Demo Input:
['Bulbbasaur\n', 'F\n', 'aBddulbasaurrgndgbualdBdsagaurrgndbb\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".
|
```python
# your code goes here
# Bulbbasaur
import sys
def main():
news = input()
# print(news)
dic = make_dict(news)
count = catch_bulb(dic)
print(count)
def make_dict(news):
dic = {}
for s in news:
if s in dic:
dic[s] += 1
else:
dic[s] = 1
return dic
def catch_bulb(dic):
# print(dic)
count = 0
while ('B' in dic and dic['B'] >= 1) and ('u' in dic and dic['u'] > 1) and ('l' in dic and dic['l'] >= 1) and ('b' in dic and dic['b'] > 1) and ('a' in dic and dic['a'] > 1) and ('s' in dic and dic['s'] >= 1) and ('r' in dic and dic['r'] >= 1):
count += 1
dic['B'] -= 1
dic['u'] -= 2
dic['l'] -= 1
dic['b'] -= 2
dic['a'] -= 2
dic['s'] -= 1
dic['r'] -= 1
return count
if __name__ == '__main__':
main()
```
| 0
|
|
687
|
A
|
NP-Hard Problem
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"graphs"
] | null | null |
Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.
Suppose the graph *G* is given. Subset *A* of its vertices is called a vertex cover of this graph, if for each edge *uv* there is at least one endpoint of it in this set, i.e. or (or both).
Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.
They have agreed to give you their graph and you need to find two disjoint subsets of its vertices *A* and *B*, such that both *A* and *B* are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).
|
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — the number of vertices and the number of edges in the prize graph, respectively.
Each of the next *m* lines contains a pair of integers *u**i* and *v**i* (1<=<=≤<=<=*u**i*,<=<=*v**i*<=<=≤<=<=*n*), denoting an undirected edge between *u**i* and *v**i*. It's guaranteed the graph won't contain any self-loops or multiple edges.
|
If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).
If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer *k* denoting the number of vertices in that vertex cover, and the second line contains *k* integers — the indices of vertices. Note that because of *m*<=≥<=1, vertex cover cannot be empty.
|
[
"4 2\n1 2\n2 3\n",
"3 3\n1 2\n2 3\n1 3\n"
] |
[
"1\n2 \n2\n1 3 \n",
"-1\n"
] |
In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).
In the second sample, there is no way to satisfy both Pari and Arya.
| 500
|
[
{
"input": "4 2\n1 2\n2 3",
"output": "1\n2 \n2\n1 3 "
},
{
"input": "3 3\n1 2\n2 3\n1 3",
"output": "-1"
},
{
"input": "5 7\n3 2\n5 4\n3 4\n1 3\n1 5\n1 4\n2 5",
"output": "-1"
},
{
"input": "10 11\n4 10\n8 10\n2 3\n2 4\n7 1\n8 5\n2 8\n7 2\n1 2\n2 9\n6 8",
"output": "-1"
},
{
"input": "10 9\n2 5\n2 4\n2 7\n2 9\n2 3\n2 8\n2 6\n2 10\n2 1",
"output": "1\n2 \n9\n1 5 4 7 9 3 8 6 10 "
},
{
"input": "10 16\n6 10\n5 2\n6 4\n6 8\n5 3\n5 4\n6 2\n5 9\n5 7\n5 1\n6 9\n5 8\n5 10\n6 1\n6 7\n6 3",
"output": "2\n5 6 \n8\n1 2 10 4 8 9 7 3 "
},
{
"input": "10 17\n5 1\n8 1\n2 1\n2 6\n3 1\n5 7\n3 7\n8 6\n4 7\n2 7\n9 7\n10 7\n3 6\n4 1\n9 1\n8 7\n10 1",
"output": "7\n5 3 2 8 4 9 10 \n3\n1 7 6 "
},
{
"input": "10 15\n5 9\n7 8\n2 9\n1 9\n3 8\n3 9\n5 8\n1 8\n6 9\n7 9\n4 8\n4 9\n10 9\n10 8\n6 8",
"output": "2\n9 8 \n8\n1 5 7 3 4 10 6 2 "
},
{
"input": "10 9\n4 9\n1 9\n10 9\n2 9\n3 9\n6 9\n5 9\n7 9\n8 9",
"output": "1\n9 \n9\n1 4 10 2 3 6 5 7 8 "
},
{
"input": "2 1\n1 2",
"output": "1\n2 \n1\n1 "
},
{
"input": "10 10\n6 4\n9 1\n3 6\n6 7\n4 2\n9 6\n8 6\n5 7\n1 4\n6 10",
"output": "6\n9 4 3 7 8 10 \n4\n1 6 2 5 "
},
{
"input": "20 22\n20 8\n1 3\n3 18\n14 7\n19 6\n7 20\n14 8\n8 10\n2 5\n11 2\n4 19\n14 2\n7 11\n15 1\n12 15\n7 6\n11 13\n1 16\n9 12\n1 19\n17 3\n11 20",
"output": "-1"
},
{
"input": "20 22\n3 18\n9 19\n6 15\n7 1\n16 8\n18 7\n12 3\n18 4\n9 15\n20 1\n4 2\n6 7\n14 2\n7 15\n7 10\n8 1\n13 6\n9 7\n11 8\n2 6\n18 5\n17 15",
"output": "-1"
},
{
"input": "1000 1\n839 771",
"output": "1\n839 \n1\n771 "
},
{
"input": "1000 1\n195 788",
"output": "1\n788 \n1\n195 "
},
{
"input": "100000 1\n42833 64396",
"output": "1\n64396 \n1\n42833 "
},
{
"input": "100000 1\n26257 21752",
"output": "1\n26257 \n1\n21752 "
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "-1"
}
] | 1,665,622,610
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 11
| 62
| 102,400
|
from collections import defaultdict, deque
n, k = list(map(int, input().split()))
graph = defaultdict(list)
for _ in range(k):
a, b = list(map(int, input().split()))
a-=1
b-=1
graph[a].append(b)
graph[b].append(a)
def dfs(node,color,u,v):
if color == 0 and node+1 in u:
return False
if color == 1 and node+1 in v:
return False
if node in visited:
return True
visited.add(node)
if color == 0: v.add(node+1)
if color == 1: u.add(node+1)
color = 0 if color == 1 else 1
for nei in graph[node]:
if not dfs(nei, color , u,v):
return False
return True
visited = set()
u = set()
v = set()
isNotPossible = True
for i in range(n):
if i not in visited and i in graph:
if dfs(i,0,u,v):
print(len(u))
print(*u)
print(len(v))
print(*v)
isNotPossible = False
break
if isNotPossible:
print("-1")
|
Title: NP-Hard Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.
Suppose the graph *G* is given. Subset *A* of its vertices is called a vertex cover of this graph, if for each edge *uv* there is at least one endpoint of it in this set, i.e. or (or both).
Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.
They have agreed to give you their graph and you need to find two disjoint subsets of its vertices *A* and *B*, such that both *A* and *B* are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — the number of vertices and the number of edges in the prize graph, respectively.
Each of the next *m* lines contains a pair of integers *u**i* and *v**i* (1<=<=≤<=<=*u**i*,<=<=*v**i*<=<=≤<=<=*n*), denoting an undirected edge between *u**i* and *v**i*. It's guaranteed the graph won't contain any self-loops or multiple edges.
Output Specification:
If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).
If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer *k* denoting the number of vertices in that vertex cover, and the second line contains *k* integers — the indices of vertices. Note that because of *m*<=≥<=1, vertex cover cannot be empty.
Demo Input:
['4 2\n1 2\n2 3\n', '3 3\n1 2\n2 3\n1 3\n']
Demo Output:
['1\n2 \n2\n1 3 \n', '-1\n']
Note:
In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).
In the second sample, there is no way to satisfy both Pari and Arya.
|
```python
from collections import defaultdict, deque
n, k = list(map(int, input().split()))
graph = defaultdict(list)
for _ in range(k):
a, b = list(map(int, input().split()))
a-=1
b-=1
graph[a].append(b)
graph[b].append(a)
def dfs(node,color,u,v):
if color == 0 and node+1 in u:
return False
if color == 1 and node+1 in v:
return False
if node in visited:
return True
visited.add(node)
if color == 0: v.add(node+1)
if color == 1: u.add(node+1)
color = 0 if color == 1 else 1
for nei in graph[node]:
if not dfs(nei, color , u,v):
return False
return True
visited = set()
u = set()
v = set()
isNotPossible = True
for i in range(n):
if i not in visited and i in graph:
if dfs(i,0,u,v):
print(len(u))
print(*u)
print(len(v))
print(*v)
isNotPossible = False
break
if isNotPossible:
print("-1")
```
| 0
|
|
40
|
A
|
Find Color
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"geometry",
"implementation",
"math"
] |
A. Find Color
|
2
|
256
|
Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture:
The picture shows only the central part of the clock. This coloring naturally extends to infinity.
The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball.
All the points located on the border of one of the areas have to be considered painted black.
|
The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000.
|
Find the required color.
All the points between which and the origin of coordinates the distance is integral-value are painted black.
|
[
"-2 1\n",
"2 1\n",
"4 3\n"
] |
[
"white\n",
"black\n",
"black\n"
] |
none
| 500
|
[
{
"input": "-2 1",
"output": "white"
},
{
"input": "2 1",
"output": "black"
},
{
"input": "4 3",
"output": "black"
},
{
"input": "3 3",
"output": "black"
},
{
"input": "4 4",
"output": "white"
},
{
"input": "-4 4",
"output": "black"
},
{
"input": "4 -4",
"output": "black"
},
{
"input": "-4 -4",
"output": "white"
},
{
"input": "0 0",
"output": "black"
},
{
"input": "0 1",
"output": "black"
},
{
"input": "0 2",
"output": "black"
},
{
"input": "0 1000",
"output": "black"
},
{
"input": "1000 0",
"output": "black"
},
{
"input": "-1000 0",
"output": "black"
},
{
"input": "0 -1000",
"output": "black"
},
{
"input": "1000 -1000",
"output": "white"
},
{
"input": "12 5",
"output": "black"
},
{
"input": "12 -5",
"output": "black"
},
{
"input": "-12 -35",
"output": "black"
},
{
"input": "20 -21",
"output": "black"
},
{
"input": "-677 492",
"output": "white"
},
{
"input": "-673 -270",
"output": "white"
},
{
"input": "-668 970",
"output": "black"
},
{
"input": "-220 208",
"output": "white"
},
{
"input": "-215 -996",
"output": "black"
},
{
"input": "-211 243",
"output": "black"
},
{
"input": "-206 -518",
"output": "white"
},
{
"input": "-201 278",
"output": "black"
},
{
"input": "-196 -484",
"output": "black"
},
{
"input": "902 479",
"output": "white"
},
{
"input": "-441 572",
"output": "white"
},
{
"input": "217 221",
"output": "white"
},
{
"input": "875 -129",
"output": "white"
},
{
"input": "-469 -36",
"output": "black"
},
{
"input": "189 -387",
"output": "white"
},
{
"input": "847 -294",
"output": "white"
},
{
"input": "-496 -644",
"output": "black"
},
{
"input": "-281 -552",
"output": "white"
},
{
"input": "377 -902",
"output": "black"
},
{
"input": "165 -738",
"output": "white"
},
{
"input": "61 -175",
"output": "black"
},
{
"input": "-42 389",
"output": "black"
},
{
"input": "-589 952",
"output": "black"
},
{
"input": "-693 -929",
"output": "white"
},
{
"input": "-796 -365",
"output": "white"
},
{
"input": "658 198",
"output": "white"
},
{
"input": "555 319",
"output": "black"
},
{
"input": "8 882",
"output": "black"
},
{
"input": "-96 -556",
"output": "black"
},
{
"input": "-129 489",
"output": "black"
},
{
"input": "207 -224",
"output": "black"
},
{
"input": "64 0",
"output": "black"
},
{
"input": "17 144",
"output": "black"
},
{
"input": "60 -448",
"output": "black"
},
{
"input": "-399 -40",
"output": "black"
},
{
"input": "128 -504",
"output": "black"
},
{
"input": "0 72",
"output": "black"
},
{
"input": "168 -26",
"output": "black"
},
{
"input": "72 -154",
"output": "black"
},
{
"input": "117 -44",
"output": "black"
},
{
"input": "-72 -646",
"output": "black"
},
{
"input": "253 -204",
"output": "black"
},
{
"input": "-40 198",
"output": "black"
},
{
"input": "-216 -90",
"output": "black"
},
{
"input": "15 -8",
"output": "black"
},
{
"input": "-180 -432",
"output": "black"
},
{
"input": "280 342",
"output": "black"
},
{
"input": "132 224",
"output": "black"
},
{
"input": "-192 -256",
"output": "black"
},
{
"input": "351 -280",
"output": "black"
}
] | 1,550,331,243
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 0
|
from math import hypot, ceil
x, y = map(int, input().split())
s = hypot(x, y)
if s != ceil(s):
print('black')
else:
print('white' if int(s) & 1 ^ x * y < 0 else 'black')
|
Title: Find Color
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture:
The picture shows only the central part of the clock. This coloring naturally extends to infinity.
The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball.
All the points located on the border of one of the areas have to be considered painted black.
Input Specification:
The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000.
Output Specification:
Find the required color.
All the points between which and the origin of coordinates the distance is integral-value are painted black.
Demo Input:
['-2 1\n', '2 1\n', '4 3\n']
Demo Output:
['white\n', 'black\n', 'black\n']
Note:
none
|
```python
from math import hypot, ceil
x, y = map(int, input().split())
s = hypot(x, y)
if s != ceil(s):
print('black')
else:
print('white' if int(s) & 1 ^ x * y < 0 else 'black')
```
| 0
|
948
|
A
|
Protect Sheep
|
PROGRAMMING
| 900
|
[
"brute force",
"dfs and similar",
"graphs",
"implementation"
] | null | null |
Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.
|
First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively.
Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.
|
If it is impossible to protect all sheep, output a single line with the word "No".
Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.
|
[
"6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n",
"1 2\nSW\n",
"5 5\n.S...\n...S.\nS....\n...S.\n.S...\n"
] |
[
"Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n",
"No\n",
"Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n"
] |
In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.
| 500
|
[
{
"input": "1 2\nSW",
"output": "No"
},
{
"input": "10 10\n....W.W.W.\n.........S\n.S.S...S..\nW.......SS\n.W..W.....\n.W...W....\nS..S...S.S\n....W...S.\n..S..S.S.S\nSS.......S",
"output": "Yes\nDDDDWDWDWD\nDDDDDDDDDS\nDSDSDDDSDD\nWDDDDDDDSS\nDWDDWDDDDD\nDWDDDWDDDD\nSDDSDDDSDS\nDDDDWDDDSD\nDDSDDSDSDS\nSSDDDDDDDS"
},
{
"input": "10 10\n....W.W.W.\n...W.....S\n.S.S...S..\nW......WSS\n.W..W.....\n.W...W....\nS..S...S.S\n...WWW..S.\n..S..S.S.S\nSS.......S",
"output": "No"
},
{
"input": "1 50\nW...S..............W.....S..S...............S...W.",
"output": "Yes\nWDDDSDDDDDDDDDDDDDDWDDDDDSDDSDDDDDDDDDDDDDDDSDDDWD"
},
{
"input": "2 4\n...S\n...W",
"output": "No"
},
{
"input": "4 2\n..\n..\n..\nSW",
"output": "No"
},
{
"input": "4 2\n..\n..\n..\nWS",
"output": "No"
},
{
"input": "2 4\n...W\n...S",
"output": "No"
},
{
"input": "50 1\nS\n.\n.\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\n.\nS\n.\nW\n.\nS\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\nW\n.\n.\n.\nW\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.",
"output": "Yes\nS\nD\nD\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nD\nS\nD\nW\nD\nS\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nW\nD\nD\nD\nW\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD"
},
{
"input": "4 4\nW..S\nW..S\nW..S\nW..S",
"output": "Yes\nWDDS\nWDDS\nWDDS\nWDDS"
},
{
"input": "4 4\nSSSS\n....\n....\nWWWW",
"output": "Yes\nSSSS\nDDDD\nDDDD\nWWWW"
},
{
"input": "4 4\nWWWW\n....\n....\nSSSS",
"output": "Yes\nWWWW\nDDDD\nDDDD\nSSSS"
},
{
"input": "4 4\nS..W\nS..W\nS..W\nS..W",
"output": "Yes\nSDDW\nSDDW\nSDDW\nSDDW"
},
{
"input": "1 1\n.",
"output": "Yes\nD"
},
{
"input": "1 1\nW",
"output": "Yes\nW"
},
{
"input": "1 1\nS",
"output": "Yes\nS"
},
{
"input": "4 2\n..\n..\n.W\n.S",
"output": "No"
},
{
"input": "4 2\n..\n..\n.S\n.W",
"output": "No"
},
{
"input": "4 2\n..\n..\nW.\nS.",
"output": "No"
},
{
"input": "4 2\n..\n..\nS.\nW.",
"output": "No"
},
{
"input": "2 4\n....\n..SW",
"output": "No"
},
{
"input": "2 4\n....\n..WS",
"output": "No"
},
{
"input": "1 2\nS.",
"output": "Yes\nSD"
}
] | 1,572,120,533
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 98
| 218
| 1,945,600
|
R,C=map(int, input().split())
m=[[str(j) for j in input()] for i in range(R)]
row=0
for i in range(R):
if row == 1:
break
for j in range(C):
if m[i][j] == ".":
m[i][j] = "D"
elif m[i][j] == "S":
if j > 0 and m[i][j-1] == "W":
row=1
break
elif j < C-1 and m[i][j+1] == "W":
row=1
break
elif i < R-1 and m[i+1][j] == "W":
row=1
break
elif i > 0 and m[i-1][j] == "W":
row=1
break
if row == 1:
print("NO")
else:
print("YES")
for i in m:
print("".join(i))
|
Title: Protect Sheep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.
Input Specification:
First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively.
Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.
Output Specification:
If it is impossible to protect all sheep, output a single line with the word "No".
Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.
Demo Input:
['6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n', '1 2\nSW\n', '5 5\n.S...\n...S.\nS....\n...S.\n.S...\n']
Demo Output:
['Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n', 'No\n', 'Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n']
Note:
In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.
|
```python
R,C=map(int, input().split())
m=[[str(j) for j in input()] for i in range(R)]
row=0
for i in range(R):
if row == 1:
break
for j in range(C):
if m[i][j] == ".":
m[i][j] = "D"
elif m[i][j] == "S":
if j > 0 and m[i][j-1] == "W":
row=1
break
elif j < C-1 and m[i][j+1] == "W":
row=1
break
elif i < R-1 and m[i+1][j] == "W":
row=1
break
elif i > 0 and m[i-1][j] == "W":
row=1
break
if row == 1:
print("NO")
else:
print("YES")
for i in m:
print("".join(i))
```
| 3
|
|
92
|
A
|
Chips
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] |
A. Chips
|
2
|
256
|
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
|
Print the number of chips the presenter ended up with.
|
[
"4 11\n",
"17 107\n",
"3 8\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
| 500
|
[
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,693,736,437
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 62
| 0
|
l=list(map(int,input().split()))
i=1
while l[1]>0:
l[1]=l[1]-i
if i==l[0]:
i=0
i=i+1
if l[1]>=0:
print(l[1])
else:
print(l[1]+i-1)
|
Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
|
```python
l=list(map(int,input().split()))
i=1
while l[1]>0:
l[1]=l[1]-i
if i==l[0]:
i=0
i=i+1
if l[1]>=0:
print(l[1])
else:
print(l[1]+i-1)
```
| 0
|
810
|
A
|
Straight <<A>>
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one.
In school, where Noora is studying, teachers are putting down marks to the online class register, which are integers from 1 to *k*. The worst mark is 1, the best is *k*. Mark that is going to the certificate, is calculated as an average of all the marks, rounded to the closest integer. If several answers are possible, rounding up is produced. For example, 7.3 is rounded to 7, but 7.5 and 7.8784 — to 8.
For instance, if Noora has marks [8,<=9], then the mark to the certificate is 9, because the average is equal to 8.5 and rounded to 9, but if the marks are [8,<=8,<=9], Noora will have graduation certificate with 8.
To graduate with «A» certificate, Noora has to have mark *k*.
Noora got *n* marks in register this year. However, she is afraid that her marks are not enough to get final mark *k*. Noora decided to ask for help in the internet, where hacker Leha immediately responded to her request. He is ready to hack class register for Noora and to add Noora any number of additional marks from 1 to *k*. At the same time, Leha want his hack be unseen to everyone, so he decided to add as less as possible additional marks. Please help Leha to calculate the minimal number of marks he has to add, so that final Noora's mark will become equal to *k*.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100) denoting the number of marks, received by Noora and the value of highest possible mark.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*k*) denoting marks received by Noora before Leha's hack.
|
Print a single integer — minimal number of additional marks, that Leha has to add in order to change Noora's final mark to *k*.
|
[
"2 10\n8 9\n",
"3 5\n4 4 4\n"
] |
[
"4",
"3"
] |
Consider the first example testcase.
Maximal mark is 10, Noora received two marks — 8 and 9, so current final mark is 9. To fix it, Leha can add marks [10, 10, 10, 10] (4 marks in total) to the registry, achieving Noora having average mark equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1b961585522f76271546da990a6228e7c666277f.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Consequently, new final mark is 10. Less number of marks won't fix the situation.
In the second example Leha can add [5, 5, 5] to the registry, so that making average mark equal to 4.5, which is enough to have 5 in the certificate.
| 500
|
[
{
"input": "2 10\n8 9",
"output": "4"
},
{
"input": "3 5\n4 4 4",
"output": "3"
},
{
"input": "3 10\n10 8 9",
"output": "3"
},
{
"input": "2 23\n21 23",
"output": "2"
},
{
"input": "5 10\n5 10 10 9 10",
"output": "7"
},
{
"input": "12 50\n18 10 26 22 22 23 14 21 27 18 25 12",
"output": "712"
},
{
"input": "38 12\n2 7 10 8 5 3 5 6 3 6 5 1 9 7 7 8 3 4 4 4 5 2 3 6 6 1 6 7 4 4 8 7 4 5 3 6 6 6",
"output": "482"
},
{
"input": "63 86\n32 31 36 29 36 26 28 38 39 32 29 26 33 38 36 38 36 28 43 48 28 33 25 39 39 27 34 25 37 28 40 26 30 31 42 32 36 44 29 36 30 35 48 40 26 34 30 33 33 46 42 24 36 38 33 51 33 41 38 29 29 32 28",
"output": "6469"
},
{
"input": "100 38\n30 24 38 31 31 33 32 32 29 34 29 22 27 23 34 25 32 30 30 26 16 27 38 33 38 38 37 34 32 27 33 23 33 32 24 24 30 36 29 30 33 30 29 30 36 33 33 35 28 24 30 32 38 29 30 36 31 30 27 38 31 36 15 37 32 27 29 24 38 33 28 29 34 21 37 35 32 31 27 25 27 28 31 31 36 38 35 35 36 29 35 22 38 31 38 28 31 27 34 31",
"output": "1340"
},
{
"input": "33 69\n60 69 68 69 69 60 64 60 62 59 54 47 60 62 69 69 69 58 67 69 62 69 68 53 69 69 66 66 57 58 65 69 61",
"output": "329"
},
{
"input": "39 92\n19 17 16 19 15 30 21 25 14 17 19 19 23 16 14 15 17 19 29 15 11 25 19 14 18 20 10 16 11 15 18 20 20 17 18 16 12 17 16",
"output": "5753"
},
{
"input": "68 29\n29 29 29 29 29 28 29 29 29 27 29 29 29 29 29 29 29 23 29 29 26 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 26 29 29 29 29 29 29 29 29 29 29 29 29 22 29 29 29 29 29 29 29 29 29 29 29 29 29 28 29 29 29 29",
"output": "0"
},
{
"input": "75 30\n22 18 21 26 23 18 28 30 24 24 19 25 28 30 23 29 18 23 23 30 26 30 17 30 18 19 25 26 26 15 27 23 30 21 19 26 25 30 25 28 20 22 22 21 26 17 23 23 24 15 25 19 18 22 30 30 29 21 30 28 28 30 27 25 24 15 22 19 30 21 20 30 18 20 25",
"output": "851"
},
{
"input": "78 43\n2 7 6 5 5 6 4 5 3 4 6 8 4 5 5 4 3 1 2 4 4 6 5 6 4 4 6 4 8 4 6 5 6 1 4 5 6 3 2 5 2 5 3 4 8 8 3 3 4 4 6 6 5 4 5 5 7 9 3 9 6 4 7 3 6 9 6 5 1 7 2 5 6 3 6 2 5 4",
"output": "5884"
},
{
"input": "82 88\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1",
"output": "14170"
},
{
"input": "84 77\n28 26 36 38 37 44 48 34 40 22 42 35 40 37 30 31 33 35 36 55 47 36 33 47 40 38 27 38 36 33 35 31 47 33 30 38 38 47 49 24 38 37 28 43 39 36 34 33 29 38 36 43 48 38 36 34 33 34 35 31 26 33 39 37 37 37 35 52 47 30 24 46 38 26 43 46 41 50 33 40 36 41 37 30",
"output": "6650"
},
{
"input": "94 80\n21 19 15 16 27 16 20 18 19 19 15 15 20 19 19 21 20 19 13 17 15 9 17 15 23 15 12 18 12 13 15 12 14 13 14 17 20 20 14 21 15 6 10 23 24 8 18 18 13 23 17 22 17 19 19 18 17 24 8 16 18 20 24 19 10 19 15 10 13 14 19 15 16 19 20 15 14 21 16 16 14 14 22 19 12 11 14 13 19 32 16 16 13 20",
"output": "11786"
},
{
"input": "96 41\n13 32 27 34 28 34 30 26 21 24 29 20 25 34 25 16 27 15 22 22 34 22 25 19 23 17 17 22 26 24 23 20 21 27 19 33 13 24 22 18 30 30 27 14 26 24 20 20 22 11 19 31 19 29 18 28 30 22 17 15 28 32 17 24 17 24 24 19 26 23 22 29 18 22 23 29 19 32 26 23 22 22 24 23 27 30 24 25 21 21 33 19 35 27 34 28",
"output": "3182"
},
{
"input": "1 26\n26",
"output": "0"
},
{
"input": "99 39\n25 28 30 28 32 34 31 28 29 28 29 30 33 19 33 31 27 33 29 24 27 30 25 38 28 34 35 31 34 37 30 22 21 24 34 27 34 33 34 33 26 26 36 19 30 22 35 30 21 28 23 35 33 29 21 22 36 31 34 32 34 32 30 32 27 33 38 25 35 26 39 27 29 29 19 33 28 29 34 38 26 30 36 26 29 30 26 34 22 32 29 38 25 27 24 17 25 28 26",
"output": "1807"
},
{
"input": "100 12\n7 6 6 3 5 5 9 8 7 7 4 7 12 6 9 5 6 3 4 7 9 10 7 7 5 3 9 6 9 9 6 7 4 10 4 8 8 6 9 8 6 5 7 4 10 7 5 6 8 9 3 4 8 5 4 8 6 10 5 8 7 5 9 8 5 8 5 6 9 11 4 9 5 5 11 4 6 6 7 3 8 9 6 7 10 4 7 6 9 4 8 11 5 4 10 8 5 10 11 4",
"output": "946"
},
{
"input": "100 18\n1 2 2 2 2 2 1 1 1 2 3 1 3 1 1 4 2 4 1 2 1 2 1 3 2 1 2 1 1 1 2 1 2 2 1 1 4 3 1 1 2 1 3 3 2 1 2 2 1 1 1 1 3 1 1 2 2 1 1 1 5 1 2 1 3 2 2 1 4 2 2 1 1 1 1 1 1 1 1 2 2 1 2 1 1 1 2 1 2 2 2 1 1 3 1 1 2 1 1 2",
"output": "3164"
},
{
"input": "100 27\n16 20 21 10 16 17 18 25 19 18 20 12 11 21 21 23 20 26 20 21 27 16 25 18 25 21 27 12 20 27 18 17 27 13 21 26 12 22 15 21 25 21 18 27 24 15 16 18 23 21 24 27 19 17 24 14 21 16 24 26 13 14 25 18 27 26 22 16 27 27 17 25 17 12 22 10 19 27 19 20 23 22 25 23 17 25 14 20 22 10 22 27 21 20 15 26 24 27 12 16",
"output": "1262"
},
{
"input": "100 29\n20 18 23 24 14 14 16 23 22 17 18 22 21 21 19 19 14 11 18 19 16 22 25 20 14 13 21 24 18 16 18 29 17 25 12 10 18 28 11 16 17 14 15 20 17 20 18 22 10 16 16 20 18 19 29 18 25 27 17 19 24 15 24 25 16 23 19 16 16 20 19 15 12 21 20 13 21 15 15 23 16 23 17 13 17 21 13 18 17 18 18 20 16 12 19 15 27 14 11 18",
"output": "2024"
},
{
"input": "100 30\n16 10 20 11 14 27 15 17 22 26 24 17 15 18 19 22 22 15 21 22 14 21 22 22 21 22 15 17 17 22 18 19 26 18 22 20 22 25 18 18 17 23 18 18 20 13 19 30 17 24 22 19 29 20 20 21 17 18 26 25 22 19 15 18 18 20 19 19 18 18 24 16 19 17 12 21 20 16 23 21 16 17 26 23 25 28 22 20 9 21 17 24 15 19 17 21 29 13 18 15",
"output": "1984"
},
{
"input": "100 59\n56 58 53 59 59 48 59 54 46 59 59 58 48 59 55 59 59 50 59 56 59 59 59 59 59 59 59 57 59 53 45 53 50 59 50 55 58 54 59 56 54 59 59 59 59 48 56 59 59 57 59 59 48 43 55 57 39 59 46 55 55 52 58 57 51 59 59 59 59 53 59 43 51 54 46 59 57 43 50 59 47 58 59 59 59 55 46 56 55 59 56 47 56 56 46 51 47 48 59 55",
"output": "740"
},
{
"input": "100 81\n6 7 6 6 7 6 6 6 3 9 4 5 4 3 4 6 6 6 1 3 9 5 2 3 8 5 6 9 6 6 6 5 4 4 7 7 3 6 11 7 6 4 8 7 12 6 4 10 2 4 9 11 7 4 7 7 8 8 6 7 9 8 4 5 8 13 6 6 6 8 6 2 5 6 7 5 4 4 4 4 2 6 4 8 3 4 7 7 6 7 7 10 5 10 6 7 4 11 8 4",
"output": "14888"
},
{
"input": "100 100\n30 35 23 43 28 49 31 32 30 44 32 37 33 34 38 28 43 32 33 32 50 32 41 38 33 20 40 36 29 21 42 25 23 34 43 32 37 31 30 27 36 32 45 37 33 29 38 34 35 33 28 19 37 33 28 41 31 29 41 27 32 39 30 34 37 40 33 38 35 32 32 34 35 34 28 39 28 34 40 45 31 25 42 28 29 31 33 21 36 33 34 37 40 42 39 30 36 34 34 40",
"output": "13118"
},
{
"input": "100 100\n71 87 100 85 89 98 90 90 71 65 76 75 85 100 81 100 91 80 73 89 86 78 82 89 77 92 78 90 100 81 85 89 73 100 66 60 72 88 91 73 93 76 88 81 86 78 83 77 74 93 97 94 85 78 82 78 91 91 100 78 89 76 78 82 81 78 83 88 87 83 78 98 85 97 98 89 88 75 76 86 74 81 70 76 86 84 99 100 89 94 72 84 82 88 83 89 78 99 87 76",
"output": "3030"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "19700"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100 100\n1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "19696"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "0"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 98 100 100 100 100 98 100 100 100 100 100 100 99 98 100 100 93 100 100 98 100 100 100 100 93 100 96 100 100 100 94 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 95 88 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100 100\n95 100 100 100 100 100 100 100 100 100 100 100 100 100 87 100 100 100 94 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 90 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 97 100 100 100 96 100 98 100 100 100 100 100 96 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 97 100 100 100 100",
"output": "2"
},
{
"input": "100 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "100 2\n2 1 1 2 1 1 1 1 2 2 2 2 1 1 1 2 1 1 1 2 2 2 2 1 1 1 1 2 2 2 1 2 2 2 2 1 2 2 1 1 1 1 1 1 2 2 1 2 1 1 1 2 1 2 2 2 2 1 1 1 2 2 1 2 1 1 1 2 1 2 2 1 1 1 2 2 1 1 2 1 1 2 1 1 1 2 1 1 1 1 2 1 1 1 1 2 1 2 1 1",
"output": "16"
},
{
"input": "3 5\n5 5 5",
"output": "0"
},
{
"input": "7 7\n1 1 1 1 1 1 1",
"output": "77"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "19700"
},
{
"input": "4 10\n10 10 10 10",
"output": "0"
},
{
"input": "1 10\n10",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3 10\n10 10 10",
"output": "0"
},
{
"input": "2 4\n3 4",
"output": "0"
},
{
"input": "1 2\n2",
"output": "0"
},
{
"input": "3 4\n4 4 4",
"output": "0"
},
{
"input": "3 2\n2 2 1",
"output": "0"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "0"
},
{
"input": "3 3\n3 3 3",
"output": "0"
},
{
"input": "2 9\n8 9",
"output": "0"
},
{
"input": "3 10\n9 10 10",
"output": "0"
},
{
"input": "1 3\n3",
"output": "0"
},
{
"input": "2 2\n1 2",
"output": "0"
},
{
"input": "2 10\n10 10",
"output": "0"
},
{
"input": "23 14\n7 11 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14",
"output": "0"
},
{
"input": "2 10\n9 10",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "0"
},
{
"input": "10 5\n5 5 5 5 5 5 5 5 5 4",
"output": "0"
},
{
"input": "3 5\n4 5 5",
"output": "0"
},
{
"input": "5 4\n4 4 4 4 4",
"output": "0"
},
{
"input": "2 10\n10 9",
"output": "0"
},
{
"input": "4 5\n3 5 5 5",
"output": "0"
},
{
"input": "10 5\n5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "3 10\n10 10 9",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "4 10\n9 10 10 10",
"output": "0"
},
{
"input": "5 2\n2 2 2 2 2",
"output": "0"
},
{
"input": "2 5\n4 5",
"output": "0"
},
{
"input": "5 10\n10 10 10 10 10",
"output": "0"
},
{
"input": "2 6\n6 6",
"output": "0"
},
{
"input": "2 9\n9 9",
"output": "0"
},
{
"input": "3 10\n10 9 10",
"output": "0"
},
{
"input": "4 40\n39 40 40 40",
"output": "0"
},
{
"input": "3 4\n3 4 4",
"output": "0"
},
{
"input": "9 9\n9 9 9 9 9 9 9 9 9",
"output": "0"
},
{
"input": "1 4\n4",
"output": "0"
},
{
"input": "4 7\n1 1 1 1",
"output": "44"
},
{
"input": "1 5\n5",
"output": "0"
},
{
"input": "3 1\n1 1 1",
"output": "0"
},
{
"input": "1 100\n100",
"output": "0"
},
{
"input": "2 7\n3 5",
"output": "10"
},
{
"input": "3 6\n6 6 6",
"output": "0"
},
{
"input": "4 2\n1 2 2 2",
"output": "0"
},
{
"input": "4 5\n4 5 5 5",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "35"
},
{
"input": "66 2\n1 2 2 2 2 1 1 2 1 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 2 1 2 2 1 1 2 1 2 2 1 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 1 2 1 1 1 2 2 1",
"output": "0"
},
{
"input": "2 2\n2 1",
"output": "0"
},
{
"input": "5 5\n5 5 5 4 5",
"output": "0"
},
{
"input": "3 7\n1 1 1",
"output": "33"
},
{
"input": "2 5\n5 5",
"output": "0"
},
{
"input": "1 7\n1",
"output": "11"
},
{
"input": "6 7\n1 1 1 1 1 1",
"output": "66"
},
{
"input": "99 97\n15 80 78 69 12 84 36 51 89 77 88 10 1 19 67 85 6 36 8 70 14 45 88 97 22 13 75 57 83 27 13 97 9 90 68 51 76 37 5 2 16 92 11 48 13 77 35 19 15 74 22 29 21 12 28 42 56 5 32 41 62 75 71 71 68 72 24 77 11 28 78 27 53 88 74 66 1 42 18 16 18 39 75 38 81 5 13 39 40 75 13 36 53 83 9 54 57 63 64",
"output": "10077"
},
{
"input": "8 7\n1 1 1 1 1 1 1 1",
"output": "88"
},
{
"input": "3 2\n2 2 2",
"output": "0"
},
{
"input": "6 5\n5 5 5 5 5 5",
"output": "0"
},
{
"input": "10 5\n5 5 5 5 5 5 5 4 1 1",
"output": "8"
},
{
"input": "1 5\n1",
"output": "7"
},
{
"input": "10 10\n10 10 10 10 10 10 10 10 10 10",
"output": "0"
},
{
"input": "2 3\n2 3",
"output": "0"
},
{
"input": "1 9\n9",
"output": "0"
},
{
"input": "74 2\n2 2 2 2 1 2 2 1 1 1 2 2 1 2 2 2 2 1 2 1 1 1 2 1 1 2 2 1 2 1 1 2 1 1 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 1 1 2 1 1 1 1 1 1 2 2 2 1 1 1 1 1 2 2 2 2 2 2 1 2",
"output": "0"
},
{
"input": "5 5\n5 5 5 5 4",
"output": "0"
}
] | 1,551,793,221
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 106
| 124
| 0
|
def rg(n):
return int(n) if n * 10 % 10 < 5 else int(n) + 1
n, k = map(int, input().split())
s = sum(map(int, input().split()))
c = 0
while rg(s / n) != k:
s += k
n += 1
c += 1
print(c)
|
Title: Straight <<A>>
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one.
In school, where Noora is studying, teachers are putting down marks to the online class register, which are integers from 1 to *k*. The worst mark is 1, the best is *k*. Mark that is going to the certificate, is calculated as an average of all the marks, rounded to the closest integer. If several answers are possible, rounding up is produced. For example, 7.3 is rounded to 7, but 7.5 and 7.8784 — to 8.
For instance, if Noora has marks [8,<=9], then the mark to the certificate is 9, because the average is equal to 8.5 and rounded to 9, but if the marks are [8,<=8,<=9], Noora will have graduation certificate with 8.
To graduate with «A» certificate, Noora has to have mark *k*.
Noora got *n* marks in register this year. However, she is afraid that her marks are not enough to get final mark *k*. Noora decided to ask for help in the internet, where hacker Leha immediately responded to her request. He is ready to hack class register for Noora and to add Noora any number of additional marks from 1 to *k*. At the same time, Leha want his hack be unseen to everyone, so he decided to add as less as possible additional marks. Please help Leha to calculate the minimal number of marks he has to add, so that final Noora's mark will become equal to *k*.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100) denoting the number of marks, received by Noora and the value of highest possible mark.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*k*) denoting marks received by Noora before Leha's hack.
Output Specification:
Print a single integer — minimal number of additional marks, that Leha has to add in order to change Noora's final mark to *k*.
Demo Input:
['2 10\n8 9\n', '3 5\n4 4 4\n']
Demo Output:
['4', '3']
Note:
Consider the first example testcase.
Maximal mark is 10, Noora received two marks — 8 and 9, so current final mark is 9. To fix it, Leha can add marks [10, 10, 10, 10] (4 marks in total) to the registry, achieving Noora having average mark equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1b961585522f76271546da990a6228e7c666277f.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Consequently, new final mark is 10. Less number of marks won't fix the situation.
In the second example Leha can add [5, 5, 5] to the registry, so that making average mark equal to 4.5, which is enough to have 5 in the certificate.
|
```python
def rg(n):
return int(n) if n * 10 % 10 < 5 else int(n) + 1
n, k = map(int, input().split())
s = sum(map(int, input().split()))
c = 0
while rg(s / n) != k:
s += k
n += 1
c += 1
print(c)
```
| 3
|
|
20
|
C
|
Dijkstra?
|
PROGRAMMING
| 1,900
|
[
"graphs",
"shortest paths"
] |
C. Dijkstra?
|
1
|
64
|
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
|
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
|
[
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] |
[
"1 4 3 5 ",
"1 4 3 5 "
] |
none
| 1,500
|
[
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,600,144,378
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 27
| 1,000
| 22,118,400
|
import queue
import sys
def get_int(): return int(sys.stdin.readline().strip())
def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
def get_strings(): return list(sys.stdin.readline().strip().split())
INF = int(1e9)
MAX = 10 ** 5 + 1
graph = [[] for _ in range(MAX)]
dist = [INF for _ in range(MAX)]
path = [-1 for _ in range(MAX)]
class Node:
def __init__(self, id, w):
self.id = id
self.w = w
def __lt__(self, other):
return self.w < other.w
def dijkstra(s):
pq = queue.PriorityQueue()
dist[s] = 0
pq.put(Node(s, 0))
while not pq.empty():
top = pq.get()
u = top.id
uw = top.w
for (v, vw) in graph[u]:
if uw + vw < dist[v]:
dist[v] = uw + vw
pq.put(Node(v, dist[v]))
path[v] = u
n, m = get_ints()
for _ in range(m):
a, b, w = get_ints()
graph[a].append((b, w))
graph[b].append((a, w))
dijkstra(1)
if dist[n] == INF:
print(-1)
else:
ans = [n]
i = n
while path[i] != -1:
ans.append(path[i])
i = path[i]
ans.reverse()
print(" ".join(list(map(str, ans))))
|
Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none
|
```python
import queue
import sys
def get_int(): return int(sys.stdin.readline().strip())
def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
def get_strings(): return list(sys.stdin.readline().strip().split())
INF = int(1e9)
MAX = 10 ** 5 + 1
graph = [[] for _ in range(MAX)]
dist = [INF for _ in range(MAX)]
path = [-1 for _ in range(MAX)]
class Node:
def __init__(self, id, w):
self.id = id
self.w = w
def __lt__(self, other):
return self.w < other.w
def dijkstra(s):
pq = queue.PriorityQueue()
dist[s] = 0
pq.put(Node(s, 0))
while not pq.empty():
top = pq.get()
u = top.id
uw = top.w
for (v, vw) in graph[u]:
if uw + vw < dist[v]:
dist[v] = uw + vw
pq.put(Node(v, dist[v]))
path[v] = u
n, m = get_ints()
for _ in range(m):
a, b, w = get_ints()
graph[a].append((b, w))
graph[b].append((a, w))
dijkstra(1)
if dist[n] == INF:
print(-1)
else:
ans = [n]
i = n
while path[i] != -1:
ans.append(path[i])
i = path[i]
ans.reverse()
print(" ".join(list(map(str, ans))))
```
| 0
|
687
|
B
|
Remainders Game
|
PROGRAMMING
| 1,800
|
[
"chinese remainder theorem",
"math",
"number theory"
] | null | null |
Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000).
|
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise.
|
[
"4 5\n2 3 5 12\n",
"2 7\n2 3\n"
] |
[
"Yes\n",
"No\n"
] |
In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.
| 1,000
|
[
{
"input": "4 5\n2 3 5 12",
"output": "Yes"
},
{
"input": "2 7\n2 3",
"output": "No"
},
{
"input": "1 6\n8",
"output": "No"
},
{
"input": "2 3\n9 4",
"output": "Yes"
},
{
"input": "4 16\n19 16 13 9",
"output": "Yes"
},
{
"input": "5 10\n5 16 19 9 17",
"output": "Yes"
},
{
"input": "11 95\n31 49 8 139 169 121 71 17 43 29 125",
"output": "No"
},
{
"input": "17 71\n173 43 139 73 169 199 49 81 11 89 131 107 23 29 125 152 17",
"output": "No"
},
{
"input": "13 86\n41 64 17 31 13 97 19 25 81 47 61 37 71",
"output": "No"
},
{
"input": "15 91\n49 121 83 67 128 125 27 113 41 169 149 19 37 29 71",
"output": "Yes"
},
{
"input": "2 4\n2 2",
"output": "No"
},
{
"input": "14 87\n1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619",
"output": "No"
},
{
"input": "12 100\n1766 1766 1766 1766 1766 1766 1766 1766 1766 1766 1766 1766",
"output": "No"
},
{
"input": "1 994619\n216000",
"output": "No"
},
{
"input": "1 651040\n911250",
"output": "No"
},
{
"input": "1 620622\n60060",
"output": "No"
},
{
"input": "1 1\n559872",
"output": "Yes"
},
{
"input": "88 935089\n967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967",
"output": "No"
},
{
"input": "93 181476\n426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426",
"output": "No"
},
{
"input": "91 4900\n630 630 70 630 910 630 630 630 770 70 770 630 630 770 70 630 70 630 70 630 70 630 630 70 910 630 630 630 770 630 630 630 70 910 70 630 70 630 770 630 630 70 630 770 70 630 70 70 630 630 70 70 70 70 630 70 70 770 910 630 70 630 770 70 910 70 630 910 630 70 770 70 70 630 770 630 70 630 70 70 630 70 630 770 630 70 630 630 70 910 630",
"output": "No"
},
{
"input": "61 531012\n698043 698043 698043 963349 698043 698043 698043 963349 698043 698043 698043 963349 698043 698043 698043 698043 966694 698043 698043 698043 698043 698043 698043 636247 698043 963349 698043 698043 698043 698043 697838 698043 963349 698043 698043 966694 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 963349 698043 698043 698043 698043 963349 698043",
"output": "No"
},
{
"input": "1 216000\n648000",
"output": "Yes"
},
{
"input": "2 8\n4 4",
"output": "No"
},
{
"input": "3 8\n4 4 4",
"output": "No"
},
{
"input": "2 8\n2 4",
"output": "No"
},
{
"input": "3 12\n2 2 3",
"output": "No"
},
{
"input": "10 4\n2 2 2 2 2 2 2 2 2 2",
"output": "No"
},
{
"input": "10 1024\n1 2 4 8 16 32 64 128 256 512",
"output": "No"
},
{
"input": "3 24\n2 2 3",
"output": "No"
},
{
"input": "1 8\n2",
"output": "No"
},
{
"input": "2 9\n3 3",
"output": "No"
},
{
"input": "3 4\n2 2 2",
"output": "No"
},
{
"input": "3 4\n1 2 2",
"output": "No"
},
{
"input": "1 4\n2",
"output": "No"
},
{
"input": "1 100003\n2",
"output": "No"
},
{
"input": "1 2\n12",
"output": "Yes"
},
{
"input": "2 988027\n989018 995006",
"output": "Yes"
},
{
"input": "3 9\n3 3 3",
"output": "No"
},
{
"input": "1 49\n7",
"output": "No"
},
{
"input": "2 600000\n200000 300000",
"output": "Yes"
},
{
"input": "3 8\n2 2 2",
"output": "No"
},
{
"input": "7 510510\n524288 531441 390625 823543 161051 371293 83521",
"output": "Yes"
},
{
"input": "2 30\n6 10",
"output": "Yes"
},
{
"input": "2 27000\n5400 4500",
"output": "Yes"
},
{
"input": "3 8\n1 2 4",
"output": "No"
},
{
"input": "4 16\n2 2 2 2",
"output": "No"
},
{
"input": "2 16\n4 8",
"output": "No"
},
{
"input": "2 8\n4 2",
"output": "No"
},
{
"input": "3 4\n2 2 3",
"output": "No"
},
{
"input": "1 8\n4",
"output": "No"
},
{
"input": "1 999983\n2",
"output": "No"
},
{
"input": "3 16\n2 4 8",
"output": "No"
},
{
"input": "2 216\n12 18",
"output": "No"
},
{
"input": "2 16\n8 8",
"output": "No"
},
{
"input": "2 36\n18 12",
"output": "Yes"
},
{
"input": "2 36\n12 18",
"output": "Yes"
},
{
"input": "2 1000000\n1000000 1000000",
"output": "Yes"
},
{
"input": "3 20\n2 2 5",
"output": "No"
},
{
"input": "1 2\n6",
"output": "Yes"
},
{
"input": "4 4\n2 3 6 5",
"output": "No"
},
{
"input": "1 2\n1",
"output": "No"
},
{
"input": "1 6\n6",
"output": "Yes"
},
{
"input": "2 16\n4 4",
"output": "No"
},
{
"input": "2 3779\n1 2",
"output": "No"
},
{
"input": "2 8\n4 12",
"output": "No"
},
{
"input": "2 24\n4 6",
"output": "No"
},
{
"input": "1 1\n5",
"output": "Yes"
},
{
"input": "10 255255\n1000000 700000 300000 110000 130000 170000 190000 230000 290000 310000",
"output": "Yes"
},
{
"input": "2 1000\n500 2",
"output": "No"
},
{
"input": "4 8\n2 2 2 2",
"output": "No"
},
{
"input": "1 666013\n1",
"output": "No"
},
{
"input": "1 999983\n1",
"output": "No"
},
{
"input": "1 125\n5",
"output": "No"
},
{
"input": "2 32\n4 8",
"output": "No"
},
{
"input": "3 32\n2 4 8",
"output": "No"
},
{
"input": "3 8\n2 4 11",
"output": "No"
},
{
"input": "1 32\n2",
"output": "No"
},
{
"input": "4 4\n2 2 2 2",
"output": "No"
},
{
"input": "8 32\n2 2 2 2 2 2 2 2",
"output": "No"
},
{
"input": "3 24\n2 4 3",
"output": "No"
},
{
"input": "2 4\n2 6",
"output": "No"
},
{
"input": "2 24\n6 4",
"output": "No"
},
{
"input": "1 20998\n2",
"output": "No"
},
{
"input": "1 1\n1",
"output": "Yes"
},
{
"input": "2 20998\n2 10499",
"output": "Yes"
},
{
"input": "1 5\n1",
"output": "No"
},
{
"input": "2 25\n5 5",
"output": "No"
},
{
"input": "5 2\n2 2 2 2 2",
"output": "Yes"
},
{
"input": "3 12\n2 3 6",
"output": "No"
},
{
"input": "1 1\n3",
"output": "Yes"
},
{
"input": "2 16\n8 4",
"output": "No"
},
{
"input": "40 10\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "No"
}
] | 1,696,303,566
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
from math import lcm
n, k = [int(s) for s in input().split(' ')]
c = [int(s) for s in input().split(' ')]
m = lcm(*c)
if m % k == 0:
print('Yes')
else:
print('No')
|
Title: Remainders Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000).
Output Specification:
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise.
Demo Input:
['4 5\n2 3 5 12\n', '2 7\n2 3\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.
|
```python
from math import lcm
n, k = [int(s) for s in input().split(' ')]
c = [int(s) for s in input().split(' ')]
m = lcm(*c)
if m % k == 0:
print('Yes')
else:
print('No')
```
| -1
|
|
219
|
A
|
k-String
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
|
The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.
|
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
|
[
"2\naazz\n",
"3\nabcabcabz\n"
] |
[
"azaz\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
},
{
"input": "5\naaaaa",
"output": "aaaaa"
},
{
"input": "3\naabaaaaabb",
"output": "-1"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "2\nbabac",
"output": "-1"
},
{
"input": "3\nbbbccc",
"output": "bcbcbc"
},
{
"input": "2\naa",
"output": "aa"
},
{
"input": "250\ncececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece",
"output": "cececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece"
},
{
"input": "15\nabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaa",
"output": "aaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbc"
},
{
"input": "1\naaa",
"output": "aaa"
},
{
"input": "1\naabaab",
"output": "aaaabb"
},
{
"input": "2\naabbbbccccccdddddddd",
"output": "abbcccddddabbcccdddd"
},
{
"input": "3\naaaaaaaaacccdddddd",
"output": "aaacddaaacddaaacdd"
},
{
"input": "3\naaaaaaaaacccbbbbbb",
"output": "aaabbcaaabbcaaabbc"
},
{
"input": "2\naazzzz",
"output": "azzazz"
},
{
"input": "2\naaaa",
"output": "aaaa"
},
{
"input": "2\naaaazz",
"output": "aazaaz"
},
{
"input": "2\naaaabb",
"output": "aabaab"
},
{
"input": "2\naabbbb",
"output": "abbabb"
},
{
"input": "2\naaaazzzz",
"output": "aazzaazz"
},
{
"input": "2\naaazz",
"output": "-1"
},
{
"input": "2\nbbbbaa",
"output": "abbabb"
},
{
"input": "2\naaaaaabbbb",
"output": "aaabbaaabb"
},
{
"input": "2\naaaaaazz",
"output": "aaazaaaz"
},
{
"input": "2\naaaabbbb",
"output": "aabbaabb"
},
{
"input": "2\naabaab",
"output": "aabaab"
},
{
"input": "1\naaaaa",
"output": "aaaaa"
},
{
"input": "2\nbbaaaa",
"output": "aabaab"
},
{
"input": "2\nacaccc",
"output": "accacc"
}
] | 1,682,592,211
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 154
| 0
|
def rearrange(k, s):
n = len(s)
#frequency of character
char_count = [0] * 26
for char in s:
char_count[ord(char) - ord('a')] += 1
for i in range(26):
if char_count[i] % k != 0:
return -1
result = ''
for i in range(26):
result += chr(ord('a') + i) * (char_count[i] // k)
result *= k
return result
#input
k = int(input())
s = input().strip()
result = rearrange(k, s)
print(result)
|
Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none
|
```python
def rearrange(k, s):
n = len(s)
#frequency of character
char_count = [0] * 26
for char in s:
char_count[ord(char) - ord('a')] += 1
for i in range(26):
if char_count[i] % k != 0:
return -1
result = ''
for i in range(26):
result += chr(ord('a') + i) * (char_count[i] // k)
result *= k
return result
#input
k = int(input())
s = input().strip()
result = rearrange(k, s)
print(result)
```
| 3
|
Subsets and Splits
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Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.