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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
418
|
D
|
Big Problems for Organizers
|
PROGRAMMING
| 2,800
|
[
"data structures",
"graphs",
"trees"
] | null | null |
The Finals of the "Russian Code Cup" 2214 will be held in *n* hotels. Two hotels (let's assume that they are the main hotels), will host all sorts of events, and the remaining hotels will accommodate the participants. The hotels are connected by *n*<=-<=1 roads, you can get from any hotel to any other one.
The organizers wonder what is the minimum time all the participants need to get to the main hotels, if each participant goes to the main hotel that is nearest to him and moving between two hotels connected by a road takes one unit of time.
The hosts consider various options for the location of the main hotels. For each option help the organizers to find minimal time.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100000) — the number of hotels. The next *n*<=-<=1 lines contain two integers each — the numbers of the hotels that have a road between them. Consider hotels are numbered from 1 to *n*.
The next line contains an integer *m* (1<=≤<=*m*<=≤<=100000) — the number of queries. The following *m* lines contains two distinct integers each — the numbers of the hotels we assume to be the main.
|
For each request of the organizers print a single integer — the time that all participants need to reach the main hotels.
|
[
"3\n2 3\n3 1\n3\n2 1\n2 3\n3 1\n",
"4\n1 4\n1 2\n2 3\n3\n1 4\n1 3\n2 3\n"
] |
[
"1\n1\n1\n",
"2\n1\n2\n"
] |
none
| 2,500
|
[] | 1,689,418,621
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1689418621.2949936")# 1689418621.2950141
|
Title: Big Problems for Organizers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Finals of the "Russian Code Cup" 2214 will be held in *n* hotels. Two hotels (let's assume that they are the main hotels), will host all sorts of events, and the remaining hotels will accommodate the participants. The hotels are connected by *n*<=-<=1 roads, you can get from any hotel to any other one.
The organizers wonder what is the minimum time all the participants need to get to the main hotels, if each participant goes to the main hotel that is nearest to him and moving between two hotels connected by a road takes one unit of time.
The hosts consider various options for the location of the main hotels. For each option help the organizers to find minimal time.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100000) — the number of hotels. The next *n*<=-<=1 lines contain two integers each — the numbers of the hotels that have a road between them. Consider hotels are numbered from 1 to *n*.
The next line contains an integer *m* (1<=≤<=*m*<=≤<=100000) — the number of queries. The following *m* lines contains two distinct integers each — the numbers of the hotels we assume to be the main.
Output Specification:
For each request of the organizers print a single integer — the time that all participants need to reach the main hotels.
Demo Input:
['3\n2 3\n3 1\n3\n2 1\n2 3\n3 1\n', '4\n1 4\n1 2\n2 3\n3\n1 4\n1 3\n2 3\n']
Demo Output:
['1\n1\n1\n', '2\n1\n2\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1689418621.2949936")# 1689418621.2950141
```
| 0
|
|
254
|
A
|
Cards with Numbers
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"sortings"
] | null | null |
Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.
|
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs.
Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.
|
[
"3\n20 30 10 30 20 10\n",
"1\n1 2\n"
] |
[
"4 2\n1 5\n6 3\n",
"-1"
] |
none
| 500
|
[
{
"input": "3\n20 30 10 30 20 10",
"output": "4 2\n1 5\n6 3"
},
{
"input": "1\n1 2",
"output": "-1"
},
{
"input": "5\n2 2 2 2 2 1 2 2 1 2",
"output": "2 1\n3 4\n7 5\n6 9\n10 8"
},
{
"input": "5\n2 1 2 2 1 1 1 1 1 2",
"output": "3 1\n2 5\n7 6\n8 9\n10 4"
},
{
"input": "5\n1 2 2 2 1 2 2 1 2 1",
"output": "3 2\n1 5\n6 4\n7 9\n10 8"
},
{
"input": "5\n3 3 1 1 1 3 2 3 1 2",
"output": "2 1\n3 4\n8 6\n5 9\n10 7"
},
{
"input": "5\n1 1 3 1 3 3 3 1 1 1",
"output": "2 1\n3 5\n7 6\n4 8\n10 9"
},
{
"input": "5\n3 1 1 1 2 3 3 3 2 1",
"output": "3 2\n1 6\n8 7\n5 9\n10 4"
},
{
"input": "5\n3 3 2 2 3 3 1 3 1 3",
"output": "2 1\n3 4\n6 5\n7 9\n10 8"
},
{
"input": "5\n4 1 3 1 4 1 2 2 3 1",
"output": "4 2\n1 5\n8 7\n3 9\n10 6"
},
{
"input": "100\n8 6 7 8 7 9 1 7 3 3 5 8 7 8 5 4 8 4 8 1 2 8 3 7 8 7 6 5 7 9 6 10 7 6 7 8 6 8 9 5 1 5 6 1 4 8 4 8 7 2 6 2 6 6 2 8 2 8 7 1 5 4 4 6 4 9 7 5 1 8 1 3 9 2 3 2 4 7 6 10 5 3 4 10 8 9 6 7 2 7 10 1 8 10 4 1 1 1 2 7 5 4 9 10 6 8 3 1 10 9 9 6 1 5 8 6 6 3 3 4 10 10 8 9 7 10 9 3 7 6 3 2 10 8 5 8 5 5 5 10 8 5 7 6 10 7 7 9 10 10 9 9 3 6 5 6 8 1 9 8 2 4 8 8 6 8 10 2 3 5 2 6 8 4 8 6 4 5 10 8 1 10 5 2 5 6 8 2 6 8 1 3 4 5 7 5 6 9 2 8",
"output": "4 1\n3 5\n10 9\n8 13\n14 12\n11 15\n18 16\n17 19\n20 7\n22 25\n26 24\n2 27\n30 6\n29 33\n34 31\n36 38\n40 28\n37 43\n44 41\n45 47\n48 46\n35 49\n50 21\n51 53\n55 52\n56 58\n61 42\n62 63\n64 54\n39 66\n67 59\n60 69\n72 23\n57 74\n77 65\n32 80\n81 68\n75 82\n85 70\n73 86\n87 79\n78 88\n89 76\n84 91\n92 71\n83 95\n97 96\n90 100\n104 94\n93 106\n108 98\n103 110\n112 105\n101 114\n117 116\n107 118\n120 102\n109 121\n123 115\n111 124\n126 122\n119 128\n129 125\n99 132\n136 134\n135 137\n139 138\n133 140\n144 130..."
},
{
"input": "100\n7 3 8 8 1 9 6 6 3 3 8 2 7 9 9 10 2 10 4 4 9 3 6 5 2 6 3 6 3 5 2 3 8 2 5 10 3 9 7 2 1 6 7 4 8 3 9 10 9 4 3 3 7 1 4 2 2 5 6 6 1 7 9 1 8 1 2 2 5 9 7 7 6 4 6 10 1 1 8 1 5 6 4 9 5 4 4 10 6 4 5 1 9 1 7 8 6 10 3 2 4 7 10 4 8 10 6 7 8 4 1 3 8 3 2 1 9 4 2 4 3 1 6 8 6 2 2 5 6 8 6 10 1 6 4 2 7 3 6 10 6 5 6 6 3 9 4 6 4 1 5 4 4 2 8 4 10 3 7 6 6 10 2 5 5 6 1 6 1 9 9 1 10 5 10 1 1 5 7 5 2 1 4 2 3 3 3 5 1 8 10 3 3 5 9 6 3 6 8 1",
"output": "4 3\n7 8\n9 2\n1 13\n14 6\n12 17\n18 16\n19 20\n21 15\n10 22\n26 23\n27 29\n30 24\n25 31\n33 11\n32 37\n40 34\n5 41\n42 28\n39 43\n47 38\n36 48\n50 44\n46 51\n57 56\n35 58\n60 59\n54 61\n62 53\n49 63\n65 45\n64 66\n68 67\n71 72\n74 55\n73 75\n78 77\n69 81\n84 70\n83 86\n88 76\n82 89\n90 87\n85 91\n92 80\n79 96\n99 52\n95 102\n103 98\n101 104\n107 97\n105 109\n111 94\n112 114\n115 100\n93 117\n118 110\n116 122\n124 113\n123 125\n126 119\n129 131\n132 106\n120 135\n136 127\n108 137\n138 121\n134 139\n142 128..."
},
{
"input": "100\n6 3 6 8 8 4 3 7 10 3 1 3 9 5 10 10 6 7 6 6 2 3 8 8 7 6 4 9 6 7 4 4 10 4 7 3 2 7 10 8 6 7 9 1 3 5 3 7 9 1 1 7 1 1 7 7 8 3 2 7 4 8 7 8 10 3 1 7 2 7 9 8 8 8 5 2 8 1 2 7 8 7 8 8 5 10 10 4 9 10 8 7 8 8 7 7 3 6 4 3 4 8 10 8 6 3 7 1 8 6 3 3 7 10 3 9 3 5 10 9 9 2 8 7 2 3 2 1 10 9 6 2 8 7 2 2 5 3 10 6 7 2 1 1 5 10 7 5 4 9 7 7 8 1 1 3 3 7 10 5 9 8 6 8 2 2 1 7 8 9 6 2 2 6 2 9 10 2 10 9 6 3 3 10 6 5 3 6 6 3 6 10 8 7 4 8 6 3 4 7",
"output": "-1"
},
{
"input": "1\n2 2",
"output": "2 1"
},
{
"input": "2\n1 2 4 7",
"output": "-1"
}
] | 1,556,375,350
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
f = open("input.txt", "rt")
R = open("output.txt", "wt")
n = int(f.readline())
A = map(int, f.readline().split())
C = {}
X = []
i = 1
O = 0
for a in A:
if a not in C or C[a]==0:
C[a] = i
O += 1
else:
X += [`C[a]`+' '+ `i` + '\n']
C[a] = 0
O -= 1
i += 1
if O:
R.write("-1")
else:
R.writelines(X)
R.close()
|
Title: Cards with Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.
Output Specification:
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs.
Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.
Demo Input:
['3\n20 30 10 30 20 10\n', '1\n1 2\n']
Demo Output:
['4 2\n1 5\n6 3\n', '-1']
Note:
none
|
```python
f = open("input.txt", "rt")
R = open("output.txt", "wt")
n = int(f.readline())
A = map(int, f.readline().split())
C = {}
X = []
i = 1
O = 0
for a in A:
if a not in C or C[a]==0:
C[a] = i
O += 1
else:
X += [`C[a]`+' '+ `i` + '\n']
C[a] = 0
O -= 1
i += 1
if O:
R.write("-1")
else:
R.writelines(X)
R.close()
```
| -1
|
|
680
|
B
|
Bear and Finding Criminals
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation"
] | null | null |
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
|
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
|
Print the number of criminals Limak will catch.
|
[
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] |
[
"3\n",
"1\n"
] |
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
| 1,000
|
[
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0 1 0 1",
"output": "5"
},
{
"input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0",
"output": "10"
},
{
"input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0",
"output": "27"
},
{
"input": "8 1\n1 0 1 1 0 0 1 0",
"output": "4"
},
{
"input": "11 11\n0 1 0 0 1 1 1 0 0 0 0",
"output": "4"
},
{
"input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1",
"output": "4"
},
{
"input": "100 38\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 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 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 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 38\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 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 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 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100 38\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 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 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 0 0 0 0 0",
"output": "3"
},
{
"input": "99 38\n0 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 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "25"
},
{
"input": "99 38\n0 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 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24"
},
{
"input": "99 38\n0 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 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24"
},
{
"input": "98 70\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 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 0 0 0 0 0",
"output": "41"
},
{
"input": "99 70\n0 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 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 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 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": "9"
},
{
"input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "34"
},
{
"input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1",
"output": "39"
},
{
"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": "100"
},
{
"input": "100 1\n0 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 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 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "2 1\n0 1",
"output": "1"
}
] | 1,604,879,184
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 0
|
# Solve for the
for _ in range(int(input())):
a,b,c,d = map(int, input().split())
print(a+c)
|
Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
|
```python
# Solve for the
for _ in range(int(input())):
a,b,c,d = map(int, input().split())
print(a+c)
```
| -1
|
|
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"
}
] | 1,653,970,365
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 30
| 0
|
n,k=map(int,input().split())
if n<k:
print(-1)
elif n//k%k==0 and n//k>1:
print(n//k)
else:
while n//k%k!=0 and n//k>1:
n+=1
print(n//k)
|
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,k=map(int,input().split())
if n<k:
print(-1)
elif n//k%k==0 and n//k>1:
print(n//k)
else:
while n//k%k!=0 and n//k>1:
n+=1
print(n//k)
```
| 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"
}
] | 1,654,181,404
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
"""
def CheckForHello (word):
hello = ["h", "e", "l", "o"]
for i in range(len(word)):
if word[i] != "h" or word[i] != "e" or word[i] != "l" or word[i] != "o":
word.replace(word[i], "")
elif word[i] in hello:
if word == "hello":
return "YES"
else:
return "NO"
"""
def CheckForLetters(word):
CorrectList = []
for i in range(len(word)):
if word[i] == "h":
CorrectList.append(-10)
if word[i] == "e":
CorrectList.append(-9)
if word[i] == "l":
CorrectList.append(-8)
#if word[i] == "l":
# CorrectList.append(-7)
if word[i] == "o":
CorrectList.append(-6)
#print(CorrectList)
return CorrectList
def CheckForAscending (AList):
flag = 0
i = 1
while i < len(AList):
if(AList[i] < AList[i - 1]):
flag = 1
i += 1
if (not flag):
return "YES"
else:
return "NO"
word = input()
print(CheckForAscending(CheckForLetters(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 CheckForHello (word):
hello = ["h", "e", "l", "o"]
for i in range(len(word)):
if word[i] != "h" or word[i] != "e" or word[i] != "l" or word[i] != "o":
word.replace(word[i], "")
elif word[i] in hello:
if word == "hello":
return "YES"
else:
return "NO"
"""
def CheckForLetters(word):
CorrectList = []
for i in range(len(word)):
if word[i] == "h":
CorrectList.append(-10)
if word[i] == "e":
CorrectList.append(-9)
if word[i] == "l":
CorrectList.append(-8)
#if word[i] == "l":
# CorrectList.append(-7)
if word[i] == "o":
CorrectList.append(-6)
#print(CorrectList)
return CorrectList
def CheckForAscending (AList):
flag = 0
i = 1
while i < len(AList):
if(AList[i] < AList[i - 1]):
flag = 1
i += 1
if (not flag):
return "YES"
else:
return "NO"
word = input()
print(CheckForAscending(CheckForLetters(word)))
```
| 0
|
36
|
A
|
Extra-terrestrial Intelligence
|
PROGRAMMING
| 1,300
|
[
"implementation"
] |
A. Extra-terrestrial Intelligence
|
2
|
64
|
Recently Vasya got interested in finding extra-terrestrial intelligence. He made a simple extra-terrestrial signals’ receiver and was keeping a record of the signals for *n* days in a row. Each of those *n* days Vasya wrote a 1 in his notebook if he had received a signal that day and a 0 if he hadn’t. Vasya thinks that he has found extra-terrestrial intelligence if there is a system in the way the signals has been received, i.e. if all the intervals between successive signals are equal. Otherwise, Vasya thinks that the signals were sent by some stupid aliens no one cares about. Help Vasya to deduce from the information given by the receiver if he has found extra-terrestrial intelligence or not.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of days during which Vasya checked if there were any signals. The second line contains *n* characters 1 or 0 — the record Vasya kept each of those *n* days. It’s guaranteed that the given record sequence contains at least three 1s.
|
If Vasya has found extra-terrestrial intelligence, output YES, otherwise output NO.
|
[
"8\n00111000\n",
"7\n1001011\n",
"7\n1010100\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "8\n00111000",
"output": "YES"
},
{
"input": "7\n1001011",
"output": "NO"
},
{
"input": "7\n1010100",
"output": "YES"
},
{
"input": "5\n10101",
"output": "YES"
},
{
"input": "3\n111",
"output": "YES"
},
{
"input": "10\n0011111011",
"output": "NO"
},
{
"input": "12\n001010101010",
"output": "YES"
},
{
"input": "25\n1000000010000000100000001",
"output": "YES"
},
{
"input": "30\n111110111101110111111111111111",
"output": "NO"
},
{
"input": "50\n00101010101010101010101010101010101010101010101010",
"output": "YES"
},
{
"input": "60\n000000000000000000100001000000001000000001000000001000000001",
"output": "NO"
},
{
"input": "66\n000000000000000000000010011110000010000010000010000000000000000000",
"output": "NO"
},
{
"input": "77\n10000100001010000000000001000000000000100000101000010010000000001100000000000",
"output": "NO"
},
{
"input": "99\n000000000000000000100000010000001000000100000010000001000000100000010000001000000100000010000001000",
"output": "YES"
},
{
"input": "100\n0000000000001000000000000000000000000000000000000010000000000000000000000000000000000000100000000000",
"output": "YES"
},
{
"input": "100\n0010000000000000000000000000000100000000000010000000000000001000000000000000000000000000010000000000",
"output": "NO"
},
{
"input": "5\n10111",
"output": "NO"
},
{
"input": "100\n0000100000000000000000000000000000000000010000000000000000000000000000000000001000000000000000000000",
"output": "YES"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111101111111111111111111111111111111111111",
"output": "NO"
}
] | 1,607,346,128
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 156
| 0
|
p=int(input().strip())
a=input().strip('0')
if a==a[::-1]:
print("YES")
exit(0)
else:
print("NO")
exit(0)
|
Title: Extra-terrestrial Intelligence
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Recently Vasya got interested in finding extra-terrestrial intelligence. He made a simple extra-terrestrial signals’ receiver and was keeping a record of the signals for *n* days in a row. Each of those *n* days Vasya wrote a 1 in his notebook if he had received a signal that day and a 0 if he hadn’t. Vasya thinks that he has found extra-terrestrial intelligence if there is a system in the way the signals has been received, i.e. if all the intervals between successive signals are equal. Otherwise, Vasya thinks that the signals were sent by some stupid aliens no one cares about. Help Vasya to deduce from the information given by the receiver if he has found extra-terrestrial intelligence or not.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of days during which Vasya checked if there were any signals. The second line contains *n* characters 1 or 0 — the record Vasya kept each of those *n* days. It’s guaranteed that the given record sequence contains at least three 1s.
Output Specification:
If Vasya has found extra-terrestrial intelligence, output YES, otherwise output NO.
Demo Input:
['8\n00111000\n', '7\n1001011\n', '7\n1010100\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
p=int(input().strip())
a=input().strip('0')
if a==a[::-1]:
print("YES")
exit(0)
else:
print("NO")
exit(0)
```
| -1
|
490
|
A
|
Team Olympiad
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
|
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
|
[
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] |
[
"2\n3 5 2\n6 7 4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "1\n1 2 3"
},
{
"input": "12\n3 3 3 3 3 3 3 3 1 3 3 2",
"output": "1\n9 12 2"
},
{
"input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2",
"output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47"
},
{
"input": "12\n3 1 1 1 1 1 1 2 1 1 1 1",
"output": "1\n3 8 1"
},
{
"input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2",
"output": "1\n18 2 11"
},
{
"input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3",
"output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36"
},
{
"input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1",
"output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200"
},
{
"input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3",
"output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64"
},
{
"input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1",
"output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44"
},
{
"input": "5\n1 2 2 3 3",
"output": "1\n1 3 4"
}
] | 1,668,726,618
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,867,200
|
n=int(input())
L=list(map(int,input().split()))
L.sort()
k1,k2,k3=0,0,0
for i in range(len(L)):
if L[i]==1:
k1+=1
elif L[i]==2:
k2+=1
elif L[i]==3:
k3+=1
x=min(k1,k2,k3)
print(x)
for i in range(x):
i1,i2,i3=L.index(1),L.index(2),L.index(3)
print(i1,i2,i3)
L.pop(i1)
L.pop(i2)
L.pop(i3)
|
Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none
|
```python
n=int(input())
L=list(map(int,input().split()))
L.sort()
k1,k2,k3=0,0,0
for i in range(len(L)):
if L[i]==1:
k1+=1
elif L[i]==2:
k2+=1
elif L[i]==3:
k3+=1
x=min(k1,k2,k3)
print(x)
for i in range(x):
i1,i2,i3=L.index(1),L.index(2),L.index(3)
print(i1,i2,i3)
L.pop(i1)
L.pop(i2)
L.pop(i3)
```
| -1
|
|
106
|
B
|
Choosing Laptop
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] |
B. Choosing Laptop
|
2
|
256
|
Vasya is choosing a laptop. The shop has *n* laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
|
The first line contains number *n* (1<=≤<=*n*<=≤<=100).
Then follow *n* lines. Each describes a laptop as *speed* *ram* *hdd* *cost*. Besides,
- *speed*, *ram*, *hdd* and *cost* are integers - 1000<=≤<=*speed*<=≤<=4200 is the processor's speed in megahertz - 256<=≤<=*ram*<=≤<=4096 the RAM volume in megabytes - 1<=≤<=*hdd*<=≤<=500 is the HDD in gigabytes - 100<=≤<=*cost*<=≤<=1000 is price in tugriks
All laptops have different prices.
|
Print a single number — the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to *n* in the order in which they are given in the input data.
|
[
"5\n2100 512 150 200\n2000 2048 240 350\n2300 1024 200 320\n2500 2048 80 300\n2000 512 180 150\n"
] |
[
"4"
] |
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
| 1,000
|
[
{
"input": "5\n2100 512 150 200\n2000 2048 240 350\n2300 1024 200 320\n2500 2048 80 300\n2000 512 180 150",
"output": "4"
},
{
"input": "2\n1500 500 50 755\n1600 600 80 700",
"output": "2"
},
{
"input": "2\n1500 512 50 567\n1600 400 70 789",
"output": "1"
},
{
"input": "4\n1000 300 5 700\n1100 400 10 600\n1200 500 15 500\n1300 600 20 400",
"output": "4"
},
{
"input": "10\n2123 389 397 747\n2705 3497 413 241\n3640 984 470 250\n3013 2004 276 905\n3658 3213 353 602\n1428 626 188 523\n2435 1140 459 824\n2927 2586 237 860\n2361 4004 386 719\n2863 2429 476 310",
"output": "2"
},
{
"input": "25\n2123 389 397 747\n2705 3497 413 241\n3640 984 470 250\n3013 2004 276 905\n3658 3213 353 602\n1428 626 188 523\n2435 1140 459 824\n2927 2586 237 860\n2361 4004 386 719\n2863 2429 476 310\n3447 3875 1 306\n3950 1901 31 526\n4130 1886 152 535\n1951 1840 122 814\n1798 3722 474 106\n2305 3979 82 971\n3656 3148 349 992\n1062 1648 320 491\n3113 3706 302 542\n3545 1317 184 853\n1277 2153 95 492\n2189 3495 427 655\n4014 3030 22 963\n1455 3840 155 485\n2760 717 309 891",
"output": "15"
},
{
"input": "1\n1200 512 300 700",
"output": "1"
},
{
"input": "1\n4200 4096 500 1000",
"output": "1"
},
{
"input": "1\n1000 256 1 100",
"output": "1"
},
{
"input": "2\n2000 500 200 100\n3000 600 100 200",
"output": "1"
},
{
"input": "2\n2000 500 200 200\n3000 600 100 100",
"output": "2"
},
{
"input": "2\n2000 600 100 100\n3000 500 200 200",
"output": "1"
},
{
"input": "2\n2000 700 100 200\n3000 500 200 100",
"output": "2"
},
{
"input": "2\n3000 500 100 100\n1500 600 200 200",
"output": "1"
},
{
"input": "2\n3000 500 100 300\n1500 600 200 200",
"output": "2"
},
{
"input": "3\n3467 1566 191 888\n3047 3917 3 849\n1795 1251 97 281",
"output": "2"
},
{
"input": "4\n3835 1035 5 848\n2222 3172 190 370\n2634 2698 437 742\n1748 3112 159 546",
"output": "2"
},
{
"input": "5\n3511 981 276 808\n3317 2320 354 878\n3089 702 20 732\n1088 2913 327 756\n3837 691 173 933",
"output": "4"
},
{
"input": "6\n1185 894 287 455\n2465 3317 102 240\n2390 2353 81 615\n2884 603 170 826\n3202 2070 320 184\n3074 3776 497 466",
"output": "5"
},
{
"input": "7\n3987 1611 470 720\n1254 4048 226 626\n1747 630 25 996\n2336 2170 402 123\n1902 3952 337 663\n1416 271 77 499\n1802 1399 419 929",
"output": "4"
},
{
"input": "10\n3888 1084 420 278\n2033 277 304 447\n1774 514 61 663\n2055 3437 67 144\n1237 1590 145 599\n3648 663 244 525\n3691 2276 332 504\n1496 2655 324 313\n2462 1930 13 644\n1811 331 390 284",
"output": "4"
},
{
"input": "13\n3684 543 70 227\n3953 1650 151 681\n2452 655 102 946\n3003 990 121 411\n2896 1936 158 155\n1972 717 366 754\n3989 2237 32 521\n2738 2140 445 965\n2884 1772 251 369\n2240 741 465 209\n4073 2812 494 414\n3392 955 425 133\n4028 717 90 123",
"output": "11"
},
{
"input": "17\n3868 2323 290 182\n1253 3599 38 217\n2372 354 332 897\n1286 649 332 495\n1642 1643 301 216\n1578 792 140 299\n3329 3039 359 525\n1362 2006 172 183\n1058 3961 423 591\n3196 914 484 675\n3032 3752 217 954\n2391 2853 171 579\n4102 3170 349 516\n1218 1661 451 354\n3375 1997 196 404\n1030 918 198 893\n2546 2029 399 647",
"output": "14"
},
{
"input": "22\n1601 1091 249 107\n2918 3830 312 767\n4140 409 393 202\n3485 2409 446 291\n2787 530 272 147\n2303 3400 265 206\n2164 1088 143 667\n1575 2439 278 863\n2874 699 369 568\n4017 1625 368 641\n3446 916 53 509\n3627 3229 328 256\n1004 2525 109 670\n2369 3299 57 351\n4147 3038 73 309\n3510 3391 390 470\n3308 3139 268 736\n3733 1054 98 809\n3967 2992 408 873\n2104 3191 83 687\n2223 2910 209 563\n1406 2428 147 673",
"output": "3"
},
{
"input": "27\n1689 1927 40 270\n3833 2570 167 134\n2580 3589 390 300\n1898 2587 407 316\n1841 2772 411 187\n1296 288 407 506\n1215 263 236 307\n2737 1427 84 992\n1107 1879 284 866\n3311 2507 475 147\n2951 2214 209 375\n1352 2582 110 324\n2082 747 289 521\n2226 1617 209 108\n2253 1993 109 835\n2866 2360 29 206\n1431 3581 185 918\n3800 1167 463 943\n4136 1156 266 490\n3511 1396 478 169\n3498 1419 493 792\n2660 2165 204 172\n3509 2358 178 469\n1568 3564 276 319\n3871 2660 472 366\n3569 2829 146 761\n1365 2943 460 611",
"output": "10"
},
{
"input": "2\n1000 2000 300 120\n1000 2000 300 130",
"output": "1"
},
{
"input": "10\n2883 1110 230 501\n2662 821 163 215\n2776 1131 276 870\n2776 1131 276 596\n2776 1131 276 981\n2662 821 163 892\n2662 821 163 997\n2883 1110 230 132\n2776 1131 276 317\n2883 1110 230 481",
"output": "8"
},
{
"input": "23\n1578 3681 380 163\n2640 3990 180 576\n3278 2311 131 386\n3900 513 443 873\n1230 1143 267 313\n2640 3990 180 501\n1230 1143 267 428\n1578 3681 380 199\n1578 3681 380 490\n3900 513 443 980\n3900 513 443 882\n3278 2311 131 951\n3278 2311 131 863\n2640 3990 180 916\n3278 2311 131 406\n3278 2311 131 455\n3278 2311 131 239\n1230 1143 267 439\n3900 513 443 438\n3900 513 443 514\n3278 2311 131 526\n1578 3681 380 123\n1578 3681 380 263",
"output": "22"
},
{
"input": "6\n2100 512 150 200\n2000 2048 240 350\n2300 1024 200 320\n2500 2048 80 300\n2000 512 180 150\n1000 256 1 100",
"output": "4"
},
{
"input": "2\n1000 256 1 100\n1000 256 1 101",
"output": "1"
},
{
"input": "2\n1500 500 300 1000\n1500 500 300 900",
"output": "2"
},
{
"input": "4\n1000 256 1 500\n1000 256 1 400\n1000 256 1 300\n1000 256 1 200",
"output": "4"
},
{
"input": "3\n1500 1024 300 150\n1200 512 150 100\n1000 256 50 200",
"output": "1"
}
] | 1,636,217,466
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 122
| 4,505,600
|
def compare(l1, l2):
for i in range(3):
if l1[i]>=l2[i]:
return False
return True
n = int(input())
laptop = []
outdated = [False]*n
for i in range(n):
laptop.append(list(map(int, input().split())))
for i in range(n):
for j in range(n):
if compare(laptop[i], laptop[j]):
outdated[i] = True
minCost, idx = 1001, -1
for i in range(n):
if (not outdated[i]) and laptop[i][3]<minCost:
minCost, idx = laptop[i][3], i+1
print(idx)
|
Title: Choosing Laptop
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is choosing a laptop. The shop has *n* laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=100).
Then follow *n* lines. Each describes a laptop as *speed* *ram* *hdd* *cost*. Besides,
- *speed*, *ram*, *hdd* and *cost* are integers - 1000<=≤<=*speed*<=≤<=4200 is the processor's speed in megahertz - 256<=≤<=*ram*<=≤<=4096 the RAM volume in megabytes - 1<=≤<=*hdd*<=≤<=500 is the HDD in gigabytes - 100<=≤<=*cost*<=≤<=1000 is price in tugriks
All laptops have different prices.
Output Specification:
Print a single number — the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to *n* in the order in which they are given in the input data.
Demo Input:
['5\n2100 512 150 200\n2000 2048 240 350\n2300 1024 200 320\n2500 2048 80 300\n2000 512 180 150\n']
Demo Output:
['4']
Note:
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
|
```python
def compare(l1, l2):
for i in range(3):
if l1[i]>=l2[i]:
return False
return True
n = int(input())
laptop = []
outdated = [False]*n
for i in range(n):
laptop.append(list(map(int, input().split())))
for i in range(n):
for j in range(n):
if compare(laptop[i], laptop[j]):
outdated[i] = True
minCost, idx = 1001, -1
for i in range(n):
if (not outdated[i]) and laptop[i][3]<minCost:
minCost, idx = laptop[i][3], i+1
print(idx)
```
| 3.961108
|
227
|
B
|
Effective Approach
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
|
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
[
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] |
[
"1 2\n",
"2 1\n",
"6 6\n"
] |
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
| 1,000
|
[
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,659,149,714
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 2,000
| 6,656,000
|
t = int(input())
arr = list(map(int, input().split()))
q = int(input())
searchq = list(map(int, input().split()))
vas = 0
pet = 0
for j in searchq:
i=0
while j!=arr[i] and i<len(arr):
vas+=1
i+=1
vas+=1
for j in searchq:
i=len(arr)-1
while j!=arr[i] and i>0:
pet+=1
i-=1
pet+=1
print(vas,pet)
|
Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
|
```python
t = int(input())
arr = list(map(int, input().split()))
q = int(input())
searchq = list(map(int, input().split()))
vas = 0
pet = 0
for j in searchq:
i=0
while j!=arr[i] and i<len(arr):
vas+=1
i+=1
vas+=1
for j in searchq:
i=len(arr)-1
while j!=arr[i] and i>0:
pet+=1
i-=1
pet+=1
print(vas,pet)
```
| 0
|
|
141
|
A
|
Amusing Joke
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings",
"strings"
] | null | null |
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
|
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
|
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
|
[
"SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n",
"PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n",
"BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L".
| 500
|
[
{
"input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS",
"output": "YES"
},
{
"input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI",
"output": "NO"
},
{
"input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER",
"output": "NO"
},
{
"input": "B\nA\nAB",
"output": "YES"
},
{
"input": "ONDOL\nJNPB\nONLNJBODP",
"output": "YES"
},
{
"input": "Y\nW\nYW",
"output": "YES"
},
{
"input": "OI\nM\nIMO",
"output": "YES"
},
{
"input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF",
"output": "YES"
},
{
"input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB",
"output": "NO"
},
{
"input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH",
"output": "NO"
},
{
"input": "IQ\nOQ\nQOQIGGKFNHJSGCGM",
"output": "NO"
},
{
"input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR",
"output": "YES"
},
{
"input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY",
"output": "YES"
},
{
"input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX",
"output": "YES"
},
{
"input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW",
"output": "YES"
},
{
"input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU",
"output": "YES"
},
{
"input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK",
"output": "YES"
},
{
"input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE",
"output": "NO"
},
{
"input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ",
"output": "NO"
},
{
"input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI",
"output": "NO"
},
{
"input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT",
"output": "NO"
},
{
"input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY",
"output": "NO"
},
{
"input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT",
"output": "NO"
},
{
"input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY",
"output": "NO"
},
{
"input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI",
"output": "NO"
},
{
"input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE",
"output": "NO"
},
{
"input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH",
"output": "NO"
},
{
"input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU",
"output": "NO"
},
{
"input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG",
"output": "NO"
},
{
"input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM",
"output": "NO"
},
{
"input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW",
"output": "NO"
},
{
"input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ",
"output": "YES"
},
{
"input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS",
"output": "YES"
},
{
"input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH",
"output": "YES"
},
{
"input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB",
"output": "YES"
},
{
"input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP",
"output": "YES"
},
{
"input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR",
"output": "YES"
},
{
"input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT",
"output": "YES"
},
{
"input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH",
"output": "YES"
},
{
"input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG",
"output": "YES"
},
{
"input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF",
"output": "YES"
},
{
"input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA",
"output": "NO"
},
{
"input": "IXFDY\nJRMOU\nDF",
"output": "NO"
},
{
"input": "JPSPZ\nUGCUB\nJMZZZZZZZZ",
"output": "NO"
},
{
"input": "AC\nA\nBBA",
"output": "NO"
},
{
"input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB",
"output": "NO"
},
{
"input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW",
"output": "NO"
},
{
"input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE",
"output": "NO"
},
{
"input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH",
"output": "NO"
},
{
"input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ",
"output": "NO"
},
{
"input": "GIO\nPRL\nPRL",
"output": "NO"
},
{
"input": "A\nB\nABC",
"output": "NO"
},
{
"input": "KKK\nKKK\nZZZZZ",
"output": "NO"
},
{
"input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA",
"output": "NO"
}
] | 1,691,585,138
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 92
| 0
|
def main():
alpha_length = 26
first = [0] * alpha_length
second = [0] * alpha_length
line = input().strip()
for char in line:
first[ord(char) - ord('A')] += 1
line = input().strip()
for char in line:
first[ord(char) - ord('A')] += 1
line = input().strip()
for char in line:
second[ord(char) - ord('A')] += 1
output = "YES"
for k in range(alpha_length):
if first[k] != second[k]:
output = "NO"
break
print(output)
if __name__ == "__main__":
main()
|
Title: Amusing Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
Input Specification:
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Output Specification:
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
Demo Input:
['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L".
|
```python
def main():
alpha_length = 26
first = [0] * alpha_length
second = [0] * alpha_length
line = input().strip()
for char in line:
first[ord(char) - ord('A')] += 1
line = input().strip()
for char in line:
first[ord(char) - ord('A')] += 1
line = input().strip()
for char in line:
second[ord(char) - ord('A')] += 1
output = "YES"
for k in range(alpha_length):
if first[k] != second[k]:
output = "NO"
break
print(output)
if __name__ == "__main__":
main()
```
| 3
|
|
877
|
B
|
Nikita and string
|
PROGRAMMING
| 1,500
|
[
"brute force",
"dp"
] | null | null |
One day Nikita found the string containing letters "a" and "b" only.
Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".
Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?
|
The first line contains a non-empty string of length not greater than 5<=000 containing only lowercase English letters "a" and "b".
|
Print a single integer — the maximum possible size of beautiful string Nikita can get.
|
[
"abba\n",
"bab\n"
] |
[
"4",
"2"
] |
It the first sample the string is already beautiful.
In the second sample he needs to delete one of "b" to make it beautiful.
| 1,000
|
[
{
"input": "abba",
"output": "4"
},
{
"input": "bab",
"output": "2"
},
{
"input": "bbabbbaabbbb",
"output": "9"
},
{
"input": "bbabbbbbaaba",
"output": "10"
},
{
"input": "bbabbbababaa",
"output": "9"
},
{
"input": "aabbaababbab",
"output": "8"
},
{
"input": "a",
"output": "1"
},
{
"input": "b",
"output": "1"
},
{
"input": "ab",
"output": "2"
},
{
"input": "ba",
"output": "2"
},
{
"input": "bb",
"output": "2"
},
{
"input": "aa",
"output": "2"
},
{
"input": "babbbaab",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaa",
"output": "3"
}
] | 1,604,178,349
| 1,549
|
PyPy 3
|
OK
|
TESTS
| 46
| 170
| 2,355,200
|
s = input()
n = len(s)
dp = []
dp.append((0, 0, 0))
for i in range(n):
if s[i] == 'a':
dp.append((dp[-1][0] + 1, dp[-1][1], max(dp[-1][1:]) + 1))
elif s[i] == 'b':
dp.append((dp[-1][0], max(dp[-1][1], dp[-1][0]) + 1, dp[-1][-1]))
print(max(dp[-1]))
|
Title: Nikita and string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Nikita found the string containing letters "a" and "b" only.
Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".
Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?
Input Specification:
The first line contains a non-empty string of length not greater than 5<=000 containing only lowercase English letters "a" and "b".
Output Specification:
Print a single integer — the maximum possible size of beautiful string Nikita can get.
Demo Input:
['abba\n', 'bab\n']
Demo Output:
['4', '2']
Note:
It the first sample the string is already beautiful.
In the second sample he needs to delete one of "b" to make it beautiful.
|
```python
s = input()
n = len(s)
dp = []
dp.append((0, 0, 0))
for i in range(n):
if s[i] == 'a':
dp.append((dp[-1][0] + 1, dp[-1][1], max(dp[-1][1:]) + 1))
elif s[i] == 'b':
dp.append((dp[-1][0], max(dp[-1][1], dp[-1][0]) + 1, dp[-1][-1]))
print(max(dp[-1]))
```
| 3
|
|
990
|
B
|
Micro-World
|
PROGRAMMING
| 1,200
|
[
"greedy",
"sortings"
] | null | null |
You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
|
The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have.
|
Print the only integer — minimal possible number of bacteria can remain.
|
[
"7 1\n101 53 42 102 101 55 54\n",
"6 5\n20 15 10 15 20 25\n",
"7 1000000\n1 1 1 1 1 1 1\n"
] |
[
"3\n",
"1\n",
"7\n"
] |
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria.
| 0
|
[
{
"input": "7 1\n101 53 42 102 101 55 54",
"output": "3"
},
{
"input": "6 5\n20 15 10 15 20 25",
"output": "1"
},
{
"input": "7 1000000\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 4\n8",
"output": "1"
},
{
"input": "10 1\n1 2 3 5 6 8 10 11 9 4",
"output": "2"
},
{
"input": "9 2\n1 6 1 5 5 8 6 8 7",
"output": "4"
},
{
"input": "15 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "15"
},
{
"input": "2 1000000\n1 1000000",
"output": "1"
},
{
"input": "7 2\n1 5 5 8 9 8 8",
"output": "4"
},
{
"input": "10 1\n2 6 3 4 2 4 4 3 2 1",
"output": "4"
},
{
"input": "4 1\n2 2 1 1",
"output": "2"
},
{
"input": "10 1\n6 3 1 3 6 4 1 3 6 4",
"output": "7"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "8 2\n3 13 9 8 3 13 9 14",
"output": "5"
},
{
"input": "8 1000000\n1 1 5 1000000 1000000 2 2 2",
"output": "2"
},
{
"input": "2 1\n999152 999153",
"output": "1"
}
] | 1,528,628,478
| 3,378
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 3
| 2,000
| 14,028,800
|
#zadb
n,K=map(int,input().split())
line=list(sorted((map(int,input().split()))))
i=1
while True:
if i>len(line)-1:break
else:
if line[i]>line[i-1]:
if (line[i-1]+K)>=line[i]:
j=line[i-1]
while j in line:
line.remove(j)
continue
i+=1
print(len(line))
|
Title: Micro-World
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
Input Specification:
The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have.
Output Specification:
Print the only integer — minimal possible number of bacteria can remain.
Demo Input:
['7 1\n101 53 42 102 101 55 54\n', '6 5\n20 15 10 15 20 25\n', '7 1000000\n1 1 1 1 1 1 1\n']
Demo Output:
['3\n', '1\n', '7\n']
Note:
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria.
|
```python
#zadb
n,K=map(int,input().split())
line=list(sorted((map(int,input().split()))))
i=1
while True:
if i>len(line)-1:break
else:
if line[i]>line[i-1]:
if (line[i-1]+K)>=line[i]:
j=line[i-1]
while j in line:
line.remove(j)
continue
i+=1
print(len(line))
```
| 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"
}
] | 1,698,396,637
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 61
| 0
|
x = int(input())
i = 1
while x//2 > 0:
if x % 2 == 1: i += 1
x = x // 2
print(i)
|
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
x = int(input())
i = 1
while x//2 > 0:
if x % 2 == 1: i += 1
x = x // 2
print(i)
```
| 3
|
|
230
|
B
|
T-primes
|
PROGRAMMING
| 1,300
|
[
"binary search",
"implementation",
"math",
"number theory"
] | null | null |
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
|
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
|
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
|
[
"3\n4 5 6\n"
] |
[
"YES\nNO\nNO\n"
] |
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
| 500
|
[
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"output": "YES"
},
{
"input": "1\n999993399999",
"output": "NO"
},
{
"input": "9\n111 121 131 111 121 131 111 121 131",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n976197352729",
"output": "NO"
},
{
"input": "1\n1000000000000",
"output": "NO"
},
{
"input": "1\n9",
"output": "YES"
},
{
"input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n223092870 6469693230 200560490130",
"output": "NO\nNO\nNO"
},
{
"input": "2\n81 25",
"output": "NO\nYES"
},
{
"input": "1\n16",
"output": "NO"
},
{
"input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81",
"output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "1\n225",
"output": "NO"
},
{
"input": "1\n1521",
"output": "NO"
},
{
"input": "1\n81",
"output": "NO"
}
] | 1,697,542,610
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 68
| 1,560
| 22,323,200
|
import math
n1 = 1000005
pa = [1]*(n1+1)
pa[0]= pa[1]= 0
for i in range(2, int(math.sqrt(n1)+1)):
if pa[i] == 1:
j =i
while i*j <= n1:
pa[i*j] = 0
j+=1
n = int(input())
arr = list(map(int, input().split()))
for i in range(n):
s = int(math.sqrt(arr[i]))
if s*s == arr[i] and pa[s] ==1:
print("YES")
else:
print("NO")
|
Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
|
```python
import math
n1 = 1000005
pa = [1]*(n1+1)
pa[0]= pa[1]= 0
for i in range(2, int(math.sqrt(n1)+1)):
if pa[i] == 1:
j =i
while i*j <= n1:
pa[i*j] = 0
j+=1
n = int(input())
arr = list(map(int, input().split()))
for i in range(n):
s = int(math.sqrt(arr[i]))
if s*s == arr[i] and pa[s] ==1:
print("YES")
else:
print("NO")
```
| 3
|
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,696,104,366
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 92
| 0
|
word = input()
if word[0] >= 'a':
print(chr(ord(word[0]) - 32) + word[1:])
else:
print(word)
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
word = input()
if word[0] >= 'a':
print(chr(ord(word[0]) - 32) + word[1:])
else:
print(word)
```
| 3
|
|
888
|
B
|
Buggy Robot
|
PROGRAMMING
| 1,000
|
[
"greedy"
] | null | null |
Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform:
- U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*).
Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!
|
The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100).
The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.
|
Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.
|
[
"4\nLDUR\n",
"5\nRRRUU\n",
"6\nLLRRRR\n"
] |
[
"4\n",
"0\n",
"4\n"
] |
none
| 0
|
[
{
"input": "4\nLDUR",
"output": "4"
},
{
"input": "5\nRRRUU",
"output": "0"
},
{
"input": "6\nLLRRRR",
"output": "4"
},
{
"input": "88\nLLUUULRDRRURDDLURRLRDRLLRULRUUDDLLLLRRDDURDURRLDURRLDRRRUULDDLRRRDDRRLUULLURDURUDDDDDLDR",
"output": "76"
},
{
"input": "89\nLDLLLDRDUDURRRRRUDULDDDLLUDLRLRLRLDLDUULRDUDLRRDLUDLURRDDRRDLDUDUUURUUUDRLUDUDLURDLDLLDDU",
"output": "80"
},
{
"input": "90\nRRRDUULLLRDUUDDRLDLRLUDURDRDUUURUURDDRRRURLDDDUUDRLLLULURDRDRURLDRRRRUULDULDDLLLRRLRDLLLLR",
"output": "84"
},
{
"input": "91\nRLDRLRRLLDLULULLURULLRRULUDUULLUDULDUULURUDRUDUURDULDUDDUUUDRRUUDLLRULRULURLDRDLDRURLLLRDDD",
"output": "76"
},
{
"input": "92\nRLRDDLULRLLUURRDDDLDDDLDDUURRRULLRDULDULLLUUULDUDLRLRRDRDRDDULDRLUDRDULDRURUDUULLRDRRLLDRLRR",
"output": "86"
},
{
"input": "93\nRLLURLULRURDDLUURLUDDRDLUURLRDLRRRDUULLRDRRLRLDURRDLLRDDLLLDDDLDRRURLLDRUDULDDRRULRRULRLDRDLR",
"output": "84"
},
{
"input": "94\nRDULDDDLULRDRUDRUUDUUDRRRULDRRUDURUULRDUUDLULLLUDURRDRDLUDRULRRRULUURUDDDDDUDLLRDLDRLLRUUURLUL",
"output": "86"
},
{
"input": "95\nRDLUUULLUURDDRLDLLRRRULRLRDULULRULRUDURLULDDDRLURLDRULDUDUUULLRDDURUULULLDDLDRDRLLLURLRDLLDDDDU",
"output": "86"
},
{
"input": "96\nRDDRLRLLDDULRLRURUDLRLDUDRURLLUUDLLURDLRRUURDRRUDRURLLDLLRDURDURLRLUDURULLLRDUURULUUULRRURRDLURL",
"output": "84"
},
{
"input": "97\nRURDDLRLLRULUDURDLRLLUUDURRLLUDLLLDUDRUULDRUUURURULRDLDRRLLUUUDLLLDDLLLLRLLDUDRRDLLUDLRURUDULRLUR",
"output": "82"
},
{
"input": "98\nRUDURLULLDDLLRDLLRDDLLLLRLDDDDRRRDDRRURLDRLLRUUUDLUUUDDDUDRUURLURUUDUUDRULRRULLRRLRULLULDLUURLULRD",
"output": "92"
},
{
"input": "99\nRRULLDULRRDRULLDUDRUDDDRLLUUDRDDUDURLDDRUUDRRUUURRRURDDLDUURDLRLURRDDLUDDLUDURDRUDDURLURURLRUDRURLD",
"output": "86"
},
{
"input": "100\nUDRLRRLLRRLRRRDDLLDDDLULLDDLURUURUULUDDDRDDLLRDLLUURLRDRLRRLRLLLULDUDDUURRLRDULDRDURRRRRRULDRRDLDRRL",
"output": "88"
},
{
"input": "1\nU",
"output": "0"
},
{
"input": "5\nUUULD",
"output": "2"
},
{
"input": "1\nD",
"output": "0"
},
{
"input": "5\nURLUL",
"output": "2"
},
{
"input": "5\nDDDRU",
"output": "2"
},
{
"input": "2\nLR",
"output": "2"
},
{
"input": "8\nDDRDLDUR",
"output": "4"
},
{
"input": "6\nLLLLUD",
"output": "2"
},
{
"input": "13\nRRRLLLLLLLLLL",
"output": "6"
}
] | 1,604,055,830
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 108
| 0
|
n = int(input())
a = list(input())
x = y = 0
count = {}
for c in a:
if c not in count:
count[c] = 0
count[c] += 1
print(2*(min(count.get('L', 0), count.get('R', 0)) + min(count.get('U', 0), count.get('D', 0))))
|
Title: Buggy Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform:
- U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*).
Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!
Input Specification:
The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100).
The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.
Output Specification:
Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.
Demo Input:
['4\nLDUR\n', '5\nRRRUU\n', '6\nLLRRRR\n']
Demo Output:
['4\n', '0\n', '4\n']
Note:
none
|
```python
n = int(input())
a = list(input())
x = y = 0
count = {}
for c in a:
if c not in count:
count[c] = 0
count[c] += 1
print(2*(min(count.get('L', 0), count.get('R', 0)) + min(count.get('U', 0), count.get('D', 0))))
```
| 3
|
|
176
|
B
|
Word Cut
|
PROGRAMMING
| 1,700
|
[
"dp"
] | null | null |
Let's consider one interesting word game. In this game you should transform one word into another through special operations.
Let's say we have word *w*, let's split this word into two non-empty parts *x* and *y* so, that *w*<==<=*xy*. A split operation is transforming word *w*<==<=*xy* into word *u*<==<=*yx*. For example, a split operation can transform word "wordcut" into word "cutword".
You are given two words *start* and *end*. Count in how many ways we can transform word *start* into word *end*, if we apply exactly *k* split operations consecutively to word *start*.
Two ways are considered different if the sequences of applied operations differ. Two operation sequences are different if exists such number *i* (1<=≤<=*i*<=≤<=*k*), that in the *i*-th operation of the first sequence the word splits into parts *x* and *y*, in the *i*-th operation of the second sequence the word splits into parts *a* and *b*, and additionally *x*<=≠<=*a* holds.
|
The first line contains a non-empty word *start*, the second line contains a non-empty word *end*. The words consist of lowercase Latin letters. The number of letters in word *start* equals the number of letters in word *end* and is at least 2 and doesn't exceed 1000 letters.
The third line contains integer *k* (0<=≤<=*k*<=≤<=105) — the required number of operations.
|
Print a single number — the answer to the problem. As this number can be rather large, print it modulo 1000000007 (109<=+<=7).
|
[
"ab\nab\n2\n",
"ababab\nababab\n1\n",
"ab\nba\n2\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
The sought way in the first sample is:
ab → a|b → ba → b|a → ab
In the second sample the two sought ways are:
- ababab → abab|ab → ababab - ababab → ab|abab → ababab
| 1,000
|
[
{
"input": "ab\nab\n2",
"output": "1"
},
{
"input": "ababab\nababab\n1",
"output": "2"
},
{
"input": "ab\nba\n2",
"output": "0"
},
{
"input": "aaa\naaa\n0",
"output": "1"
},
{
"input": "hi\nhi\n1",
"output": "0"
},
{
"input": "abcd\ncbad\n5",
"output": "0"
},
{
"input": "ab\nba\n10",
"output": "0"
},
{
"input": "voodoo\ndoovoo\n100000",
"output": "792428974"
},
{
"input": "ababab\nbababa\n100000",
"output": "377286908"
},
{
"input": "abcdefgh\ncdefghab\n666",
"output": "83913683"
},
{
"input": "aaaabaaaaaaaaaaabaaaaaaa\naaaaaaaaaabaaaaaaaaabaaa\n7477",
"output": "0"
},
{
"input": "ssgqcodnqgfbhqsgineioafkkhcmmmihbiefialidgkffrhaiekebpieqgpplmsgmghphjsfgpscrbcgrssbccqroffnfgkfohljdarbpqmkolldcjcfhpodeqmgbdddlgoolesecdqsochdfgjsmorbnmiinjlpda\nljdarbpqmkolldcjcfhpodeqmgbdddlgoolesecdqsochdfgjsmorbnmiinjlpdassgqcodnqgfbhqsgineioafkkhcmmmihbiefialidgkffrhaiekebpieqgpplmsgmghphjsfgpscrbcgrssbccqroffnfgkfoh\n50897",
"output": "222669762"
},
{
"input": "jfemedqrsqaopiekdosgjnhbshanggdqqpkhepjfrkgkshepbmkdnidmpgfojjjbeddkelccoqapnpkqbimlbgagllioqbdgnsejqcbicjbbijjlrjmkkarjdoganmfsmfohlspbsoldfspdacasgsrcndlhg\nhepbmkdnidmpgfojjjbeddkelccoqapnpkqbimlbgagllioqbdgnsejqcbicjbbijjlrjmkkarjdoganmfsmfohlspbsoldfspdacasgsrcndlhgjfemedqrsqaopiekdosgjnhbshanggdqqpkhepjfrkgks\n6178",
"output": "568786732"
},
{
"input": "aaeddddadbcacbdccaeeeddecadbecbbcebdcdbcddcadcadccecccecdbabd\nadbecbbcebdcdbcddcadcadccecccecdbabdaaeddddadbcacbdccaeeeddec\n55400",
"output": "471327413"
},
{
"input": "chajciihijjbjcgaedebdcjaaeaiffiggfdfbdjhikhbiijhbjciebgkadbbekijadafhjhgiidfjkjbgcdfdgjjfficbagghkdgdhdedihifcfkedcefcdfjaagiehccjbjhihcbdakbjfjdgakkfagddhekccbdjhejhakfccgghkdc\ndafhjhgiidfjkjbgcdfdgjjfficbagghkdgdhdedihifcfkedcefcdfjaagiehccjbjhihcbdakbjfjdgakkfagddhekccbdjhejhakfccgghkdcchajciihijjbjcgaedebdcjaaeaiffiggfdfbdjhikhbiijhbjciebgkadbbekija\n67572",
"output": "18146811"
},
{
"input": "dkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjijdkjij\ndddkikjjidkkidijjjjkkjjikjdikiidijjikikjijjiijdikkjjjiddjjijkkkjkiijijkijdjjikikdjjjijdddjkjjdijjjjjjjddkjjkjjjdjjdkijjkijkkjkkkiiijdjijkkdjdjjjkkjkdddjidjjijdddkijididjdddidijjjjjdkidijjkkjiijkjjjjjjdjkkdddkjj\n31328",
"output": "0"
},
{
"input": "lnxjsxjjfbrhijshlptgogxstvaybbueilocmaqwhlauulubpwtjtsdktgrowgbkhdotemhstqtxnpspmxjxrqymonnpcvrendrfxbwqxeedyamwvyyytxtmorkovoeklkabyexrfseimlaqlibaoaaeae\nalxckgrkneuqrqstpxormswxsyhaaoebqhfspyqjbxgytrpdoybxmmenllxxrrjtxtvhaoidjmaekiflxykmpnwageytemoewalwihbunbnvomqbututijjvdsbelrxestajokwysctfptalvegdahlboo\n44031",
"output": "0"
},
{
"input": "cfacacbfaeadfdbedfdccdccdddaaa\ncbfaeadfdbedfdccdccdddaaacfaca\n15215",
"output": "668863397"
},
{
"input": "xxx\nyyy\n0",
"output": "0"
},
{
"input": "abc\ncba\n0",
"output": "0"
},
{
"input": "ab\nba\n0",
"output": "0"
},
{
"input": "abc\nabc\n1",
"output": "0"
},
{
"input": "ab\nba\n9",
"output": "1"
},
{
"input": "ab\nba\n10",
"output": "0"
},
{
"input": "aa\naa\n1",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n100000",
"output": "486606280"
},
{
"input": "aaaba\naaaad\n0",
"output": "0"
}
] | 1,687,496,487
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 90
| 810
| 2,662,400
|
import sys
input = lambda:sys.stdin.readline().strip()
M = lambda:map(int,input().split())
from collections import deque
inf = float('inf')
mod = 10**9+7
a = input()
a += a
b = input()
n = len(b)
k = int(input())
bo = [0]*n
for i in range(n):
if a[i:i+n] == b:
bo[i] = 1
res = [0]*n
res[0] = 1
for i in range(k):
sm = sum(res)%mod
for j in range(n):
res[j] = (sm-res[j])%mod
ans = 0
for i in range(n):
ans += bo[i]*res[i]
ans %= mod
print(ans)
|
Title: Word Cut
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider one interesting word game. In this game you should transform one word into another through special operations.
Let's say we have word *w*, let's split this word into two non-empty parts *x* and *y* so, that *w*<==<=*xy*. A split operation is transforming word *w*<==<=*xy* into word *u*<==<=*yx*. For example, a split operation can transform word "wordcut" into word "cutword".
You are given two words *start* and *end*. Count in how many ways we can transform word *start* into word *end*, if we apply exactly *k* split operations consecutively to word *start*.
Two ways are considered different if the sequences of applied operations differ. Two operation sequences are different if exists such number *i* (1<=≤<=*i*<=≤<=*k*), that in the *i*-th operation of the first sequence the word splits into parts *x* and *y*, in the *i*-th operation of the second sequence the word splits into parts *a* and *b*, and additionally *x*<=≠<=*a* holds.
Input Specification:
The first line contains a non-empty word *start*, the second line contains a non-empty word *end*. The words consist of lowercase Latin letters. The number of letters in word *start* equals the number of letters in word *end* and is at least 2 and doesn't exceed 1000 letters.
The third line contains integer *k* (0<=≤<=*k*<=≤<=105) — the required number of operations.
Output Specification:
Print a single number — the answer to the problem. As this number can be rather large, print it modulo 1000000007 (109<=+<=7).
Demo Input:
['ab\nab\n2\n', 'ababab\nababab\n1\n', 'ab\nba\n2\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
The sought way in the first sample is:
ab → a|b → ba → b|a → ab
In the second sample the two sought ways are:
- ababab → abab|ab → ababab - ababab → ab|abab → ababab
|
```python
import sys
input = lambda:sys.stdin.readline().strip()
M = lambda:map(int,input().split())
from collections import deque
inf = float('inf')
mod = 10**9+7
a = input()
a += a
b = input()
n = len(b)
k = int(input())
bo = [0]*n
for i in range(n):
if a[i:i+n] == b:
bo[i] = 1
res = [0]*n
res[0] = 1
for i in range(k):
sm = sum(res)%mod
for j in range(n):
res[j] = (sm-res[j])%mod
ans = 0
for i in range(n):
ans += bo[i]*res[i]
ans %= mod
print(ans)
```
| 3
|
|
734
|
A
|
Anton and Danik
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.
|
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
|
[
"6\nADAAAA\n",
"7\nDDDAADA\n",
"6\nDADADA\n"
] |
[
"Anton\n",
"Danik\n",
"Friendship\n"
] |
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
| 500
|
[
{
"input": "6\nADAAAA",
"output": "Anton"
},
{
"input": "7\nDDDAADA",
"output": "Danik"
},
{
"input": "6\nDADADA",
"output": "Friendship"
},
{
"input": "10\nDDDDADDADD",
"output": "Danik"
},
{
"input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA",
"output": "Anton"
},
{
"input": "200\nDDDDDDDADDDDDDAADADAADAAADAADADAAADDDADDDDDDADDDAADDDAADADDDDDADDDAAAADAAADDDDDAAADAADDDAAAADDADADDDAADDAADAAADAADAAAADDAADDADAAAADADDDAAAAAADDAADAADAADADDDAAADAAAADADDADAAAAAADADADDDADDDAADDADDDAAAAD",
"output": "Friendship"
},
{
"input": "1\nA",
"output": "Anton"
},
{
"input": "1\nD",
"output": "Danik"
},
{
"input": "2\nDA",
"output": "Friendship"
},
{
"input": "4\nDADA",
"output": "Friendship"
},
{
"input": "4\nDAAD",
"output": "Friendship"
},
{
"input": "3\nADD",
"output": "Danik"
},
{
"input": "3\nDAD",
"output": "Danik"
},
{
"input": "2\nDA",
"output": "Friendship"
},
{
"input": "379\nAADAAAAAADDAAAAAADAADADADDAAAAADADDAADAAAADDDADAAAAAAADAADAAAAAAADAAAAAAAAADAAAAAAADAAAAAAAAAAADDDADAAAAAAAADAADADAAAADAAAAAAAAAAAAAAAAADAAAADDDAADAAAAAAADAAADAAADAADDDADDAAADAAAAAADDDADDDAAADAAAADAAAAAAAAADAAADAAAAAAAAADAAAAAAAAAAAAAAAAAADADAAAAAAAAAAADAAAAADAAAADAAAAAAAAAAAAADADAADAAAAAAAADAADAAAAAAAADAAAAAAAADDDAAAAAADAAADAAAAAADAADAAAAAADAAAADADAADAAAAAADAAAADAADDAADAADAAA",
"output": "Anton"
}
] | 1,699,720,612
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 93
| 512,000
|
n = int(input())
str = input().strip()
a = str.count("A")
d = str.count("D")
if (a>d):
print("Anton")
elif(d>a):
print("Danik")
else:
print("Friendship")
|
Title: Anton and Danik
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.
Output Specification:
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Demo Input:
['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n']
Demo Output:
['Anton\n', 'Danik\n', 'Friendship\n']
Note:
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
|
```python
n = int(input())
str = input().strip()
a = str.count("A")
d = str.count("D")
if (a>d):
print("Anton")
elif(d>a):
print("Danik")
else:
print("Friendship")
```
| 3
|
|
315
|
A
|
Sereja and Bottles
|
PROGRAMMING
| 1,400
|
[
"brute force"
] | null | null |
Sereja and his friends went to a picnic. The guys had *n* soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the *i*-th bottle is from brand *a**i*, besides, you can use it to open other bottles of brand *b**i*. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles. The next *n* lines contain the bottles' description. The *i*-th line contains two integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the description of the *i*-th bottle.
|
In a single line print a single integer — the answer to the problem.
|
[
"4\n1 1\n2 2\n3 3\n4 4\n",
"4\n1 2\n2 3\n3 4\n4 1\n"
] |
[
"4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "4\n1 1\n2 2\n3 3\n4 4",
"output": "4"
},
{
"input": "4\n1 2\n2 3\n3 4\n4 1",
"output": "0"
},
{
"input": "3\n2 828\n4 392\n4 903",
"output": "3"
},
{
"input": "4\n2 3\n1 772\n3 870\n3 668",
"output": "2"
},
{
"input": "5\n1 4\n6 6\n4 3\n3 4\n4 758",
"output": "2"
},
{
"input": "6\n4 843\n2 107\n10 943\n9 649\n7 806\n6 730",
"output": "6"
},
{
"input": "7\n351 955\n7 841\n102 377\n394 102\n549 440\n630 324\n624 624",
"output": "6"
},
{
"input": "8\n83 978\n930 674\n542 22\n834 116\n116 271\n640 930\n659 930\n705 987",
"output": "6"
},
{
"input": "9\n162 942\n637 967\n356 108\n768 53\n656 656\n575 32\n32 575\n53 53\n351 222",
"output": "6"
},
{
"input": "10\n423 360\n947 538\n507 484\n31 947\n414 351\n169 901\n901 21\n592 22\n763 200\n656 485",
"output": "8"
},
{
"input": "1\n1000 1000",
"output": "1"
},
{
"input": "1\n500 1000",
"output": "1"
},
{
"input": "11\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11",
"output": "11"
},
{
"input": "49\n1 758\n5 3\n5 3\n4 2\n4 36\n3 843\n5 107\n1 943\n1 649\n2 806\n3 730\n2 351\n2 102\n1 4\n3 4\n3 955\n2 841\n2 377\n5 2\n3 440\n4 324\n3 3\n3 83\n2 2\n2 1\n4 1\n1 931\n3 4\n2 5\n2 5\n4 73\n5 830\n3 4\n3 5\n5 291\n1 2\n5 3\n4 4\n2 3\n3 151\n4 2\n4 431\n5 1\n2 5\n2 4\n4 2\n4 4\n3 1\n5 2",
"output": "0"
},
{
"input": "50\n507 31\n31 250\n414 763\n169 304\n901 9\n592 610\n763 414\n656 789\n411 422\n360 468\n625 504\n538 201\n549 619\n484 797\n596 282\n42 310\n603 656\n351 623\n292 293\n837 180\n375 658\n21 192\n597 729\n22 512\n349 635\n200 56\n669 647\n485 887\n282 939\n735 808\n54 417\n1000 310\n419 652\n939 617\n901 669\n789 390\n128 549\n468 511\n729 837\n894 729\n649 894\n484 22\n808 586\n422 286\n311 427\n618 656\n814 933\n515 901\n310 894\n617 330",
"output": "30"
},
{
"input": "2\n7 7\n5 359",
"output": "2"
},
{
"input": "2\n465 706\n706 706",
"output": "1"
},
{
"input": "2\n1 1\n1 1",
"output": "0"
},
{
"input": "3\n1 1\n1 1\n2 2",
"output": "1"
},
{
"input": "5\n1 1\n1 2\n2 3\n3 4\n4 3",
"output": "1"
},
{
"input": "3\n1 2\n1 2\n1 1",
"output": "1"
}
] | 1,666,937,068
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 0
|
import numpy as np
n=int(input())
a=[eval(input().replace(' ',',')) for i in range(n)]
a=np.array(a)
out=n
for i,ai in enumerate(a):
if (ai[0] in a[:,1])and(ai[0]!=ai[1]) :
out -= 1
print(out)
|
Title: Sereja and Bottles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and his friends went to a picnic. The guys had *n* soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the *i*-th bottle is from brand *a**i*, besides, you can use it to open other bottles of brand *b**i*. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles. The next *n* lines contain the bottles' description. The *i*-th line contains two integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the description of the *i*-th bottle.
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['4\n1 1\n2 2\n3 3\n4 4\n', '4\n1 2\n2 3\n3 4\n4 1\n']
Demo Output:
['4\n', '0\n']
Note:
none
|
```python
import numpy as np
n=int(input())
a=[eval(input().replace(' ',',')) for i in range(n)]
a=np.array(a)
out=n
for i,ai in enumerate(a):
if (ai[0] in a[:,1])and(ai[0]!=ai[1]) :
out -= 1
print(out)
```
| -1
|
|
901
|
A
|
Hashing Trees
|
PROGRAMMING
| 1,500
|
[
"constructive algorithms",
"trees"
] | null | null |
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence *a*0,<=*a*1,<=...,<=*a**h*, where *h* is the height of the tree, and *a**i* equals to the number of vertices that are at distance of *i* edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence *a**i*, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
|
The first line contains a single integer *h* (2<=≤<=*h*<=≤<=105) — the height of the tree.
The second line contains *h*<=+<=1 integers — the sequence *a*0,<=*a*1,<=...,<=*a**h* (1<=≤<=*a**i*<=≤<=2·105). The sum of all *a**i* does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
|
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print integers, the *k*-th of them should be the parent of vertex *k* or be equal to zero, if the *k*-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
|
[
"2\n1 1 1\n",
"2\n1 2 2\n"
] |
[
"perfect\n",
"ambiguous\n0 1 1 3 3\n0 1 1 3 2\n"
] |
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<img class="tex-graphics" src="https://espresso.codeforces.com/ae5d1889e09854f9d8ad6e29ab7afbe690ca4702.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "2\n1 1 1",
"output": "perfect"
},
{
"input": "2\n1 2 2",
"output": "ambiguous\n0 1 1 3 3\n0 1 1 3 2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1 1",
"output": "perfect"
},
{
"input": "10\n1 1 1 1 1 2 1 1 1 1 1",
"output": "perfect"
},
{
"input": "10\n1 1 1 1 2 2 1 1 1 1 1",
"output": "ambiguous\n0 1 2 3 4 4 6 6 8 9 10 11 12\n0 1 2 3 4 4 6 5 8 9 10 11 12"
},
{
"input": "10\n1 1 1 1 1 1 1 2 1 1 2",
"output": "perfect"
},
{
"input": "10\n1 1 1 3 2 1 2 4 1 3 1",
"output": "ambiguous\n0 1 2 3 3 3 6 6 8 9 9 11 11 11 11 15 16 16 16 19\n0 1 2 3 3 3 6 5 8 9 9 11 10 10 10 15 16 16 16 19"
},
{
"input": "10\n1 1 1 4 1 1 2 1 5 1 2",
"output": "perfect"
},
{
"input": "10\n1 1 11 12 12 11 15 13 8 8 8",
"output": "ambiguous\n0 1 2 2 2 2 2 2 2 2 2 2 2 13 13 13 13 13 13 13 13 13 13 13 13 25 25 25 25 25 25 25 25 25 25 25 25 37 37 37 37 37 37 37 37 37 37 37 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 63 63 63 63 63 63 63 63 63 63 63 63 63 76 76 76 76 76 76 76 76 84 84 84 84 84 84 84 84 92 92 92 92 92 92 92 92\n0 1 2 2 2 2 2 2 2 2 2 2 2 13 12 12 12 12 12 12 12 12 12 12 12 25 24 24 24 24 24 24 24 24 24 24 24 37 36 36 36 36 36 36 36 36 36 36 48 47 47 47 47 47 47 47 47 47 47 47 47 47 47 63 62 62 62 62 62 62 62 62 62 62 62 ..."
},
{
"input": "10\n1 1 21 1 20 1 14 1 19 1 20",
"output": "perfect"
},
{
"input": "10\n1 1 93 121 112 103 114 112 112 122 109",
"output": "ambiguous\n0 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 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 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 9..."
},
{
"input": "10\n1 1 262 1 232 1 245 1 1 254 1",
"output": "perfect"
},
{
"input": "2\n1 1 199998",
"output": "perfect"
},
{
"input": "3\n1 1 199997 1",
"output": "perfect"
},
{
"input": "3\n1 1 100009 99989",
"output": "ambiguous\n0 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 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 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 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 2 2 2 2 ..."
},
{
"input": "123\n1 1 1 3714 1 3739 1 3720 1 1 3741 1 1 3726 1 3836 1 3777 1 1 3727 1 1 3866 1 3799 1 3785 1 3693 1 1 3667 1 3930 1 3849 1 1 3767 1 3792 1 3792 1 3808 1 3680 1 3798 1 3817 1 3636 1 3833 1 1 3765 1 3774 1 3747 1 1 3897 1 3773 1 3814 1 3739 1 1 3852 1 3759 1 3783 1 1 3836 1 3787 1 3752 1 1 3818 1 3794 1 3745 1 3785 1 3784 1 1 3765 1 3750 1 3690 1 1 3806 1 3781 1 3680 1 1 3748 1 3709 1 3793 1 3618 1 1 3893 1",
"output": "perfect"
},
{
"input": "13\n1 1 16677 16757 16710 16596 16512 16762 16859 16750 16658 16489 16594 16634",
"output": "ambiguous\n0 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 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 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 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 2 2 2 2 ..."
},
{
"input": "13\n1 1 40049 1 1 39777 1 1 40008 1 40060 1 40097 1",
"output": "perfect"
},
{
"input": "4\n1 2 1 2 2",
"output": "ambiguous\n0 1 1 3 4 4 6 6\n0 1 1 3 4 4 6 5"
},
{
"input": "16\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536",
"output": "ambiguous\n0 1 1 3 3 3 3 7 7 7 7 7 7 7 7 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 127 12..."
},
{
"input": "4\n1 2 1 2 3",
"output": "ambiguous\n0 1 1 3 4 4 6 6 6\n0 1 1 3 4 4 6 5 5"
},
{
"input": "2\n1 3 199969",
"output": "ambiguous\n0 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 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 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 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 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 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 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 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 4 4 4 4 4 4 4 4 4 4 4 4 4 4 ..."
},
{
"input": "2\n1 99999 99999",
"output": "ambiguous\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "2\n1 3 2",
"output": "ambiguous\n0 1 1 1 4 4\n0 1 1 1 4 3"
}
] | 1,586,987,153
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
import sys
import bisect
# from collections import deque
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 30
MOD = 998244353
n = int(ri())
a = Ri()
ite= 2
par = 1
ans = [0]*(sum(a)+3)
flag = -1
for i in range(1,len(a)):
if a[i] != a[i-1]:
flag = i
break
ans[ite] = par
ite+=1
par+=1
if flag == -1 or flag == len(a):
print("perfect")
else:
i = flag
nextpar = par
pflag = False
nextsidepar = -1
sidechild = -1
cflag = False
for j in range(flag,len(a)):
par = nextpar
for i in range(a[j]):
if i== 0:
nextpar = ite
if a[j] >= 2 and pflag == False:
nextsidepar =ite+1
pflag = True
if par == nextsidepar-1 and cflag == False:
sidechild = ite
cflag = True
ans[ite] = par
ite+=1
print("ambiguous")
print(*ans[1:sum(a)+1])
ans[sidechild] = nextsidepar
print(*ans[1:sum(a)+1])
|
Title: Hashing Trees
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence *a*0,<=*a*1,<=...,<=*a**h*, where *h* is the height of the tree, and *a**i* equals to the number of vertices that are at distance of *i* edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence *a**i*, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input Specification:
The first line contains a single integer *h* (2<=≤<=*h*<=≤<=105) — the height of the tree.
The second line contains *h*<=+<=1 integers — the sequence *a*0,<=*a*1,<=...,<=*a**h* (1<=≤<=*a**i*<=≤<=2·105). The sum of all *a**i* does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output Specification:
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print integers, the *k*-th of them should be the parent of vertex *k* or be equal to zero, if the *k*-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Demo Input:
['2\n1 1 1\n', '2\n1 2 2\n']
Demo Output:
['perfect\n', 'ambiguous\n0 1 1 3 3\n0 1 1 3 2\n']
Note:
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<img class="tex-graphics" src="https://espresso.codeforces.com/ae5d1889e09854f9d8ad6e29ab7afbe690ca4702.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
import sys
import bisect
# from collections import deque
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 30
MOD = 998244353
n = int(ri())
a = Ri()
ite= 2
par = 1
ans = [0]*(sum(a)+3)
flag = -1
for i in range(1,len(a)):
if a[i] != a[i-1]:
flag = i
break
ans[ite] = par
ite+=1
par+=1
if flag == -1 or flag == len(a):
print("perfect")
else:
i = flag
nextpar = par
pflag = False
nextsidepar = -1
sidechild = -1
cflag = False
for j in range(flag,len(a)):
par = nextpar
for i in range(a[j]):
if i== 0:
nextpar = ite
if a[j] >= 2 and pflag == False:
nextsidepar =ite+1
pflag = True
if par == nextsidepar-1 and cflag == False:
sidechild = ite
cflag = True
ans[ite] = par
ite+=1
print("ambiguous")
print(*ans[1:sum(a)+1])
ans[sidechild] = nextsidepar
print(*ans[1:sum(a)+1])
```
| 0
|
|
984
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
|
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
|
Print one number that will be left on the board.
|
[
"3\n2 1 3\n",
"3\n2 2 2\n"
] |
[
"2",
"2"
] |
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
| 500
|
[
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,577,013,094
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 108
| 0
|
number = int(input())
data = list(map(int, input().split()))
for m in range(int((number - 1)/2)):
data.remove(max(data))
data.remove(min(data))
print(data[0])
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
|
```python
number = int(input())
data = list(map(int, input().split()))
for m in range(int((number - 1)/2)):
data.remove(max(data))
data.remove(min(data))
print(data[0])
```
| 0
|
|
16
|
A
|
Flag
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Flag
|
2
|
64
|
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
|
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
|
Output YES, if the flag meets the new ISO standard, and NO otherwise.
|
[
"3 3\n000\n111\n222\n",
"3 3\n000\n000\n111\n",
"3 3\n000\n111\n002\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "3 3\n000\n111\n222",
"output": "YES"
},
{
"input": "3 3\n000\n000\n111",
"output": "NO"
},
{
"input": "3 3\n000\n111\n002",
"output": "NO"
},
{
"input": "10 10\n2222222222\n5555555555\n0000000000\n4444444444\n1111111111\n3333333393\n3333333333\n5555555555\n0000000000\n8888888888",
"output": "NO"
},
{
"input": "10 13\n4442444444444\n8888888888888\n6666666666666\n0000000000000\n3333333333333\n4444444444444\n7777777777777\n8388888888888\n1111111111111\n5555555555555",
"output": "NO"
},
{
"input": "10 8\n33333333\n44444444\n11111115\n81888888\n44444444\n11111111\n66666666\n33330333\n33333333\n33333333",
"output": "NO"
},
{
"input": "5 5\n88888\n44444\n66666\n55555\n88888",
"output": "YES"
},
{
"input": "20 19\n1111111111111111111\n5555555555555555555\n0000000000000000000\n3333333333333333333\n1111111111111111111\n2222222222222222222\n4444444444444444444\n5555555555555555555\n0000000000000000000\n4444444444444444444\n0000000000000000000\n5555555555555555555\n7777777777777777777\n9999999999999999999\n2222222222222222222\n4444444444444444444\n1111111111111111111\n6666666666666666666\n7777777777777777777\n2222222222222222222",
"output": "YES"
},
{
"input": "1 100\n8888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888",
"output": "YES"
},
{
"input": "100 1\n5\n7\n9\n4\n7\n2\n5\n1\n6\n7\n2\n7\n6\n8\n7\n4\n0\n2\n9\n8\n9\n1\n6\n4\n3\n4\n7\n1\n9\n3\n0\n8\n3\n1\n7\n5\n3\n9\n5\n1\n3\n5\n8\n1\n9\n3\n9\n0\n6\n0\n7\n6\n5\n2\n8\n3\n7\n6\n5\n1\n8\n3\n6\n9\n6\n0\n5\n8\n5\n2\n9\n1\n0\n1\n8\n3\n2\n1\n0\n3\n9\n0\n5\n1\n0\n4\n9\n3\n0\n4\n8\n4\n8\n6\n3\n0\n4\n6\n8\n4",
"output": "YES"
},
{
"input": "1 1\n2",
"output": "YES"
},
{
"input": "1 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111181111111111111111111111",
"output": "NO"
},
{
"input": "100 1\n3\n6\n4\n3\n0\n2\n8\n7\n3\n2\n1\n7\n1\n3\n2\n3\n6\n9\n0\n8\n5\n9\n7\n9\n2\n1\n4\n5\n1\n9\n2\n5\n1\n4\n6\n4\n9\n1\n0\n2\n1\n4\n7\n1\n4\n8\n0\n9\n2\n1\n6\n2\n8\n6\n9\n5\n8\n6\n4\n5\n9\n2\n7\n4\n1\n5\n8\n0\n9\n5\n4\n6\n5\n0\n6\n3\n6\n9\n7\n2\n0\n9\n7\n3\n2\n4\n9\n4\n7\n1\n2\n3\n1\n7\n9\n1\n9\n0\n4\n0",
"output": "YES"
}
] | 1,643,724,781
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
def solve():
n, m = map(int,input().split())
arr = [set(input()) for x in range(n)]
curr = '#'
for x in arr:
if len(x) != 1 or x == curr:
return 'NO'
curr = x
return 'YES'
print(solve())
|
Title: Flag
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
Input Specification:
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output Specification:
Output YES, if the flag meets the new ISO standard, and NO otherwise.
Demo Input:
['3 3\n000\n111\n222\n', '3 3\n000\n000\n111\n', '3 3\n000\n111\n002\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
def solve():
n, m = map(int,input().split())
arr = [set(input()) for x in range(n)]
curr = '#'
for x in arr:
if len(x) != 1 or x == curr:
return 'NO'
curr = x
return 'YES'
print(solve())
```
| 3.977
|
707
|
B
|
Bakery
|
PROGRAMMING
| 1,300
|
[
"graphs"
] | null | null |
Masha wants to open her own bakery and bake muffins in one of the *n* cities numbered from 1 to *n*. There are *m* bidirectional roads, each of whose connects some pair of cities.
To bake muffins in her bakery, Masha needs to establish flour supply from some storage. There are only *k* storages, located in different cities numbered *a*1,<=*a*2,<=...,<=*a**k*.
Unforunately the law of the country Masha lives in prohibits opening bakery in any of the cities which has storage located in it. She can open it only in one of another *n*<=-<=*k* cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1 ruble.
Formally, Masha will pay *x* roubles, if she will open the bakery in some city *b* (*a**i*<=≠<=*b* for every 1<=≤<=*i*<=≤<=*k*) and choose a storage in some city *s* (*s*<==<=*a**j* for some 1<=≤<=*j*<=≤<=*k*) and *b* and *s* are connected by some path of roads of summary length *x* (if there are more than one path, Masha is able to choose which of them should be used).
Masha is very thrifty and rational. She is interested in a city, where she can open her bakery (and choose one of *k* storages and one of the paths between city with bakery and city with storage) and pay minimum possible amount of rubles for flour delivery. Please help Masha find this amount.
|
The first line of the input contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.
Then *m* lines follow. Each of them contains three integers *u*, *v* and *l* (1<=≤<=*u*,<=*v*<=≤<=*n*, 1<=≤<=*l*<=≤<=109, *u*<=≠<=*v*) meaning that there is a road between cities *u* and *v* of length of *l* kilometers .
If *k*<=><=0, then the last line of the input contains *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*) — the number of cities having flour storage located in. If *k*<==<=0 then this line is not presented in the input.
|
Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.
If the bakery can not be opened (while satisfying conditions) in any of the *n* cities, print <=-<=1 in the only line.
|
[
"5 4 2\n1 2 5\n1 2 3\n2 3 4\n1 4 10\n1 5\n",
"3 1 1\n1 2 3\n3\n"
] |
[
"3",
"-1"
] |
Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.
| 1,000
|
[
{
"input": "5 4 2\n1 2 5\n1 2 3\n2 3 4\n1 4 10\n1 5",
"output": "3"
},
{
"input": "3 1 1\n1 2 3\n3",
"output": "-1"
},
{
"input": "2 3 1\n1 2 3\n1 2 18\n1 2 13\n2",
"output": "3"
},
{
"input": "3 7 0\n1 3 9\n1 2 5\n1 2 21\n1 2 12\n1 2 13\n2 3 19\n2 3 8",
"output": "-1"
},
{
"input": "4 13 1\n1 4 10\n1 3 6\n1 4 3\n3 4 1\n1 3 2\n1 2 15\n1 4 21\n1 4 20\n2 4 13\n1 4 7\n2 4 2\n1 2 8\n1 3 17\n1",
"output": "2"
},
{
"input": "5 7 3\n2 3 20\n1 2 10\n1 2 11\n4 5 15\n2 3 3\n1 5 19\n1 2 3\n5 3 2",
"output": "3"
},
{
"input": "6 7 4\n5 6 21\n3 6 18\n1 6 5\n4 6 4\n1 2 13\n3 4 7\n1 2 15\n6 1 3 2",
"output": "4"
},
{
"input": "7 39 2\n2 7 10\n5 6 18\n2 7 13\n4 5 11\n3 6 14\n1 2 16\n3 4 2\n1 3 13\n1 5 1\n1 2 20\n1 5 11\n1 4 14\n3 6 21\n1 2 18\n1 4 13\n2 3 4\n3 6 12\n2 5 18\n4 7 17\n1 2 3\n2 3 6\n1 2 21\n1 7 18\n4 6 13\n1 2 13\n1 7 17\n2 3 16\n5 6 5\n2 4 17\n1 2 9\n1 2 21\n4 5 9\n1 2 18\n2 6 6\n2 3 9\n1 4 7\n2 5 7\n3 7 21\n4 5 2\n6 2",
"output": "3"
},
{
"input": "8 57 3\n1 3 15\n2 3 1\n1 7 21\n1 2 8\n2 5 16\n1 6 4\n1 3 2\n3 7 17\n5 8 3\n1 3 18\n1 4 3\n1 2 1\n2 8 14\n1 4 17\n4 5 21\n2 3 6\n3 5 11\n2 8 11\n3 4 1\n1 3 9\n1 4 3\n2 3 12\n1 5 9\n2 3 15\n1 2 14\n1 2 10\n1 4 19\n5 7 7\n5 8 20\n5 8 1\n1 4 3\n4 5 8\n5 7 2\n1 2 14\n4 5 9\n6 7 2\n2 6 9\n2 6 4\n3 7 4\n3 5 11\n4 8 19\n3 7 15\n1 8 21\n6 7 11\n4 6 2\n2 3 21\n6 7 2\n6 8 4\n1 3 21\n3 4 1\n4 5 15\n4 7 21\n2 6 2\n5 6 16\n5 8 9\n2 5 6\n1 7 17\n1 4 8",
"output": "1"
},
{
"input": "350 10 39\n2 13 693\n6 31 482\n72 312 617\n183 275 782\n81 123 887\n26 120 1205\n135 185 822\n64 219 820\n74 203 874\n19 167 1422\n252 332 204 334 100 350 26 14 134 213 32 84 331 215 181 158 99 190 206 265 343 241 287 74 113 15 12 338 27 110 98 132 35 95 51 315 297 69 163",
"output": "874"
},
{
"input": "7 7 3\n1 2 1\n2 4 1\n3 4 1\n1 3 1\n5 7 2\n6 7 10\n5 6 5\n5 6 7",
"output": "-1"
},
{
"input": "7 8 3\n1 2 1\n2 4 1\n3 4 1\n1 3 1\n5 7 2\n6 7 10\n5 6 5\n2 5 31246\n5 6 7",
"output": "31246"
},
{
"input": "5 5 5\n1 2 1\n2 3 2\n3 4 3\n4 5 5\n1 5 6\n1 2 3 4 5",
"output": "-1"
},
{
"input": "10 10 3\n1 2 1000000000\n2 3 1000000000\n3 4 1000000000\n4 5 1000000000\n5 6 1000000000\n6 7 1000000000\n7 8 1000000000\n8 9 1000000000\n9 10 1000000000\n10 1 1000000000\n1 2 3",
"output": "1000000000"
},
{
"input": "2 1 1\n1 2 1000000000\n1",
"output": "1000000000"
},
{
"input": "99999 1 0\n1 2 3",
"output": "-1"
},
{
"input": "99999 1 2\n1 2 3\n2 4",
"output": "3"
},
{
"input": "2 1 1\n1 2 99999999\n1",
"output": "99999999"
},
{
"input": "2 1 1\n1 2 999999\n2",
"output": "999999"
}
] | 1,681,303,793
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 16
| 889
| 819,200
|
def quick_sort(li, left, right): #框架
if left < right: #至少两个元素
mid = partition(li, left, right) #归位
quick_sort(li, left, mid - 1)
quick_sort(li, mid + 1, right)
def partition(li, left, right):
tmp = li[left]
while left < right:
while left < right and li[right] >= tmp: #找比左边小的，从右边找
right -= 1 #右指针往左走
li[left] = li[right] #把右边的值放入左边空位
while left < right and li[left] <= tmp:
left += 1
li[right] = li[left] #把左边的值放入右边空位
li[left] = tmp #把tmp归位
return left
n, m, k = map(int, input().split())
if k == 0:
print(-1)
else:
ul, vl, ll = [], [], []
for line in range(m):
u, v, l = map(int, input().split())
ul.append(u)
vl.append(v)
ll.append(l)
kl = list(map(int, input().split()))
nl = []
for i in range(m):
if ((ul[i] in kl) and(vl[i] not in kl)) or ((ul[i] not in kl)and(vl[i] in kl)):
nl.append(ll[i])
if nl == []:
print(-1)
else:
quick_sort(nl,0,len(nl)-1)
print(nl[0])
|
Title: Bakery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Masha wants to open her own bakery and bake muffins in one of the *n* cities numbered from 1 to *n*. There are *m* bidirectional roads, each of whose connects some pair of cities.
To bake muffins in her bakery, Masha needs to establish flour supply from some storage. There are only *k* storages, located in different cities numbered *a*1,<=*a*2,<=...,<=*a**k*.
Unforunately the law of the country Masha lives in prohibits opening bakery in any of the cities which has storage located in it. She can open it only in one of another *n*<=-<=*k* cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1 ruble.
Formally, Masha will pay *x* roubles, if she will open the bakery in some city *b* (*a**i*<=≠<=*b* for every 1<=≤<=*i*<=≤<=*k*) and choose a storage in some city *s* (*s*<==<=*a**j* for some 1<=≤<=*j*<=≤<=*k*) and *b* and *s* are connected by some path of roads of summary length *x* (if there are more than one path, Masha is able to choose which of them should be used).
Masha is very thrifty and rational. She is interested in a city, where she can open her bakery (and choose one of *k* storages and one of the paths between city with bakery and city with storage) and pay minimum possible amount of rubles for flour delivery. Please help Masha find this amount.
Input Specification:
The first line of the input contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.
Then *m* lines follow. Each of them contains three integers *u*, *v* and *l* (1<=≤<=*u*,<=*v*<=≤<=*n*, 1<=≤<=*l*<=≤<=109, *u*<=≠<=*v*) meaning that there is a road between cities *u* and *v* of length of *l* kilometers .
If *k*<=><=0, then the last line of the input contains *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*) — the number of cities having flour storage located in. If *k*<==<=0 then this line is not presented in the input.
Output Specification:
Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.
If the bakery can not be opened (while satisfying conditions) in any of the *n* cities, print <=-<=1 in the only line.
Demo Input:
['5 4 2\n1 2 5\n1 2 3\n2 3 4\n1 4 10\n1 5\n', '3 1 1\n1 2 3\n3\n']
Demo Output:
['3', '-1']
Note:
Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.
|
```python
def quick_sort(li, left, right): #框架
if left < right: #至少两个元素
mid = partition(li, left, right) #归位
quick_sort(li, left, mid - 1)
quick_sort(li, mid + 1, right)
def partition(li, left, right):
tmp = li[left]
while left < right:
while left < right and li[right] >= tmp: #找比左边小的，从右边找
right -= 1 #右指针往左走
li[left] = li[right] #把右边的值放入左边空位
while left < right and li[left] <= tmp:
left += 1
li[right] = li[left] #把左边的值放入右边空位
li[left] = tmp #把tmp归位
return left
n, m, k = map(int, input().split())
if k == 0:
print(-1)
else:
ul, vl, ll = [], [], []
for line in range(m):
u, v, l = map(int, input().split())
ul.append(u)
vl.append(v)
ll.append(l)
kl = list(map(int, input().split()))
nl = []
for i in range(m):
if ((ul[i] in kl) and(vl[i] not in kl)) or ((ul[i] not in kl)and(vl[i] in kl)):
nl.append(ll[i])
if nl == []:
print(-1)
else:
quick_sort(nl,0,len(nl)-1)
print(nl[0])
```
| -1
|
|
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"
}
] | 1,657,200,652
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
import random
roll_dice = random.randint(1, 6)
print(roll_dice)
|
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
import random
roll_dice = random.randint(1, 6)
print(roll_dice)
```
| 0
|
832
|
B
|
Petya and Exam
|
PROGRAMMING
| 1,600
|
[
"implementation",
"strings"
] | null | null |
It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one..
There is a glob pattern in the statements (a string consisting of lowercase English letters, characters "?" and "*"). It is known that character "*" occurs no more than once in the pattern.
Also, *n* query strings are given, it is required to determine for each of them if the pattern matches it or not.
Everything seemed easy to Petya, but then he discovered that the special pattern characters differ from their usual meaning.
A pattern matches a string if it is possible to replace each character "?" with one good lowercase English letter, and the character "*" (if there is one) with any, including empty, string of bad lowercase English letters, so that the resulting string is the same as the given string.
The good letters are given to Petya. All the others are bad.
|
The first line contains a string with length from 1 to 26 consisting of distinct lowercase English letters. These letters are good letters, all the others are bad.
The second line contains the pattern — a string *s* of lowercase English letters, characters "?" and "*" (1<=≤<=|*s*|<=≤<=105). It is guaranteed that character "*" occurs in *s* no more than once.
The third line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of query strings.
*n* lines follow, each of them contains single non-empty string consisting of lowercase English letters — a query string.
It is guaranteed that the total length of all query strings is not greater than 105.
|
Print *n* lines: in the *i*-th of them print "YES" if the pattern matches the *i*-th query string, and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrary.
|
[
"ab\na?a\n2\naaa\naab\n",
"abc\na?a?a*\n4\nabacaba\nabaca\napapa\naaaaax\n"
] |
[
"YES\nNO\n",
"NO\nYES\nNO\nYES\n"
] |
In the first example we can replace "?" with good letters "a" and "b", so we can see that the answer for the first query is "YES", and the answer for the second query is "NO", because we can't match the third letter.
Explanation of the second example.
- The first query: "NO", because character "*" can be replaced with a string of bad letters only, but the only way to match the query string is to replace it with the string "ba", in which both letters are good. - The second query: "YES", because characters "?" can be replaced with corresponding good letters, and character "*" can be replaced with empty string, and the strings will coincide. - The third query: "NO", because characters "?" can't be replaced with bad letters. - The fourth query: "YES", because characters "?" can be replaced with good letters "a", and character "*" can be replaced with a string of bad letters "x".
| 1,000
|
[
{
"input": "ab\na?a\n2\naaa\naab",
"output": "YES\nNO"
},
{
"input": "abc\na?a?a*\n4\nabacaba\nabaca\napapa\naaaaax",
"output": "NO\nYES\nNO\nYES"
},
{
"input": "s\nc*?cb\n26\nbbaa\nb\ncc\ncbaab\nacacc\nca\na\nc\ncb\nabb\nba\nb\nba\ncac\nccccb\nccb\nbbbc\nabbcb\na\nbc\nc\na\nabb\nca\ncacb\nac",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "o\n*\n28\nbac\nbcc\ncbcb\ncaabc\ncb\nacab\ncbccb\ncbccc\nc\nbbaa\ncaaaa\nbbc\nba\nc\ncacbc\ncbab\naa\nac\nacc\na\nac\nbac\naaac\nba\nabbbb\nbbcc\nbaacb\naabaa",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "u\n*b??c\n23\na\nbcbcc\nacb\na\nbacaa\nbb\nb\nbcba\ncbbcc\nb\nabbb\nbcacb\nabcb\ncbca\nb\ncba\ncabcb\nbc\ncc\naaacc\nccac\ncc\nccbcb",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "g\nc?*\n58\nb\ncaac\nbbc\nabb\ncaccc\ncb\naba\nbcaa\ncca\ncbbcb\ncac\nbbaca\nbcba\nbba\nabbab\nccc\nc\nbcb\naac\nbcbbc\nbca\nc\ncbb\nccabb\naaccc\nccaa\nc\nc\nbcca\naa\nccb\ncb\ncbcb\ncc\nab\ncccc\nbbbab\nbab\na\nc\ncbba\nbbacb\naa\nb\nbaab\nacabb\nbcbab\ncbbcb\nbc\ncccba\naa\ncccca\ncacc\naacbb\na\nc\nab\nccca",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "g\nbc*a\n40\nbabac\nccbb\ncacbc\nc\na\naba\nbc\na\nba\nbbcca\nccbac\na\nc\nbabc\ncccbc\nab\nabca\nccb\nacbbb\nb\nbbac\naa\nb\nca\nbc\naaba\nbaaaa\nbcc\nab\na\naba\nb\nc\nba\nbc\nca\nbb\nc\nc\nca",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "g\n*cc?\n93\nac\ncaab\nacaca\ncccc\nbcc\nbab\nbc\nc\nc\nbbaa\nb\ncc\ncb\naa\nabcbb\nbccc\nc\ncbcbc\nac\nca\nbcba\nbb\nbab\nba\nb\nbbba\nbabbc\nbacab\nbc\na\ncbccc\nbbaac\ncbab\ncab\ncc\ncbbcb\nc\nc\ncbaa\nca\nbacab\nc\nbcac\nbbbc\nc\nac\nccab\nccccb\ncccab\nc\nacb\nac\nbccba\nca\nbbbbc\naaca\naa\na\nbabac\nbb\nc\ncac\naca\naacb\naacbb\na\nacaab\ncbb\nbcc\ncb\nbcbaa\ncca\nb\nbaac\nbcca\nc\ncbb\nac\nc\naccc\naac\nbcbc\nabc\nbacab\nb\na\na\nbbacc\ncb\na\nccac\nb\nbbc",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "c\n*\n83\nbbc\ncacb\nbcbc\naca\nba\nc\nccac\nab\nab\nbacba\nbb\nc\nbcc\nc\ncbc\ncbbb\nac\nb\nacbcb\nbccc\ncccb\nb\na\nca\nc\nccaa\naa\ncacb\nccc\na\nccc\nababb\nbab\ncaa\nbaa\na\ncc\ncbbbc\naaaa\nabbab\naabac\nbcbab\nbcb\nacaa\nbcb\na\ncca\na\nbacc\nacacb\nc\nc\ncba\nbcaca\na\ncaac\na\nb\na\nccc\naabca\nbbab\nb\nac\nbabc\nc\nac\nba\nbbcb\nc\naaab\ncab\nacb\nbba\nbbcba\nc\na\naccbb\naaccc\nac\nbaa\nbaabb\nabca",
"output": "NO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nYES\nYES\nNO\nYES\nYES\nNO\nNO\nYES\nYES\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nYES\nYES\nNO\nNO\nYES\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO"
},
{
"input": "s\n*cb\n70\nab\nccb\naaab\nb\nab\ncba\na\nbbaca\nac\nccacb\nbaabb\naaab\nccca\ncb\nba\nbccac\nc\ncc\ncbbbb\ncab\nabbb\ncbb\naabc\ncac\nacb\na\nc\nc\ncbbbb\nbaaca\ncbcc\nbc\naa\nabcb\nacbbc\nbaaa\naa\ncc\ncc\nb\nb\nbcba\ncbacc\nbcb\ncaabc\nacaac\ncb\ncba\ncbaaa\nbcaaa\naccbb\naccac\nca\nacaa\ncc\nc\nb\nbac\nb\nbab\nb\ncca\naacc\nacb\nccc\nbc\nb\naab\naaca\naac",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "k\n*b\n70\ncbc\nb\ncca\nacbc\nca\nab\nc\nbbb\nbaa\nbbb\nac\nbaacc\nbaabc\naac\na\nba\nb\nc\nc\nba\ncacbb\nabb\nbc\nabcb\nca\na\nbbbbb\ncca\nccacc\ncbaca\nba\ncbcca\ncb\nc\nbbbba\ncca\nabaac\na\nac\nc\nccbc\nbcac\nbcb\na\nc\nabbca\nbaacb\ncc\nacba\nc\nbcc\ncbba\nccba\na\na\ncbb\ncba\nb\naaaac\ncb\nbaacb\nab\nc\ncbbcb\nbab\nac\nca\nc\nac\nb",
"output": "NO\nYES\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\nNO\nNO\nNO\nNO\nYES"
},
{
"input": "l\na*\n40\nacbb\naba\nb\naab\nbb\nbbab\ncaba\naab\naaab\nacac\nacbaa\nbca\nac\nbb\na\nba\naaa\nbc\nbba\ncca\naacab\na\nc\nca\naacaa\nbaac\nbb\nc\nba\nc\nbab\nb\na\ncabaa\nccacc\ncbbab\nbaaca\ncabb\naaccc\nbcbac",
"output": "YES\nYES\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES\nNO"
},
{
"input": "u\ncba*\n26\ncaa\ncccb\nbc\nbacb\nca\nccaaa\nb\naaca\nba\ncacc\ncccac\nabba\nbabc\na\nac\nca\nbbba\na\naa\naaabb\nb\nc\nbba\nbbba\nacaa\nba",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "cba\n?*cbc\n88\ncccca\ncbc\nb\nbcb\naaa\ncaac\nbacb\nacbb\na\nab\ncbcca\nbccc\nabcc\naca\nba\nbbac\nacc\ncba\nbcba\nbc\naa\nab\ncaba\ncccab\ncba\ncbcc\nba\ncacbb\nabcc\na\nc\nbac\nccaba\nb\nac\nbbb\nac\nccaca\na\nba\nacbcc\nbbc\nacbc\nbbabc\nccbb\nb\nacaa\na\nba\nacb\na\nab\naa\nbbbb\naabb\nbcbc\nb\nca\nb\nccab\nab\nc\nb\naabab\nc\ncbbbc\nacbbb\nbacaa\nbcccc\ncbac\nc\nac\nb\nca\ncbb\nccbc\nc\nc\nbcb\nc\nbaaba\nc\nbac\nb\nba\ncb\ncc\nbaaca",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "a\naa\n1\naaa",
"output": "NO"
},
{
"input": "a\naaa\n1\naaaa",
"output": "NO"
},
{
"input": "a\naa*aa\n1\naaa",
"output": "NO"
},
{
"input": "a\nbb*bb\n1\nbbbbbbbbbbbbbbbb",
"output": "YES"
},
{
"input": "a\na*\n1\nabbbbbbb",
"output": "YES"
},
{
"input": "a\na?a\n1\naaab",
"output": "NO"
},
{
"input": "xy\ncab*aba\n1\ncaba",
"output": "NO"
},
{
"input": "a\n*\n4\nb\na\nab\nba",
"output": "YES\nNO\nNO\nNO"
},
{
"input": "abc\na?a?*a\n3\nababxa\nababca\nababa",
"output": "YES\nNO\nYES"
},
{
"input": "abc\n??a*df?c\n6\nabadfcc\naaaadfac\nbbagthfac\nacadddfac\ndaagdffc\naaaadfcc",
"output": "YES\nNO\nNO\nYES\nNO\nNO"
},
{
"input": "abc\nabc*a\n1\nabckka",
"output": "YES"
},
{
"input": "b\n*a\n1\naba",
"output": "NO"
},
{
"input": "a\nabc*g\n1\nabcdefg",
"output": "YES"
},
{
"input": "a\nab\n1\na",
"output": "NO"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\n*a\n1\na",
"output": "YES"
},
{
"input": "as\naba*aba\n1\naba",
"output": "NO"
},
{
"input": "ab\naweerrtab\n4\naw\naweerrtabwqeqrw\naweerrtabxcvxcbcxbdsfdsfewrewrqweq\naweerrtabaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO\nNO\nNO\nNO"
},
{
"input": "a\na\n1\nab",
"output": "NO"
},
{
"input": "a\na*b\n1\nabb",
"output": "YES"
},
{
"input": "a\nb*a\n1\nbbadd",
"output": "NO"
},
{
"input": "a\naaaa\n1\naaa",
"output": "NO"
},
{
"input": "z\n*abcd\n1\nggggggg",
"output": "NO"
},
{
"input": "abc\n*??\n1\nqqqqqqqqab",
"output": "YES"
},
{
"input": "b\naa\n1\na",
"output": "NO"
},
{
"input": "ab\na*pa\n1\nappppa",
"output": "YES"
},
{
"input": "a\nbbb\n1\nbbbbb",
"output": "NO"
},
{
"input": "ab\nabcd?\n1\nabcd",
"output": "NO"
},
{
"input": "c\na\n1\nab",
"output": "NO"
}
] | 1,599,664,975
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 11
| 420
| 17,920,000
|
alpha = "abcdefghijklmnopqrstuvwxyz"
good = [i for i in input()]
bad = ["-1"]
for i in alpha:
if(i not in good):
bad.append(i)
s = [i for i in input()]
k = int(input())
n = len(s)
for i in range(k):
ok = "YES"
s1 = [i for i in input()]
while(len(s1)<n):
s1.append("-1")
for i in range(n):
if(s[i]==s1[i]):
continue
elif(s[i]=="?"):
if(s1[i] in bad):
ok = "NO"
break
elif(s[i]=="*"):
if(s1[i] in good):
ok="NO"
break
else:
ok="NO"
break
print(ok)
|
Title: Petya and Exam
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one..
There is a glob pattern in the statements (a string consisting of lowercase English letters, characters "?" and "*"). It is known that character "*" occurs no more than once in the pattern.
Also, *n* query strings are given, it is required to determine for each of them if the pattern matches it or not.
Everything seemed easy to Petya, but then he discovered that the special pattern characters differ from their usual meaning.
A pattern matches a string if it is possible to replace each character "?" with one good lowercase English letter, and the character "*" (if there is one) with any, including empty, string of bad lowercase English letters, so that the resulting string is the same as the given string.
The good letters are given to Petya. All the others are bad.
Input Specification:
The first line contains a string with length from 1 to 26 consisting of distinct lowercase English letters. These letters are good letters, all the others are bad.
The second line contains the pattern — a string *s* of lowercase English letters, characters "?" and "*" (1<=≤<=|*s*|<=≤<=105). It is guaranteed that character "*" occurs in *s* no more than once.
The third line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of query strings.
*n* lines follow, each of them contains single non-empty string consisting of lowercase English letters — a query string.
It is guaranteed that the total length of all query strings is not greater than 105.
Output Specification:
Print *n* lines: in the *i*-th of them print "YES" if the pattern matches the *i*-th query string, and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['ab\na?a\n2\naaa\naab\n', 'abc\na?a?a*\n4\nabacaba\nabaca\napapa\naaaaax\n']
Demo Output:
['YES\nNO\n', 'NO\nYES\nNO\nYES\n']
Note:
In the first example we can replace "?" with good letters "a" and "b", so we can see that the answer for the first query is "YES", and the answer for the second query is "NO", because we can't match the third letter.
Explanation of the second example.
- The first query: "NO", because character "*" can be replaced with a string of bad letters only, but the only way to match the query string is to replace it with the string "ba", in which both letters are good. - The second query: "YES", because characters "?" can be replaced with corresponding good letters, and character "*" can be replaced with empty string, and the strings will coincide. - The third query: "NO", because characters "?" can't be replaced with bad letters. - The fourth query: "YES", because characters "?" can be replaced with good letters "a", and character "*" can be replaced with a string of bad letters "x".
|
```python
alpha = "abcdefghijklmnopqrstuvwxyz"
good = [i for i in input()]
bad = ["-1"]
for i in alpha:
if(i not in good):
bad.append(i)
s = [i for i in input()]
k = int(input())
n = len(s)
for i in range(k):
ok = "YES"
s1 = [i for i in input()]
while(len(s1)<n):
s1.append("-1")
for i in range(n):
if(s[i]==s1[i]):
continue
elif(s[i]=="?"):
if(s1[i] in bad):
ok = "NO"
break
elif(s[i]=="*"):
if(s1[i] in good):
ok="NO"
break
else:
ok="NO"
break
print(ok)
```
| 0
|
|
142
|
B
|
Help General
|
PROGRAMMING
| 1,800
|
[
"constructive algorithms",
"greedy",
"implementation"
] | null | null |
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular *n*<=×<=*m* field, split into *nm* square 1<=×<=1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (*x*1,<=*y*1) and (*x*2,<=*y*2) equals exactly (*x*1<=-<=*x*2)2<=+<=(*y*1<=-<=*y*2)2. Now not all *nm* squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2,<=2), then he cannot put soldiers in the squares (1,<=4), (3,<=4), (4,<=1) and (4,<=3) — each of them will conflict with the soldier in the square (2,<=2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
|
The single line contains space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) that represent the size of the drill exercise field.
|
Print the desired maximum number of warriors.
|
[
"2 4\n",
"3 4\n"
] |
[
"4",
"6"
] |
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
| 1,000
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "4 3",
"output": "6"
},
{
"input": "4 2",
"output": "4"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "3 5",
"output": "8"
},
{
"input": "5 3",
"output": "8"
},
{
"input": "506 44",
"output": "11132"
},
{
"input": "555 349",
"output": "96848"
},
{
"input": "757 210",
"output": "79485"
},
{
"input": "419 503",
"output": "105379"
},
{
"input": "515 19",
"output": "4893"
},
{
"input": "204 718",
"output": "73236"
},
{
"input": "862 330",
"output": "142230"
},
{
"input": "494 982",
"output": "242554"
},
{
"input": "967 4",
"output": "1934"
},
{
"input": "449 838",
"output": "188131"
},
{
"input": "635 458",
"output": "145415"
},
{
"input": "156 911",
"output": "71058"
},
{
"input": "409 295",
"output": "60328"
},
{
"input": "755 458",
"output": "172895"
},
{
"input": "936 759",
"output": "355212"
},
{
"input": "771 460",
"output": "177330"
},
{
"input": "563 802",
"output": "225763"
},
{
"input": "953 874",
"output": "416461"
},
{
"input": "354 720",
"output": "127440"
},
{
"input": "915 72",
"output": "32940"
},
{
"input": "860 762",
"output": "327660"
},
{
"input": "396 387",
"output": "76626"
},
{
"input": "675 710",
"output": "239625"
},
{
"input": "728 174",
"output": "63336"
},
{
"input": "883 312",
"output": "137748"
},
{
"input": "701 600",
"output": "210300"
},
{
"input": "824 729",
"output": "300348"
},
{
"input": "886 80",
"output": "35440"
},
{
"input": "762 742",
"output": "282702"
},
{
"input": "781 586",
"output": "228833"
},
{
"input": "44 343",
"output": "7546"
},
{
"input": "847 237",
"output": "100370"
},
{
"input": "169 291",
"output": "24590"
},
{
"input": "961 61",
"output": "29311"
},
{
"input": "695 305",
"output": "105988"
},
{
"input": "854 503",
"output": "214781"
},
{
"input": "1 744",
"output": "744"
},
{
"input": "1 383",
"output": "383"
},
{
"input": "1 166",
"output": "166"
},
{
"input": "557 1",
"output": "557"
},
{
"input": "650 1",
"output": "650"
},
{
"input": "1 995",
"output": "995"
},
{
"input": "1 865",
"output": "865"
},
{
"input": "1 393",
"output": "393"
},
{
"input": "363 1",
"output": "363"
},
{
"input": "1 506",
"output": "506"
},
{
"input": "2 348",
"output": "348"
},
{
"input": "583 2",
"output": "584"
},
{
"input": "2 89",
"output": "90"
},
{
"input": "576 2",
"output": "576"
},
{
"input": "180 2",
"output": "180"
},
{
"input": "719 2",
"output": "720"
},
{
"input": "2 951",
"output": "952"
},
{
"input": "313 2",
"output": "314"
},
{
"input": "433 2",
"output": "434"
},
{
"input": "804 2",
"output": "804"
},
{
"input": "1 991",
"output": "991"
},
{
"input": "1 992",
"output": "992"
},
{
"input": "1 993",
"output": "993"
},
{
"input": "994 1",
"output": "994"
},
{
"input": "995 1",
"output": "995"
},
{
"input": "996 1",
"output": "996"
},
{
"input": "997 1",
"output": "997"
},
{
"input": "1 998",
"output": "998"
},
{
"input": "1 999",
"output": "999"
},
{
"input": "1 1000",
"output": "1000"
},
{
"input": "991 2",
"output": "992"
},
{
"input": "2 992",
"output": "992"
},
{
"input": "993 2",
"output": "994"
},
{
"input": "994 2",
"output": "996"
},
{
"input": "995 2",
"output": "996"
},
{
"input": "2 996",
"output": "996"
},
{
"input": "997 2",
"output": "998"
},
{
"input": "2 998",
"output": "1000"
},
{
"input": "2 999",
"output": "1000"
},
{
"input": "2 1000",
"output": "1000"
},
{
"input": "997 997",
"output": "497005"
},
{
"input": "997 998",
"output": "497503"
},
{
"input": "997 999",
"output": "498002"
},
{
"input": "997 1000",
"output": "498500"
},
{
"input": "998 997",
"output": "497503"
},
{
"input": "998 998",
"output": "498002"
},
{
"input": "998 999",
"output": "498501"
},
{
"input": "998 1000",
"output": "499000"
},
{
"input": "999 997",
"output": "498002"
},
{
"input": "999 998",
"output": "498501"
},
{
"input": "999 999",
"output": "499001"
},
{
"input": "999 1000",
"output": "499500"
},
{
"input": "1000 997",
"output": "498500"
},
{
"input": "1000 998",
"output": "499000"
},
{
"input": "1000 999",
"output": "499500"
},
{
"input": "1000 1000",
"output": "500000"
},
{
"input": "3 3",
"output": "5"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "2 2",
"output": "4"
}
] | 1,514,028,752
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 103
| 218
| 23,142,400
|
n, m = sorted(map(int, input().split()))
k = 4 * (m >> 2)
print(m if n == 1 else k + 2 * min(2, m - k) if n == 2 else (m * n + 1 >> 1))
|
Title: Help General
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular *n*<=×<=*m* field, split into *nm* square 1<=×<=1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (*x*1,<=*y*1) and (*x*2,<=*y*2) equals exactly (*x*1<=-<=*x*2)2<=+<=(*y*1<=-<=*y*2)2. Now not all *nm* squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2,<=2), then he cannot put soldiers in the squares (1,<=4), (3,<=4), (4,<=1) and (4,<=3) — each of them will conflict with the soldier in the square (2,<=2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input Specification:
The single line contains space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) that represent the size of the drill exercise field.
Output Specification:
Print the desired maximum number of warriors.
Demo Input:
['2 4\n', '3 4\n']
Demo Output:
['4', '6']
Note:
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
|
```python
n, m = sorted(map(int, input().split()))
k = 4 * (m >> 2)
print(m if n == 1 else k + 2 * min(2, m - k) if n == 2 else (m * n + 1 >> 1))
```
| 3
|
|
859
|
E
|
Desk Disorder
|
PROGRAMMING
| 2,100
|
[
"combinatorics",
"dfs and similar",
"dsu",
"graphs",
"trees"
] | null | null |
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007<==<=109<=+<=7.
|
Input will begin with a line containing *N* (1<=≤<=*N*<=≤<=100000), the number of engineers.
*N* lines follow, each containing exactly two integers. The *i*-th line contains the number of the current desk of the *i*-th engineer and the number of the desk the *i*-th engineer wants to move to. Desks are numbered from 1 to 2·*N*. It is guaranteed that no two engineers sit at the same desk.
|
Print the number of possible assignments, modulo 1000000007<==<=109<=+<=7.
|
[
"4\n1 5\n5 2\n3 7\n7 3\n",
"5\n1 10\n2 10\n3 10\n4 10\n5 5\n"
] |
[
"6\n",
"5\n"
] |
These are the possible assignments for the first example:
- 1 5 3 7 - 1 2 3 7 - 5 2 3 7 - 1 5 7 3 - 1 2 7 3 - 5 2 7 3
| 2,000
|
[
{
"input": "4\n1 5\n5 2\n3 7\n7 3",
"output": "6"
},
{
"input": "5\n1 10\n2 10\n3 10\n4 10\n5 5",
"output": "5"
},
{
"input": "1\n1 2",
"output": "2"
},
{
"input": "30\n22 37\n12 37\n37 58\n29 57\n43 57\n57 58\n58 53\n45 4\n1 4\n4 51\n35 31\n21 31\n31 51\n51 53\n53 48\n60 55\n52 55\n55 33\n36 9\n10 9\n9 33\n33 19\n5 23\n47 23\n23 32\n50 44\n26 44\n44 32\n32 19\n19 48",
"output": "31"
},
{
"input": "50\n73 1\n65 73\n16 65\n57 65\n33 16\n34 57\n98 16\n84 98\n55 34\n64 84\n80 55\n75 64\n28 75\n20 75\n42 75\n88 42\n50 20\n48 28\n32 48\n58 88\n92 76\n76 53\n53 15\n15 1\n1 10\n10 71\n71 37\n37 95\n95 63\n63 92\n45 97\n97 51\n51 96\n96 12\n12 62\n62 31\n31 5\n5 29\n29 19\n19 49\n49 6\n6 40\n40 18\n18 22\n22 17\n17 46\n46 72\n72 82\n82 14\n14 14",
"output": "2"
},
{
"input": "10\n15 8\n8 13\n13 3\n1 4\n14 3\n11 17\n9 10\n10 18\n19 20\n17 20",
"output": "120"
},
{
"input": "4\n5 6\n6 7\n7 8\n8 5",
"output": "2"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "2"
}
] | 1,505,587,282
| 3,982
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 3
| 2,000
| 6,758,400
|
def num_combos(n, arr):
combos = 0
prev_combos = set()
decisions = [0] * n
seats = {}
for i in arr:
seats[i[0]] = True
seats[i[1]] = True
curr_seat = [None] * n
i = 0
while True:
if i == n:
# at the end
combos += 1
i -= 1
elif decisions[i] == 0:
if seats[arr[i][0]]:
seats[arr[i][0]] = False
curr_seat[i] = arr[i][0]
decisions[i] += 1
i += 1
else:
decisions[i] += 1
elif decisions[i] == 1:
if curr_seat[i] is not None:
seats[curr_seat[i]] = True
curr_seat[i] = None
if seats[arr[i][1]] and arr[i][0] != arr[i][1]:
seats[arr[i][1]] = False
curr_seat[i] = arr[i][1]
decisions[i] += 1
i += 1
else:
decisions[i] += 1
else:
# go up a level
if curr_seat[i] is not None:
seats[curr_seat[i]] = True
curr_seat[i] = None
decisions[i] = 0
i -= 1
if i < 0:
return combos
k = int(input().strip())
a = []
for _ in range(k):
a.append(list(map(int, input().strip().split())))
print(num_combos(k, a))
|
Title: Desk Disorder
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007<==<=109<=+<=7.
Input Specification:
Input will begin with a line containing *N* (1<=≤<=*N*<=≤<=100000), the number of engineers.
*N* lines follow, each containing exactly two integers. The *i*-th line contains the number of the current desk of the *i*-th engineer and the number of the desk the *i*-th engineer wants to move to. Desks are numbered from 1 to 2·*N*. It is guaranteed that no two engineers sit at the same desk.
Output Specification:
Print the number of possible assignments, modulo 1000000007<==<=109<=+<=7.
Demo Input:
['4\n1 5\n5 2\n3 7\n7 3\n', '5\n1 10\n2 10\n3 10\n4 10\n5 5\n']
Demo Output:
['6\n', '5\n']
Note:
These are the possible assignments for the first example:
- 1 5 3 7 - 1 2 3 7 - 5 2 3 7 - 1 5 7 3 - 1 2 7 3 - 5 2 7 3
|
```python
def num_combos(n, arr):
combos = 0
prev_combos = set()
decisions = [0] * n
seats = {}
for i in arr:
seats[i[0]] = True
seats[i[1]] = True
curr_seat = [None] * n
i = 0
while True:
if i == n:
# at the end
combos += 1
i -= 1
elif decisions[i] == 0:
if seats[arr[i][0]]:
seats[arr[i][0]] = False
curr_seat[i] = arr[i][0]
decisions[i] += 1
i += 1
else:
decisions[i] += 1
elif decisions[i] == 1:
if curr_seat[i] is not None:
seats[curr_seat[i]] = True
curr_seat[i] = None
if seats[arr[i][1]] and arr[i][0] != arr[i][1]:
seats[arr[i][1]] = False
curr_seat[i] = arr[i][1]
decisions[i] += 1
i += 1
else:
decisions[i] += 1
else:
# go up a level
if curr_seat[i] is not None:
seats[curr_seat[i]] = True
curr_seat[i] = None
decisions[i] = 0
i -= 1
if i < 0:
return combos
k = int(input().strip())
a = []
for _ in range(k):
a.append(list(map(int, input().strip().split())))
print(num_combos(k, a))
```
| 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"
},
{
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] | 1,540,631,890
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 216
| 0
|
n , m = map(int,input().split())
l = sorted(list(map(int,input().split())))
ans = 0
h = 0
for f in range(m) :
ans += (-1 * l[f])
if ans < h :
print(ans + l[f])
break
else : h = ans
print(ans)
|
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 = sorted(list(map(int,input().split())))
ans = 0
h = 0
for f in range(m) :
ans += (-1 * l[f])
if ans < h :
print(ans + l[f])
break
else : h = ans
print(ans)
```
| 0
|
851
|
A
|
Arpa and a research in Mexican wave
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Arpa is researching the Mexican wave.
There are *n* spectators in the stadium, labeled from 1 to *n*. They start the Mexican wave at time 0.
- At time 1, the first spectator stands. - At time 2, the second spectator stands. - ... - At time *k*, the *k*-th spectator stands. - At time *k*<=+<=1, the (*k*<=+<=1)-th spectator stands and the first spectator sits. - At time *k*<=+<=2, the (*k*<=+<=2)-th spectator stands and the second spectator sits. - ... - At time *n*, the *n*-th spectator stands and the (*n*<=-<=*k*)-th spectator sits. - At time *n*<=+<=1, the (*n*<=+<=1<=-<=*k*)-th spectator sits. - ... - At time *n*<=+<=*k*, the *n*-th spectator sits.
Arpa wants to know how many spectators are standing at time *t*.
|
The first line contains three integers *n*, *k*, *t* (1<=≤<=*n*<=≤<=109, 1<=≤<=*k*<=≤<=*n*, 1<=≤<=*t*<=<<=*n*<=+<=*k*).
|
Print single integer: how many spectators are standing at time *t*.
|
[
"10 5 3\n",
"10 5 7\n",
"10 5 12\n"
] |
[
"3\n",
"5\n",
"3\n"
] |
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At *t* = 0 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0. - At *t* = 1 ^--------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 2 ^^-------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 3 ^^^------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 4 ^^^^------ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 5 ^^^^^----- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 6 -^^^^^---- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 7 --^^^^^--- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 8 ---^^^^^-- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 9 ----^^^^^- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 10 -----^^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 11 ------^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 12 -------^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 13 --------^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 14 ---------^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 15 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0.
| 500
|
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{
"input": "100 5 50",
"output": "5"
},
{
"input": "10000 10 10000",
"output": "10"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "6 4 4",
"output": "4"
},
{
"input": "9979797 555554 10101010",
"output": "434341"
},
{
"input": "13 5 12",
"output": "5"
},
{
"input": "9 4 10",
"output": "3"
},
{
"input": "7 5 10",
"output": "2"
},
{
"input": "100000000 10000000 100005000",
"output": "9995000"
},
{
"input": "100000 50000 100001",
"output": "49999"
},
{
"input": "15 10 20",
"output": "5"
},
{
"input": "4 4 5",
"output": "3"
},
{
"input": "5 3 3",
"output": "3"
},
{
"input": "30 5 30",
"output": "5"
},
{
"input": "200000 10 200005",
"output": "5"
},
{
"input": "10 9 12",
"output": "7"
},
{
"input": "10 6 15",
"output": "1"
},
{
"input": "1000000000 10 1000000000",
"output": "10"
},
{
"input": "7 5 11",
"output": "1"
},
{
"input": "9 4 4",
"output": "4"
},
{
"input": "14 3 15",
"output": "2"
},
{
"input": "1000000000 100000000 1000000000",
"output": "100000000"
},
{
"input": "40 10 22",
"output": "10"
},
{
"input": "50 10 51",
"output": "9"
},
{
"input": "999999997 999999995 1999999991",
"output": "1"
},
{
"input": "92 79 144",
"output": "27"
},
{
"input": "8 4 4",
"output": "4"
}
] | 1,585,719,888
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 166
| 140
| 0
|
#-------------Program--------------
#----Kuzlyaev-Nikita-Codeforces----
#-------------Training-------------
#----------------------------------
n,k,t=map(int,input().split())
if t<=k:print(t)
elif t>=n:print(n+k-t)
else:
print(k)
|
Title: Arpa and a research in Mexican wave
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arpa is researching the Mexican wave.
There are *n* spectators in the stadium, labeled from 1 to *n*. They start the Mexican wave at time 0.
- At time 1, the first spectator stands. - At time 2, the second spectator stands. - ... - At time *k*, the *k*-th spectator stands. - At time *k*<=+<=1, the (*k*<=+<=1)-th spectator stands and the first spectator sits. - At time *k*<=+<=2, the (*k*<=+<=2)-th spectator stands and the second spectator sits. - ... - At time *n*, the *n*-th spectator stands and the (*n*<=-<=*k*)-th spectator sits. - At time *n*<=+<=1, the (*n*<=+<=1<=-<=*k*)-th spectator sits. - ... - At time *n*<=+<=*k*, the *n*-th spectator sits.
Arpa wants to know how many spectators are standing at time *t*.
Input Specification:
The first line contains three integers *n*, *k*, *t* (1<=≤<=*n*<=≤<=109, 1<=≤<=*k*<=≤<=*n*, 1<=≤<=*t*<=<<=*n*<=+<=*k*).
Output Specification:
Print single integer: how many spectators are standing at time *t*.
Demo Input:
['10 5 3\n', '10 5 7\n', '10 5 12\n']
Demo Output:
['3\n', '5\n', '3\n']
Note:
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At *t* = 0 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0. - At *t* = 1 ^--------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 2 ^^-------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 3 ^^^------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 4 ^^^^------ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 5 ^^^^^----- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 6 -^^^^^---- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 7 --^^^^^--- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 8 ---^^^^^-- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 9 ----^^^^^- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 10 -----^^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 11 ------^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 12 -------^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 13 --------^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 14 ---------^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 15 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0.
|
```python
#-------------Program--------------
#----Kuzlyaev-Nikita-Codeforces----
#-------------Training-------------
#----------------------------------
n,k,t=map(int,input().split())
if t<=k:print(t)
elif t>=n:print(n+k-t)
else:
print(k)
```
| 3
|
|
479
|
A
|
Expression
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
|
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
|
Print the maximum value of the expression that you can obtain.
|
[
"1\n2\n3\n",
"2\n10\n3\n"
] |
[
"9\n",
"60\n"
] |
none
| 500
|
[
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{
"input": "3\n1\n5",
"output": "20"
},
{
"input": "6\n7\n1",
"output": "48"
},
{
"input": "1\n8\n3",
"output": "27"
},
{
"input": "9\n7\n2",
"output": "126"
},
{
"input": "1\n1\n10",
"output": "20"
},
{
"input": "9\n1\n1",
"output": "18"
},
{
"input": "10\n5\n6",
"output": "300"
},
{
"input": "8\n9\n7",
"output": "504"
},
{
"input": "4\n2\n10",
"output": "80"
},
{
"input": "2\n8\n3",
"output": "48"
},
{
"input": "3\n5\n7",
"output": "105"
},
{
"input": "1\n10\n1",
"output": "12"
},
{
"input": "2\n2\n2",
"output": "8"
},
{
"input": "5\n6\n1",
"output": "35"
},
{
"input": "10\n1\n1",
"output": "20"
},
{
"input": "1\n6\n1",
"output": "8"
},
{
"input": "1\n9\n1",
"output": "11"
},
{
"input": "2\n1\n2",
"output": "6"
},
{
"input": "2\n6\n1",
"output": "14"
},
{
"input": "9\n2\n1",
"output": "27"
},
{
"input": "1\n9\n2",
"output": "20"
},
{
"input": "1\n3\n1",
"output": "5"
},
{
"input": "2\n1\n1",
"output": "4"
}
] | 1,689,002,835
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
x=int(input())
y=int(input())
z=int(input())
values=[x+y+z,
x*y*z,
x+y*z
(x+y)*z,
x*(y+z)
]
value=max(values)
print(value)
|
Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none
|
```python
x=int(input())
y=int(input())
z=int(input())
values=[x+y+z,
x*y*z,
x+y*z
(x+y)*z,
x*(y+z)
]
value=max(values)
print(value)
```
| -1
|
|
299
|
B
|
Ksusha the Squirrel
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Ksusha the Squirrel is standing at the beginning of a straight road, divided into *n* sectors. The sectors are numbered 1 to *n*, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector *n*. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector *i* to any of the sectors *i*<=+<=1,<=*i*<=+<=2,<=...,<=*i*<=+<=*k*.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
|
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=3·105,<=1<=≤<=*k*<=≤<=3·105). The next line contains *n* characters — the description of the road: the *i*-th character equals ".", if the *i*-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
|
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
|
[
"2 1\n..\n",
"5 2\n.#.#.\n",
"7 3\n.#.###.\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 1,000
|
[
{
"input": "2 1\n..",
"output": "YES"
},
{
"input": "5 2\n.#.#.",
"output": "YES"
},
{
"input": "7 3\n.#.###.",
"output": "NO"
},
{
"input": "2 200\n..",
"output": "YES"
},
{
"input": "2 1\n..",
"output": "YES"
},
{
"input": "2 2\n..",
"output": "YES"
},
{
"input": "2 100000\n..",
"output": "YES"
},
{
"input": "3 1\n.#.",
"output": "NO"
},
{
"input": "3 2\n.#.",
"output": "YES"
},
{
"input": "3 10000\n.#.",
"output": "YES"
}
] | 1,660,297,965
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 218
| 819,200
|
n, m = map(int, input(). split())
s = input()
ans = 'YES'
d = 0
for i in range(n):
if s[i] == '#':
d += 1
else:
if d >= m:
ans = 'NO'
break
d = 0
print(ans)
|
Title: Ksusha the Squirrel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksusha the Squirrel is standing at the beginning of a straight road, divided into *n* sectors. The sectors are numbered 1 to *n*, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector *n*. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector *i* to any of the sectors *i*<=+<=1,<=*i*<=+<=2,<=...,<=*i*<=+<=*k*.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=3·105,<=1<=≤<=*k*<=≤<=3·105). The next line contains *n* characters — the description of the road: the *i*-th character equals ".", if the *i*-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
Output Specification:
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
Demo Input:
['2 1\n..\n', '5 2\n.#.#.\n', '7 3\n.#.###.\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n, m = map(int, input(). split())
s = input()
ans = 'YES'
d = 0
for i in range(n):
if s[i] == '#':
d += 1
else:
if d >= m:
ans = 'NO'
break
d = 0
print(ans)
```
| 3
|
|
1,008
|
A
|
Romaji
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
|
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
|
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
|
[
"sumimasen\n",
"ninja\n",
"codeforces\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
| 500
|
[
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"input": "n",
"output": "YES"
},
{
"input": "necnei",
"output": "NO"
},
{
"input": "nternn",
"output": "NO"
},
{
"input": "aucunuohja",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "b",
"output": "NO"
},
{
"input": "nn",
"output": "YES"
},
{
"input": "nnnzaaa",
"output": "YES"
},
{
"input": "zn",
"output": "NO"
},
{
"input": "ab",
"output": "NO"
},
{
"input": "aaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaab",
"output": "NO"
},
{
"input": "aaaaaaaaan",
"output": "YES"
},
{
"input": "baaaaaaaaa",
"output": "YES"
},
{
"input": "naaaaaaaaa",
"output": "YES"
},
{
"input": "nbaaaaaaaa",
"output": "YES"
},
{
"input": "bbaaaaaaaa",
"output": "NO"
},
{
"input": "bnaaaaaaaa",
"output": "NO"
},
{
"input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu",
"output": "YES"
},
{
"input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino",
"output": "NO"
},
{
"input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul",
"output": "NO"
},
{
"input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu",
"output": "YES"
},
{
"input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo",
"output": "NO"
},
{
"input": "y",
"output": "NO"
},
{
"input": "by",
"output": "NO"
},
{
"input": "yy",
"output": "NO"
},
{
"input": "nbn",
"output": "NO"
},
{
"input": "nz",
"output": "NO"
},
{
"input": "king",
"output": "NO"
},
{
"input": "g",
"output": "NO"
},
{
"input": "az",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "yn",
"output": "NO"
},
{
"input": "aeo",
"output": "YES"
},
{
"input": "nb",
"output": "NO"
},
{
"input": "npn",
"output": "NO"
},
{
"input": "kini",
"output": "YES"
},
{
"input": "pya",
"output": "NO"
},
{
"input": "m",
"output": "NO"
},
{
"input": "p",
"output": "NO"
},
{
"input": "aaaaaaaak",
"output": "NO"
},
{
"input": "aab",
"output": "NO"
},
{
"input": "d",
"output": "NO"
},
{
"input": "at",
"output": "NO"
},
{
"input": "aaaaaak",
"output": "NO"
},
{
"input": "aaz",
"output": "NO"
},
{
"input": "aaab",
"output": "NO"
},
{
"input": "s",
"output": "NO"
},
{
"input": "nzzen",
"output": "NO"
},
{
"input": "aeionnhhhn",
"output": "NO"
},
{
"input": "h",
"output": "NO"
},
{
"input": "ny",
"output": "NO"
}
] | 1,637,341,793
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 62
| 0
|
s=input()
L=len(s)
result = 'YES'
v=['a','e','i','o','u']
for i in range(L-1):
if s[i] in v :
continue
elif s[i] in 'n' :
continue
elif s[i+1] not in v :
result = 'NO'
print(result)
|
Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
|
```python
s=input()
L=len(s)
result = 'YES'
v=['a','e','i','o','u']
for i in range(L-1):
if s[i] in v :
continue
elif s[i] in 'n' :
continue
elif s[i+1] not in v :
result = 'NO'
print(result)
```
| 0
|
|
928
|
A
|
Login Verification
|
PROGRAMMING
| 1,200
|
[
"*special",
"strings"
] | null | null |
When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc.
Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types:
- transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other.
For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not.
You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones.
|
The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50 — the login itself.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins.
The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar.
|
Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it.
Otherwise print «No» (without quotes).
|
[
"1_wat\n2\n2_wat\nwat_1\n",
"000\n3\n00\nooA\noOo\n",
"_i_\n3\n__i_\n_1_\nI\n",
"La0\n3\n2a0\nLa1\n1a0\n",
"abc\n1\naBc\n",
"0Lil\n2\nLIL0\n0Ril\n"
] |
[
"Yes\n",
"No\n",
"No\n",
"No\n",
"No\n",
"Yes\n"
] |
In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing.
In the third sample case the new login is similar with the second one.
| 500
|
[
{
"input": "1_wat\n2\n2_wat\nwat_1",
"output": "Yes"
},
{
"input": "000\n3\n00\nooA\noOo",
"output": "No"
},
{
"input": "_i_\n3\n__i_\n_1_\nI",
"output": "No"
},
{
"input": "La0\n3\n2a0\nLa1\n1a0",
"output": "No"
},
{
"input": "abc\n1\naBc",
"output": "No"
},
{
"input": "0Lil\n2\nLIL0\n0Ril",
"output": "Yes"
},
{
"input": "iloO\n3\niIl0\noIl0\nIooO",
"output": "Yes"
},
{
"input": "L1il0o1L1\n5\niLLoLL\noOI1Io10il\nIoLLoO\nO01ilOoI\nI10l0o",
"output": "Yes"
},
{
"input": "ELioO1lOoOIOiLoooi1iolul1O\n7\nOoEIuOIl1ui1010uiooOoi0Oio001L0EoEolO0\nOLIoOEuoE11u1u1iLOI0oO\nuEOuO0uIOOlO01OlEI0E1Oo0IO1LI0uE0LILO0\nEOo0Il11iIOOOIiuOiIiiLOLEOOII001EE\niOoO0LOulioE0OLIIIulli01OoiuOOOoOlEiI0EiiElIIu0\nlE1LOE1Oil\n1u0EOliIiIOl1u110il0l1O0u",
"output": "Yes"
},
{
"input": "0blo7X\n20\n1oobb6\nXIXIO2X\n2iYI2\n607XXol\n2I6io22\nOl10I\nbXX0Lo\nolOOb7X\n07LlXL\nlXY17\n12iIX2\n7lL70\nbOo11\n17Y6b62\n0O6L7\n1lX2L\n2iYl6lI\n7bXIi1o\niLIY2\n0OIo1X",
"output": "Yes"
},
{
"input": "lkUL\n25\nIIfL\nokl\nfoo\ni0U\noko\niIoU\nUUv\nvli\nv0Uk\n0Of\niill\n1vkl\nUIf\nUfOO\nlvLO\nUUo0\nIOf1\nlovL\nIkk\noIv\nLvfU\n0UI\nkol\n1OO0\n1OOi",
"output": "Yes"
},
{
"input": "L1lo\n3\nOOo1\nL1lo\n0lOl",
"output": "No"
},
{
"input": "LIoooiLO\n5\nLIoooiLO\nl0o01I00\n0OOl0lLO01\nil10i0\noiloi",
"output": "No"
},
{
"input": "1i1lQI\n7\nuLg1uLLigIiOLoggu\nLLLgIuQIQIIloiQuIIoIO0l0o000\n0u1LQu11oIuooIl0OooLg0i0IQu1O1lloI1\nQuQgIQi0LOIliLOuuuioLQou1l\nlLIO00QLi01LogOliOIggII1\no0Ll1uIOQl10IL0IILQ\n1i1lQI",
"output": "No"
},
{
"input": "oIzz1\n20\n1TTl0O\nloF0LT\n1lLzo\noi0Ov\nFlIF1zT\nzoITzx\n0TIFlT\nl1vllil\nOviix1F\nLFvI1lL\nLIl0loz\nixz1v\n1i1vFi\nTIFTol\noIzz1\nIvTl0o\nxv1U0O\niiiioF\n1oiLUlO\nxToxv1",
"output": "No"
},
{
"input": "00L0\n25\n0il\nIlkZ\nL0I\n00L0\nBd0\nZLd\n0d1k\nddk\nIdl\nkBd\nkBOL\nZ1lI\nkBL\nLOko\noZ0i\nZ1lO\nLiOk\niBld\nLO0d\ndIo\nZ10\n1k1i\n0o0L\nIoBd\ni0B0",
"output": "No"
},
{
"input": "Z\n1\nz",
"output": "No"
},
{
"input": "0\n1\no",
"output": "No"
},
{
"input": "0\n1\nO",
"output": "No"
},
{
"input": "o\n1\n0",
"output": "No"
},
{
"input": "o\n1\nO",
"output": "No"
},
{
"input": "o\n1\no",
"output": "No"
},
{
"input": "O\n1\no",
"output": "No"
},
{
"input": "O\n1\n0",
"output": "No"
},
{
"input": "1\n1\nl",
"output": "No"
},
{
"input": "1\n1\nL",
"output": "No"
},
{
"input": "1\n1\ni",
"output": "No"
},
{
"input": "1\n1\nI",
"output": "No"
},
{
"input": "1\n1\no",
"output": "Yes"
},
{
"input": "i\n1\n1",
"output": "No"
},
{
"input": "i\n1\nL",
"output": "No"
},
{
"input": "i\n1\nl",
"output": "No"
},
{
"input": "I\n1\nL",
"output": "No"
},
{
"input": "I\n1\nl",
"output": "No"
},
{
"input": "I\n1\n1",
"output": "No"
},
{
"input": "l\n1\n1",
"output": "No"
},
{
"input": "l\n1\nL",
"output": "No"
},
{
"input": "l\n1\ni",
"output": "No"
},
{
"input": "l\n1\nI",
"output": "No"
},
{
"input": "L\n1\nl",
"output": "No"
},
{
"input": "L\n1\n1",
"output": "No"
},
{
"input": "L\n1\ni",
"output": "No"
},
{
"input": "L\n1\nI",
"output": "No"
}
] | 1,519,489,781
| 3,281
|
Python 3
|
OK
|
TESTS
| 73
| 77
| 5,632,000
|
def checkEquals( a , b):
if a == b :
return True
if a == b.upper() or b == a.upper():
return True
if (a == '0' or a == 'O' or a == 'o') and (b == '0' or b == 'O' or b == 'o') :
return True
if (a == '1' or a == 'i' or a == 'I' or a == 'l' or a == 'L') and (b == '1' or b == 'i' or b == 'I' or b == 'l' or b == 'L'):
return True
return False
def cheakName( name1 , name2):
if len(name1) != len(name2) : return False
for i in range(len(name1)):
if not checkEquals(name1[i] , name2[i]):
return False
return True
newName = input()
N = int(input())
mass = []
for i in range(N):
mass.append(input())
flag = False
for name in mass :
if cheakName(newName , name) : flag = True
if flag : print('No')
else: print('Yes')
|
Title: Login Verification
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc.
Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types:
- transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other.
For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not.
You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones.
Input Specification:
The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50 — the login itself.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins.
The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar.
Output Specification:
Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it.
Otherwise print «No» (without quotes).
Demo Input:
['1_wat\n2\n2_wat\nwat_1\n', '000\n3\n00\nooA\noOo\n', '_i_\n3\n__i_\n_1_\nI\n', 'La0\n3\n2a0\nLa1\n1a0\n', 'abc\n1\naBc\n', '0Lil\n2\nLIL0\n0Ril\n']
Demo Output:
['Yes\n', 'No\n', 'No\n', 'No\n', 'No\n', 'Yes\n']
Note:
In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing.
In the third sample case the new login is similar with the second one.
|
```python
def checkEquals( a , b):
if a == b :
return True
if a == b.upper() or b == a.upper():
return True
if (a == '0' or a == 'O' or a == 'o') and (b == '0' or b == 'O' or b == 'o') :
return True
if (a == '1' or a == 'i' or a == 'I' or a == 'l' or a == 'L') and (b == '1' or b == 'i' or b == 'I' or b == 'l' or b == 'L'):
return True
return False
def cheakName( name1 , name2):
if len(name1) != len(name2) : return False
for i in range(len(name1)):
if not checkEquals(name1[i] , name2[i]):
return False
return True
newName = input()
N = int(input())
mass = []
for i in range(N):
mass.append(input())
flag = False
for name in mass :
if cheakName(newName , name) : flag = True
if flag : print('No')
else: print('Yes')
```
| 3
|
|
268
|
A
|
Games
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
|
In a single line print the number of games where the host team is going to play in the guest uniform.
|
[
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] |
[
"1\n",
"5\n",
"0\n"
] |
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
| 500
|
[
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28",
"output": "1"
},
{
"input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77",
"output": "6"
},
{
"input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35",
"output": "4"
},
{
"input": "2\n46 6\n6 46",
"output": "2"
},
{
"input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11",
"output": "10"
},
{
"input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33",
"output": "6"
},
{
"input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73",
"output": "0"
},
{
"input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2",
"output": "1"
},
{
"input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100",
"output": "8"
},
{
"input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "450"
},
{
"input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3",
"output": "0"
},
{
"input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5",
"output": "20"
},
{
"input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2",
"output": "108"
},
{
"input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2",
"output": "312"
},
{
"input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6",
"output": "96"
},
{
"input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97",
"output": "210"
},
{
"input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78",
"output": "73"
},
{
"input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80",
"output": "277"
},
{
"input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21",
"output": "154"
},
{
"input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17",
"output": "418"
},
{
"input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80",
"output": "74"
},
{
"input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9",
"output": "59"
},
{
"input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67",
"output": "100"
},
{
"input": "4\n8 7\n8 7\n7 8\n7 8",
"output": "8"
},
{
"input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1",
"output": "10"
},
{
"input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "72"
},
{
"input": "4\n1 2\n1 2\n2 1\n2 1",
"output": "8"
}
] | 1,695,898,708
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 31
| 124
| 0
|
import sys
n = int(sys.stdin.readline().strip())
h_count = {}
g_count = {}
for _ in range(n):
h, g = list(map(int, sys.stdin.readline().strip().split()))
h_count[h] = h_count.get(h, 0) + 1
g_count[g] = g_count.get(g, 0) + 1
ans = 0
for h, h_num in h_count.items():
if h in g_count:
ans += h_num * g_count[h]
print(ans)
|
Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
|
```python
import sys
n = int(sys.stdin.readline().strip())
h_count = {}
g_count = {}
for _ in range(n):
h, g = list(map(int, sys.stdin.readline().strip().split()))
h_count[h] = h_count.get(h, 0) + 1
g_count[g] = g_count.get(g, 0) + 1
ans = 0
for h, h_num in h_count.items():
if h in g_count:
ans += h_num * g_count[h]
print(ans)
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,680,081,871
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
def change_case(word):
upper_count = 0
lower_count = 0
for letter in word:
if letter.isupper():
upper_count += 1
else:
lower_count += 1
if upper_count > lower_count:
return word.upper()
else:
return word.lower()
word = "HoUse"
print(change_case(word))
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
def change_case(word):
upper_count = 0
lower_count = 0
for letter in word:
if letter.isupper():
upper_count += 1
else:
lower_count += 1
if upper_count > lower_count:
return word.upper()
else:
return word.lower()
word = "HoUse"
print(change_case(word))
```
| 0
|
509
|
A
|
Maximum in Table
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
|
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
|
Print a single line containing a positive integer *m* — the maximum value in the table.
|
[
"1\n",
"5\n"
] |
[
"1",
"70"
] |
In the second test the rows of the table look as follows:
| 0
|
[
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
},
{
"input": "9",
"output": "12870"
},
{
"input": "10",
"output": "48620"
},
{
"input": "8",
"output": "3432"
}
] | 1,621,354,791
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 46
| 0
|
n=int(input())
lst=[]
final=[]
for i in range(n):
ls=[]
for j in range(n):
if i==0 and j==0:
ls.append(1)
else:
temp=lst[i-1][j]+ls[j-1]
ls.append(temp)
lst.append(ls)
final.extend(ls)
print(max(final))
|
Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows:
|
```python
n=int(input())
lst=[]
final=[]
for i in range(n):
ls=[]
for j in range(n):
if i==0 and j==0:
ls.append(1)
else:
temp=lst[i-1][j]+ls[j-1]
ls.append(temp)
lst.append(ls)
final.extend(ls)
print(max(final))
```
| -1
|
|
598
|
F
|
Cut Length
|
PROGRAMMING
| 2,900
|
[
"geometry"
] | null | null |
Given simple (without self-intersections) *n*-gon. It is not necessary convex. Also you are given *m* lines. For each line find the length of common part of the line and the *n*-gon.
The boundary of *n*-gon belongs to polygon. It is possible that *n*-gon contains 180-degree angles.
|
The first line contains integers *n* and *m* (3<=≤<=*n*<=≤<=1000;1<=≤<=*m*<=≤<=100). The following *n* lines contain coordinates of polygon vertices (in clockwise or counterclockwise direction). All vertices are distinct.
The following *m* lines contain line descriptions. Each of them contains two distict points of a line by their coordinates.
All given in the input coordinates are real numbers, given with at most two digits after decimal point. They do not exceed 105 by absolute values.
|
Print *m* lines, the *i*-th line should contain the length of common part of the given *n*-gon and the *i*-th line. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"4 3\n0 0\n1 0\n1 1\n0 1\n0 0 1 1\n0 0 0 1\n0 0 1 -1\n"
] |
[
"1.41421356237309514547\n1.00000000000000000000\n0.00000000000000000000\n"
] |
none
| 0
|
[
{
"input": "4 3\n0 0\n1 0\n1 1\n0 1\n0 0 1 1\n0 0 0 1\n0 0 1 -1",
"output": "1.41421356237309514547\n1.00000000000000000000\n0.00000000000000000000"
},
{
"input": "4 9\n0 0\n0 1\n1 1\n1 0\n0 0 1 1\n1 1 0 0\n0 0 1 0\n0 0 0.5 0\n0 0.5 1 0.5\n0 1 1 1\n1 1 1 0\n0.75 0.75 0.75 0.25\n0 0.25 1 0.75",
"output": "1.41421356237309514547\n1.41421356237309514547\n1.00000000000000000000\n1.00000000000000000000\n1.00000000000000000000\n1.00000000000000000000\n1.00000000000000000000\n1.00000000000000000000\n1.11803398874989490253"
},
{
"input": "9 7\n0 0\n0 2\n1 1\n2 2\n3 1\n4 2\n5 1\n6 2\n6 0\n0 1 6 1\n0 1.5 6 1.5\n0 2 6 1\n0 1 6 2\n0 2 6 2\n0 0 0 2\n0 0 6 0",
"output": "6.00000000000000000000\n3.00000000000000000000\n2.95448465757342138005\n2.95448465757342093596\n0.00000000000000000000\n2.00000000000000000000\n6.00000000000000000000"
},
{
"input": "12 8\n0 0\n10000 0\n10000 5002\n2 5002\n2 9998\n10000 9998\n10000 10000\n0 10000\n0 5000\n9998 5000\n9998 2\n0 2\n5000 -1 5000 10001\n0 0 10000 10000\n0 0 1 1\n0 0 2 2\n0 5000 5000 5000\n-2 0 10002 10000\n-2 10000 10002 0\n-10000 0 20000 10000",
"output": "6.00000000000000000000\n8.48528137423856954058\n8.48528137423856954058\n8.48528137423856954058\n10000.00000000000000000000\n2.83125515610529765453\n8.48471631080716015560\n6.32455532033675815740"
},
{
"input": "12 7\n0 2\n9998 2\n9998 5000\n0 5000\n0 10000\n10000 10000\n10000 9998\n2 9998\n2 5002\n10000 5002\n10000 0\n0 0\n5000 -1 5000 10001\n0 0 10000 10000\n0 0 1 1\n0 5000 5000 5000\n-2 0 10002 10000\n-2 10000 10002 0\n-10000 0 20000 10000",
"output": "6.00000000000000000000\n8.48528137423856954058\n8.48528137423856954058\n10000.00000000000000000000\n2.83125515610529765453\n8.48471631080716015560\n6.32455532033675815740"
},
{
"input": "9 5\n0 0\n6 0\n6 2\n5 1\n4 2\n3 1\n2 2\n1 1\n0 2\n-1 2 7.5 1\n0 1 6 1\n0 1.5 6 1.5\n0 2 6 1\n0 0 0 2",
"output": "2.79811274979359403403\n6.00000000000000000000\n3.00000000000000000000\n2.95448465757342138005\n2.00000000000000000000"
},
{
"input": "4 5\n0 0\n1 0\n1 1\n0 1\n0 0 1 1\n0 0 0 1\n0 0 1 -1\n0 0 100000 99999\n-100000 -100000 100000 99999",
"output": "1.41421356237309514547\n1.00000000000000000000\n0.00000000000000000000\n1.41420649132296061268\n0.70710147787905963668"
},
{
"input": "9 5\n0 0\n0 2\n1 1\n2 2\n3 1\n4 2\n5 1\n6 2\n6 0\n-1 2 7.5 1\n0 1 6 1\n0 1.5 6 1.5\n0 2 6 1\n0 0 0 2",
"output": "2.79811274979359403403\n6.00000000000000000000\n3.00000000000000000000\n2.95448465757342138005\n2.00000000000000000000"
},
{
"input": "5 1\n0 0\n-1 1\n2 1\n2 -1\n1 0\n100 0 101 0",
"output": "2.00000000000000000000"
},
{
"input": "3 1\n100000 100000\n-100000 -100000\n100000 -100000\n1.15 99999.99 1.16 99999.99",
"output": "0.01000000000000000021"
},
{
"input": "3 1\n100000 100000\n-100000 -100000\n100000 -100000\n0 99999.99 0.01 99999.99",
"output": "0.01000000000000000021"
},
{
"input": "3 1\n99999.99 -100000\n-100000 100000\n100000 -100000\n99999.99 -100000 -100000 100000",
"output": "282842.70540355128468945622"
},
{
"input": "3 3\n100000 100000\n-99999.99 -100000\n-100000 -100000\n100000 100000 -99999.99 -100000\n-99999.99 -100000 -100000 -100000\n-100000 -100000 -100000 -99999.99",
"output": "282842.70540355128468945622\n0.01000000000000000021\n0.00000000000000000000"
},
{
"input": "3 7\n99999.99 -100000\n-100000 100000\n-99999.99 100000\n1 1 1.01 1.01\n99999.99 -100000 -100000 100000\n-99999.99 100000 -100000 100000\n99999.99 -100000 -99999.99 100000\n-100000 100000 99999.99 -100000\n1 1 2 2\n-100000 100000 100000 -100000",
"output": "0.00353553399432175773\n282842.70540355128468945622\n0.01000000000000000021\n282842.69833248376380652189\n282842.70540355128468945622\n0.00353553399432175860\n141421.35623730949009768665"
},
{
"input": "3 1\n-99999.99 100000\n100000 -100000\n100000 -99999.99\n-99999.99 100000 100000 -100000",
"output": "282842.70540355128468945622"
},
{
"input": "3 1\n2 2\n100000 100000\n-99999.99 -100000\n100000 100000 -99999.99 -100000",
"output": "282842.70540355128468945622"
},
{
"input": "3 1\n-100000 -100000\n100000 -100000\n0 1\n-100000 -100000 100000 -99999",
"output": "199999.00001749978400766850"
},
{
"input": "3 8\n-100000 -100000\n100000 100000\n0 1\n-100000 -100000 100000 -80000\n-100000 -100000 100000 -99000\n-100000 -100000 100000 -99990\n-100000 -100000 100000 -99999\n-100000 -100000 100000 -99999.99\n-100000 -100000 100000 100000\n-100000 -100000 100000 99999.99\n-100000 100000 100000 -100000",
"output": "0.00000000000000000000\n0.00000000000000000000\n0.00000000000000000000\n0.00000000000000000000\n0.00000000000000000000\n282842.71247461898019537330\n0.00000000000000000000\n0.70710324567031901388"
},
{
"input": "3 8\n-100000 -100000\n100000 100000\n1 0\n-100000 -100000 100000 -80000\n-100000 -100000 100000 -99000\n-100000 -100000 100000 -99990\n-100000 -100000 100000 -99999\n-100000 -100000 100000 -99999.99\n-100000 -100000 100000 100000\n-100000 -100000 100000 99999.99\n-100000 100000 100000 -100000",
"output": "0.00000000000000000000\n0.00000000000000000000\n0.00000000000000000000\n0.00000000000000000000\n0.00000000000000000000\n282842.71247461898019537330\n281435.54176649585133418441\n0.70710324567031901388"
}
] | 1,680,278,469
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
from sys import stdin
from itertools import cycle
import math
def read_input():
n, m = map(int, input().split())
vertices = [tuple(map(float, input().split())) for _ in range(n)]
lines = [tuple(map(float, input().split())) for _ in range(m * 2)]
lines = [(lines[2 * i], lines[2 * i + 1]) for i in range(m)]
return n, m, vertices, lines
def intersection(p1, p2, p3, p4):
d = (p2[0] - p1[0]) * (p4[1] - p3[1]) - (p2[1] - p1[1]) * (p4[0] - p3[0])
if d == 0:
return None
u = ((p3[0] - p1[0]) * (p4[1] - p3[1]) - (p3[1] - p1[1]) * (p4[0] - p3[0])) / d
v = ((p3[0] - p1[0]) * (p2[1] - p1[1]) - (p3[1] - p1[1]) * (p2[0] - p1[0])) / d
if 0 <= u <= 1 and 0 <= v <= 1:
return (p1[0] + u * (p2[0] - p1[0]), p1[1] + u * (p2[1] - p1[1]))
return None
def is_inside(p, vertices):
count = 0
n = len(vertices)
for i in range(n):
p1, p2 = vertices[i], vertices[(i + 1) % n]
if p1[1] <= p[1] and p2[1] > p[1] or p1[1] > p[1] and p2[1] <= p[1]:
if p1[0] + (p[1] - p1[1]) / (p2[1] - p1[1]) * (p2[0] - p1[0]) < p[0]:
count += 1
return count % 2 == 1
def main():
n, m, vertices, lines = read_input()
for line in lines:
intersections = []
for p1, p2 in zip(vertices, cycle(vertices[1:])):
p = intersection(p1, p2, *line)
if p is not None:
intersections.append(p)
intersections.sort(key=lambda p: (p[0] - line[0][0]) * (line[1][1] - line[0][1]) - (p[1] - line[0][1]) * (line[1][0] - line[0][0]))
total_length = 0
for p1, p2 in zip(intersections[::2], intersections[1::2]):
total_length += math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2)
print("{:.15f}".format(total_length))
if __name__ == "__main__":
main()
|
Title: Cut Length
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Given simple (without self-intersections) *n*-gon. It is not necessary convex. Also you are given *m* lines. For each line find the length of common part of the line and the *n*-gon.
The boundary of *n*-gon belongs to polygon. It is possible that *n*-gon contains 180-degree angles.
Input Specification:
The first line contains integers *n* and *m* (3<=≤<=*n*<=≤<=1000;1<=≤<=*m*<=≤<=100). The following *n* lines contain coordinates of polygon vertices (in clockwise or counterclockwise direction). All vertices are distinct.
The following *m* lines contain line descriptions. Each of them contains two distict points of a line by their coordinates.
All given in the input coordinates are real numbers, given with at most two digits after decimal point. They do not exceed 105 by absolute values.
Output Specification:
Print *m* lines, the *i*-th line should contain the length of common part of the given *n*-gon and the *i*-th line. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['4 3\n0 0\n1 0\n1 1\n0 1\n0 0 1 1\n0 0 0 1\n0 0 1 -1\n']
Demo Output:
['1.41421356237309514547\n1.00000000000000000000\n0.00000000000000000000\n']
Note:
none
|
```python
from sys import stdin
from itertools import cycle
import math
def read_input():
n, m = map(int, input().split())
vertices = [tuple(map(float, input().split())) for _ in range(n)]
lines = [tuple(map(float, input().split())) for _ in range(m * 2)]
lines = [(lines[2 * i], lines[2 * i + 1]) for i in range(m)]
return n, m, vertices, lines
def intersection(p1, p2, p3, p4):
d = (p2[0] - p1[0]) * (p4[1] - p3[1]) - (p2[1] - p1[1]) * (p4[0] - p3[0])
if d == 0:
return None
u = ((p3[0] - p1[0]) * (p4[1] - p3[1]) - (p3[1] - p1[1]) * (p4[0] - p3[0])) / d
v = ((p3[0] - p1[0]) * (p2[1] - p1[1]) - (p3[1] - p1[1]) * (p2[0] - p1[0])) / d
if 0 <= u <= 1 and 0 <= v <= 1:
return (p1[0] + u * (p2[0] - p1[0]), p1[1] + u * (p2[1] - p1[1]))
return None
def is_inside(p, vertices):
count = 0
n = len(vertices)
for i in range(n):
p1, p2 = vertices[i], vertices[(i + 1) % n]
if p1[1] <= p[1] and p2[1] > p[1] or p1[1] > p[1] and p2[1] <= p[1]:
if p1[0] + (p[1] - p1[1]) / (p2[1] - p1[1]) * (p2[0] - p1[0]) < p[0]:
count += 1
return count % 2 == 1
def main():
n, m, vertices, lines = read_input()
for line in lines:
intersections = []
for p1, p2 in zip(vertices, cycle(vertices[1:])):
p = intersection(p1, p2, *line)
if p is not None:
intersections.append(p)
intersections.sort(key=lambda p: (p[0] - line[0][0]) * (line[1][1] - line[0][1]) - (p[1] - line[0][1]) * (line[1][0] - line[0][0]))
total_length = 0
for p1, p2 in zip(intersections[::2], intersections[1::2]):
total_length += math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2)
print("{:.15f}".format(total_length))
if __name__ == "__main__":
main()
```
| -1
|
|
389
|
A
|
Fox and Number Game
|
PROGRAMMING
| 1,000
|
[
"greedy",
"math"
] | null | null |
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
|
Output a single integer — the required minimal sum.
|
[
"2\n1 2\n",
"3\n2 4 6\n",
"2\n12 18\n",
"5\n45 12 27 30 18\n"
] |
[
"2\n",
"6\n",
"12\n",
"15\n"
] |
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
| 500
|
[
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 4 6",
"output": "6"
},
{
"input": "2\n12 18",
"output": "12"
},
{
"input": "5\n45 12 27 30 18",
"output": "15"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n100 100",
"output": "200"
},
{
"input": "2\n87 58",
"output": "58"
},
{
"input": "39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52",
"output": "2028"
},
{
"input": "59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96",
"output": "5664"
},
{
"input": "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": "10000"
},
{
"input": "100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98",
"output": "700"
},
{
"input": "39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42",
"output": "819"
},
{
"input": "59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70",
"output": "4130"
},
{
"input": "87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22",
"output": "1914"
},
{
"input": "15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63",
"output": "945"
},
{
"input": "39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91",
"output": "273"
},
{
"input": "18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36",
"output": "324"
},
{
"input": "46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71",
"output": "3266"
},
{
"input": "70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33",
"output": "770"
},
{
"input": "10\n60 12 96 48 60 24 60 36 60 60",
"output": "120"
},
{
"input": "20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "1020"
},
{
"input": "50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58",
"output": "2900"
},
{
"input": "98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90",
"output": "980"
},
{
"input": "100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82",
"output": "8200"
},
{
"input": "100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70",
"output": "100"
},
{
"input": "2\n100 1",
"output": "2"
},
{
"input": "2\n18 30",
"output": "12"
},
{
"input": "2\n3 5",
"output": "2"
},
{
"input": "2\n1 10",
"output": "2"
},
{
"input": "2\n8 5",
"output": "2"
},
{
"input": "5\n2 3 5 8 18",
"output": "5"
},
{
"input": "5\n2 4 1 6 8",
"output": "5"
},
{
"input": "3\n12 10 5",
"output": "3"
},
{
"input": "3\n6 10 15",
"output": "3"
}
] | 1,391,534,146
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
def solve():
n = int(raw_input())
inp = raw_input().split()
arr = [int(w) for w in inp]
arr.sort()
ans = reduce(gcd, arr)
ans *= n
print ans
solve()
|
Title: Fox and Number Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output Specification:
Output a single integer — the required minimal sum.
Demo Input:
['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n']
Demo Output:
['2\n', '6\n', '12\n', '15\n']
Note:
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
|
```python
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
def solve():
n = int(raw_input())
inp = raw_input().split()
arr = [int(w) for w in inp]
arr.sort()
ans = reduce(gcd, arr)
ans *= n
print ans
solve()
```
| -1
|
|
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"
}
] | 1,589,200,134
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 248
| 7,065,600
|
import sys
import collections
import math
import itertools as it
def readArray(type= int):
line = input()
return [type(x) for x in line.split()]
def solve():
n = int(input())
points = []
for x in range(n):
points.append(readArray())
xfreq = collections.defaultdict(list)
yfreq = collections.defaultdict(list)
for p in points:
xfreq[p[0]].append(p)
yfreq[p[1]].append(p)
cc = 0
for p in points:
nb = 0
# upper neighbours
for q in xfreq[p[0]]:
if(q[1] > p[1]):
nb+= 1
break
# lower neighbours
for q in xfreq[p[0]]:
if (q[1] < p[1]):
nb += 1
break
# left neighbours
for q in yfreq[p[1]]:
if (q[0] < p[0]):
nb += 1
break
# right neighbours
for q in yfreq[p[1]]:
if (q[0] > p[0]):
nb += 1
break
if nb == 4:
cc+= 1
print(cc)
if __name__ == '__main__':
solve()
|
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
import sys
import collections
import math
import itertools as it
def readArray(type= int):
line = input()
return [type(x) for x in line.split()]
def solve():
n = int(input())
points = []
for x in range(n):
points.append(readArray())
xfreq = collections.defaultdict(list)
yfreq = collections.defaultdict(list)
for p in points:
xfreq[p[0]].append(p)
yfreq[p[1]].append(p)
cc = 0
for p in points:
nb = 0
# upper neighbours
for q in xfreq[p[0]]:
if(q[1] > p[1]):
nb+= 1
break
# lower neighbours
for q in xfreq[p[0]]:
if (q[1] < p[1]):
nb += 1
break
# left neighbours
for q in yfreq[p[1]]:
if (q[0] < p[0]):
nb += 1
break
# right neighbours
for q in yfreq[p[1]]:
if (q[0] > p[0]):
nb += 1
break
if nb == 4:
cc+= 1
print(cc)
if __name__ == '__main__':
solve()
```
| 3
|
|
950
|
A
|
Left-handers, Right-handers and Ambidexters
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
|
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
|
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
|
[
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] |
[
"6\n",
"14\n",
"0\n"
] |
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
| 500
|
[
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "100 100 100",
"output": "300"
},
{
"input": "1 1 1",
"output": "2"
},
{
"input": "30 70 35",
"output": "130"
},
{
"input": "89 44 76",
"output": "208"
},
{
"input": "0 100 100",
"output": "200"
},
{
"input": "100 0 100",
"output": "200"
},
{
"input": "100 1 100",
"output": "200"
},
{
"input": "1 100 100",
"output": "200"
},
{
"input": "100 100 0",
"output": "200"
},
{
"input": "100 100 1",
"output": "200"
},
{
"input": "1 2 1",
"output": "4"
},
{
"input": "0 0 100",
"output": "100"
},
{
"input": "0 100 0",
"output": "0"
},
{
"input": "100 0 0",
"output": "0"
},
{
"input": "10 8 7",
"output": "24"
},
{
"input": "45 47 16",
"output": "108"
},
{
"input": "59 43 100",
"output": "202"
},
{
"input": "34 1 30",
"output": "62"
},
{
"input": "14 81 1",
"output": "30"
},
{
"input": "53 96 94",
"output": "242"
},
{
"input": "62 81 75",
"output": "218"
},
{
"input": "21 71 97",
"output": "188"
},
{
"input": "49 82 73",
"output": "204"
},
{
"input": "88 19 29",
"output": "96"
},
{
"input": "89 4 62",
"output": "132"
},
{
"input": "58 3 65",
"output": "126"
},
{
"input": "27 86 11",
"output": "76"
},
{
"input": "35 19 80",
"output": "134"
},
{
"input": "4 86 74",
"output": "156"
},
{
"input": "32 61 89",
"output": "182"
},
{
"input": "68 60 98",
"output": "226"
},
{
"input": "37 89 34",
"output": "142"
},
{
"input": "92 9 28",
"output": "74"
},
{
"input": "79 58 98",
"output": "234"
},
{
"input": "35 44 88",
"output": "166"
},
{
"input": "16 24 19",
"output": "58"
},
{
"input": "74 71 75",
"output": "220"
},
{
"input": "83 86 99",
"output": "268"
},
{
"input": "97 73 15",
"output": "176"
},
{
"input": "77 76 73",
"output": "226"
},
{
"input": "48 85 55",
"output": "188"
},
{
"input": "1 2 2",
"output": "4"
},
{
"input": "2 2 2",
"output": "6"
},
{
"input": "2 1 2",
"output": "4"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "3 2 1",
"output": "6"
},
{
"input": "1 2 3",
"output": "6"
},
{
"input": "1 3 2",
"output": "6"
},
{
"input": "2 1 3",
"output": "6"
},
{
"input": "2 3 1",
"output": "6"
},
{
"input": "3 1 2",
"output": "6"
},
{
"input": "99 99 99",
"output": "296"
},
{
"input": "99 99 100",
"output": "298"
},
{
"input": "99 100 99",
"output": "298"
},
{
"input": "99 100 100",
"output": "298"
},
{
"input": "100 99 99",
"output": "298"
},
{
"input": "100 99 100",
"output": "298"
},
{
"input": "100 100 99",
"output": "298"
},
{
"input": "89 32 23",
"output": "110"
},
{
"input": "4 5 0",
"output": "8"
},
{
"input": "3 0 3",
"output": "6"
},
{
"input": "0 0 2",
"output": "2"
},
{
"input": "97 97 0",
"output": "194"
},
{
"input": "1 4 0",
"output": "2"
},
{
"input": "5 2 0",
"output": "4"
},
{
"input": "0 5 10",
"output": "14"
},
{
"input": "0 1 2",
"output": "2"
},
{
"input": "5 2 3",
"output": "10"
},
{
"input": "5 5 0",
"output": "10"
},
{
"input": "0 0 10",
"output": "10"
},
{
"input": "0 1 1",
"output": "2"
},
{
"input": "0 0 1",
"output": "0"
}
] | 1,650,397,589
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 1
| 1,000
| 0
|
lst = [int(item) for item in input().split()]
l, r, amb = lst
flag = True
while(amb != 0):
if l < r:
l += 1
amb -= 1
flag = False
elif l > r:
r += 1
amb -= 1
if l != r and not flag:
print(l+r - abs(l-r))
else:
print(0)
|
Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
|
```python
lst = [int(item) for item in input().split()]
l, r, amb = lst
flag = True
while(amb != 0):
if l < r:
l += 1
amb -= 1
flag = False
elif l > r:
r += 1
amb -= 1
if l != r and not flag:
print(l+r - abs(l-r))
else:
print(0)
```
| 0
|
|
934
|
B
|
A Prosperous Lot
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"implementation"
] | null | null |
Apart from Nian, there is a daemon named Sui, which terrifies children and causes them to become sick. Parents give their children money wrapped in red packets and put them under the pillow, so that when Sui tries to approach them, it will be driven away by the fairies inside.
Big Banban is hesitating over the amount of money to give out. He considers loops to be lucky since it symbolizes unity and harmony.
He would like to find a positive integer *n* not greater than 1018, such that there are exactly *k* loops in the decimal representation of *n*, or determine that such *n* does not exist.
A loop is a planar area enclosed by lines in the digits' decimal representation written in Arabic numerals. For example, there is one loop in digit 4, two loops in 8 and no loops in 5. Refer to the figure below for all exact forms.
|
The first and only line contains an integer *k* (1<=≤<=*k*<=≤<=106) — the desired number of loops.
|
Output an integer — if no such *n* exists, output -1; otherwise output any such *n*. In the latter case, your output should be a positive decimal integer not exceeding 1018.
|
[
"2\n",
"6\n"
] |
[
"462",
"8080"
] |
none
| 1,000
|
[
{
"input": "2",
"output": "8"
},
{
"input": "6",
"output": "888"
},
{
"input": "3",
"output": "86"
},
{
"input": "4",
"output": "88"
},
{
"input": "5",
"output": "886"
},
{
"input": "1000000",
"output": "-1"
},
{
"input": "1",
"output": "6"
},
{
"input": "7",
"output": "8886"
},
{
"input": "8",
"output": "8888"
},
{
"input": "9",
"output": "88886"
},
{
"input": "10",
"output": "88888"
},
{
"input": "11",
"output": "888886"
},
{
"input": "12",
"output": "888888"
},
{
"input": "13",
"output": "8888886"
},
{
"input": "14",
"output": "8888888"
},
{
"input": "15",
"output": "88888886"
},
{
"input": "16",
"output": "88888888"
},
{
"input": "17",
"output": "888888886"
},
{
"input": "18",
"output": "888888888"
},
{
"input": "19",
"output": "8888888886"
},
{
"input": "20",
"output": "8888888888"
},
{
"input": "21",
"output": "88888888886"
},
{
"input": "22",
"output": "88888888888"
},
{
"input": "23",
"output": "888888888886"
},
{
"input": "24",
"output": "888888888888"
},
{
"input": "25",
"output": "8888888888886"
},
{
"input": "26",
"output": "8888888888888"
},
{
"input": "27",
"output": "88888888888886"
},
{
"input": "28",
"output": "88888888888888"
},
{
"input": "29",
"output": "888888888888886"
},
{
"input": "30",
"output": "888888888888888"
},
{
"input": "31",
"output": "8888888888888886"
},
{
"input": "32",
"output": "8888888888888888"
},
{
"input": "33",
"output": "88888888888888886"
},
{
"input": "34",
"output": "88888888888888888"
},
{
"input": "35",
"output": "888888888888888886"
},
{
"input": "36",
"output": "888888888888888888"
},
{
"input": "37",
"output": "-1"
},
{
"input": "38",
"output": "-1"
},
{
"input": "39",
"output": "-1"
},
{
"input": "40",
"output": "-1"
},
{
"input": "462",
"output": "-1"
},
{
"input": "1317",
"output": "-1"
},
{
"input": "88",
"output": "-1"
},
{
"input": "87",
"output": "-1"
}
] | 1,649,584,047
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 5
| 296
| 9,523,200
|
n = int(input())
h = n//2
r = n%2
a = 8
b = 4
for i in range(h):
print(a,end="")
if r:
print(b)
|
Title: A Prosperous Lot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Apart from Nian, there is a daemon named Sui, which terrifies children and causes them to become sick. Parents give their children money wrapped in red packets and put them under the pillow, so that when Sui tries to approach them, it will be driven away by the fairies inside.
Big Banban is hesitating over the amount of money to give out. He considers loops to be lucky since it symbolizes unity and harmony.
He would like to find a positive integer *n* not greater than 1018, such that there are exactly *k* loops in the decimal representation of *n*, or determine that such *n* does not exist.
A loop is a planar area enclosed by lines in the digits' decimal representation written in Arabic numerals. For example, there is one loop in digit 4, two loops in 8 and no loops in 5. Refer to the figure below for all exact forms.
Input Specification:
The first and only line contains an integer *k* (1<=≤<=*k*<=≤<=106) — the desired number of loops.
Output Specification:
Output an integer — if no such *n* exists, output -1; otherwise output any such *n*. In the latter case, your output should be a positive decimal integer not exceeding 1018.
Demo Input:
['2\n', '6\n']
Demo Output:
['462', '8080']
Note:
none
|
```python
n = int(input())
h = n//2
r = n%2
a = 8
b = 4
for i in range(h):
print(a,end="")
if r:
print(b)
```
| 0
|
|
887
|
A
|
Div. 64
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.
|
In the only line given a non-empty binary string *s* with length up to 100.
|
Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.
|
[
"100010001\n",
"100\n"
] |
[
"yes",
"no"
] |
In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system)
| 500
|
[
{
"input": "100010001",
"output": "yes"
},
{
"input": "100",
"output": "no"
},
{
"input": "0000001000000",
"output": "yes"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "no"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111111111",
"output": "no"
},
{
"input": "0111111101111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "no"
},
{
"input": "1111011111111111111111111111110111110111111111111111111111011111111111111110111111111111111111111111",
"output": "no"
},
{
"input": "1111111111101111111111111111111111111011111111111111111111111101111011111101111111111101111111111111",
"output": "yes"
},
{
"input": "0110111111111111111111011111111110110111110111111111111111111111111111111111111110111111111111111111",
"output": "yes"
},
{
"input": "1100110001111011001101101000001110111110011110111110010100011000100101000010010111100000010001001101",
"output": "yes"
},
{
"input": "000000",
"output": "no"
},
{
"input": "0001000",
"output": "no"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "1000000",
"output": "yes"
},
{
"input": "0",
"output": "no"
},
{
"input": "1",
"output": "no"
},
{
"input": "10000000000",
"output": "yes"
},
{
"input": "0000000000",
"output": "no"
},
{
"input": "0010000",
"output": "no"
},
{
"input": "000000011",
"output": "no"
},
{
"input": "000000000",
"output": "no"
},
{
"input": "00000000",
"output": "no"
},
{
"input": "000000000011",
"output": "no"
},
{
"input": "0000000",
"output": "no"
},
{
"input": "00000000011",
"output": "no"
},
{
"input": "000000001",
"output": "no"
},
{
"input": "000000000000000000000000000",
"output": "no"
},
{
"input": "0000001",
"output": "no"
},
{
"input": "00000001",
"output": "no"
},
{
"input": "00000000100",
"output": "no"
},
{
"input": "00000000000000000000",
"output": "no"
},
{
"input": "0000000000000000000",
"output": "no"
},
{
"input": "00001000",
"output": "no"
},
{
"input": "0000000000010",
"output": "no"
},
{
"input": "000000000010",
"output": "no"
},
{
"input": "000000000000010",
"output": "no"
},
{
"input": "0100000",
"output": "no"
},
{
"input": "00010000",
"output": "no"
},
{
"input": "00000000000000000",
"output": "no"
},
{
"input": "00000000000",
"output": "no"
},
{
"input": "000001000",
"output": "no"
},
{
"input": "000000000000",
"output": "no"
},
{
"input": "100000000000000",
"output": "yes"
},
{
"input": "000010000",
"output": "no"
},
{
"input": "00000100",
"output": "no"
},
{
"input": "0001100000",
"output": "no"
},
{
"input": "000000000000000000000000001",
"output": "no"
},
{
"input": "000000100",
"output": "no"
},
{
"input": "0000000000001111111111",
"output": "no"
},
{
"input": "00000010",
"output": "no"
},
{
"input": "0001110000",
"output": "no"
},
{
"input": "0000000000000000000000",
"output": "no"
},
{
"input": "000000010010",
"output": "no"
},
{
"input": "0000100",
"output": "no"
},
{
"input": "0000000001",
"output": "no"
},
{
"input": "000000111",
"output": "no"
},
{
"input": "0000000000000",
"output": "no"
},
{
"input": "000000000000000000",
"output": "no"
},
{
"input": "0000000000000000000000000",
"output": "no"
},
{
"input": "000000000000000",
"output": "no"
},
{
"input": "0010000000000100",
"output": "yes"
},
{
"input": "0000001000",
"output": "no"
},
{
"input": "00000000000000000001",
"output": "no"
},
{
"input": "100000000",
"output": "yes"
},
{
"input": "000000000001",
"output": "no"
},
{
"input": "0000011001",
"output": "no"
},
{
"input": "000",
"output": "no"
},
{
"input": "000000000000000000000",
"output": "no"
},
{
"input": "0000000000011",
"output": "no"
},
{
"input": "0000000000000000",
"output": "no"
},
{
"input": "00000000000000001",
"output": "no"
},
{
"input": "00000000000000",
"output": "no"
},
{
"input": "0000000000000000010",
"output": "no"
},
{
"input": "00000000000000000000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "000011000",
"output": "no"
},
{
"input": "00000011",
"output": "no"
},
{
"input": "0000000000001100",
"output": "no"
},
{
"input": "00000",
"output": "no"
},
{
"input": "000000000000000000000000000111111111111111",
"output": "no"
},
{
"input": "000000010",
"output": "no"
},
{
"input": "00000000111",
"output": "no"
},
{
"input": "000000000000001",
"output": "no"
},
{
"input": "0000000000000011111111111111111",
"output": "no"
},
{
"input": "0000000010",
"output": "no"
},
{
"input": "0000000000000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "00000000010",
"output": "no"
},
{
"input": "101000000000",
"output": "yes"
},
{
"input": "00100000",
"output": "no"
},
{
"input": "00000000000001",
"output": "no"
},
{
"input": "0000000000100",
"output": "no"
},
{
"input": "0000",
"output": "no"
},
{
"input": "00000000000111",
"output": "no"
},
{
"input": "0000000000000011",
"output": "no"
},
{
"input": "0000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "0000000000000010",
"output": "no"
},
{
"input": "0010101010",
"output": "no"
},
{
"input": "0000000000000001",
"output": "no"
},
{
"input": "1010101",
"output": "no"
}
] | 1,510,915,337
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 98
| 62
| 102,400
|
n = input()
if '1' in n and n[n.index('1'):].count('0') >= 6:
print('yes')
else:
print('no')
|
Title: Div. 64
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.
Input Specification:
In the only line given a non-empty binary string *s* with length up to 100.
Output Specification:
Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.
Demo Input:
['100010001\n', '100\n']
Demo Output:
['yes', 'no']
Note:
In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system)
|
```python
n = input()
if '1' in n and n[n.index('1'):].count('0') >= 6:
print('yes')
else:
print('no')
```
| 3
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,677,006,015
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
x ='WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB'
r = x.replace('WUB',' ').split()
print(*r)
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
x ='WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB'
r = x.replace('WUB',' ').split()
print(*r)
```
| 0
|
|
982
|
B
|
Bus of Characters
|
PROGRAMMING
| 1,300
|
[
"data structures",
"greedy",
"implementation"
] | null | null |
In the Bus of Characters there are $n$ rows of seat, each having $2$ seats. The width of both seats in the $i$-th row is $w_i$ centimeters. All integers $w_i$ are distinct.
Initially the bus is empty. On each of $2n$ stops one passenger enters the bus. There are two types of passengers:
- an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; - an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it.
You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take.
|
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of rows in the bus.
The second line contains the sequence of integers $w_1, w_2, \dots, w_n$ ($1 \le w_i \le 10^{9}$), where $w_i$ is the width of each of the seats in the $i$-th row. It is guaranteed that all $w_i$ are distinct.
The third line contains a string of length $2n$, consisting of digits '0' and '1' — the description of the order the passengers enter the bus. If the $j$-th character is '0', then the passenger that enters the bus on the $j$-th stop is an introvert. If the $j$-th character is '1', the the passenger that enters the bus on the $j$-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal $n$), and for each extrovert there always is a suitable row.
|
Print $2n$ integers — the rows the passengers will take. The order of passengers should be the same as in input.
|
[
"2\n3 1\n0011\n",
"6\n10 8 9 11 13 5\n010010011101\n"
] |
[
"2 1 1 2 \n",
"6 6 2 3 3 1 4 4 1 2 5 5 \n"
] |
In the first example the first passenger (introvert) chooses the row $2$, because it has the seats with smallest width. The second passenger (introvert) chooses the row $1$, because it is the only empty row now. The third passenger (extrovert) chooses the row $1$, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row $2$, because it is the only row with an empty place.
| 1,000
|
[
{
"input": "2\n3 1\n0011",
"output": "2 1 1 2 "
},
{
"input": "6\n10 8 9 11 13 5\n010010011101",
"output": "6 6 2 3 3 1 4 4 1 2 5 5 "
},
{
"input": "1\n1\n01",
"output": "1 1 "
},
{
"input": "1\n1000000\n01",
"output": "1 1 "
},
{
"input": "2\n1 1000000\n0011",
"output": "1 2 2 1 "
},
{
"input": "2\n1000000000 1\n0101",
"output": "2 2 1 1 "
},
{
"input": "2\n1000000000 999999999\n0011",
"output": "2 1 1 2 "
},
{
"input": "10\n24 53 10 99 83 9 15 62 33 47\n00100000000111111111",
"output": "6 3 3 7 1 9 10 2 8 5 4 4 5 8 2 10 9 1 7 6 "
}
] | 1,624,894,334
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 49
| 1,185
| 30,720,000
|
n=int(input())
i=iter(sorted(zip(map(int,input().split()),range(1,n+1))))
s,o=[],[]
for c in input():
if c=='0':
x=next(i)[1];o+=[x];s+=[x]
else:o.append(s.pop())
print(*o)
|
Title: Bus of Characters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the Bus of Characters there are $n$ rows of seat, each having $2$ seats. The width of both seats in the $i$-th row is $w_i$ centimeters. All integers $w_i$ are distinct.
Initially the bus is empty. On each of $2n$ stops one passenger enters the bus. There are two types of passengers:
- an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; - an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it.
You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of rows in the bus.
The second line contains the sequence of integers $w_1, w_2, \dots, w_n$ ($1 \le w_i \le 10^{9}$), where $w_i$ is the width of each of the seats in the $i$-th row. It is guaranteed that all $w_i$ are distinct.
The third line contains a string of length $2n$, consisting of digits '0' and '1' — the description of the order the passengers enter the bus. If the $j$-th character is '0', then the passenger that enters the bus on the $j$-th stop is an introvert. If the $j$-th character is '1', the the passenger that enters the bus on the $j$-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal $n$), and for each extrovert there always is a suitable row.
Output Specification:
Print $2n$ integers — the rows the passengers will take. The order of passengers should be the same as in input.
Demo Input:
['2\n3 1\n0011\n', '6\n10 8 9 11 13 5\n010010011101\n']
Demo Output:
['2 1 1 2 \n', '6 6 2 3 3 1 4 4 1 2 5 5 \n']
Note:
In the first example the first passenger (introvert) chooses the row $2$, because it has the seats with smallest width. The second passenger (introvert) chooses the row $1$, because it is the only empty row now. The third passenger (extrovert) chooses the row $1$, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row $2$, because it is the only row with an empty place.
|
```python
n=int(input())
i=iter(sorted(zip(map(int,input().split()),range(1,n+1))))
s,o=[],[]
for c in input():
if c=='0':
x=next(i)[1];o+=[x];s+=[x]
else:o.append(s.pop())
print(*o)
```
| 3
|
|
483
|
A
|
Counterexample
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math",
"number theory"
] | null | null |
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
|
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
|
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
|
[
"2 4\n",
"10 11\n",
"900000000000000009 900000000000000029\n"
] |
[
"2 3 4\n",
"-1\n",
"900000000000000009 900000000000000010 900000000000000021\n"
] |
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
| 500
|
[
{
"input": "2 4",
"output": "2 3 4"
},
{
"input": "10 11",
"output": "-1"
},
{
"input": "900000000000000009 900000000000000029",
"output": "900000000000000009 900000000000000010 900000000000000021"
},
{
"input": "640097987171091791 640097987171091835",
"output": "640097987171091792 640097987171091793 640097987171091794"
},
{
"input": "19534350415104721 19534350415104725",
"output": "19534350415104722 19534350415104723 19534350415104724"
},
{
"input": "933700505788726243 933700505788726280",
"output": "933700505788726244 933700505788726245 933700505788726246"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "2 3 4"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "266540997167959130 266540997167959164",
"output": "266540997167959130 266540997167959131 266540997167959132"
},
{
"input": "267367244641009850 267367244641009899",
"output": "267367244641009850 267367244641009851 267367244641009852"
},
{
"input": "268193483524125978 268193483524125993",
"output": "268193483524125978 268193483524125979 268193483524125980"
},
{
"input": "269019726702209402 269019726702209432",
"output": "269019726702209402 269019726702209403 269019726702209404"
},
{
"input": "269845965585325530 269845965585325576",
"output": "269845965585325530 269845965585325531 269845965585325532"
},
{
"input": "270672213058376250 270672213058376260",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492378",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608523",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691962",
"output": "273150934002691930 273150934002691931 273150934002691932"
},
{
"input": "996517375802030516 996517375802030524",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146694",
"output": "997343614685146644 997343614685146645 997343614685146646"
},
{
"input": "998169857863230068 998169857863230083",
"output": "998169857863230068 998169857863230069 998169857863230070"
},
{
"input": "998996101041313492 998996101041313522",
"output": "998996101041313492 998996101041313493 998996101041313494"
},
{
"input": "999822344219396916 999822344219396961",
"output": "999822344219396916 999822344219396917 999822344219396918"
},
{
"input": "648583102513043 648583102513053",
"output": "648583102513044 648583102513045 648583102513046"
},
{
"input": "266540997167959130 266540997167959131",
"output": "-1"
},
{
"input": "267367244641009850 267367244641009850",
"output": "-1"
},
{
"input": "268193483524125978 268193483524125979",
"output": "-1"
},
{
"input": "269019726702209402 269019726702209402",
"output": "-1"
},
{
"input": "269845965585325530 269845965585325530",
"output": "-1"
},
{
"input": "270672213058376250 270672213058376254",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492379",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608508",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691931",
"output": "-1"
},
{
"input": "996517375802030516 996517375802030518",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146644",
"output": "-1"
},
{
"input": "2147483647 2147483649",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "1 7",
"output": "2 3 4"
},
{
"input": "9 12",
"output": "9 11 12"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "11 13",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
}
] | 1,624,281,464
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 62
| 0
|
a, c = map(int, input().split(" "))
if a & 1 == 0:
if a+1 >= c:
print(-1)
else:
print(a, a+1, a+2)
else:
if a+2 >= c:
print(-1)
else:
print(a+1, a+2, a+3)
|
Title: Counterexample
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
Input Specification:
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
Output Specification:
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
Demo Input:
['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n']
Demo Output:
['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n']
Note:
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
|
```python
a, c = map(int, input().split(" "))
if a & 1 == 0:
if a+1 >= c:
print(-1)
else:
print(a, a+1, a+2)
else:
if a+2 >= c:
print(-1)
else:
print(a+1, a+2, a+3)
```
| 3
|
|
478
|
B
|
Random Teams
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"constructive algorithms",
"greedy",
"math"
] | null | null |
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
|
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
|
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
|
[
"5 1\n",
"3 2\n",
"6 3\n"
] |
[
"10 10\n",
"1 1\n",
"3 6\n"
] |
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
| 1,000
|
[
{
"input": "5 1",
"output": "10 10"
},
{
"input": "3 2",
"output": "1 1"
},
{
"input": "6 3",
"output": "3 6"
},
{
"input": "5 3",
"output": "2 3"
},
{
"input": "10 2",
"output": "20 36"
},
{
"input": "10 6",
"output": "4 10"
},
{
"input": "1000000000 1",
"output": "499999999500000000 499999999500000000"
},
{
"input": "5000000 12",
"output": "1041664166668 12499942500066"
},
{
"input": "1833 195",
"output": "7722 1342341"
},
{
"input": "1000000000 1000000000",
"output": "0 0"
},
{
"input": "1000000000 1000000",
"output": "499500000000 499000500499500000"
},
{
"input": "1000000000 32170",
"output": "15541930838100 499967831017438365"
},
{
"input": "1000000 1000",
"output": "499500000 499000999500"
},
{
"input": "1234 1123",
"output": "111 6216"
},
{
"input": "599222887 298488",
"output": "601178656545 179355218158217800"
},
{
"input": "999999999 500000000",
"output": "499999999 124999999750000000"
},
{
"input": "1000000000 384842119",
"output": "845473643 189209609585784021"
},
{
"input": "1000000000 384842119",
"output": "845473643 189209609585784021"
},
{
"input": "1000000000 2",
"output": "249999999500000000 499999998500000001"
},
{
"input": "1000000000 999999999",
"output": "1 1"
},
{
"input": "38447 383",
"output": "1910550 724453080"
},
{
"input": "100000000 99999799",
"output": "201 20301"
},
{
"input": "1 1",
"output": "0 0"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "2 2",
"output": "0 0"
},
{
"input": "10 10",
"output": "0 0"
}
] | 1,662,302,934
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 6
| 46
| 512,000
|
from math import sqrt
def integral(n):
if n == 0:
return 1
return n * integral(n - 1)
def pair_combs(n):
if n == 1:
return 0
if n == 2:
return 1
else:
n2 = integral(n - 2)
return int(n2 * n * (n - 1) / 2 / n2)
wah = list(map(int, input().split(' ')))
n, m = wah[0], wah[1]
ost = n % m
team = n // m
k_min = ost * pair_combs(team + 1) + (m - ost) * pair_combs(team)
k_max = pair_combs(n - (m - 1))
print(k_min, k_max)
|
Title: Random Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input Specification:
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
Output Specification:
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Demo Input:
['5 1\n', '3 2\n', '6 3\n']
Demo Output:
['10 10\n', '1 1\n', '3 6\n']
Note:
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
|
```python
from math import sqrt
def integral(n):
if n == 0:
return 1
return n * integral(n - 1)
def pair_combs(n):
if n == 1:
return 0
if n == 2:
return 1
else:
n2 = integral(n - 2)
return int(n2 * n * (n - 1) / 2 / n2)
wah = list(map(int, input().split(' ')))
n, m = wah[0], wah[1]
ost = n % m
team = n // m
k_min = ost * pair_combs(team + 1) + (m - ost) * pair_combs(team)
k_max = pair_combs(n - (m - 1))
print(k_min, k_max)
```
| -1
|
|
920
|
C
|
Swap Adjacent Elements
|
PROGRAMMING
| 1,400
|
[
"dfs and similar",
"greedy",
"math",
"sortings",
"two pointers"
] | null | null |
You have an array *a* consisting of *n* integers. Each integer from 1 to *n* appears exactly once in this array.
For some indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1) it is possible to swap *i*-th element with (*i*<=+<=1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap *i*-th element with (*i*<=+<=1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations?
|
The first line contains one integer *n* (2<=≤<=*n*<=≤<=200000) — the number of elements in the array.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200000) — the elements of the array. Each integer from 1 to *n* appears exactly once.
The third line contains a string of *n*<=-<=1 characters, each character is either 0 or 1. If *i*-th character is 1, then you can swap *i*-th element with (*i*<=+<=1)-th any number of times, otherwise it is forbidden to swap *i*-th element with (*i*<=+<=1)-th.
|
If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.
|
[
"6\n1 2 5 3 4 6\n01110\n",
"6\n1 2 5 3 4 6\n01010\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example you may swap *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, and then swap *a*<sub class="lower-index">4</sub> and *a*<sub class="lower-index">5</sub>.
| 0
|
[
{
"input": "6\n1 2 5 3 4 6\n01110",
"output": "YES"
},
{
"input": "6\n1 2 5 3 4 6\n01010",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01101",
"output": "NO"
},
{
"input": "6\n2 3 1 4 5 6\n01111",
"output": "NO"
},
{
"input": "4\n2 3 1 4\n011",
"output": "NO"
},
{
"input": "2\n2 1\n0",
"output": "NO"
},
{
"input": "5\n1 2 4 5 3\n0101",
"output": "NO"
},
{
"input": "5\n1 2 4 5 3\n0001",
"output": "NO"
},
{
"input": "5\n1 4 5 2 3\n0110",
"output": "NO"
},
{
"input": "5\n4 5 1 2 3\n0111",
"output": "NO"
},
{
"input": "3\n3 1 2\n10",
"output": "NO"
},
{
"input": "5\n2 3 4 5 1\n0011",
"output": "NO"
},
{
"input": "16\n3 4 14 16 11 7 13 9 10 8 6 5 15 12 1 2\n111111101111111",
"output": "NO"
},
{
"input": "5\n1 5 3 4 2\n1101",
"output": "NO"
},
{
"input": "6\n6 1 2 3 4 5\n11101",
"output": "NO"
},
{
"input": "3\n2 3 1\n01",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01110",
"output": "NO"
},
{
"input": "7\n1 7 3 4 5 6 2\n010001",
"output": "NO"
},
{
"input": "5\n5 2 3 4 1\n1001",
"output": "NO"
},
{
"input": "4\n1 3 4 2\n001",
"output": "NO"
},
{
"input": "5\n4 5 1 2 3\n1011",
"output": "NO"
},
{
"input": "6\n1 5 3 4 2 6\n11011",
"output": "NO"
},
{
"input": "5\n1 4 2 5 3\n1101",
"output": "NO"
},
{
"input": "5\n3 2 4 1 5\n1010",
"output": "NO"
},
{
"input": "6\n1 4 3 5 6 2\n01101",
"output": "NO"
},
{
"input": "6\n2 3 4 5 1 6\n00010",
"output": "NO"
},
{
"input": "10\n5 2 7 9 1 10 3 4 6 8\n111101000",
"output": "NO"
},
{
"input": "5\n2 4 3 1 5\n0110",
"output": "NO"
},
{
"input": "4\n3 1 2 4\n100",
"output": "NO"
},
{
"input": "6\n1 5 3 4 2 6\n01010",
"output": "NO"
},
{
"input": "4\n3 1 2 4\n101",
"output": "NO"
},
{
"input": "4\n2 4 3 1\n011",
"output": "NO"
},
{
"input": "4\n2 3 4 1\n001",
"output": "NO"
},
{
"input": "4\n3 4 1 2\n011",
"output": "NO"
},
{
"input": "5\n2 4 1 3 5\n0110",
"output": "NO"
},
{
"input": "4\n1 3 4 2\n101",
"output": "NO"
},
{
"input": "20\n20 19 18 17 16 15 1 2 3 4 5 14 13 12 11 10 9 8 7 6\n1111111011111111111",
"output": "NO"
},
{
"input": "6\n6 5 4 1 2 3\n11100",
"output": "NO"
},
{
"input": "5\n2 3 5 1 4\n0011",
"output": "NO"
},
{
"input": "4\n1 4 2 3\n010",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01001",
"output": "NO"
},
{
"input": "7\n1 7 2 4 3 5 6\n011110",
"output": "NO"
},
{
"input": "5\n1 3 4 2 5\n0010",
"output": "NO"
},
{
"input": "5\n5 4 3 1 2\n1110",
"output": "NO"
},
{
"input": "5\n2 5 4 3 1\n0111",
"output": "NO"
},
{
"input": "4\n2 3 4 1\n101",
"output": "NO"
},
{
"input": "5\n1 4 5 2 3\n1011",
"output": "NO"
},
{
"input": "5\n1 3 2 5 4\n1110",
"output": "NO"
},
{
"input": "6\n3 2 4 1 5 6\n10111",
"output": "NO"
},
{
"input": "7\n3 1 7 4 5 2 6\n101110",
"output": "NO"
},
{
"input": "10\n5 4 10 9 2 1 6 7 3 8\n011111111",
"output": "NO"
},
{
"input": "5\n1 5 3 2 4\n1110",
"output": "NO"
},
{
"input": "4\n2 3 4 1\n011",
"output": "NO"
},
{
"input": "5\n5 4 3 2 1\n0000",
"output": "NO"
},
{
"input": "12\n6 9 11 1 12 7 5 8 10 4 3 2\n11111111110",
"output": "NO"
},
{
"input": "5\n3 1 5 2 4\n1011",
"output": "NO"
},
{
"input": "5\n4 5 1 2 3\n1110",
"output": "NO"
},
{
"input": "10\n1 2 3 4 5 6 8 9 7 10\n000000000",
"output": "NO"
},
{
"input": "6\n5 6 3 2 4 1\n01111",
"output": "NO"
},
{
"input": "5\n1 3 4 2 5\n0100",
"output": "NO"
},
{
"input": "4\n2 1 4 3\n100",
"output": "NO"
},
{
"input": "6\n1 2 3 4 6 5\n00000",
"output": "NO"
},
{
"input": "6\n4 6 5 3 2 1\n01111",
"output": "NO"
},
{
"input": "5\n3 1 4 5 2\n1001",
"output": "NO"
},
{
"input": "5\n5 2 3 1 4\n1011",
"output": "NO"
},
{
"input": "3\n2 3 1\n10",
"output": "NO"
},
{
"input": "10\n6 5 9 4 3 2 8 10 7 1\n111111110",
"output": "NO"
},
{
"input": "7\n1 2 7 3 4 5 6\n111101",
"output": "NO"
},
{
"input": "6\n5 6 1 2 4 3\n11101",
"output": "NO"
},
{
"input": "6\n4 6 3 5 2 1\n11110",
"output": "NO"
},
{
"input": "5\n5 4 2 3 1\n1110",
"output": "NO"
},
{
"input": "2\n2 1\n1",
"output": "YES"
},
{
"input": "3\n1 3 2\n10",
"output": "NO"
},
{
"input": "5\n3 4 5 1 2\n1110",
"output": "NO"
},
{
"input": "5\n3 4 2 1 5\n0110",
"output": "NO"
},
{
"input": "6\n6 1 2 3 4 5\n10001",
"output": "NO"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10\n000000000",
"output": "YES"
},
{
"input": "3\n3 2 1\n00",
"output": "NO"
},
{
"input": "5\n5 4 3 2 1\n1110",
"output": "NO"
},
{
"input": "6\n3 1 2 5 6 4\n10011",
"output": "NO"
},
{
"input": "6\n3 2 1 6 5 4\n11000",
"output": "NO"
},
{
"input": "2\n1 2\n0",
"output": "YES"
},
{
"input": "2\n1 2\n1",
"output": "YES"
},
{
"input": "11\n1 2 3 4 5 6 7 8 9 10 11\n0000000000",
"output": "YES"
},
{
"input": "4\n2 4 3 1\n101",
"output": "NO"
},
{
"input": "4\n3 4 1 2\n101",
"output": "NO"
},
{
"input": "3\n1 3 2\n01",
"output": "YES"
},
{
"input": "6\n6 2 3 1 4 5\n11110",
"output": "NO"
},
{
"input": "3\n2 1 3\n01",
"output": "NO"
},
{
"input": "5\n1 5 4 3 2\n0111",
"output": "YES"
},
{
"input": "6\n1 2 6 3 4 5\n11110",
"output": "NO"
},
{
"input": "7\n2 3 1 7 6 5 4\n011111",
"output": "NO"
},
{
"input": "6\n5 6 1 2 3 4\n01111",
"output": "NO"
},
{
"input": "4\n1 2 4 3\n001",
"output": "YES"
},
{
"input": "6\n1 2 3 6 4 5\n11001",
"output": "NO"
},
{
"input": "11\n9 8 10 11 1 2 3 4 5 6 7\n1101111111",
"output": "NO"
},
{
"input": "5\n1 5 3 4 2\n0101",
"output": "NO"
},
{
"input": "10\n9 1 2 3 7 8 5 6 4 10\n110111100",
"output": "NO"
},
{
"input": "7\n1 2 7 3 4 5 6\n111011",
"output": "NO"
},
{
"input": "10\n3 10 1 2 6 4 5 7 8 9\n111111001",
"output": "NO"
},
{
"input": "10\n1 3 6 5 2 9 7 8 4 10\n001101111",
"output": "NO"
},
{
"input": "10\n1 8 9 7 6 10 4 2 3 5\n111111101",
"output": "NO"
},
{
"input": "7\n1 2 5 3 6 4 7\n111011",
"output": "NO"
},
{
"input": "4\n2 4 3 1\n100",
"output": "NO"
},
{
"input": "6\n1 2 3 4 6 5\n00001",
"output": "YES"
},
{
"input": "6\n2 1 3 4 5 6\n10000",
"output": "YES"
},
{
"input": "5\n3 2 1 5 4\n1100",
"output": "NO"
},
{
"input": "9\n2 1 3 6 5 4 7 9 8\n10011001",
"output": "YES"
},
{
"input": "8\n2 6 4 1 5 7 3 8\n1010010",
"output": "NO"
},
{
"input": "5\n1 2 4 5 3\n1101",
"output": "NO"
},
{
"input": "6\n1 3 5 2 4 6\n00110",
"output": "NO"
},
{
"input": "6\n1 3 6 2 4 5\n10111",
"output": "NO"
},
{
"input": "9\n9 8 7 6 5 4 3 1 2\n11111110",
"output": "NO"
},
{
"input": "10\n6 7 8 9 10 1 2 3 4 5\n111111110",
"output": "NO"
},
{
"input": "8\n6 1 7 8 3 2 5 4\n1011111",
"output": "NO"
},
{
"input": "70\n4 65 66 30 67 16 39 35 57 14 42 51 5 21 61 53 63 13 60 29 68 70 69 46 20 2 43 47 49 52 26 44 54 62 25 19 12 28 27 24 18 36 6 33 7 8 11 1 45 32 64 38 23 22 56 59 15 9 41 37 40 55 3 31 34 48 50 10 17 58\n111111101101111111111110101111111111111101101111010010110011011110010",
"output": "NO"
},
{
"input": "5\n5 3 2 4 1\n0100",
"output": "NO"
},
{
"input": "6\n3 2 6 5 1 4\n11011",
"output": "NO"
},
{
"input": "6\n1 2 4 5 6 3\n10011",
"output": "NO"
},
{
"input": "7\n1 7 3 2 5 6 4\n111001",
"output": "NO"
}
] | 1,567,994,363
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 171
| 14,131,200
|
import sys
n = int(input())
l = list(map(int,input().split()))
s = input()
i = 0
while i < n-1 :
if s[i] == '1':
k = i
while k < n - 1 and s[k] == '1':
k += 1
for j in range(i,k+1):
if not ( i+1 <= l[j] <= k+1):
print("NO")
sys.exit()
i = k
else:
k = i
while k < n - 1 and s[k] == '0':
k += 1
if k - i > 1:
for j in range(i,k):
if l[j] != j+1:
print("NO")
sys.exit()
i = k
print("YES")
|
Title: Swap Adjacent Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have an array *a* consisting of *n* integers. Each integer from 1 to *n* appears exactly once in this array.
For some indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1) it is possible to swap *i*-th element with (*i*<=+<=1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap *i*-th element with (*i*<=+<=1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations?
Input Specification:
The first line contains one integer *n* (2<=≤<=*n*<=≤<=200000) — the number of elements in the array.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200000) — the elements of the array. Each integer from 1 to *n* appears exactly once.
The third line contains a string of *n*<=-<=1 characters, each character is either 0 or 1. If *i*-th character is 1, then you can swap *i*-th element with (*i*<=+<=1)-th any number of times, otherwise it is forbidden to swap *i*-th element with (*i*<=+<=1)-th.
Output Specification:
If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.
Demo Input:
['6\n1 2 5 3 4 6\n01110\n', '6\n1 2 5 3 4 6\n01010\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example you may swap *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, and then swap *a*<sub class="lower-index">4</sub> and *a*<sub class="lower-index">5</sub>.
|
```python
import sys
n = int(input())
l = list(map(int,input().split()))
s = input()
i = 0
while i < n-1 :
if s[i] == '1':
k = i
while k < n - 1 and s[k] == '1':
k += 1
for j in range(i,k+1):
if not ( i+1 <= l[j] <= k+1):
print("NO")
sys.exit()
i = k
else:
k = i
while k < n - 1 and s[k] == '0':
k += 1
if k - i > 1:
for j in range(i,k):
if l[j] != j+1:
print("NO")
sys.exit()
i = k
print("YES")
```
| 0
|
|
864
|
B
|
Polycarp and Letters
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters.
Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met:
- letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=<<=*j*<=<<=*a*2 for some *a*1 and *a*2 from *A*).
Write a program that will determine the maximum number of elements in a pretty set of positions.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*.
The second line contains a string *s* consisting of lowercase and uppercase Latin letters.
|
Print maximum number of elements in pretty set of positions for string *s*.
|
[
"11\naaaaBaabAbA\n",
"12\nzACaAbbaazzC\n",
"3\nABC\n"
] |
[
"2\n",
"3\n",
"0\n"
] |
In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.
In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.
In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.
| 1,000
|
[
{
"input": "11\naaaaBaabAbA",
"output": "2"
},
{
"input": "12\nzACaAbbaazzC",
"output": "3"
},
{
"input": "3\nABC",
"output": "0"
},
{
"input": "1\na",
"output": "1"
},
{
"input": "2\naz",
"output": "2"
},
{
"input": "200\nXbTJZqcbpYuZQEoUrbxlPXAPCtVLrRExpQzxzqzcqsqzsiisswqitswzCtJQxOavicSdBIodideVRKHPojCNHmbnrLgwJlwOpyrJJIhrUePszxSjJGeUgTtOfewPQnPVWhZAtogRPrJLwyShNQaeNsvrJwjuuBOMPCeSckBMISQzGngfOmeyfDObncyeNsihYVtQbSEh",
"output": "8"
},
{
"input": "2\nAZ",
"output": "0"
},
{
"input": "28\nAabcBabcCBNMaaaaabbbbbcccccc",
"output": "3"
},
{
"input": "200\nrsgraosldglhdoorwhkrsehjpuxrjuwgeanjgezhekprzarelduuaxdnspzjuooguuwnzkowkuhzduakdrzpnslauejhrrkalwpurpuuswdgeadlhjwzjgegwpknepazwwleulppwrlgrgedlwdzuodzropsrrkxusjnuzshdkjrxxpgzanzdrpnggdwxarpwohxdepJ",
"output": "17"
},
{
"input": "1\nk",
"output": "1"
},
{
"input": "1\nH",
"output": "0"
},
{
"input": "2\nzG",
"output": "1"
},
{
"input": "2\ngg",
"output": "1"
},
{
"input": "2\nai",
"output": "2"
},
{
"input": "20\npEjVrKWLIFCZjIHgggVU",
"output": "1"
},
{
"input": "20\niFSiiigiYFSKmDnMGcgM",
"output": "2"
},
{
"input": "20\nedxedxxxCQiIVmYEUtLi",
"output": "3"
},
{
"input": "20\nprnchweyabjvzkoqiltm",
"output": "20"
},
{
"input": "35\nQLDZNKFXKVSVLUVHRTDPQYMSTDXBELXBOTS",
"output": "0"
},
{
"input": "35\nbvZWiitgxodztelnYUyljYGnCoWluXTvBLp",
"output": "10"
},
{
"input": "35\nBTexnaeplecllxwlanarpcollawHLVMHIIF",
"output": "10"
},
{
"input": "35\nhhwxqysolegsthsvfcqiryenbujbrrScobu",
"output": "20"
},
{
"input": "26\npbgfqosklxjuzmdheyvawrictn",
"output": "26"
},
{
"input": "100\nchMRWwymTDuZDZuSTvUmmuxvSscnTasyjlwwodhzcoifeahnbmcifyeobbydwparebduoLDCgHlOsPtVRbYGGQXfnkdvrWKIwCRl",
"output": "20"
},
{
"input": "100\nhXYLXKUMBrGkjqQJTGbGWAfmztqqapdbjbhcualhypgnaieKXmhzGMnqXVlcPesskfaEVgvWQTTShRRnEtFahWDyuBzySMpugxCM",
"output": "19"
},
{
"input": "100\nucOgELrgjMrFOgtHzqgvUgtHngKJxdMFKBjfcCppciqmGZXXoiSZibgpadshyljqrwxbomzeutvnhTLGVckZUmyiFPLlwuLBFito",
"output": "23"
},
{
"input": "200\nWTCKAKLVGXSYFVMVJDUYERXNMVNTGWXUGRFCGMYXJQGLODYZTUIDENHYEGFKXFIEUILAMESAXAWZXVCZPJPEYUXBITHMTZOTMKWITGRSFHODKVJHPAHVVWTCTHIVAWAREQXWMPUWQSTPPJFHKGKELBTPUYDAVIUMGASPUEDIODRYXIWCORHOSLIBLOZUNJPHHMXEXOAY",
"output": "0"
},
{
"input": "200\neLCCuYMPPwQoNlCpPOtKWJaQJmWfHeZCKiMSpILHSKjFOYGpRMzMCfMXdDuQdBGNsCNrHIVJzEFfBZcNMwNcFjOFVJvEtUQmLbFNKVHgNDyFkFVQhUTUQDgXhMjJZgFSSiHhMKuTgZQYJqAqKBpHoHddddddddddddddddXSSYNKNnRrKuOjAVKZlRLzCjExPdHaDHBT",
"output": "1"
},
{
"input": "200\nitSYxgOLlwOoAkkkkkzzzzzzzzkzkzkzkkkkkzkzzkzUDJSKybRPBvaIDsNuWImPJvrHkKiMeYukWmtHtgZSyQsgYanZvXNbKXBlFLSUcqRnGWSriAvKxsTkDJfROqaKdzXhvJsPEDATueCraWOGEvRDWjPwXuiNpWsEnCuhDcKWOQxjBkdBqmFatWFkgKsbZuLtRGtY",
"output": "2"
},
{
"input": "200\noggqoqqogoqoggggoggqgooqggogogooogqqgggoqgggqoqogogggogggqgooqgqggqqqoqgqgoooqgqogqoggoqqgqoqgoooqoogooqoogqoqoqqgoqgoqgggogqqqoqoggoqoqqoqggqoggooqqqoqggoggqqqqqqqqqgogqgggggooogogqgggqogqgoqoqogoooq",
"output": "3"
},
{
"input": "200\nCtclUtUnmqFniaLqGRmMoUMeLyFfAgWxIZxdrBarcRQprSOGcdUYsmDbooSuOvBLgrYlgaIjJtFgcxJKHGkCXpYfVKmUbouuIqGstFrrwJzYQqjjqqppqqqqqpqqqjpjjpjqjXRYkfPhGAatOigFuItkKxkjCBLdiNMVGjmdWNMgOOvmaJEdGsWNoaERrINNKqKeQajv",
"output": "3"
},
{
"input": "200\nmeZNrhqtSTSmktGQnnNOTcnyAMTKSixxKQKiagrMqRYBqgbRlsbJhvtNeHVUuMCyZLCnsIixRYrYEAkfQOxSVqXkrPqeCZQksInzRsRKBgvIqlGVPxPQnypknSXjgMjsjElcqGsaJRbegJVAKtWcHoOnzHqzhoKReqBBsOhZYLaYJhmqOMQsizdCsQfjUDHcTtHoeYwu",
"output": "4"
},
{
"input": "200\nvFAYTHJLZaivWzSYmiuDBDUFACDSVbkImnVaXBpCgrbgmTfXKJfoglIkZxWPSeVSFPnHZDNUAqLyhjLXSuAqGLskBlDxjxGPJyGdwzlPfIekwsblIrkxzfhJeNoHywdfAGlJzqXOfQaKceSqViVFTRJEGfACnsFeSFpOYisIHJciqTMNAmgeXeublTvfWoPnddtvKIyF",
"output": "6"
},
{
"input": "200\ngnDdkqJjYvduVYDSsswZDvoCouyaYZTfhmpSakERWLhufZtthWsfbQdTGwhKYjEcrqWBOyxBbiFhdLlIjChLOPiOpYmcrJgDtXsJfmHtLrabyGKOfHQRukEtTzwoqBHfmyVXPebfcpGQacLkGWFwerszjdHpTBXGssYXmGHlcCBgBXyGJqxbVhvDffLyCrZnxonABEXV",
"output": "7"
},
{
"input": "200\nBmggKNRZBXPtJqlJaXLdKKQLDJvXpDuQGupiRQfDwCJCJvAlDDGpPZNOvXkrdKOFOEFBVfrsZjWyHPoKGzXmTAyPJGEmxCyCXpeAdTwbrMtWLmlmGNqxvuxmqpmtpuhrmxxtrquSLFYVlnSYgRJDYHWgHBbziBLZRwCIJNvbtsEdLLxmTbnjkoqSPAuzEeTYLlmejOUH",
"output": "9"
},
{
"input": "200\nMkuxcDWdcnqsrlTsejehQKrTwoOBRCUAywqSnZkDLRmVBDVoOqdZHbrInQQyeRFAjiYYmHGrBbWgWstCPfLPRdNVDXBdqFJsGQfSXbufsiogybEhKDlWfPazIuhpONwGzZWaQNwVnmhTqWdewaklgjwaumXYDGwjSeEcYXjkVtLiYSWULEnTFukIlWQGWsXwWRMJGTcI",
"output": "10"
},
{
"input": "200\nOgMBgYeuMJdjPtLybvwmGDrQEOhliaabEtwulzNEjsfnaznXUMoBbbxkLEwSQzcLrlJdjJCLGVNBxorghPxTYCoqniySJMcilpsqpBAbqdzqRUDVaYOgqGhGrxlIJkyYgkOdTUgRZwpgIkeZFXojLXpDilzirHVVadiHaMrxhzodzpdvhvrzdzxbhmhdpxqqpoDegfFQ",
"output": "11"
},
{
"input": "200\nOLaJOtwultZLiZPSYAVGIbYvbIuZkqFZXwfsqpsavCDmBMStAuUFLBVknWDXNzmiuUYIsUMGxtoadWlPYPqvqSvpYdOiJRxFzGGnnmstniltvitnrmyrblnqyruylummmlsqtqitlbulvtuitiqimuintbimqyurviuntqnnvslynlNYMpYVKYwKVTbIUVdlNGrcFZON",
"output": "12"
},
{
"input": "200\nGAcmlaqfjSAQLvXlkhxujXgSbxdFAwnoxDuldDvYmpUhTWJdcEQSdARLrozJzIgFVCkzPUztWIpaGfiKeqzoXinEjVuoKqyBHmtFjBWcRdBmyjviNlGAIkpikjAimmBgayfphrstfbjexjbttzfzfzaysxfyrjazfhtpghnbbeffjhxrjxpttesgzrnrfbgzzsRsCgmz",
"output": "15"
},
{
"input": "200\nYRvIopNqSTYDhViTqCLMwEbTTIdHkoeuBmAJWhgtOgVxlcHSsavDNzMfpwTghkBvYEtCYQxicLUxdgAcaCzOOgbQYsfnaTXFlFxbeEiGwdNvxwHzkTdKtWlqzalwniDDBDipkxfflpaqkfkgfezbkxdvzemlfohwtgytzzywmwhvzUgPlPdeAVqTPAUZbogQheRXetvT",
"output": "20"
},
{
"input": "200\nNcYVomemswLCUqVRSDKHCknlBmqeSWhVyRzQrnZaOANnTGqsRFMjpczllcEVebqpxdavzppvztxsnfmtcharzqlginndyjkawzurqkxJLXiXKNZTIIxhSQghDpjwzatEqnLMTLxwoEKpHytvWkKFDUcZjLShCiVdocxRvvJtbXHCDGpJvMwRKWLhcTFtswdLUHkbhfau",
"output": "25"
},
{
"input": "200\nDxNZuvkTkQEqdWIkLzcKAwfqvZQiptnTazydSCTIfGjDhLMrlPZiKEsqIdDhgKPAlEvXyzNwWtYorotgkcwydpabjqnzubaksdchucxtkmjzfretdmvlxgklyvicrtftvztsbiUaQorfNIYUOdwQDRsKpxLUiLknbLbinilpPXPTTwLAnXVpMHBaAcKWgDBeOFabPtXU",
"output": "26"
},
{
"input": "4\nabbc",
"output": "3"
},
{
"input": "3\naaa",
"output": "1"
},
{
"input": "3\naba",
"output": "2"
},
{
"input": "3\nabb",
"output": "2"
},
{
"input": "3\nbba",
"output": "2"
},
{
"input": "3\nAaa",
"output": "1"
},
{
"input": "3\nAba",
"output": "2"
},
{
"input": "3\naBa",
"output": "1"
},
{
"input": "3\naAa",
"output": "1"
},
{
"input": "3\naAb",
"output": "1"
},
{
"input": "3\nAaA",
"output": "1"
},
{
"input": "5\naBacd",
"output": "3"
},
{
"input": "5\naAabc",
"output": "3"
}
] | 1,521,486,387
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 93
| 7,065,600
|
IL = lambda: list(map(int, input().split()))
IS = lambda: input().split()
I = lambda: int(input())
S = lambda: input()
n = I()
s = S()
for l in "QWERTYUIOPASDFGHJKLZXCVBNM":
s = s.replace(l, '#')
print(max([len(set(i)) for i in s.split('#')]))
|
Title: Polycarp and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters.
Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met:
- letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=<<=*j*<=<<=*a*2 for some *a*1 and *a*2 from *A*).
Write a program that will determine the maximum number of elements in a pretty set of positions.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*.
The second line contains a string *s* consisting of lowercase and uppercase Latin letters.
Output Specification:
Print maximum number of elements in pretty set of positions for string *s*.
Demo Input:
['11\naaaaBaabAbA\n', '12\nzACaAbbaazzC\n', '3\nABC\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.
In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.
In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.
|
```python
IL = lambda: list(map(int, input().split()))
IS = lambda: input().split()
I = lambda: int(input())
S = lambda: input()
n = I()
s = S()
for l in "QWERTYUIOPASDFGHJKLZXCVBNM":
s = s.replace(l, '#')
print(max([len(set(i)) for i in s.split('#')]))
```
| 3
|
|
1,011
|
B
|
Planning The Expedition
|
PROGRAMMING
| 1,200
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
Natasha is planning an expedition to Mars for $n$ people. One of the important tasks is to provide food for each participant.
The warehouse has $m$ daily food packages. Each package has some food type $a_i$.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant $j$ Natasha should select his food type $b_j$ and each day $j$-th participant will eat one food package of type $b_j$. The values $b_j$ for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above?
|
The first line contains two integers $n$ and $m$ ($1 \le n \le 100$, $1 \le m \le 100$) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le 100$), where $a_i$ is the type of $i$-th food package.
|
Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.
|
[
"4 10\n1 5 2 1 1 1 2 5 7 2\n",
"100 1\n1\n",
"2 5\n5 4 3 2 1\n",
"3 9\n42 42 42 42 42 42 42 42 42\n"
] |
[
"2\n",
"0\n",
"1\n",
"3\n"
] |
In the first example, Natasha can assign type $1$ food to the first participant, the same type $1$ to the second, type $5$ to the third and type $2$ to the fourth. In this case, the expedition can last for $2$ days, since each participant can get two food packages of his food type (there will be used $4$ packages of type $1$, two packages of type $2$ and two packages of type $5$).
In the second example, there are $100$ participants and only $1$ food package. In this case, the expedition can't last even $1$ day.
| 1,000
|
[
{
"input": "4 10\n1 5 2 1 1 1 2 5 7 2",
"output": "2"
},
{
"input": "100 1\n1",
"output": "0"
},
{
"input": "2 5\n5 4 3 2 1",
"output": "1"
},
{
"input": "3 9\n42 42 42 42 42 42 42 42 42",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "4 100\n84 99 66 69 86 94 89 96 98 93 93 82 87 93 91 100 69 99 93 81 99 84 75 100 86 88 98 100 84 96 44 70 94 91 85 78 86 79 45 88 91 78 98 94 81 87 93 72 96 88 96 97 96 62 86 72 94 84 80 98 88 90 93 73 73 98 78 50 91 96 97 82 85 90 87 41 97 82 97 77 100 100 92 83 98 81 70 81 74 78 84 79 98 98 55 99 97 99 79 98",
"output": "5"
},
{
"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": "1"
},
{
"input": "1 100\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": "100"
},
{
"input": "6 100\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "15"
},
{
"input": "1 1\n59",
"output": "1"
},
{
"input": "1 50\n39 1 46 21 23 28 100 32 63 63 18 15 40 29 34 49 56 74 47 42 96 97 59 62 76 62 69 61 36 21 66 18 92 58 63 85 5 6 77 75 91 66 38 10 66 43 20 74 37 83",
"output": "3"
},
{
"input": "1 100\n83 72 21 55 49 5 61 60 87 21 89 88 3 75 49 81 36 25 50 61 96 19 36 55 48 8 97 69 50 24 23 39 26 25 41 90 69 20 19 62 38 52 60 6 66 31 9 45 36 12 69 94 22 60 91 65 35 58 13 85 33 87 83 11 95 20 20 85 13 21 57 69 17 94 78 37 59 45 60 7 64 51 60 89 91 22 6 58 95 96 51 53 89 22 28 16 27 56 1 54",
"output": "5"
},
{
"input": "50 1\n75",
"output": "0"
},
{
"input": "50 50\n85 20 12 73 52 78 70 95 88 43 31 88 81 41 80 99 16 11 97 11 21 44 2 34 47 38 87 2 32 47 97 93 52 14 35 37 97 48 58 19 52 55 97 72 17 25 16 85 90 58",
"output": "1"
},
{
"input": "50 100\n2 37 74 32 99 75 73 86 67 33 62 30 15 21 51 41 73 75 67 39 90 10 56 74 72 26 38 65 75 55 46 99 34 49 92 82 11 100 15 71 75 12 22 56 47 74 20 98 59 65 14 76 1 40 89 36 43 93 83 73 75 100 50 95 27 10 72 51 25 69 15 3 57 60 84 99 31 44 12 61 69 95 51 31 28 36 57 35 31 52 44 19 79 12 27 27 7 81 68 1",
"output": "1"
},
{
"input": "100 1\n26",
"output": "0"
},
{
"input": "100 50\n8 82 62 11 85 57 5 32 99 92 77 2 61 86 8 88 10 28 83 4 68 79 8 64 56 98 4 88 22 54 30 60 62 79 72 38 17 28 32 16 62 26 56 44 72 33 22 84 77 45",
"output": "0"
},
{
"input": "100 100\n13 88 64 65 78 10 61 97 16 32 76 9 60 1 40 35 90 61 60 85 26 16 38 36 33 95 24 55 82 88 13 9 47 34 94 2 90 74 11 81 46 70 94 11 55 32 19 36 97 16 17 35 38 82 89 16 74 94 97 79 9 94 88 12 28 2 4 25 72 95 49 31 88 82 6 77 70 98 90 57 57 33 38 61 26 75 2 66 22 44 13 35 16 4 33 16 12 66 32 86",
"output": "1"
},
{
"input": "34 64\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",
"output": "1"
},
{
"input": "53 98\n1 1 2 2 2 2 2 1 2 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 1 2 1 1 1 2 1 2 1 1 1 2 2 1 2 1 1 1 2 2 1 2 1 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 1 1 2 2 1 2 1 2 1 2 1 1 2 2 2 1 1 2 1 2 1 1 1 1 2 2 2 2 2 1 1 2 2 2 1 1",
"output": "1"
},
{
"input": "17 8\n2 5 3 4 3 2 2 2",
"output": "0"
},
{
"input": "24 77\n8 6 10 4 6 6 4 10 9 7 7 5 5 4 6 7 10 6 3 4 6 6 4 9 4 6 2 5 3 4 4 1 4 6 6 8 1 1 6 4 6 2 5 7 7 2 4 4 10 1 10 9 2 3 8 1 10 4 3 9 3 8 3 5 6 3 4 9 5 3 4 1 1 6 1 2 1",
"output": "2"
},
{
"input": "65 74\n7 19 2 38 28 44 34 49 14 13 30 22 11 4 4 12 8 1 40 8 34 31 44 38 21 35 13 7 19 32 37 5 36 26 7 2 15 11 47 45 48 2 49 10 10 42 42 31 50 24 29 34 31 38 39 48 43 47 32 46 10 1 33 21 12 50 13 44 38 11 41 41 10 7",
"output": "1"
},
{
"input": "37 71\n50 93 15 80 82 23 35 90 70 73 55 23 23 6 86 63 38 70 38 52 88 34 25 75 32 19 6 98 31 38 21 8 66 8 59 71 7 80 69 23 17 70 6 40 72 5 48 59 18 1 48 91 17 41 11 27 53 95 87 31 62 94 94 60 38 99 70 50 81 86 44",
"output": "1"
},
{
"input": "35 4\n100 100 100 100",
"output": "0"
},
{
"input": "68 12\n100 100 100 99 99 100 100 100 99 99 99 99",
"output": "0"
},
{
"input": "91 33\n97 100 96 96 97 100 97 97 96 96 99 99 98 97 97 99 99 98 100 96 96 99 100 96 97 100 97 97 99 98 96 98 97",
"output": "0"
},
{
"input": "59 72\n99 96 96 93 94 97 93 96 100 99 93 99 96 92 92 97 93 100 92 100 91 96 99 100 97 98 93 93 96 100 97 99 100 100 91 93 96 91 93 100 95 96 92 97 91 91 99 100 91 95 95 100 92 93 98 93 95 92 95 100 91 95 99 95 92 91 100 93 98 98 96 99",
"output": "1"
},
{
"input": "9 35\n95 54 78 94 56 63 68 73 61 85 73 62 61 94 67 91 66 54 89 69 92 60 91 87 63 62 85 70 62 68 97 88 96 55 62",
"output": "2"
},
{
"input": "99 100\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": "1"
},
{
"input": "1 50\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",
"output": "50"
}
] | 1,696,055,970
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 46
| 0
|
# LUOGU_RID: 126684742
a,b=map(int,input().split());b=input().split();c={}
for i in b:
if i in c:c[i]+=1
else:c[i]=1
b=1
while 1:
if sum(c[j]//b for j in c)<a:print(b-1);break
b+=1
|
Title: Planning The Expedition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is planning an expedition to Mars for $n$ people. One of the important tasks is to provide food for each participant.
The warehouse has $m$ daily food packages. Each package has some food type $a_i$.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant $j$ Natasha should select his food type $b_j$ and each day $j$-th participant will eat one food package of type $b_j$. The values $b_j$ for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above?
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n \le 100$, $1 \le m \le 100$) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le 100$), where $a_i$ is the type of $i$-th food package.
Output Specification:
Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.
Demo Input:
['4 10\n1 5 2 1 1 1 2 5 7 2\n', '100 1\n1\n', '2 5\n5 4 3 2 1\n', '3 9\n42 42 42 42 42 42 42 42 42\n']
Demo Output:
['2\n', '0\n', '1\n', '3\n']
Note:
In the first example, Natasha can assign type $1$ food to the first participant, the same type $1$ to the second, type $5$ to the third and type $2$ to the fourth. In this case, the expedition can last for $2$ days, since each participant can get two food packages of his food type (there will be used $4$ packages of type $1$, two packages of type $2$ and two packages of type $5$).
In the second example, there are $100$ participants and only $1$ food package. In this case, the expedition can't last even $1$ day.
|
```python
# LUOGU_RID: 126684742
a,b=map(int,input().split());b=input().split();c={}
for i in b:
if i in c:c[i]+=1
else:c[i]=1
b=1
while 1:
if sum(c[j]//b for j in c)<a:print(b-1);break
b+=1
```
| 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"
}
] | 1,696,009,912
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
import math
n = int(input())
print(int(math.ceil(n/5)))
|
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
n = int(input())
print(int(math.ceil(n/5)))
```
| 3
|
|
1,011
|
A
|
Stages
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
|
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
|
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
|
[
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] |
[
"29",
"34",
"-1",
"1"
] |
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
| 500
|
[
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "169"
},
{
"input": "50 14\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "50 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "1"
},
{
"input": "50 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "13 13\nuwgmkyqeiaocs",
"output": "169"
},
{
"input": "13 13\nhzdxpbfvrltnj",
"output": "182"
},
{
"input": "1 1\nn",
"output": "14"
},
{
"input": "10 8\nsmzeblyjqw",
"output": "113"
},
{
"input": "20 20\ntzmvhskkyugkuuxpvtbh",
"output": "-1"
},
{
"input": "30 15\nwjzolzzkfulwgioksfxmcxmnnjtoav",
"output": "-1"
},
{
"input": "40 30\nxumfrflllrrgswehqtsskefixhcxjrxbjmrpsshv",
"output": "-1"
},
{
"input": "50 31\nahbyyoxltryqdmvenemaqnbakglgqolxnaifnqtoclnnqiabpz",
"output": "-1"
},
{
"input": "10 7\niuiukrxcml",
"output": "99"
},
{
"input": "38 2\nvjzarfykmrsrvwbwfwldsulhxtykmjbnwmdufa",
"output": "5"
},
{
"input": "12 6\nfwseyrarkwcd",
"output": "61"
},
{
"input": "2 2\nac",
"output": "4"
},
{
"input": "1 1\nc",
"output": "3"
},
{
"input": "2 2\nad",
"output": "5"
},
{
"input": "2 1\nac",
"output": "1"
},
{
"input": "4 3\nadjz",
"output": "15"
},
{
"input": "3 3\naoz",
"output": "42"
},
{
"input": "3 1\nzzz",
"output": "26"
},
{
"input": "2 1\nxz",
"output": "24"
},
{
"input": "5 1\naaddd",
"output": "1"
}
] | 1,564,940,744
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 124
| 0
|
n,k=map(int,input().split())
s=list(map(ord,sorted(input())))
last=0
total=0
i=0
while i<n:
if s[i]>last+1:
total += s[i]-96
last = s[i]
k -= 1
if k==0:
break
i+=1
if i<n:
print(total)
else:
print(-1)
|
Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
|
```python
n,k=map(int,input().split())
s=list(map(ord,sorted(input())))
last=0
total=0
i=0
while i<n:
if s[i]>last+1:
total += s[i]-96
last = s[i]
k -= 1
if k==0:
break
i+=1
if i<n:
print(total)
else:
print(-1)
```
| 3
|
|
858
|
A
|
k-rounding
|
PROGRAMMING
| 1,100
|
[
"brute force",
"math",
"number theory"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 750
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,615,269,152
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 96
| 62
| 0
|
import math
a,b=input().split()
a=int(a)
b=int(b)
ab=1
for i in range(b):
ab=ab*10
print((ab//math.gcd(ab,a))*a)
|
Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
import math
a,b=input().split()
a=int(a)
b=int(b)
ab=1
for i in range(b):
ab=ab*10
print((ab//math.gcd(ab,a))*a)
```
| 3
|
|
672
|
B
|
Different is Good
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation",
"strings"
] | null | null |
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
|
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
|
[
"2\naa\n",
"4\nkoko\n",
"5\nmurat\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
| 1,000
|
[
{
"input": "2\naa",
"output": "1"
},
{
"input": "4\nkoko",
"output": "2"
},
{
"input": "5\nmurat",
"output": "0"
},
{
"input": "6\nacbead",
"output": "1"
},
{
"input": "7\ncdaadad",
"output": "4"
},
{
"input": "25\npeoaicnbisdocqofsqdpgobpn",
"output": "12"
},
{
"input": "25\ntcqpchnqskqjacruoaqilgebu",
"output": "7"
},
{
"input": "13\naebaecedabbee",
"output": "8"
},
{
"input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "10\nbababbdaee",
"output": "6"
},
{
"input": "11\ndbadcdbdbca",
"output": "7"
},
{
"input": "12\nacceaabddaaa",
"output": "7"
},
{
"input": "13\nabddfbfaeecfa",
"output": "7"
},
{
"input": "14\neeceecacdbcbbb",
"output": "9"
},
{
"input": "15\ndcbceaaggabaheb",
"output": "8"
},
{
"input": "16\nhgiegfbadgcicbhd",
"output": "7"
},
{
"input": "17\nabhfibbdddfghgfdi",
"output": "10"
},
{
"input": "26\nbbbbbabbaababaaabaaababbaa",
"output": "24"
},
{
"input": "26\nahnxdnbfbcrirerssyzydihuee",
"output": "11"
},
{
"input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc",
"output": "8"
},
{
"input": "26\nrvxmulriorilidecqwmfaemifj",
"output": "10"
},
{
"input": "26\naowpmreooavnmamogdoopuisge",
"output": "12"
},
{
"input": "26\ninimevtuefhvuefirdehmmfudh",
"output": "15"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "27\nqdcfjtblgglnilgassirrjekcjt",
"output": "-1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyza",
"output": "-1"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "0"
},
{
"input": "5\nzzzzz",
"output": "4"
},
{
"input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa",
"output": "-1"
},
{
"input": "1\nq",
"output": "0"
},
{
"input": "27\nqwertyuioplkjhgfdsazxcvbnmm",
"output": "-1"
},
{
"input": "9\nxxxyyyzzz",
"output": "6"
},
{
"input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk",
"output": "-1"
},
{
"input": "27\nqwertyuiopasdfghjklzxcvbnmm",
"output": "-1"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyz",
"output": "0"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxya",
"output": "1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyzz",
"output": "-1"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "26\nqwertyuioplkjhgfdsazxcvbnm",
"output": "0"
},
{
"input": "10\nzzzzzzzzzz",
"output": "9"
},
{
"input": "1\na",
"output": "0"
},
{
"input": "30\nabcdefghtyabcdefghtyabcdefghty",
"output": "-1"
}
] | 1,547,222,809
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 7
| 2,000
| 204,800
|
n = int(input())
ph = input()
tab = []
sum = 0
for i in ph:
if ph.count(i) > 1 and i not in tab:
sum += ph.count(i) - 1
tab.append(i)
print(sum)
|
Title: Different is Good
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
Output Specification:
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
Demo Input:
['2\naa\n', '4\nkoko\n', '5\nmurat\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
|
```python
n = int(input())
ph = input()
tab = []
sum = 0
for i in ph:
if ph.count(i) > 1 and i not in tab:
sum += ph.count(i) - 1
tab.append(i)
print(sum)
```
| 0
|
|
455
|
A
|
Boredom
|
PROGRAMMING
| 1,500
|
[
"dp"
] | null | null |
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
|
Print a single integer — the maximum number of points that Alex can earn.
|
[
"2\n1 2\n",
"3\n1 2 3\n",
"9\n1 2 1 3 2 2 2 2 3\n"
] |
[
"2\n",
"4\n",
"10\n"
] |
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
| 500
|
[
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "4"
},
{
"input": "9\n1 2 1 3 2 2 2 2 3",
"output": "10"
},
{
"input": "5\n3 3 4 5 4",
"output": "11"
},
{
"input": "5\n5 3 5 3 4",
"output": "16"
},
{
"input": "5\n4 2 3 2 5",
"output": "9"
},
{
"input": "10\n10 5 8 9 5 6 8 7 2 8",
"output": "46"
},
{
"input": "10\n1 1 1 1 1 1 2 3 4 4",
"output": "14"
},
{
"input": "100\n6 6 8 9 7 9 6 9 5 7 7 4 5 3 9 1 10 3 4 5 8 9 6 5 6 4 10 9 1 4 1 7 1 4 9 10 8 2 9 9 10 5 8 9 5 6 8 7 2 8 7 6 2 6 10 8 6 2 5 5 3 2 8 8 5 3 6 2 1 4 7 2 7 3 7 4 10 10 7 5 4 7 5 10 7 1 1 10 7 7 7 2 3 4 2 8 4 7 4 4",
"output": "296"
},
{
"input": "100\n6 1 5 7 10 10 2 7 3 7 2 10 7 6 3 5 5 5 3 7 2 4 2 7 7 4 2 8 2 10 4 7 9 1 1 7 9 7 1 10 10 9 5 6 10 1 7 5 8 1 1 5 3 10 2 4 3 5 2 7 4 9 5 10 1 3 7 6 6 9 3 6 6 10 1 10 6 1 10 3 4 1 7 9 2 7 8 9 3 3 2 4 6 6 1 2 9 4 1 2",
"output": "313"
},
{
"input": "100\n7 6 3 8 8 3 10 5 3 8 6 4 6 9 6 7 3 9 10 7 5 5 9 10 7 2 3 8 9 5 4 7 9 3 6 4 9 10 7 6 8 7 6 6 10 3 7 4 5 7 7 5 1 5 4 8 7 3 3 4 7 8 5 9 2 2 3 1 6 4 6 6 6 1 7 10 7 4 5 3 9 2 4 1 5 10 9 3 9 6 8 5 2 1 10 4 8 5 10 9",
"output": "298"
},
{
"input": "100\n2 10 9 1 2 6 7 2 2 8 9 9 9 5 6 2 5 1 1 10 7 4 5 5 8 1 9 4 10 1 9 3 1 8 4 10 8 8 2 4 6 5 1 4 2 2 1 2 8 5 3 9 4 10 10 7 8 6 1 8 2 6 7 1 6 7 3 10 10 3 7 7 6 9 6 8 8 10 4 6 4 3 3 3 2 3 10 6 8 5 5 10 3 7 3 1 1 1 5 5",
"output": "312"
},
{
"input": "100\n4 9 7 10 4 7 2 6 1 9 1 8 7 5 5 7 6 7 9 8 10 5 3 5 7 10 3 2 1 3 8 9 4 10 4 7 6 4 9 6 7 1 9 4 3 5 8 9 2 7 10 5 7 5 3 8 10 3 8 9 3 4 3 10 6 5 1 8 3 2 5 8 4 7 5 3 3 2 6 9 9 8 2 7 6 3 2 2 8 8 4 5 6 9 2 3 2 2 5 2",
"output": "287"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n10 5 8 4 4 4 1 4 5 8 3 10 2 4 1 10 8 1 1 6 8 4 2 9 1 3 1 7 7 9 3 5 5 8 6 9 9 4 8 1 3 3 2 6 1 5 4 5 3 5 5 6 7 5 7 9 3 5 4 9 2 6 8 1 1 7 7 3 8 9 8 7 3 2 4 1 6 1 3 9 4 2 2 8 5 10 1 8 8 5 1 5 6 9 4 5 6 5 10 2",
"output": "265"
},
{
"input": "100\n7 5 1 8 5 6 6 2 6 2 7 7 3 6 2 4 4 2 10 2 2 2 10 6 6 1 5 10 9 1 5 9 8 9 4 1 10 5 7 5 7 6 4 8 8 1 7 8 3 8 2 1 8 4 10 3 5 6 6 10 9 6 5 1 10 7 6 9 9 2 10 10 9 1 2 1 7 7 4 10 1 10 5 5 3 8 9 8 1 4 10 2 4 5 4 4 1 6 2 9",
"output": "328"
},
{
"input": "100\n5 6 10 7 1 7 10 1 9 1 5 1 4 1 3 3 7 9 1 6 1 6 5 7 1 6 3 1 3 6 3 8 2 4 1 5 2 10 7 3 10 4 10 1 5 4 2 9 7 9 5 7 10 4 1 4 8 9 3 1 3 7 7 4 3 7 7 10 6 9 5 5 6 5 3 9 8 8 5 5 4 10 9 4 10 4 1 8 3 5 4 10 9 3 10 4 10 7 10 9",
"output": "324"
},
{
"input": "10\n7 4 5 3 9 1 10 3 4 5",
"output": "34"
},
{
"input": "10\n8 9 6 5 6 4 10 9 1 4",
"output": "39"
},
{
"input": "10\n1 7 1 4 9 10 8 2 9 9",
"output": "40"
},
{
"input": "1\n100000",
"output": "100000"
}
] | 1,699,687,583
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 2,867,200
|
from collections import Counter
import sys
from pkg_resources import yield_lines
input = lambda:sys.stdin.readline().rstrip("\r\n")
def I(): return input()
def II(): return int(input())
def MII(): return map(int, input().split())
def LI(): return list(input().split())
def LII(): return list(map(int, input().split()))
inf = float("inf")
mod = 10**9 + 7
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
# for _ in range(II()):
n = II()
a = LII()
mx = max(a)
cnt = Counter(a)
@bootstrap
def dfs(i):
if i < 0: yield 0
res1 = yield dfs(i-2)
res1 += i * cnt[i]
res2 = yield dfs(i-1)
yield max(res1, res2)
print(dfs(mx))
|
Title: Boredom
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a single integer — the maximum number of points that Alex can earn.
Demo Input:
['2\n1 2\n', '3\n1 2 3\n', '9\n1 2 1 3 2 2 2 2 3\n']
Demo Output:
['2\n', '4\n', '10\n']
Note:
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
|
```python
from collections import Counter
import sys
from pkg_resources import yield_lines
input = lambda:sys.stdin.readline().rstrip("\r\n")
def I(): return input()
def II(): return int(input())
def MII(): return map(int, input().split())
def LI(): return list(input().split())
def LII(): return list(map(int, input().split()))
inf = float("inf")
mod = 10**9 + 7
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
# for _ in range(II()):
n = II()
a = LII()
mx = max(a)
cnt = Counter(a)
@bootstrap
def dfs(i):
if i < 0: yield 0
res1 = yield dfs(i-2)
res1 += i * cnt[i]
res2 = yield dfs(i-1)
yield max(res1, res2)
print(dfs(mx))
```
| -1
|
|
215
|
A
|
Bicycle Chain
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
|
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
|
[
"2\n4 5\n3\n12 13 15\n",
"4\n1 2 3 4\n5\n10 11 12 13 14\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
| 500
|
[
{
"input": "2\n4 5\n3\n12 13 15",
"output": "2"
},
{
"input": "4\n1 2 3 4\n5\n10 11 12 13 14",
"output": "1"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n2\n1 2",
"output": "1"
},
{
"input": "4\n3 7 11 13\n4\n51 119 187 221",
"output": "4"
},
{
"input": "4\n2 3 4 5\n3\n1 2 3",
"output": "2"
},
{
"input": "10\n6 12 13 20 48 53 74 92 96 97\n10\n1 21 32 36 47 54 69 75 95 97",
"output": "1"
},
{
"input": "10\n5 9 10 14 15 17 19 22 24 26\n10\n2 11 17 19 21 22 24 25 27 28",
"output": "1"
},
{
"input": "10\n24 53 56 126 354 432 442 740 795 856\n10\n273 438 494 619 689 711 894 947 954 958",
"output": "1"
},
{
"input": "10\n3 4 6 7 8 10 14 16 19 20\n10\n3 4 5 7 8 10 15 16 18 20",
"output": "1"
},
{
"input": "10\n1 6 8 14 15 17 25 27 34 39\n10\n1 8 16 17 19 22 32 39 44 50",
"output": "1"
},
{
"input": "10\n5 21 22 23 25 32 35 36 38 39\n10\n3 7 8 9 18 21 23 24 36 38",
"output": "4"
},
{
"input": "50\n5 8 13 16 19 20 21 22 24 27 28 29 30 32 33 34 35 43 45 48 50 51 54 55 58 59 60 61 62 65 70 71 72 76 78 79 80 81 83 84 85 87 89 91 92 94 97 98 99 100\n50\n2 3 5 6 7 10 15 16 17 20 23 28 29 30 31 34 36 37 40 42 45 46 48 54 55 56 58 59 61 62 69 70 71 72 75 76 78 82 84 85 86 87 88 89 90 91 92 97 99 100",
"output": "1"
},
{
"input": "50\n3 5 6 8 9 11 13 19 21 23 24 32 34 35 42 50 51 52 56 58 59 69 70 72 73 75 76 77 78 80 83 88 90 95 96 100 101 102 108 109 113 119 124 135 138 141 142 143 145 150\n50\n5 8 10 11 18 19 23 30 35 43 51 53 55 58 63 68 69 71 77 78 79 82 83 86 88 89 91 92 93 94 96 102 103 105 109 110 113 114 116 123 124 126 127 132 133 135 136 137 142 149",
"output": "1"
},
{
"input": "50\n6 16 24 25 27 33 36 40 51 60 62 65 71 72 75 77 85 87 91 93 98 102 103 106 117 118 120 121 122 123 125 131 134 136 143 148 155 157 160 161 164 166 170 178 184 187 188 192 194 197\n50\n5 9 17 23 27 34 40 44 47 59 62 70 81 82 87 88 89 90 98 101 102 110 113 114 115 116 119 122 124 128 130 137 138 140 144 150 152 155 159 164 166 169 171 175 185 186 187 189 190 193",
"output": "1"
},
{
"input": "50\n14 22 23 31 32 35 48 63 76 79 88 97 101 102 103 104 106 113 114 115 116 126 136 138 145 152 155 156 162 170 172 173 179 180 182 203 208 210 212 222 226 229 231 232 235 237 245 246 247 248\n50\n2 5 6 16 28 44 45 46 54 55 56 63 72 80 87 93 94 96 97 100 101 103 132 135 140 160 164 165 167 168 173 180 182 185 186 192 194 198 199 202 203 211 213 216 217 227 232 233 236 245",
"output": "1"
},
{
"input": "50\n14 19 33 35 38 41 51 54 69 70 71 73 76 80 84 94 102 104 105 106 107 113 121 128 131 168 180 181 187 191 195 201 205 207 210 216 220 238 249 251 263 271 272 275 281 283 285 286 291 294\n50\n2 3 5 20 21 35 38 40 43 48 49 52 55 64 73 77 82 97 109 113 119 121 125 132 137 139 145 146 149 180 182 197 203 229 234 241 244 251 264 271 274 281 284 285 287 291 292 293 294 298",
"output": "1"
},
{
"input": "50\n2 4 5 16 18 19 22 23 25 26 34 44 48 54 67 79 80 84 92 110 116 133 138 154 163 171 174 202 205 218 228 229 234 245 247 249 250 263 270 272 274 275 277 283 289 310 312 334 339 342\n50\n1 5 17 18 25 37 46 47 48 59 67 75 80 83 84 107 115 122 137 141 159 162 175 180 184 204 221 224 240 243 247 248 249 258 259 260 264 266 269 271 274 293 294 306 329 330 334 335 342 350",
"output": "1"
},
{
"input": "50\n6 9 11 21 28 39 42 56 60 63 81 88 91 95 105 110 117 125 149 165 174 176 185 189 193 196 205 231 233 268 278 279 281 286 289 292 298 303 305 306 334 342 350 353 361 371 372 375 376 378\n50\n6 17 20 43 45 52 58 59 82 83 88 102 111 118 121 131 145 173 190 191 200 216 224 225 232 235 243 256 260 271 290 291 321 322 323 329 331 333 334 341 343 348 351 354 356 360 366 379 387 388",
"output": "1"
},
{
"input": "10\n17 239 443 467 661 1069 1823 2333 3767 4201\n20\n51 83 97 457 593 717 997 1329 1401 1459 1471 1983 2371 2539 3207 3251 3329 5469 6637 6999",
"output": "8"
},
{
"input": "20\n179 359 401 467 521 601 919 941 1103 1279 1709 1913 1949 2003 2099 2143 2179 2213 2399 4673\n20\n151 181 191 251 421 967 1109 1181 1249 1447 1471 1553 1619 2327 2551 2791 3049 3727 6071 7813",
"output": "3"
},
{
"input": "20\n79 113 151 709 809 983 1291 1399 1409 1429 2377 2659 2671 2897 3217 3511 3557 3797 3823 4363\n10\n19 101 659 797 1027 1963 2129 2971 3299 9217",
"output": "3"
},
{
"input": "30\n19 47 109 179 307 331 389 401 461 509 547 569 617 853 883 1249 1361 1381 1511 1723 1741 1783 2459 2531 2621 3533 3821 4091 5557 6217\n20\n401 443 563 941 967 997 1535 1567 1655 1747 1787 1945 1999 2251 2305 2543 2735 4415 6245 7555",
"output": "8"
},
{
"input": "30\n3 43 97 179 257 313 353 359 367 389 397 457 547 599 601 647 1013 1021 1063 1433 1481 1531 1669 3181 3373 3559 3769 4157 4549 5197\n50\n13 15 17 19 29 79 113 193 197 199 215 223 271 293 359 485 487 569 601 683 895 919 941 967 1283 1285 1289 1549 1565 1765 1795 1835 1907 1931 1945 1985 1993 2285 2731 2735 2995 3257 4049 4139 5105 5315 7165 7405 7655 8345",
"output": "20"
},
{
"input": "50\n11 17 23 53 59 109 137 149 173 251 353 379 419 421 439 503 593 607 661 773 821 877 941 997 1061 1117 1153 1229 1289 1297 1321 1609 1747 2311 2389 2543 2693 3041 3083 3137 3181 3209 3331 3373 3617 3767 4201 4409 4931 6379\n50\n55 59 67 73 85 89 101 115 211 263 295 353 545 599 607 685 739 745 997 1031 1255 1493 1523 1667 1709 1895 1949 2161 2195 2965 3019 3035 3305 3361 3373 3673 3739 3865 3881 4231 4253 4385 4985 5305 5585 5765 6145 6445 8045 8735",
"output": "23"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "47\n66 262 357 457 513 530 538 540 592 691 707 979 1015 1242 1246 1667 1823 1886 1963 2133 2649 2679 2916 2949 3413 3523 3699 3958 4393 4922 5233 5306 5799 6036 6302 6629 7208 7282 7315 7822 7833 7927 8068 8150 8870 8962 9987\n39\n167 199 360 528 1515 1643 1986 1988 2154 2397 2856 3552 3656 3784 3980 4096 4104 4240 4320 4736 4951 5266 5656 5849 5850 6169 6517 6875 7244 7339 7689 7832 8120 8716 9503 9509 9933 9936 9968",
"output": "12"
},
{
"input": "1\n94\n50\n423 446 485 1214 1468 1507 1853 1930 1999 2258 2271 2285 2425 2543 2715 2743 2992 3196 4074 4108 4448 4475 4652 5057 5250 5312 5356 5375 5731 5986 6298 6501 6521 7146 7255 7276 7332 7481 7998 8141 8413 8665 8908 9221 9336 9491 9504 9677 9693 9706",
"output": "1"
},
{
"input": "50\n51 67 75 186 194 355 512 561 720 876 1077 1221 1503 1820 2153 2385 2568 2608 2937 2969 3271 3311 3481 4081 4093 4171 4255 4256 4829 5020 5192 5636 5817 6156 6712 6717 7153 7436 7608 7612 7866 7988 8264 8293 8867 9311 9879 9882 9889 9908\n1\n5394",
"output": "1"
},
{
"input": "50\n26 367 495 585 675 789 855 1185 1312 1606 2037 2241 2587 2612 2628 2807 2873 2924 3774 4067 4376 4668 4902 5001 5082 5100 5104 5209 5345 5515 5661 5777 5902 5907 6155 6323 6675 6791 7503 8159 8207 8254 8740 8848 8855 8933 9069 9164 9171 9586\n5\n1557 6246 7545 8074 8284",
"output": "1"
},
{
"input": "5\n25 58 91 110 2658\n50\n21 372 909 1172 1517 1554 1797 1802 1843 1977 2006 2025 2137 2225 2317 2507 2645 2754 2919 3024 3202 3212 3267 3852 4374 4487 4553 4668 4883 4911 4916 5016 5021 5068 5104 5162 5683 5856 6374 6871 7333 7531 8099 8135 8173 8215 8462 8776 9433 9790",
"output": "4"
},
{
"input": "45\n37 48 56 59 69 70 79 83 85 86 99 114 131 134 135 145 156 250 1739 1947 2116 2315 2449 3104 3666 4008 4406 4723 4829 5345 5836 6262 6296 6870 7065 7110 7130 7510 7595 8092 8442 8574 9032 9091 9355\n50\n343 846 893 1110 1651 1837 2162 2331 2596 3012 3024 3131 3294 3394 3528 3717 3997 4125 4347 4410 4581 4977 5030 5070 5119 5229 5355 5413 5418 5474 5763 5940 6151 6161 6164 6237 6506 6519 6783 7182 7413 7534 8069 8253 8442 8505 9135 9308 9828 9902",
"output": "17"
},
{
"input": "50\n17 20 22 28 36 38 46 47 48 50 52 57 58 62 63 69 70 74 75 78 79 81 82 86 87 90 93 95 103 202 292 442 1756 1769 2208 2311 2799 2957 3483 4280 4324 4932 5109 5204 6225 6354 6561 7136 8754 9670\n40\n68 214 957 1649 1940 2078 2134 2716 3492 3686 4462 4559 4656 4756 4850 5044 5490 5529 5592 5626 6014 6111 6693 6790 7178 7275 7566 7663 7702 7857 7954 8342 8511 8730 8957 9021 9215 9377 9445 9991",
"output": "28"
},
{
"input": "39\n10 13 21 25 36 38 47 48 58 64 68 69 73 79 86 972 2012 2215 2267 2503 3717 3945 4197 4800 5266 6169 6612 6824 7023 7322 7582 7766 8381 8626 8879 9079 9088 9838 9968\n50\n432 877 970 1152 1202 1223 1261 1435 1454 1578 1843 1907 2003 2037 2183 2195 2215 2425 3065 3492 3615 3637 3686 3946 4189 4415 4559 4656 4665 4707 4886 4887 5626 5703 5955 6208 6521 6581 6596 6693 6985 7013 7081 7343 7663 8332 8342 8637 9207 9862",
"output": "15"
},
{
"input": "50\n7 144 269 339 395 505 625 688 709 950 1102 1152 1350 1381 1641 1830 1977 1999 2093 2180 2718 3308 3574 4168 4232 4259 4393 4689 4982 5154 5476 5581 5635 5721 6159 6302 6741 7010 7152 7315 7417 7482 8116 8239 8640 9347 9395 9614 9661 9822\n20\n84 162 292 1728 1866 2088 3228 3470 4068 5318 5470 6060 6380 6929 7500 8256 8399 8467 8508 9691",
"output": "8"
},
{
"input": "50\n159 880 1070 1139 1358 1608 1691 1841 2073 2171 2213 2597 2692 2759 2879 2931 3173 3217 3441 4201 4878 5106 5129 5253 5395 5647 5968 6019 6130 6276 6286 6330 6409 6728 7488 7713 7765 7828 7899 8064 8264 8457 8483 8685 8900 8946 8965 9133 9187 9638\n45\n57 159 1070 1139 1391 1608 1691 1841 2171 2213 2692 2759 2931 3173 3217 3441 4201 4878 5106 5129 5253 5647 5968 6130 6276 6286 6409 7488 7694 7713 7765 7828 7899 8003 8064 8081 8244 8264 8685 8900 8946 8965 9133 9638 9673",
"output": "38"
},
{
"input": "3\n3 4 5\n3\n6 20 25",
"output": "2"
},
{
"input": "4\n2 3 5 8\n4\n2 6 8 10",
"output": "1"
},
{
"input": "4\n3 5 7 11\n4\n3 5 7 22",
"output": "1"
},
{
"input": "2\n2 3\n3\n20 30 50",
"output": "1"
},
{
"input": "3\n1 2 3\n4\n2 4 6 49",
"output": "1"
},
{
"input": "2\n4 5\n3\n12 15 20",
"output": "1"
},
{
"input": "3\n2 5 7\n3\n4 5 7",
"output": "1"
},
{
"input": "3\n3 5 8\n3\n6 8 10",
"output": "2"
},
{
"input": "2\n2 3\n4\n4 6 9 33",
"output": "1"
},
{
"input": "2\n2 3\n4\n4 6 21 40",
"output": "1"
},
{
"input": "3\n4 9 10\n3\n8 9 10",
"output": "1"
},
{
"input": "5\n1 5 6 9 51\n5\n5 12 18 27 10000",
"output": "1"
},
{
"input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13\n1\n14",
"output": "1"
}
] | 1,660,669,330
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 57
| 154
| 1,536,000
|
import math
n=int(input())
l1=[int(i) for i in input().split()]
m=int(input())
l2=[int(j) for j in input().split()]
l3=[]
for x in range(m):
for y in range(n):
k=l2[x]/l1[y]
if k-math.floor(k)==0:
l3.append(k)
a=max(l3)
print(l3.count(a))
|
Title: Bicycle Chain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Output Specification:
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
Demo Input:
['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
|
```python
import math
n=int(input())
l1=[int(i) for i in input().split()]
m=int(input())
l2=[int(j) for j in input().split()]
l3=[]
for x in range(m):
for y in range(n):
k=l2[x]/l1[y]
if k-math.floor(k)==0:
l3.append(k)
a=max(l3)
print(l3.count(a))
```
| 3
|
|
747
|
A
|
Display Size
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] | null | null |
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
|
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
|
Print two integers — the number of rows and columns on the display.
|
[
"8\n",
"64\n",
"5\n",
"999999\n"
] |
[
"2 4\n",
"8 8\n",
"1 5\n",
"999 1001\n"
] |
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
| 500
|
[
{
"input": "8",
"output": "2 4"
},
{
"input": "64",
"output": "8 8"
},
{
"input": "5",
"output": "1 5"
},
{
"input": "999999",
"output": "999 1001"
},
{
"input": "716539",
"output": "97 7387"
},
{
"input": "1",
"output": "1 1"
},
{
"input": "2",
"output": "1 2"
},
{
"input": "3",
"output": "1 3"
},
{
"input": "4",
"output": "2 2"
},
{
"input": "6",
"output": "2 3"
},
{
"input": "7",
"output": "1 7"
},
{
"input": "9",
"output": "3 3"
},
{
"input": "10",
"output": "2 5"
},
{
"input": "11",
"output": "1 11"
},
{
"input": "12",
"output": "3 4"
},
{
"input": "15",
"output": "3 5"
},
{
"input": "100",
"output": "10 10"
},
{
"input": "101",
"output": "1 101"
},
{
"input": "169",
"output": "13 13"
},
{
"input": "179",
"output": "1 179"
},
{
"input": "190",
"output": "10 19"
},
{
"input": "1000",
"output": "25 40"
},
{
"input": "10000",
"output": "100 100"
},
{
"input": "10001",
"output": "73 137"
},
{
"input": "100000",
"output": "250 400"
},
{
"input": "100001",
"output": "11 9091"
},
{
"input": "1000000",
"output": "1000 1000"
},
{
"input": "999983",
"output": "1 999983"
},
{
"input": "524288",
"output": "512 1024"
},
{
"input": "954493",
"output": "971 983"
},
{
"input": "966289",
"output": "983 983"
},
{
"input": "944663",
"output": "961 983"
},
{
"input": "912673",
"output": "97 9409"
},
{
"input": "732641",
"output": "679 1079"
},
{
"input": "232897",
"output": "343 679"
},
{
"input": "16807",
"output": "49 343"
},
{
"input": "999958",
"output": "2 499979"
},
{
"input": "990151",
"output": "1 990151"
},
{
"input": "997002",
"output": "998 999"
},
{
"input": "20",
"output": "4 5"
},
{
"input": "20261",
"output": "1 20261"
},
{
"input": "999123",
"output": "3 333041"
},
{
"input": "901841",
"output": "1 901841"
}
] | 1,591,420,623
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 155
| 0
|
from math import*
#n,k=map(int, input().split())
n=int(input())
#l=list(map(int, input().split()))
#s=input()
k=int(pow(n,1/2))
for i in range(k,0,-1):
if n%i==0:
print(i,n//i)
break
|
Title: Display Size
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
Output Specification:
Print two integers — the number of rows and columns on the display.
Demo Input:
['8\n', '64\n', '5\n', '999999\n']
Demo Output:
['2 4\n', '8 8\n', '1 5\n', '999 1001\n']
Note:
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
|
```python
from math import*
#n,k=map(int, input().split())
n=int(input())
#l=list(map(int, input().split()))
#s=input()
k=int(pow(n,1/2))
for i in range(k,0,-1):
if n%i==0:
print(i,n//i)
break
```
| 3
|
|
248
|
A
|
Cupboards
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
|
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
|
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
|
[
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] |
[
"3\n"
] |
none
| 500
|
[
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 1\n1 1\n0 1",
"output": "6"
},
{
"input": "15\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "5\n1 0\n1 0\n1 0\n0 1\n0 1",
"output": "4"
}
] | 1,586,867,679
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 216
| 0
|
n = int(input())
l = 0
r = 0
for i in range(0, n):
inp = [int(x) for x in input().split()]
l += inp[0]
r += inp[1]
print(n - min(l, r))
|
Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none
|
```python
n = int(input())
l = 0
r = 0
for i in range(0, n):
inp = [int(x) for x in input().split()]
l += inp[0]
r += inp[1]
print(n - min(l, r))
```
| 0
|
|
106
|
A
|
Card Game
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Card Game
|
2
|
256
|
There is a card game called "Durak", which means "Fool" in Russian. The game is quite popular in the countries that used to form USSR. The problem does not state all the game's rules explicitly — you can find them later yourselves if you want.
To play durak you need a pack of 36 cards. Each card has a suit ("S", "H", "D" and "C") and a rank (in the increasing order "6", "7", "8", "9", "T", "J", "Q", "K" and "A"). At the beginning of the game one suit is arbitrarily chosen as trump.
The players move like that: one player puts one or several of his cards on the table and the other one should beat each of them with his cards.
A card beats another one if both cards have similar suits and the first card has a higher rank then the second one. Besides, a trump card can beat any non-trump card whatever the cards’ ranks are. In all other cases you can not beat the second card with the first one.
You are given the trump suit and two different cards. Determine whether the first one beats the second one or not.
|
The first line contains the tramp suit. It is "S", "H", "D" or "C".
The second line contains the description of the two different cards. Each card is described by one word consisting of two symbols. The first symbol stands for the rank ("6", "7", "8", "9", "T", "J", "Q", "K" and "A"), and the second one stands for the suit ("S", "H", "D" and "C").
|
Print "YES" (without the quotes) if the first cards beats the second one. Otherwise, print "NO" (also without the quotes).
|
[
"H\nQH 9S\n",
"S\n8D 6D\n",
"C\n7H AS\n"
] |
[
"YES\n",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "H\nQH 9S",
"output": "YES"
},
{
"input": "S\n8D 6D",
"output": "YES"
},
{
"input": "C\n7H AS",
"output": "NO"
},
{
"input": "C\nKC 9C",
"output": "YES"
},
{
"input": "D\n7D KD",
"output": "NO"
},
{
"input": "H\n7H KD",
"output": "YES"
},
{
"input": "D\nAS AH",
"output": "NO"
},
{
"input": "H\nKH KS",
"output": "YES"
},
{
"input": "C\n9H 6C",
"output": "NO"
},
{
"input": "C\n9H JC",
"output": "NO"
},
{
"input": "D\nTD JD",
"output": "NO"
},
{
"input": "H\n6S 7S",
"output": "NO"
},
{
"input": "D\n7S 8S",
"output": "NO"
},
{
"input": "S\n8H 9H",
"output": "NO"
},
{
"input": "C\n9D TD",
"output": "NO"
},
{
"input": "H\nTC JC",
"output": "NO"
},
{
"input": "C\nJH QH",
"output": "NO"
},
{
"input": "H\nQD KD",
"output": "NO"
},
{
"input": "D\nKS AS",
"output": "NO"
},
{
"input": "S\nAH 6H",
"output": "YES"
},
{
"input": "H\n7D 6D",
"output": "YES"
},
{
"input": "S\n8H 7H",
"output": "YES"
},
{
"input": "D\n9S 8S",
"output": "YES"
},
{
"input": "S\nTC 9C",
"output": "YES"
},
{
"input": "H\nJS TS",
"output": "YES"
},
{
"input": "S\nQD JD",
"output": "YES"
},
{
"input": "D\nKH QH",
"output": "YES"
},
{
"input": "H\nAD KD",
"output": "YES"
},
{
"input": "H\nQS QD",
"output": "NO"
},
{
"input": "C\nTS TH",
"output": "NO"
},
{
"input": "C\n6C 6D",
"output": "YES"
},
{
"input": "H\n8H 8D",
"output": "YES"
},
{
"input": "S\n7D 7S",
"output": "NO"
},
{
"input": "H\nJC JH",
"output": "NO"
},
{
"input": "H\n8H 9C",
"output": "YES"
},
{
"input": "D\n9D 6S",
"output": "YES"
},
{
"input": "C\nJC AH",
"output": "YES"
},
{
"input": "S\nAS KD",
"output": "YES"
},
{
"input": "S\n7S JS",
"output": "NO"
},
{
"input": "H\nTH 8H",
"output": "YES"
},
{
"input": "S\n7S QS",
"output": "NO"
},
{
"input": "C\nKC QC",
"output": "YES"
},
{
"input": "S\nAD 9S",
"output": "NO"
},
{
"input": "D\n7H 8D",
"output": "NO"
},
{
"input": "H\nJC 9H",
"output": "NO"
},
{
"input": "C\n7S AC",
"output": "NO"
},
{
"input": "C\n8C 7C",
"output": "YES"
},
{
"input": "H\n9D 8S",
"output": "NO"
},
{
"input": "D\nAC KS",
"output": "NO"
},
{
"input": "H\n8C QH",
"output": "NO"
},
{
"input": "S\n7S TS",
"output": "NO"
},
{
"input": "C\nAH 6S",
"output": "NO"
},
{
"input": "S\nKS QS",
"output": "YES"
},
{
"input": "H\nAC QC",
"output": "YES"
},
{
"input": "S\n9H 8D",
"output": "NO"
},
{
"input": "S\nTS JS",
"output": "NO"
},
{
"input": "S\n8H 7C",
"output": "NO"
},
{
"input": "C\nAH 6S",
"output": "NO"
},
{
"input": "S\n7S QS",
"output": "NO"
},
{
"input": "C\nAH 6S",
"output": "NO"
},
{
"input": "S\nTS KS",
"output": "NO"
},
{
"input": "C\nTH KH",
"output": "NO"
},
{
"input": "H\n9C 6D",
"output": "NO"
},
{
"input": "H\n9C 8D",
"output": "NO"
},
{
"input": "H\nTH AH",
"output": "NO"
},
{
"input": "H\nTH JH",
"output": "NO"
},
{
"input": "H\nQS 9C",
"output": "NO"
},
{
"input": "H\nKC AC",
"output": "NO"
},
{
"input": "H\nAH KH",
"output": "YES"
},
{
"input": "H\nKS QS",
"output": "YES"
},
{
"input": "C\nAD KS",
"output": "NO"
},
{
"input": "H\nQS 9C",
"output": "NO"
},
{
"input": "H\n9D 7S",
"output": "NO"
},
{
"input": "D\n6D 9S",
"output": "YES"
},
{
"input": "H\nAH KH",
"output": "YES"
},
{
"input": "H\nKC AC",
"output": "NO"
},
{
"input": "D\n8S 6C",
"output": "NO"
},
{
"input": "S\nAC KC",
"output": "YES"
}
] | 1,684,728,882
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
# /**
# * author: brownfox2k6
# * created: 22/05/2023 11:10:46 Hanoi, Vietnam
# **/
rank = "6789TJQKA"
trump = input()
f, s = input().split()
if f[1] == s[1]:
if rank.index(f[0]) > rank.index(s[0]):
print("YES")
else:
print("NO")
elif f[1] == trump:
print("YES")
elif s[1] == trump:
print("NO")
|
Title: Card Game
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a card game called "Durak", which means "Fool" in Russian. The game is quite popular in the countries that used to form USSR. The problem does not state all the game's rules explicitly — you can find them later yourselves if you want.
To play durak you need a pack of 36 cards. Each card has a suit ("S", "H", "D" and "C") and a rank (in the increasing order "6", "7", "8", "9", "T", "J", "Q", "K" and "A"). At the beginning of the game one suit is arbitrarily chosen as trump.
The players move like that: one player puts one or several of his cards on the table and the other one should beat each of them with his cards.
A card beats another one if both cards have similar suits and the first card has a higher rank then the second one. Besides, a trump card can beat any non-trump card whatever the cards’ ranks are. In all other cases you can not beat the second card with the first one.
You are given the trump suit and two different cards. Determine whether the first one beats the second one or not.
Input Specification:
The first line contains the tramp suit. It is "S", "H", "D" or "C".
The second line contains the description of the two different cards. Each card is described by one word consisting of two symbols. The first symbol stands for the rank ("6", "7", "8", "9", "T", "J", "Q", "K" and "A"), and the second one stands for the suit ("S", "H", "D" and "C").
Output Specification:
Print "YES" (without the quotes) if the first cards beats the second one. Otherwise, print "NO" (also without the quotes).
Demo Input:
['H\nQH 9S\n', 'S\n8D 6D\n', 'C\n7H AS\n']
Demo Output:
['YES\n', 'YES', 'NO']
Note:
none
|
```python
# /**
# * author: brownfox2k6
# * created: 22/05/2023 11:10:46 Hanoi, Vietnam
# **/
rank = "6789TJQKA"
trump = input()
f, s = input().split()
if f[1] == s[1]:
if rank.index(f[0]) > rank.index(s[0]):
print("YES")
else:
print("NO")
elif f[1] == trump:
print("YES")
elif s[1] == trump:
print("NO")
```
| 0
|
39
|
J
|
Spelling Check
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation",
"strings"
] |
J. Spelling Check
|
2
|
256
|
Petya has noticed that when he types using a keyboard, he often presses extra buttons and adds extra letters to the words. Of course, the spell-checking system underlines the words for him and he has to click every word and choose the right variant. Petya got fed up with correcting his mistakes himself, that’s why he decided to invent the function that will correct the words itself. Petya started from analyzing the case that happens to him most of the time, when all one needs is to delete one letter for the word to match a word from the dictionary. Thus, Petya faces one mini-task: he has a printed word and a word from the dictionary, and he should delete one letter from the first word to get the second one. And now the very non-trivial question that Petya faces is: which letter should he delete?
|
The input data contains two strings, consisting of lower-case Latin letters. The length of each string is from 1 to 106 symbols inclusive, the first string contains exactly 1 symbol more than the second one.
|
In the first line output the number of positions of the symbols in the first string, after the deleting of which the first string becomes identical to the second one. In the second line output space-separated positions of these symbols in increasing order. The positions are numbered starting from 1. If it is impossible to make the first string identical to the second string by deleting one symbol, output one number 0.
|
[
"abdrakadabra\nabrakadabra\n",
"aa\na\n",
"competition\ncodeforces\n"
] |
[
"1\n3\n",
"2\n1 2\n",
"0\n"
] |
none
| 0
|
[
{
"input": "abdrakadabra\nabrakadabra",
"output": "1\n3 "
},
{
"input": "aa\na",
"output": "2\n1 2 "
},
{
"input": "competition\ncodeforces",
"output": "0"
},
{
"input": "ab\na",
"output": "1\n2 "
},
{
"input": "bb\nb",
"output": "2\n1 2 "
},
{
"input": "aab\nab",
"output": "2\n1 2 "
},
{
"input": "aabb\nabb",
"output": "2\n1 2 "
},
{
"input": "babaacaacaa\nbbaacaacaa",
"output": "1\n2 "
},
{
"input": "bccaabbcccc\nbccaabcccc",
"output": "2\n6 7 "
},
{
"input": "ababcaabaaa\nabacaabaaa",
"output": "1\n4 "
},
{
"input": "cccacaccacb\ncccacaccac",
"output": "1\n11 "
},
{
"input": "aaaaaaaaaaa\naaaaaaaaaa",
"output": "11\n1 2 3 4 5 6 7 8 9 10 11 "
},
{
"input": "lcaaxcbcjca\nccaaacccca",
"output": "0"
},
{
"input": "babbbtaamba\nbabbbaabba",
"output": "0"
},
{
"input": "xdfxmcnzpch\nazvotghvtk",
"output": "0"
},
{
"input": "ki\nb",
"output": "0"
},
{
"input": "vct\nie",
"output": "0"
},
{
"input": "feee\nsnl",
"output": "0"
},
{
"input": "cbxxxxzvks\ncbxxxzvks",
"output": "4\n3 4 5 6 "
},
{
"input": "qybldcgfhdhhhhhhhhhhopqkhuczzytzluiahwbqjltgafvvoecititchjwdoljiehubngmtjckqymldhoncgtqhxnqvoagnrmur\nqybldcgfhdhhhhhhhhhopqkhuczzytzluiahwbqjltgafvvoecititchjwdoljiehubngmtjckqymldhoncgtqhxnqvoagnrmur",
"output": "10\n11 12 13 14 15 16 17 18 19 20 "
}
] | 1,553,170,560
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 1,716
| 48,742,400
|
s, t = input(), input()
n = j = len(t)
for i in range(n):
if s[i] != t[i]:
j = i
break
for i in range(j + 1, n + 1):
if s[i] != t[i - 1]:
print(0)
break
else:
i = j
while i > 0 and s[i - 1] == s[j]:
i -= 1
print(j - i + 1)
print(*range(i + 1, j + 2))
|
Title: Spelling Check
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya has noticed that when he types using a keyboard, he often presses extra buttons and adds extra letters to the words. Of course, the spell-checking system underlines the words for him and he has to click every word and choose the right variant. Petya got fed up with correcting his mistakes himself, that’s why he decided to invent the function that will correct the words itself. Petya started from analyzing the case that happens to him most of the time, when all one needs is to delete one letter for the word to match a word from the dictionary. Thus, Petya faces one mini-task: he has a printed word and a word from the dictionary, and he should delete one letter from the first word to get the second one. And now the very non-trivial question that Petya faces is: which letter should he delete?
Input Specification:
The input data contains two strings, consisting of lower-case Latin letters. The length of each string is from 1 to 106 symbols inclusive, the first string contains exactly 1 symbol more than the second one.
Output Specification:
In the first line output the number of positions of the symbols in the first string, after the deleting of which the first string becomes identical to the second one. In the second line output space-separated positions of these symbols in increasing order. The positions are numbered starting from 1. If it is impossible to make the first string identical to the second string by deleting one symbol, output one number 0.
Demo Input:
['abdrakadabra\nabrakadabra\n', 'aa\na\n', 'competition\ncodeforces\n']
Demo Output:
['1\n3\n', '2\n1 2\n', '0\n']
Note:
none
|
```python
s, t = input(), input()
n = j = len(t)
for i in range(n):
if s[i] != t[i]:
j = i
break
for i in range(j + 1, n + 1):
if s[i] != t[i - 1]:
print(0)
break
else:
i = j
while i > 0 and s[i - 1] == s[j]:
i -= 1
print(j - i + 1)
print(*range(i + 1, j + 2))
```
| 3.48021
|
716
|
A
|
Crazy Computer
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
|
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
|
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
|
[
"6 5\n1 3 8 14 19 20\n",
"6 1\n1 3 5 7 9 10\n"
] |
[
"3",
"2"
] |
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.
| 500
|
[
{
"input": "6 5\n1 3 8 14 19 20",
"output": "3"
},
{
"input": "6 1\n1 3 5 7 9 10",
"output": "2"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 7 12 13 14",
"output": "4"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "2"
},
{
"input": "3 5\n1 10 20",
"output": "1"
},
{
"input": "3 10\n1 2 3",
"output": "3"
},
{
"input": "2 1\n1 100",
"output": "1"
},
{
"input": "3 1\n1 2 10",
"output": "1"
},
{
"input": "2 1\n1 2",
"output": "2"
}
] | 1,632,740,618
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n, c = map(int, input().split())
ls = list(map(int, input().split()))
k = sum((b-a)<=c for a,b in zip(ls, ls[1:])
print(k)
|
Title: Crazy Computer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Output Specification:
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
Demo Input:
['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n']
Demo Output:
['3', '2']
Note:
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.
|
```python
n, c = map(int, input().split())
ls = list(map(int, input().split()))
k = sum((b-a)<=c for a,b in zip(ls, ls[1:])
print(k)
```
| -1
|
|
236
|
A
|
Boy or Girl
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
|
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
|
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
|
[
"wjmzbmr\n",
"xiaodao\n",
"sevenkplus\n"
] |
[
"CHAT WITH HER!\n",
"IGNORE HIM!\n",
"CHAT WITH HER!\n"
] |
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
| 500
|
[
{
"input": "wjmzbmr",
"output": "CHAT WITH HER!"
},
{
"input": "xiaodao",
"output": "IGNORE HIM!"
},
{
"input": "sevenkplus",
"output": "CHAT WITH HER!"
},
{
"input": "pezu",
"output": "CHAT WITH HER!"
},
{
"input": "wnemlgppy",
"output": "CHAT WITH HER!"
},
{
"input": "zcinitufxoldnokacdvtmdohsfdjepyfioyvclhmujiqwvmudbfjzxjfqqxjmoiyxrfsbvseawwoyynn",
"output": "IGNORE HIM!"
},
{
"input": "qsxxuoynwtebujwpxwpajitiwxaxwgbcylxneqiebzfphugwkftpaikixmumkhfbjiswmvzbtiyifbx",
"output": "CHAT WITH HER!"
},
{
"input": "qwbdfzfylckctudyjlyrtmvbidfatdoqfmrfshsqqmhzohhsczscvwzpwyoyswhktjlykumhvaounpzwpxcspxwlgt",
"output": "IGNORE HIM!"
},
{
"input": "nuezoadauueermoeaabjrkxttkatspjsjegjcjcdmcxgodowzbwuqncfbeqlhkk",
"output": "IGNORE HIM!"
},
{
"input": "lggvdmulrsvtuagoavstuyufhypdxfomjlzpnduulukszqnnwfvxbvxyzmleocmofwclmzz",
"output": "IGNORE HIM!"
},
{
"input": "tgcdptnkc",
"output": "IGNORE HIM!"
},
{
"input": "wvfgnfrzabgibzxhzsojskmnlmrokydjoexnvi",
"output": "IGNORE HIM!"
},
{
"input": "sxtburpzskucowowebgrbovhadrrayamuwypmmxhscrujkmcgvyinp",
"output": "IGNORE HIM!"
},
{
"input": "pjqxhvxkyeqqvyuujxhmbspatvrckhhkfloottuybjivkkhpyivcighxumavrxzxslfpggnwbtalmhysyfllznphzia",
"output": "IGNORE HIM!"
},
{
"input": "fpellxwskyekoyvrfnuf",
"output": "CHAT WITH HER!"
},
{
"input": "xninyvkuvakfbs",
"output": "IGNORE HIM!"
},
{
"input": "vnxhrweyvhqufpfywdwftoyrfgrhxuamqhblkvdpxmgvphcbeeqbqssresjifwyzgfhurmamhkwupymuomak",
"output": "CHAT WITH HER!"
},
{
"input": "kmsk",
"output": "IGNORE HIM!"
},
{
"input": "lqonogasrkzhryjxppjyriyfxmdfubieglthyswz",
"output": "CHAT WITH HER!"
},
{
"input": "ndormkufcrkxlihdhmcehzoimcfhqsmombnfjrlcalffq",
"output": "CHAT WITH HER!"
},
{
"input": "zqzlnnuwcfufwujygtczfakhcpqbtxtejrbgoodychepzdphdahtxyfpmlrycyicqthsgm",
"output": "IGNORE HIM!"
},
{
"input": "ppcpbnhwoizajrl",
"output": "IGNORE HIM!"
},
{
"input": "sgubujztzwkzvztitssxxxwzanfmddfqvv",
"output": "CHAT WITH HER!"
},
{
"input": "ptkyaxycecpbrjnvxcjtbqiocqcswnmicxbvhdsptbxyxswbw",
"output": "IGNORE HIM!"
},
{
"input": "yhbtzfppwcycxqjpqdfmjnhwaogyuaxamwxpnrdrnqsgdyfvxu",
"output": "CHAT WITH HER!"
},
{
"input": "ojjvpnkrxibyevxk",
"output": "CHAT WITH HER!"
},
{
"input": "wjweqcrqfuollfvfbiyriijovweg",
"output": "IGNORE HIM!"
},
{
"input": "hkdbykboclchfdsuovvpknwqr",
"output": "IGNORE HIM!"
},
{
"input": "stjvyfrfowopwfjdveduedqylerqugykyu",
"output": "IGNORE HIM!"
},
{
"input": "rafcaanqytfclvfdegak",
"output": "CHAT WITH HER!"
},
{
"input": "xczn",
"output": "CHAT WITH HER!"
},
{
"input": "arcoaeozyeawbveoxpmafxxzdjldsielp",
"output": "IGNORE HIM!"
},
{
"input": "smdfafbyehdylhaleevhoggiurdgeleaxkeqdixyfztkuqsculgslheqfafxyghyuibdgiuwrdxfcitojxika",
"output": "CHAT WITH HER!"
},
{
"input": "vbpfgjqnhfazmvtkpjrdasfhsuxnpiepxfrzvoh",
"output": "CHAT WITH HER!"
},
{
"input": "dbdokywnpqnotfrhdbrzmuyoxfdtrgrzcccninbtmoqvxfatcqg",
"output": "CHAT WITH HER!"
},
{
"input": "udlpagtpq",
"output": "CHAT WITH HER!"
},
{
"input": "zjurevbytijifnpfuyswfchdzelxheboruwjqijxcucylysmwtiqsqqhktexcynquvcwhbjsipy",
"output": "CHAT WITH HER!"
},
{
"input": "qagzrqjomdwhagkhrjahhxkieijyten",
"output": "CHAT WITH HER!"
},
{
"input": "achhcfjnnfwgoufxamcqrsontgjjhgyfzuhklkmiwybnrlsvblnsrjqdytglipxsulpnphpjpoewvlusalsgovwnsngb",
"output": "CHAT WITH HER!"
},
{
"input": "qbkjsdwpahdbbohggbclfcufqelnojoehsxxkr",
"output": "CHAT WITH HER!"
},
{
"input": "cpvftiwgyvnlmbkadiafddpgfpvhqqvuehkypqjsoibpiudfvpkhzlfrykc",
"output": "IGNORE HIM!"
},
{
"input": "lnpdosnceumubvk",
"output": "IGNORE HIM!"
},
{
"input": "efrk",
"output": "CHAT WITH HER!"
},
{
"input": "temnownneghnrujforif",
"output": "IGNORE HIM!"
},
{
"input": "ottnneymszwbumgobazfjyxewkjakglbfflsajuzescplpcxqta",
"output": "IGNORE HIM!"
},
{
"input": "eswpaclodzcwhgixhpyzvhdwsgneqidanbzdzszquefh",
"output": "IGNORE HIM!"
},
{
"input": "gwntwbpj",
"output": "IGNORE HIM!"
},
{
"input": "wuqvlbblkddeindiiswsinkfrnkxghhwunzmmvyovpqapdfbolyim",
"output": "IGNORE HIM!"
},
{
"input": "swdqsnzmzmsyvktukaoyqsqzgfmbzhezbfaqeywgwizrwjyzquaahucjchegknqaioliqd",
"output": "CHAT WITH HER!"
},
{
"input": "vlhrpzezawyolhbmvxbwhtjustdbqggexmzxyieihjlelvwjosmkwesfjmramsikhkupzvfgezmrqzudjcalpjacmhykhgfhrjx",
"output": "IGNORE HIM!"
},
{
"input": "lxxwbkrjgnqjwsnflfnsdyxihmlspgivirazsbveztnkuzpaxtygidniflyjheejelnjyjvgkgvdqks",
"output": "CHAT WITH HER!"
},
{
"input": "wpxbxzfhtdecetpljcrvpjjnllosdqirnkzesiqeukbedkayqx",
"output": "CHAT WITH HER!"
},
{
"input": "vmzxgacicvweclaodrunmjnfwtimceetsaoickarqyrkdghcmyjgmtgsqastcktyrjgvjqimdc",
"output": "CHAT WITH HER!"
},
{
"input": "yzlzmesxdttfcztooypjztlgxwcr",
"output": "IGNORE HIM!"
},
{
"input": "qpbjwzwgdzmeluheirjrvzrhbmagfsjdgvzgwumjtjzecsfkrfqjasssrhhtgdqqfydlmrktlgfc",
"output": "IGNORE HIM!"
},
{
"input": "aqzftsvezdgouyrirsxpbuvdjupnzvbhguyayeqozfzymfnepvwgblqzvmxxkxcilmsjvcgyqykpoaktjvsxbygfgsalbjoq",
"output": "CHAT WITH HER!"
},
{
"input": "znicjjgijhrbdlnwmtjgtdgziollrfxroabfhadygnomodaembllreorlyhnehijfyjbfxucazellblegyfrzuraogadj",
"output": "IGNORE HIM!"
},
{
"input": "qordzrdiknsympdrkgapjxokbldorpnmnpucmwakklmqenpmkom",
"output": "CHAT WITH HER!"
},
{
"input": "wqfldgihuxfktzanyycluzhtewmwvnawqlfoavuguhygqrrxtstxwouuzzsryjqtfqo",
"output": "CHAT WITH HER!"
},
{
"input": "vujtrrpshinkskgyknlcfckmqdrwtklkzlyipmetjvaqxdsslkskschbalmdhzsdrrjmxdltbtnxbh",
"output": "IGNORE HIM!"
},
{
"input": "zioixjibuhrzyrbzqcdjbbhhdmpgmqykixcxoqupggaqajuzonrpzihbsogjfsrrypbiphehonyhohsbybnnukqebopppa",
"output": "CHAT WITH HER!"
},
{
"input": "oh",
"output": "CHAT WITH HER!"
},
{
"input": "kxqthadqesbpgpsvpbcbznxpecqrzjoilpauttzlnxvaczcqwuri",
"output": "IGNORE HIM!"
},
{
"input": "zwlunigqnhrwirkvufqwrnwcnkqqonebrwzcshcbqqwkjxhymjjeakuzjettebciadjlkbfp",
"output": "CHAT WITH HER!"
},
{
"input": "fjuldpuejgmggvvigkwdyzytfxzwdlofrpifqpdnhfyroginqaufwgjcbgshyyruwhofctsdaisqpjxqjmtpp",
"output": "CHAT WITH HER!"
},
{
"input": "xiwntnheuitbtqxrmzvxmieldudakogealwrpygbxsbluhsqhtwmdlpjwzyafckrqrdduonkgo",
"output": "CHAT WITH HER!"
},
{
"input": "mnmbupgo",
"output": "IGNORE HIM!"
},
{
"input": "mcjehdiygkbmrbfjqwpwxidbdfelifwhstaxdapigbymmsgrhnzsdjhsqchl",
"output": "IGNORE HIM!"
},
{
"input": "yocxrzspinchmhtmqo",
"output": "CHAT WITH HER!"
},
{
"input": "vasvvnpymtgjirnzuynluluvmgpquskuaafwogeztfnvybblajvuuvfomtifeuzpikjrolzeeoftv",
"output": "CHAT WITH HER!"
},
{
"input": "ecsdicrznvglwggrdbrvehwzaenzjutjydhvimtqegweurpxtjkmpcznshtrvotkvrghxhacjkedidqqzrduzad",
"output": "IGNORE HIM!"
},
{
"input": "ubvhyaebyxoghakajqrpqpctwbrfqzli",
"output": "CHAT WITH HER!"
},
{
"input": "gogbxfeqylxoummvgxpkoqzsmobasesxbqjjktqbwqxeiaagnnhbvepbpy",
"output": "IGNORE HIM!"
},
{
"input": "nheihhxkbbrmlpxpxbhnpofcjmxemyvqqdbanwd",
"output": "IGNORE HIM!"
},
{
"input": "acrzbavz",
"output": "CHAT WITH HER!"
},
{
"input": "drvzznznvrzskftnrhvvzxcalwutxmdza",
"output": "IGNORE HIM!"
},
{
"input": "oacwxipdfcoabhkwxqdbtowiekpnflnqhlrkustgzryvws",
"output": "CHAT WITH HER!"
},
{
"input": "tpnwfmfsibnccvdwjvzviyvjfljupinfigfunyff",
"output": "CHAT WITH HER!"
},
{
"input": "gavaihhamfolcndgytcsgucqdqngxkrlovpthvteacmmthoglxu",
"output": "CHAT WITH HER!"
},
{
"input": "hsfcfvameeupldgvchmogrvwxrvsmnwxxkxoawwodtsahqvehlcw",
"output": "IGNORE HIM!"
},
{
"input": "sbkydrscoojychxchqsuciperfroumenelgiyiwlqfwximrgdbyvkmacy",
"output": "CHAT WITH HER!"
},
{
"input": "rhh",
"output": "CHAT WITH HER!"
},
{
"input": "zhdouqfmlkenjzdijxdfxnlegxeuvhelo",
"output": "IGNORE HIM!"
},
{
"input": "yufkkfwyhhvcjntsgsvpzbhqtmtgyxifqoewmuplphykmptfdebjxuaxigomjtwgtljwdjhjernkitifbomifbhysnmadtnyn",
"output": "CHAT WITH HER!"
},
{
"input": "urigreuzpxnej",
"output": "CHAT WITH HER!"
}
] | 1,697,154,509
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 85
| 124
| 0
|
name = input()
nameSet = set()
for char in name:
nameSet.add(char)
if len(nameSet) % 2 == 0:
print('CHAT WITH HER!')
else:
print('IGNORE HIM!')
|
Title: Boy or Girl
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input Specification:
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output Specification:
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Demo Input:
['wjmzbmr\n', 'xiaodao\n', 'sevenkplus\n']
Demo Output:
['CHAT WITH HER!\n', 'IGNORE HIM!\n', 'CHAT WITH HER!\n']
Note:
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
|
```python
name = input()
nameSet = set()
for char in name:
nameSet.add(char)
if len(nameSet) % 2 == 0:
print('CHAT WITH HER!')
else:
print('IGNORE HIM!')
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,686,765,603
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
from math import floor
a=list(map(int,input().split()))
print(floor(a[0]*a[1])/2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
from math import floor
a=list(map(int,input().split()))
print(floor(a[0]*a[1])/2)
```
| 0
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,648,455,004
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 62
| 0
|
s = input()
upper = ''
lower = ''
for i in s:
if i.isupper():
upper += i
else:
lower += i
if len(upper) > len(lower):
print(s.upper())
else:
print(s.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
s = input()
upper = ''
lower = ''
for i in s:
if i.isupper():
upper += i
else:
lower += i
if len(upper) > len(lower):
print(s.upper())
else:
print(s.lower())
```
| 3.9845
|
375
|
A
|
Divisible by Seven
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
|
The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters.
|
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number *a* in the required manner, print 0.
|
[
"1689\n",
"18906\n"
] |
[
"1869\n",
"18690\n"
] |
none
| 500
|
[
{
"input": "1689",
"output": "1869"
},
{
"input": "18906",
"output": "18690"
},
{
"input": "2419323689",
"output": "2432391689"
},
{
"input": "8589157262",
"output": "5857221986"
},
{
"input": "2717172350336955863014903670481525170997949309274087058935108848979319747543008692128164875210350026",
"output": "2771723503355630149036704815251709979493092740870589351088489793197475430086921281648752103500261986"
},
{
"input": "9825995656040286793128006047268547610068699214477842995873286607346639816314908021369221299622234988",
"output": "2599556040286793280060472685476100686992144778429958732866073466398163149080213692212996222349881968"
},
{
"input": "100000000689",
"output": "186900000000"
},
{
"input": "16891",
"output": "16198"
},
{
"input": "16892",
"output": "21896"
},
{
"input": "16893",
"output": "31689"
},
{
"input": "16894",
"output": "41986"
},
{
"input": "16895",
"output": "51968"
},
{
"input": "16896",
"output": "61698"
},
{
"input": "16897",
"output": "71869"
},
{
"input": "16898",
"output": "86198"
},
{
"input": "16899",
"output": "91896"
},
{
"input": "4048169",
"output": "4041968"
},
{
"input": "10994168",
"output": "94116890"
},
{
"input": "168903",
"output": "316890"
},
{
"input": "11689",
"output": "16198"
},
{
"input": "91111168",
"output": "11111968"
},
{
"input": "16890000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "18690000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1689999999999",
"output": "9999999991968"
},
{
"input": "9883291673084",
"output": "8329730841698"
},
{
"input": "6198",
"output": "1869"
}
] | 1,664,910,511
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 5,734,400
|
a=input()
cnt=[0,0,0,0,0,0,0,0,0,0]
for i in a:
cnt[int(i)]=cnt[int(i)]+1
cnt[1]=cnt[1]-1
cnt[6]=cnt[6]-1
cnt[8]=cnt[8]-1
cnt[9]=cnt[9]-1
u=""
for i in range(0,10):
for j in range(0,cnt[10-i-1]):
u=u+str(10-i-1)
print(u,end="")
if int(u)%7==1:
print(1896)
if int(u)%7==0:
print(1869)
if int(u)%7==2:
print(1986)
if int(u)%7==6:
print(1968)
if int(u)%7==3:
print(1698)
if int(u)%7==5:
print(1689)
if int(u)%7==4:
print(6891)
|
Title: Divisible by Seven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input Specification:
The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters.
Output Specification:
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number *a* in the required manner, print 0.
Demo Input:
['1689\n', '18906\n']
Demo Output:
['1869\n', '18690\n']
Note:
none
|
```python
a=input()
cnt=[0,0,0,0,0,0,0,0,0,0]
for i in a:
cnt[int(i)]=cnt[int(i)]+1
cnt[1]=cnt[1]-1
cnt[6]=cnt[6]-1
cnt[8]=cnt[8]-1
cnt[9]=cnt[9]-1
u=""
for i in range(0,10):
for j in range(0,cnt[10-i-1]):
u=u+str(10-i-1)
print(u,end="")
if int(u)%7==1:
print(1896)
if int(u)%7==0:
print(1869)
if int(u)%7==2:
print(1986)
if int(u)%7==6:
print(1968)
if int(u)%7==3:
print(1698)
if int(u)%7==5:
print(1689)
if int(u)%7==4:
print(6891)
```
| -1
|
|
593
|
A
|
2Char
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
|
The first line of the input contains number *n* (1<=≤<=*n*<=≤<=100) — the number of words in the article chosen by Andrew. Following are *n* lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
|
Print a single integer — the maximum possible total length of words in Andrew's article.
|
[
"4\nabb\ncacc\naaa\nbbb\n",
"5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa\n"
] |
[
"9",
"6"
] |
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
| 250
|
[
{
"input": "4\nabb\ncacc\naaa\nbbb",
"output": "9"
},
{
"input": "5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa",
"output": "6"
},
{
"input": "1\na",
"output": "1"
},
{
"input": "2\nz\nz",
"output": "2"
},
{
"input": "5\nabcde\nfghij\nklmno\npqrst\nuvwxy",
"output": "0"
},
{
"input": "6\ngggggg\ngggggg\ngggggg\ngggggg\ngggggg\ngggggg",
"output": "36"
},
{
"input": "6\naaaaaa\naaaaaa\nbbbbbb\nbbbbbb\naaabbb\nababab",
"output": "36"
},
{
"input": "1\nabc",
"output": "0"
},
{
"input": "2\nabc\nbca",
"output": "0"
},
{
"input": "3\nab\nba\nzzz",
"output": "4"
},
{
"input": "3\nab\nba\nzzzzz",
"output": "5"
},
{
"input": "5\nzzz\nzzzz\nzz\nz\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "37"
},
{
"input": "26\nq\nw\ne\nr\nt\ny\nu\ni\no\np\na\ns\nd\nf\ng\nh\nj\nk\nl\nz\nx\nc\nv\nb\nn\nm",
"output": "2"
},
{
"input": "5\nzzz\nzzzz\nzz\nz\naaaaaaaaaaaaaaaaaaaaaaaaaaaf",
"output": "28"
},
{
"input": "7\npavel\nerika\nalexxxxxxx\ngracio\nzhenya\nsudarev\nchelyaba",
"output": "0"
},
{
"input": "31\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml",
"output": "0"
},
{
"input": "5\nzloyfreid\ngraciocode\nschooldiary\nkazakov\nevgesha",
"output": "0"
},
{
"input": "4\nurkop\nvisualac\ngnutl\nwtf",
"output": "0"
},
{
"input": "3\naa\nb\nccc",
"output": "5"
},
{
"input": "3\na\nbd\ncaaaaaaa",
"output": "9"
},
{
"input": "4\naa\nax\nay\nxxxx",
"output": "8"
},
{
"input": "5\nc\nbb\ne\ndd\nf",
"output": "4"
},
{
"input": "2\naaaaa\naaaaa",
"output": "10"
}
] | 1,547,298,266
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 0
|
a='abcdefghijklmnopqrstuvwxyz'
n=int(input())
l=[]
ma=0
for i in range(n) :
l.append(input())
for i in range(len(a)) :
for j in range(len(a)) :
s=0
for x in l :
l1=set(x)
if len(l1)==2 :
if a[i] in l1 and a[j] in l1 :
s+=len(x)
if len(l1)==1 :
if a[i] in l1 or a[j] in l1 :
s+=len(x)
ma=max(ma,s)
print(ma)
|
Title: 2Char
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input Specification:
The first line of the input contains number *n* (1<=≤<=*n*<=≤<=100) — the number of words in the article chosen by Andrew. Following are *n* lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output Specification:
Print a single integer — the maximum possible total length of words in Andrew's article.
Demo Input:
['4\nabb\ncacc\naaa\nbbb\n', '5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa\n']
Demo Output:
['9', '6']
Note:
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
|
```python
a='abcdefghijklmnopqrstuvwxyz'
n=int(input())
l=[]
ma=0
for i in range(n) :
l.append(input())
for i in range(len(a)) :
for j in range(len(a)) :
s=0
for x in l :
l1=set(x)
if len(l1)==2 :
if a[i] in l1 and a[j] in l1 :
s+=len(x)
if len(l1)==1 :
if a[i] in l1 or a[j] in l1 :
s+=len(x)
ma=max(ma,s)
print(ma)
```
| 0
|
|
656
|
A
|
Da Vinci Powers
|
PROGRAMMING
| 1,900
|
[
"*special"
] | null | null |
The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
Output a single integer.
|
The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
|
Output a single integer.
|
[
"3\n",
"10\n"
] |
[
"8\n",
"1024\n"
] |
none
| 0
|
[
{
"input": "3",
"output": "8"
},
{
"input": "10",
"output": "1024"
},
{
"input": "35",
"output": "33940307968"
},
{
"input": "0",
"output": "1"
},
{
"input": "1",
"output": "2"
},
{
"input": "2",
"output": "4"
},
{
"input": "4",
"output": "16"
},
{
"input": "5",
"output": "32"
},
{
"input": "6",
"output": "64"
},
{
"input": "7",
"output": "128"
},
{
"input": "8",
"output": "256"
},
{
"input": "9",
"output": "512"
},
{
"input": "11",
"output": "2048"
},
{
"input": "12",
"output": "4096"
},
{
"input": "13",
"output": "8092"
},
{
"input": "14",
"output": "16184"
},
{
"input": "15",
"output": "32368"
},
{
"input": "16",
"output": "64736"
},
{
"input": "17",
"output": "129472"
},
{
"input": "18",
"output": "258944"
},
{
"input": "19",
"output": "517888"
},
{
"input": "20",
"output": "1035776"
},
{
"input": "21",
"output": "2071552"
},
{
"input": "22",
"output": "4143104"
},
{
"input": "23",
"output": "8286208"
},
{
"input": "24",
"output": "16572416"
},
{
"input": "25",
"output": "33144832"
},
{
"input": "26",
"output": "66289664"
},
{
"input": "27",
"output": "132579328"
},
{
"input": "28",
"output": "265158656"
},
{
"input": "29",
"output": "530317312"
},
{
"input": "30",
"output": "1060634624"
},
{
"input": "31",
"output": "2121269248"
},
{
"input": "32",
"output": "4242538496"
},
{
"input": "33",
"output": "8485076992"
},
{
"input": "34",
"output": "16970153984"
}
] | 1,491,020,756
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
//c++
#include<bits/stdc++.h>
using namespace std;
int main()
{
unsigned short int a;
cin>>a;
long long arr[36]={1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8092, 16184, 32368, 64736, 129472, 258944, 517888, 1035776, 2071552, 4143104, 8286208, 16572416, 33144832, 66289664, 132579328, 265158656, 530317312, 1060634624, 2121269248, 4242538496, 8485076992, 16970153984, 33940307968 };
cout<<arr[a];
return 0;
}
|
Title: Da Vinci Powers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
Output a single integer.
Input Specification:
The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
Output Specification:
Output a single integer.
Demo Input:
['3\n', '10\n']
Demo Output:
['8\n', '1024\n']
Note:
none
|
```python
//c++
#include<bits/stdc++.h>
using namespace std;
int main()
{
unsigned short int a;
cin>>a;
long long arr[36]={1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8092, 16184, 32368, 64736, 129472, 258944, 517888, 1035776, 2071552, 4143104, 8286208, 16572416, 33144832, 66289664, 132579328, 265158656, 530317312, 1060634624, 2121269248, 4242538496, 8485076992, 16970153984, 33940307968 };
cout<<arr[a];
return 0;
}
```
| -1
|
|
761
|
A
|
Dasha and Stairs
|
PROGRAMMING
| 1,000
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
On her way to programming school tiger Dasha faced her first test — a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct.
|
In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly.
|
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
|
[
"2 3\n",
"3 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.
| 500
|
[
{
"input": "2 3",
"output": "YES"
},
{
"input": "3 1",
"output": "NO"
},
{
"input": "5 4",
"output": "YES"
},
{
"input": "9 9",
"output": "YES"
},
{
"input": "85 95",
"output": "NO"
},
{
"input": "0 1",
"output": "YES"
},
{
"input": "89 25",
"output": "NO"
},
{
"input": "74 73",
"output": "YES"
},
{
"input": "62 39",
"output": "NO"
},
{
"input": "57 57",
"output": "YES"
},
{
"input": "100 99",
"output": "YES"
},
{
"input": "0 0",
"output": "NO"
},
{
"input": "98 100",
"output": "NO"
},
{
"input": "99 100",
"output": "YES"
},
{
"input": "1 0",
"output": "YES"
},
{
"input": "100 0",
"output": "NO"
},
{
"input": "0 100",
"output": "NO"
},
{
"input": "100 98",
"output": "NO"
},
{
"input": "100 100",
"output": "YES"
},
{
"input": "0 5",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
}
] | 1,485,890,957
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 93
| 4,608,000
|
n,m = map(int,input().split())
if (n+m)> 0 and abs(n-m)<= 1 :
print("YES")
else :
print("NO")
|
Title: Dasha and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On her way to programming school tiger Dasha faced her first test — a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct.
Input Specification:
In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly.
Output Specification:
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
Demo Input:
['2 3\n', '3 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.
|
```python
n,m = map(int,input().split())
if (n+m)> 0 and abs(n-m)<= 1 :
print("YES")
else :
print("NO")
```
| 3
|
|
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"
}
] | 1,655,878,113
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
x=input()
if x.count("h")>0 and x.count("e")>0 and x.count("l")>1 and x.count("o")>0:
print("YES")
else:
print("NO")
|
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
x=input()
if x.count("h")>0 and x.count("e")>0 and x.count("l")>1 and x.count("o")>0:
print("YES")
else:
print("NO")
```
| 0
|
31
|
A
|
Worms Evolution
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
A. Worms Evolution
|
2
|
256
|
Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.
|
Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.
|
[
"5\n1 2 3 5 7\n",
"5\n1 8 1 5 1\n"
] |
[
"3 2 1\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5\n1 2 3 5 7",
"output": "3 2 1"
},
{
"input": "5\n1 8 1 5 1",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "6\n86 402 133 524 405 610",
"output": "6 4 1"
},
{
"input": "8\n217 779 418 895 996 473 3 22",
"output": "5 2 1"
},
{
"input": "10\n858 972 670 15 662 114 33 273 53 310",
"output": "2 6 1"
},
{
"input": "100\n611 697 572 770 603 870 128 245 49 904 468 982 788 943 549 288 668 796 803 515 999 735 912 49 298 80 412 841 494 434 543 298 17 571 271 105 70 313 178 755 194 279 585 766 412 164 907 841 776 556 731 268 735 880 176 267 287 65 239 588 155 658 821 47 783 595 585 69 226 906 429 161 999 148 7 484 362 585 952 365 92 749 904 525 307 626 883 367 450 755 564 950 728 724 69 106 119 157 96 290",
"output": "1 38 25"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "1 63 61"
},
{
"input": "100\n156 822 179 298 981 82 610 345 373 378 895 734 768 15 78 335 764 608 932 297 717 553 916 367 425 447 361 195 66 70 901 236 905 744 919 564 296 610 963 628 840 52 100 750 345 308 37 687 192 704 101 815 10 990 216 358 823 546 578 821 706 148 182 582 421 482 829 425 121 337 500 301 402 868 66 935 625 527 746 585 308 523 488 914 608 709 875 252 151 781 447 2 756 176 976 302 450 35 680 791",
"output": "1 98 69"
},
{
"input": "100\n54 947 785 838 359 647 92 445 48 465 323 486 101 86 607 31 860 420 709 432 435 372 272 37 903 814 309 197 638 58 259 822 793 564 309 22 522 907 101 853 486 824 614 734 630 452 166 532 256 499 470 9 933 452 256 450 7 26 916 406 257 285 895 117 59 369 424 133 16 417 352 440 806 236 478 34 889 469 540 806 172 296 73 655 261 792 868 380 204 454 330 53 136 629 236 850 134 560 264 291",
"output": "2 29 27"
},
{
"input": "99\n175 269 828 129 499 890 127 263 995 807 508 289 996 226 437 320 365 642 757 22 190 8 345 499 834 713 962 889 336 171 608 492 320 257 472 801 176 325 301 306 198 729 933 4 640 322 226 317 567 586 249 237 202 633 287 128 911 654 719 988 420 855 361 574 716 899 317 356 581 440 284 982 541 111 439 29 37 560 961 224 478 906 319 416 736 603 808 87 762 697 392 713 19 459 262 238 239 599 997",
"output": "1 44 30"
},
{
"input": "98\n443 719 559 672 16 69 529 632 953 999 725 431 54 22 346 968 558 696 48 669 963 129 257 712 39 870 498 595 45 821 344 925 179 388 792 346 755 213 423 365 344 659 824 356 773 637 628 897 841 155 243 536 951 361 192 105 418 431 635 596 150 162 145 548 473 531 750 306 377 354 450 975 79 743 656 733 440 940 19 139 237 346 276 227 64 799 479 633 199 17 796 362 517 234 729 62 995 535",
"output": "2 70 40"
},
{
"input": "97\n359 522 938 862 181 600 283 1000 910 191 590 220 761 818 903 264 751 751 987 316 737 898 168 925 244 674 34 950 754 472 81 6 37 520 112 891 981 454 897 424 489 238 363 709 906 951 677 828 114 373 589 835 52 89 97 435 277 560 551 204 879 469 928 523 231 163 183 609 821 915 615 969 616 23 874 437 844 321 78 53 643 786 585 38 744 347 150 179 988 985 200 11 15 9 547 886 752",
"output": "1 23 10"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "100\n328 397 235 453 188 254 879 225 423 36 384 296 486 592 231 849 856 255 213 898 234 800 701 529 951 693 507 326 15 905 618 348 967 927 28 979 752 850 343 35 84 302 36 390 482 826 249 918 91 289 973 457 557 348 365 239 709 565 320 560 153 130 647 708 483 469 788 473 322 844 830 562 611 961 397 673 69 960 74 703 369 968 382 451 328 160 211 230 566 208 7 545 293 73 806 375 157 410 303 58",
"output": "1 79 6"
},
{
"input": "33\n52 145 137 734 180 847 178 286 716 134 181 630 358 764 593 762 785 28 1 468 189 540 764 485 165 656 114 58 628 108 605 584 257",
"output": "8 30 7"
},
{
"input": "57\n75 291 309 68 444 654 985 158 514 204 116 918 374 806 176 31 49 455 269 66 722 713 164 818 317 295 546 564 134 641 28 13 987 478 146 219 213 940 289 173 157 666 168 391 392 71 870 477 446 988 414 568 964 684 409 671 454",
"output": "2 41 29"
},
{
"input": "88\n327 644 942 738 84 118 981 686 530 404 137 197 434 16 693 183 423 325 410 345 941 329 7 106 79 867 584 358 533 675 192 718 641 329 900 768 404 301 101 538 954 590 401 954 447 14 559 337 756 586 934 367 538 928 945 936 770 641 488 579 206 869 902 139 216 446 723 150 829 205 373 578 357 368 960 40 121 206 503 385 521 161 501 694 138 370 709 308",
"output": "1 77 61"
},
{
"input": "100\n804 510 266 304 788 625 862 888 408 82 414 470 777 991 729 229 933 406 601 1 596 720 608 706 432 361 527 548 59 548 474 515 4 991 263 568 681 24 117 563 576 587 281 643 904 521 891 106 842 884 943 54 605 815 504 757 311 374 335 192 447 652 633 410 455 402 382 150 432 836 413 819 669 875 638 925 217 805 632 520 605 266 728 795 162 222 603 159 284 790 914 443 775 97 789 606 859 13 851 47",
"output": "1 77 42"
},
{
"input": "100\n449 649 615 713 64 385 927 466 138 126 143 886 80 199 208 43 196 694 92 89 264 180 617 970 191 196 910 150 275 89 693 190 191 99 542 342 45 592 114 56 451 170 64 589 176 102 308 92 402 153 414 675 352 157 69 150 91 288 163 121 816 184 20 234 836 12 593 150 793 439 540 93 99 663 186 125 349 247 476 106 77 523 215 7 363 278 441 745 337 25 148 384 15 915 108 211 240 58 23 408",
"output": "1 6 5"
},
{
"input": "90\n881 436 52 308 97 261 153 931 670 538 702 156 114 445 154 685 452 76 966 790 93 42 547 65 736 364 136 489 719 322 239 628 696 735 55 703 622 375 100 188 804 341 546 474 484 446 729 290 974 301 602 225 996 244 488 983 882 460 962 754 395 617 61 640 534 292 158 375 632 902 420 979 379 38 100 67 963 928 190 456 545 571 45 716 153 68 844 2 102 116",
"output": "1 14 2"
},
{
"input": "80\n313 674 262 240 697 146 391 221 793 504 896 818 92 899 86 370 341 339 306 887 937 570 830 683 729 519 240 833 656 847 427 958 435 704 853 230 758 347 660 575 843 293 649 396 437 787 654 599 35 103 779 783 447 379 444 585 902 713 791 150 851 228 306 721 996 471 617 403 102 168 197 741 877 481 968 545 331 715 236 654",
"output": "1 13 8"
},
{
"input": "70\n745 264 471 171 946 32 277 511 269 469 89 831 69 2 369 407 583 602 646 633 429 747 113 302 722 321 344 824 241 372 263 287 822 24 652 758 246 967 219 313 882 597 752 965 389 775 227 556 95 904 308 340 899 514 400 187 275 318 621 546 659 488 199 154 811 1 725 79 925 82",
"output": "1 63 60"
},
{
"input": "60\n176 502 680 102 546 917 516 801 392 435 635 492 398 456 653 444 472 513 634 378 273 276 44 920 68 124 800 167 825 250 452 264 561 344 98 933 381 939 426 51 568 548 206 887 342 763 151 514 156 354 486 546 998 649 356 438 295 570 450 589",
"output": "2 26 20"
},
{
"input": "50\n608 92 889 33 146 803 402 91 868 400 828 505 375 558 584 129 361 776 974 123 765 804 326 186 61 927 904 511 762 775 640 593 300 664 897 461 869 911 986 789 607 500 309 457 294 104 724 471 216 155",
"output": "3 25 11"
},
{
"input": "40\n40 330 98 612 747 336 640 381 991 366 22 167 352 12 868 166 603 40 313 869 609 981 609 804 54 729 8 854 347 300 828 922 39 633 695 988 4 530 545 176",
"output": "5 10 8"
},
{
"input": "30\n471 920 308 544 347 222 878 671 467 332 215 180 681 114 151 203 492 951 653 614 453 510 540 422 399 532 113 198 932 825",
"output": "2 21 9"
},
{
"input": "20\n551 158 517 475 595 108 764 961 590 297 761 841 659 568 82 888 733 214 993 359",
"output": "3 20 2"
},
{
"input": "10\n983 748 726 406 196 993 2 251 66 263",
"output": "-1"
},
{
"input": "9\n933 266 457 863 768 257 594 136 145",
"output": "-1"
},
{
"input": "8\n537 198 48 771 944 868 700 163",
"output": "7 8 1"
},
{
"input": "7\n140 779 639 679 768 479 158",
"output": "2 3 1"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "-1"
},
{
"input": "5\n700 939 173 494 120",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "3\n907 452 355",
"output": "-1"
},
{
"input": "3\n963 630 333",
"output": "1 3 2"
},
{
"input": "3\n2 2 4",
"output": "3 2 1"
},
{
"input": "3\n2 4 100",
"output": "-1"
}
] | 1,693,408,561
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 372
| 0
|
n=int(input())
arr=[int(i) for i in input().split()]
ansi=-1
ansj=-1
ansk=-1
for i in range(n):
for j in range(n):
for k in range(n):
if arr[i]== arr[j]+ arr[k]and j!=k:
ansi= i
ansj= j
ansk= k
if ansi==-1:
print(-1)
else:
print(ansi+1,ansj+1,ansk+1)
|
Title: Worms Evolution
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.
Output Specification:
Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.
Demo Input:
['5\n1 2 3 5 7\n', '5\n1 8 1 5 1\n']
Demo Output:
['3 2 1\n', '-1\n']
Note:
none
|
```python
n=int(input())
arr=[int(i) for i in input().split()]
ansi=-1
ansj=-1
ansk=-1
for i in range(n):
for j in range(n):
for k in range(n):
if arr[i]== arr[j]+ arr[k]and j!=k:
ansi= i
ansj= j
ansk= k
if ansi==-1:
print(-1)
else:
print(ansi+1,ansj+1,ansk+1)
```
| 3.907
|
837
|
C
|
Two Seals
|
PROGRAMMING
| 1,500
|
[
"brute force",
"implementation"
] | null | null |
One very important person has a piece of paper in the form of a rectangle *a*<=×<=*b*.
Also, he has *n* seals. Each seal leaves an impression on the paper in the form of a rectangle of the size *x**i*<=×<=*y**i*. Each impression must be parallel to the sides of the piece of paper (but seal can be rotated by 90 degrees).
A very important person wants to choose two different seals and put them two impressions. Each of the selected seals puts exactly one impression. Impressions should not overlap (but they can touch sides), and the total area occupied by them should be the largest possible. What is the largest area that can be occupied by two seals?
|
The first line contains three integer numbers *n*, *a* and *b* (1<=≤<=*n*,<=*a*,<=*b*<=≤<=100).
Each of the next *n* lines contain two numbers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=100).
|
Print the largest total area that can be occupied by two seals. If you can not select two seals, print 0.
|
[
"2 2 2\n1 2\n2 1\n",
"4 10 9\n2 3\n1 1\n5 10\n9 11\n",
"3 10 10\n6 6\n7 7\n20 5\n"
] |
[
"4\n",
"56\n",
"0\n"
] |
In the first example you can rotate the second seal by 90 degrees. Then put impression of it right under the impression of the first seal. This will occupy all the piece of paper.
In the second example you can't choose the last seal because it doesn't fit. By choosing the first and the third seals you occupy the largest area.
In the third example there is no such pair of seals that they both can fit on a piece of paper.
| 0
|
[
{
"input": "2 2 2\n1 2\n2 1",
"output": "4"
},
{
"input": "4 10 9\n2 3\n1 1\n5 10\n9 11",
"output": "56"
},
{
"input": "3 10 10\n6 6\n7 7\n20 5",
"output": "0"
},
{
"input": "2 1 1\n1 1\n1 1",
"output": "0"
},
{
"input": "2 1 2\n1 1\n1 1",
"output": "2"
},
{
"input": "2 100 100\n100 100\n1 1",
"output": "0"
},
{
"input": "2 100 100\n50 100\n100 50",
"output": "10000"
},
{
"input": "2 100 100\n100 100\n87 72",
"output": "0"
},
{
"input": "5 100 100\n100 100\n100 100\n100 100\n100 100\n100 100",
"output": "0"
},
{
"input": "15 50 50\n9 36\n28 14\n77 74\n35 2\n20 32\n83 85\n47 3\n41 50\n21 7\n38 46\n17 6\n79 90\n91 83\n9 33\n24 11",
"output": "2374"
},
{
"input": "15 100 100\n100 100\n100 100\n100 100\n42 58\n80 22\n100 100\n100 100\n100 100\n100 100\n100 100\n48 42\n100 100\n100 100\n100 100\n100 100",
"output": "4452"
},
{
"input": "30 100 100\n60 34\n29 82\n89 77\n39 1\n100 100\n82 12\n57 87\n93 43\n78 50\n38 55\n37 9\n67 5\n100 100\n100 100\n82 47\n3 71\n100 100\n19 26\n25 94\n89 5\n100 100\n32 1\n100 100\n34 3\n40 99\n100 100\n36 12\n100 100\n100 100\n100 100",
"output": "8958"
},
{
"input": "3 100 1\n1 50\n1 60\n1 30",
"output": "90"
},
{
"input": "3 1 60\n1 40\n2 2\n20 1",
"output": "60"
},
{
"input": "4 1 100\n1 25\n25 1\n1 25\n2 100",
"output": "50"
},
{
"input": "1 100 50\n4 20",
"output": "0"
},
{
"input": "2 2 4\n3 1\n2 2",
"output": "0"
},
{
"input": "2 2 4\n2 3\n2 1",
"output": "8"
},
{
"input": "2 4 2\n1 2\n2 3",
"output": "8"
},
{
"input": "2 1 4\n1 2\n1 2",
"output": "4"
},
{
"input": "2 4 5\n2 4\n4 3",
"output": "20"
},
{
"input": "2 1 4\n1 1\n3 3",
"output": "0"
},
{
"input": "6 9 5\n4 5\n6 2\n1 4\n5 6\n3 7\n6 5",
"output": "34"
},
{
"input": "6 8 5\n4 1\n3 3\n5 3\n6 7\n2 2\n5 4",
"output": "35"
},
{
"input": "6 7 5\n6 4\n5 7\n4 7\n5 4\n1 1\n3 6",
"output": "29"
},
{
"input": "6 9 7\n1 2\n1 5\n4 3\n4 7\n3 5\n6 7",
"output": "57"
},
{
"input": "6 5 9\n2 3\n7 4\n1 5\n1 7\n2 5\n7 1",
"output": "38"
},
{
"input": "2 4 2\n2 2\n1 3",
"output": "0"
},
{
"input": "2 3 2\n3 2\n1 1",
"output": "0"
},
{
"input": "6 7 5\n6 6\n4 7\n6 1\n4 1\n4 6\n1 5",
"output": "34"
},
{
"input": "2 2 3\n1 2\n2 3",
"output": "0"
},
{
"input": "2 2 2\n2 1\n1 1",
"output": "3"
},
{
"input": "5 9 7\n6 7\n4 5\n2 7\n4 2\n5 8",
"output": "56"
},
{
"input": "2 11 51\n1 10\n11 50",
"output": "560"
},
{
"input": "5 9 7\n3 8\n7 6\n4 1\n5 8\n7 8",
"output": "60"
},
{
"input": "2 4 6\n4 4\n4 2",
"output": "24"
},
{
"input": "5 9 7\n1 6\n7 9\n1 5\n1 5\n7 3",
"output": "27"
},
{
"input": "5 9 7\n5 2\n6 9\n1 4\n7 7\n6 4",
"output": "59"
},
{
"input": "5 9 7\n6 7\n4 1\n1 2\n4 7\n5 6",
"output": "58"
},
{
"input": "5 9 7\n2 8\n3 8\n2 8\n4 4\n2 2",
"output": "40"
},
{
"input": "2 2 3\n1 4\n2 1",
"output": "0"
},
{
"input": "5 9 7\n4 7\n3 9\n5 4\n3 4\n3 8",
"output": "55"
},
{
"input": "5 9 7\n7 4\n6 9\n4 3\n7 5\n2 3",
"output": "63"
},
{
"input": "2 2 3\n1 2\n2 2",
"output": "6"
},
{
"input": "2 4 3\n2 1\n1 2",
"output": "4"
},
{
"input": "2 4 6\n4 2\n4 4",
"output": "24"
},
{
"input": "2 1 4\n3 2\n3 3",
"output": "0"
}
] | 1,501,777,149
| 3,849
|
Python 3
|
OK
|
TESTS
| 51
| 93
| 4,915,200
|
def main():
n, a, b = [int(t) for t in input().split()]
points = []
for _ in range(n):
point = [int(t) for t in input().split()]
points.append(point)
largest_area = 0
for i, (x1, y1) in enumerate(points):
for j, (x2, y2) in enumerate(points):
if i == j:
continue
if satisfy(x1, y1, x2, y2, a, b):
area = x1 * y1 + x2 * y2
largest_area = max(area, largest_area)
print(largest_area)
def satisfy(x1, y1, x2, y2, a, b):
if satisfy_single(max((x1, x2)), sum((y1, y2)), a, b):
return True
if satisfy_single(sum((x1, x2)), max((y1, y2)), a, b):
return True
if satisfy_single(max((x1, y2)), sum((y1, x2)), a, b):
return True
if satisfy_single(sum((x1, y2)), max((y1, x2)), a, b):
return True
if satisfy_single(max((y1, x2)), sum((x1, y2)), a, b):
return True
if satisfy_single(sum((y1, x2)), max((x1, y2)), a, b):
return True
return False
def satisfy_single(x, y, a, b):
if x <= a and y <= b:
return True
if x <= b and y <= a:
return True
return False
if __name__ == '__main__':
main()
|
Title: Two Seals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One very important person has a piece of paper in the form of a rectangle *a*<=×<=*b*.
Also, he has *n* seals. Each seal leaves an impression on the paper in the form of a rectangle of the size *x**i*<=×<=*y**i*. Each impression must be parallel to the sides of the piece of paper (but seal can be rotated by 90 degrees).
A very important person wants to choose two different seals and put them two impressions. Each of the selected seals puts exactly one impression. Impressions should not overlap (but they can touch sides), and the total area occupied by them should be the largest possible. What is the largest area that can be occupied by two seals?
Input Specification:
The first line contains three integer numbers *n*, *a* and *b* (1<=≤<=*n*,<=*a*,<=*b*<=≤<=100).
Each of the next *n* lines contain two numbers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=100).
Output Specification:
Print the largest total area that can be occupied by two seals. If you can not select two seals, print 0.
Demo Input:
['2 2 2\n1 2\n2 1\n', '4 10 9\n2 3\n1 1\n5 10\n9 11\n', '3 10 10\n6 6\n7 7\n20 5\n']
Demo Output:
['4\n', '56\n', '0\n']
Note:
In the first example you can rotate the second seal by 90 degrees. Then put impression of it right under the impression of the first seal. This will occupy all the piece of paper.
In the second example you can't choose the last seal because it doesn't fit. By choosing the first and the third seals you occupy the largest area.
In the third example there is no such pair of seals that they both can fit on a piece of paper.
|
```python
def main():
n, a, b = [int(t) for t in input().split()]
points = []
for _ in range(n):
point = [int(t) for t in input().split()]
points.append(point)
largest_area = 0
for i, (x1, y1) in enumerate(points):
for j, (x2, y2) in enumerate(points):
if i == j:
continue
if satisfy(x1, y1, x2, y2, a, b):
area = x1 * y1 + x2 * y2
largest_area = max(area, largest_area)
print(largest_area)
def satisfy(x1, y1, x2, y2, a, b):
if satisfy_single(max((x1, x2)), sum((y1, y2)), a, b):
return True
if satisfy_single(sum((x1, x2)), max((y1, y2)), a, b):
return True
if satisfy_single(max((x1, y2)), sum((y1, x2)), a, b):
return True
if satisfy_single(sum((x1, y2)), max((y1, x2)), a, b):
return True
if satisfy_single(max((y1, x2)), sum((x1, y2)), a, b):
return True
if satisfy_single(sum((y1, x2)), max((x1, y2)), a, b):
return True
return False
def satisfy_single(x, y, a, b):
if x <= a and y <= b:
return True
if x <= b and y <= a:
return True
return False
if __name__ == '__main__':
main()
```
| 3
|
|
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"
}
] | 1,589,022,177
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 13
| 109
| 6,656,000
|
s = input()
l = list('hello#')
for x in s:
if x == l[0]:
l.pop(0)
if len(set(s)) > 4 and len(l) == 1:
print("YES")
else:
print("NO")
|
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
s = input()
l = list('hello#')
for x in s:
if x == l[0]:
l.pop(0)
if len(set(s)) > 4 and len(l) == 1:
print("YES")
else:
print("NO")
```
| 0
|
983
|
A
|
Finite or not?
|
PROGRAMMING
| 1,700
|
[
"implementation",
"math"
] | null | null |
You are given several queries. Each query consists of three integers $p$, $q$ and $b$. You need to answer whether the result of $p/q$ in notation with base $b$ is a finite fraction.
A fraction in notation with base $b$ is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
|
The first line contains a single integer $n$ ($1 \le n \le 10^5$) — the number of queries.
Next $n$ lines contain queries, one per line. Each line contains three integers $p$, $q$, and $b$ ($0 \le p \le 10^{18}$, $1 \le q \le 10^{18}$, $2 \le b \le 10^{18}$). All numbers are given in notation with base $10$.
|
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
|
[
"2\n6 12 10\n4 3 10\n",
"4\n1 1 2\n9 36 2\n4 12 3\n3 5 4\n"
] |
[
"Finite\nInfinite\n",
"Finite\nFinite\nFinite\nInfinite\n"
] |
$\frac{6}{12} = \frac{1}{2} = 0,5_{10}$
$\frac{4}{3} = 1,(3)_{10}$
$\frac{9}{36} = \frac{1}{4} = 0,01_2$
$\frac{4}{12} = \frac{1}{3} = 0,1_3$
| 500
|
[
{
"input": "2\n6 12 10\n4 3 10",
"output": "Finite\nInfinite"
},
{
"input": "4\n1 1 2\n9 36 2\n4 12 3\n3 5 4",
"output": "Finite\nFinite\nFinite\nInfinite"
},
{
"input": "10\n10 5 3\n1 7 10\n7 5 7\n4 4 9\n6 5 2\n6 7 5\n9 9 7\n7 5 5\n6 6 4\n10 8 2",
"output": "Finite\nInfinite\nInfinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nFinite"
},
{
"input": "10\n1 3 10\n6 2 6\n2 3 9\n7 8 4\n5 6 10\n1 2 7\n0 3 6\n9 3 4\n4 4 9\n10 9 10",
"output": "Infinite\nFinite\nFinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nInfinite"
},
{
"input": "10\n10 8 5\n0 6 9\n0 7 6\n5 7 3\n7 6 8\n0 4 8\n2 6 3\n10 2 9\n6 7 9\n9 1 4",
"output": "Infinite\nFinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nInfinite\nFinite"
},
{
"input": "10\n5 8 2\n0 5 8\n5 9 7\n0 7 2\n6 7 2\n10 3 7\n8 1 10\n9 1 8\n0 7 10\n9 1 4",
"output": "Finite\nFinite\nInfinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nFinite"
},
{
"input": "1\n1 864691128455135232 2",
"output": "Infinite"
},
{
"input": "11\n1 1000000000000000000 10000000\n2 999 9\n2 999 333111\n0 9 7\n17 128 2\n13 311992186885373952 18\n1971402979058461 750473176484995605 75\n14 19 23\n3 21914624432020321 23\n3 21914624432020321 46\n3 21914624432020321 47",
"output": "Finite\nInfinite\nFinite\nFinite\nFinite\nFinite\nFinite\nInfinite\nFinite\nFinite\nInfinite"
},
{
"input": "1\n1 100000000000000000 10000000000000000",
"output": "Finite"
},
{
"input": "1\n1 4294967297 4294967296",
"output": "Infinite"
},
{
"input": "1\n1 5244319080000 30030",
"output": "Finite"
}
] | 1,526,408,444
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 93
| 7,065,600
|
from math import*
n=int(input())
i=0
for i in range(n):
p,q,b=map(int,input().split())
x=(p/q)*b
while x!=int(x) and i!=13:
x=x*b
i+=1
if i==13:
print("Infinite")
else:
print("Finite")
|
Title: Finite or not?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given several queries. Each query consists of three integers $p$, $q$ and $b$. You need to answer whether the result of $p/q$ in notation with base $b$ is a finite fraction.
A fraction in notation with base $b$ is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 10^5$) — the number of queries.
Next $n$ lines contain queries, one per line. Each line contains three integers $p$, $q$, and $b$ ($0 \le p \le 10^{18}$, $1 \le q \le 10^{18}$, $2 \le b \le 10^{18}$). All numbers are given in notation with base $10$.
Output Specification:
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Demo Input:
['2\n6 12 10\n4 3 10\n', '4\n1 1 2\n9 36 2\n4 12 3\n3 5 4\n']
Demo Output:
['Finite\nInfinite\n', 'Finite\nFinite\nFinite\nInfinite\n']
Note:
$\frac{6}{12} = \frac{1}{2} = 0,5_{10}$
$\frac{4}{3} = 1,(3)_{10}$
$\frac{9}{36} = \frac{1}{4} = 0,01_2$
$\frac{4}{12} = \frac{1}{3} = 0,1_3$
|
```python
from math import*
n=int(input())
i=0
for i in range(n):
p,q,b=map(int,input().split())
x=(p/q)*b
while x!=int(x) and i!=13:
x=x*b
i+=1
if i==13:
print("Infinite")
else:
print("Finite")
```
| 0
|
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,552,796,053
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 108
| 0
|
a,b = list(map(int, input().strip().split(' ')))
mmin=a
for x in range(a,b-1):
if mmin==1 :
break
ax=x
bx=x+1
while bx!=0 :
tmp=ax
ax=bx
bx=tmp%bx
mmin=min(mmin,ax)
print(mmin)
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
a,b = list(map(int, input().strip().split(' ')))
mmin=a
for x in range(a,b-1):
if mmin==1 :
break
ax=x
bx=x+1
while bx!=0 :
tmp=ax
ax=bx
bx=tmp%bx
mmin=min(mmin,ax)
print(mmin)
```
| 0
|
|
835
|
C
|
Star sky
|
PROGRAMMING
| 1,600
|
[
"dp",
"implementation"
] | null | null |
The Cartesian coordinate system is set in the sky. There you can see *n* stars, the *i*-th has coordinates (*x**i*, *y**i*), a maximum brightness *c*, equal for all stars, and an initial brightness *s**i* (0<=≤<=*s**i*<=≤<=*c*).
Over time the stars twinkle. At moment 0 the *i*-th star has brightness *s**i*. Let at moment *t* some star has brightness *x*. Then at moment (*t*<=+<=1) this star will have brightness *x*<=+<=1, if *x*<=+<=1<=≤<=*c*, and 0, otherwise.
You want to look at the sky *q* times. In the *i*-th time you will look at the moment *t**i* and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (*x*1*i*, *y*1*i*) and the upper right — (*x*2*i*, *y*2*i*). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it.
|
The first line contains three integers *n*, *q*, *c* (1<=≤<=*n*,<=*q*<=≤<=105, 1<=≤<=*c*<=≤<=10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next *n* lines contain the stars description. The *i*-th from these lines contains three integers *x**i*, *y**i*, *s**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=100, 0<=≤<=*s**i*<=≤<=*c*<=≤<=10) — the coordinates of *i*-th star and its initial brightness.
The next *q* lines contain the views description. The *i*-th from these lines contains five integers *t**i*, *x*1*i*, *y*1*i*, *x*2*i*, *y*2*i* (0<=≤<=*t**i*<=≤<=109, 1<=≤<=*x*1*i*<=<<=*x*2*i*<=≤<=100, 1<=≤<=*y*1*i*<=<<=*y*2*i*<=≤<=100) — the moment of the *i*-th view and the coordinates of the viewed rectangle.
|
For each view print the total brightness of the viewed stars.
|
[
"2 3 3\n1 1 1\n3 2 0\n2 1 1 2 2\n0 2 1 4 5\n5 1 1 5 5\n",
"3 4 5\n1 1 2\n2 3 0\n3 3 1\n0 1 1 100 100\n1 2 2 4 4\n2 2 1 4 7\n1 50 50 51 51\n"
] |
[
"3\n0\n3\n",
"3\n3\n5\n0\n"
] |
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
| 1,250
|
[
{
"input": "2 3 3\n1 1 1\n3 2 0\n2 1 1 2 2\n0 2 1 4 5\n5 1 1 5 5",
"output": "3\n0\n3"
},
{
"input": "3 4 5\n1 1 2\n2 3 0\n3 3 1\n0 1 1 100 100\n1 2 2 4 4\n2 2 1 4 7\n1 50 50 51 51",
"output": "3\n3\n5\n0"
}
] | 1,621,146,133
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 264
| 1,024,000
|
n, q, c = list(map(int, input().split(' ')))
rng, C = 100, 10
sky = [[[None]*(rng+1) for _ in range(rng+1)] for _ in range(C+1)]
for i in range(n):
x, y, si = list(map(int, input().split(' ')))
for k in range(c+1):
sky[k][x][y] = (si+k)%(c+1)
for k in range(c+1):
# Cumulative row sum
for i in range(rng+1):
for j in range(1, rng+1):
if sky[k][i][j] is not None and sky[k][i][j-1] is not None:
sky[k][i][j] += sky[k][i][j-1]
elif sky[k][i][j-1] is not None:
sky[k][i][j] = sky[k][i][j-1]
# Cumulative column sum
for i in range(1, rng+1):
for j in range(rng+1):
if sky[k][i][j] is not None and sky[k][i-1][j] is not None:
sky[k][i][j] += sky[k][i-1][j]
elif sky[k][i-1][j] is not None:
sky[k][i][j] = sky[k][i-1][j]
# solve queries
for _ in range(q):
t, x1, y1, x2, y2 = list(map(int, input().split(' ')))
nt = t%(c+1)
ans = sky[nt][x2][y2]
ans -= (sky[nt][x1-1][y2] if x1>0 and sky[nt][x1-1][y2] is not None else 0)
ans -= (sky[nt][x2][y1-1] if y1>0 and sky[nt][x2][y1-1] is not None else 0)
ans += (sky[nt][x1-1][y1-1] if x1>0 and y1>0 and sky[nt][x1-1][y1-1] is not None else 0)
print(ans)
|
Title: Star sky
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Cartesian coordinate system is set in the sky. There you can see *n* stars, the *i*-th has coordinates (*x**i*, *y**i*), a maximum brightness *c*, equal for all stars, and an initial brightness *s**i* (0<=≤<=*s**i*<=≤<=*c*).
Over time the stars twinkle. At moment 0 the *i*-th star has brightness *s**i*. Let at moment *t* some star has brightness *x*. Then at moment (*t*<=+<=1) this star will have brightness *x*<=+<=1, if *x*<=+<=1<=≤<=*c*, and 0, otherwise.
You want to look at the sky *q* times. In the *i*-th time you will look at the moment *t**i* and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (*x*1*i*, *y*1*i*) and the upper right — (*x*2*i*, *y*2*i*). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it.
Input Specification:
The first line contains three integers *n*, *q*, *c* (1<=≤<=*n*,<=*q*<=≤<=105, 1<=≤<=*c*<=≤<=10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next *n* lines contain the stars description. The *i*-th from these lines contains three integers *x**i*, *y**i*, *s**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=100, 0<=≤<=*s**i*<=≤<=*c*<=≤<=10) — the coordinates of *i*-th star and its initial brightness.
The next *q* lines contain the views description. The *i*-th from these lines contains five integers *t**i*, *x*1*i*, *y*1*i*, *x*2*i*, *y*2*i* (0<=≤<=*t**i*<=≤<=109, 1<=≤<=*x*1*i*<=<<=*x*2*i*<=≤<=100, 1<=≤<=*y*1*i*<=<<=*y*2*i*<=≤<=100) — the moment of the *i*-th view and the coordinates of the viewed rectangle.
Output Specification:
For each view print the total brightness of the viewed stars.
Demo Input:
['2 3 3\n1 1 1\n3 2 0\n2 1 1 2 2\n0 2 1 4 5\n5 1 1 5 5\n', '3 4 5\n1 1 2\n2 3 0\n3 3 1\n0 1 1 100 100\n1 2 2 4 4\n2 2 1 4 7\n1 50 50 51 51\n']
Demo Output:
['3\n0\n3\n', '3\n3\n5\n0\n']
Note:
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
|
```python
n, q, c = list(map(int, input().split(' ')))
rng, C = 100, 10
sky = [[[None]*(rng+1) for _ in range(rng+1)] for _ in range(C+1)]
for i in range(n):
x, y, si = list(map(int, input().split(' ')))
for k in range(c+1):
sky[k][x][y] = (si+k)%(c+1)
for k in range(c+1):
# Cumulative row sum
for i in range(rng+1):
for j in range(1, rng+1):
if sky[k][i][j] is not None and sky[k][i][j-1] is not None:
sky[k][i][j] += sky[k][i][j-1]
elif sky[k][i][j-1] is not None:
sky[k][i][j] = sky[k][i][j-1]
# Cumulative column sum
for i in range(1, rng+1):
for j in range(rng+1):
if sky[k][i][j] is not None and sky[k][i-1][j] is not None:
sky[k][i][j] += sky[k][i-1][j]
elif sky[k][i-1][j] is not None:
sky[k][i][j] = sky[k][i-1][j]
# solve queries
for _ in range(q):
t, x1, y1, x2, y2 = list(map(int, input().split(' ')))
nt = t%(c+1)
ans = sky[nt][x2][y2]
ans -= (sky[nt][x1-1][y2] if x1>0 and sky[nt][x1-1][y2] is not None else 0)
ans -= (sky[nt][x2][y1-1] if y1>0 and sky[nt][x2][y1-1] is not None else 0)
ans += (sky[nt][x1-1][y1-1] if x1>0 and y1>0 and sky[nt][x1-1][y1-1] is not None else 0)
print(ans)
```
| 0
|
|
424
|
B
|
Megacity
|
PROGRAMMING
| 1,200
|
[
"binary search",
"greedy",
"implementation",
"sortings"
] | null | null |
The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.
The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with *n* other locations, the *i*-th one has coordinates (*x**i*, *y**i*) with the population of *k**i* people. You can widen the city boundaries to a circle of radius *r*. In such case all locations inside the circle and on its border are included into the city.
Your goal is to write a program that will determine the minimum radius *r*, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.
|
The first line of the input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=103; 1<=≤<=*s*<=<<=106) — the number of locatons around Tomsk city and the population of the city. Then *n* lines follow. The *i*-th line contains three integers — the *x**i* and *y**i* coordinate values of the *i*-th location and the number *k**i* of people in it (1<=≤<=*k**i*<=<<=106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.
It is guaranteed that no two locations are at the same point and no location is at point (0; 0).
|
In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.
The answer is considered correct if the absolute or relative error don't exceed 10<=-<=6.
|
[
"4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1\n",
"4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1\n",
"2 1\n1 1 999997\n2 2 1\n"
] |
[
"2.8284271\n",
"1.4142136\n",
"-1"
] |
none
| 1,000
|
[
{
"input": "4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1",
"output": "2.8284271"
},
{
"input": "4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1",
"output": "1.4142136"
},
{
"input": "2 1\n1 1 999997\n2 2 1",
"output": "-1"
},
{
"input": "4 999998\n3 3 10\n-3 3 10\n3 -3 10\n-3 -3 10",
"output": "4.2426407"
},
{
"input": "15 95473\n-9 6 199715\n0 -8 110607\n0 2 6621\n-3 -2 59894\n-10 -8 175440\n-2 0 25814\n10 -4 68131\n7 1 9971\n6 7 821\n6 5 20208\n6 2 68468\n0 7 37427\n1 -3 13337\n-10 7 113041\n-6 -2 44028",
"output": "12.8062485"
},
{
"input": "20 93350\n13 -28 486\n26 -26 48487\n5 -23 143368\n-23 -25 10371\n-2 -7 75193\n0 -8 3\n-6 -11 5015\n-19 -18 315278\n28 -15 45801\n21 8 4590\n-4 -28 12926\n-16 17 9405\n-28 -23 222092\n1 -10 1857\n14 -28 35170\n-4 -22 22036\n-2 -10 1260\n-1 12 375745\n-19 -24 38845\n10 -25 9256",
"output": "26.1725047"
},
{
"input": "30 505231\n-18 16 88130\n-10 16 15693\n16 -32 660\n-27 17 19042\n30 -37 6680\n36 19 299674\n-45 21 3300\n11 27 76\n-49 -34 28649\n-1 11 31401\n25 42 20858\n-40 6 455660\n-29 43 105001\n-38 10 6042\n19 -45 65551\n20 -9 148533\n-5 -24 393442\n-43 2 8577\n-39 18 97059\n12 28 39189\n35 23 28178\n40 -34 51687\n23 41 219028\n21 -44 927\n47 8 13206\n33 41 97342\n10 18 24895\n0 12 288\n0 -44 1065\n-25 43 44231",
"output": "24.5153013"
},
{
"input": "2 500000\n936 1000 500000\n961 976 500000",
"output": "1369.7065379"
},
{
"input": "10 764008\n959 32 23049\n-513 797 38979\n-603 -838 24916\n598 -430 25414\n-280 -624 18714\n330 891 21296\n-347 -68 27466\n650 -842 30125\n-314 889 35394\n275 969 5711",
"output": "1063.7029661"
},
{
"input": "30 295830\n1 -4 24773\n4 3 26175\n-2 -3 14789\n2 -1 46618\n-2 -2 52997\n-3 0 517\n-2 0 18173\n-4 -3 54465\n2 4 63579\n4 -4 41821\n2 2 11018\n0 4 42856\n0 -1 51885\n-3 4 57137\n3 0 4688\n0 2 60137\n-4 4 33484\n-1 3 66196\n3 -1 53634\n0 -2 41630\n-2 1 54606\n2 -2 2978\n2 -3 23733\n1 -2 35248\n-3 -3 15124\n-2 -4 26518\n4 0 28151\n4 -1 18348\n3 3 16914\n-4 2 26013",
"output": "4.4721360"
},
{
"input": "10 511500\n-5129 -3858 76954\n1296 1130 36126\n1219 6732 102003\n-8026 -178 4150\n-3261 1342 105429\n7965 -3013 62561\n5607 8963 53539\n-9044 -3999 16509\n1406 4103 115667\n-3716 2522 110626",
"output": "6841.4753526"
},
{
"input": "20 39342\n2 0 36476\n-3 1 136925\n1 3 31234\n0 -3 23785\n-1 3 77700\n-3 -1 50490\n-1 -3 13965\n-3 2 121093\n3 0 118933\n-3 0 125552\n-3 3 54779\n-2 0 96250\n1 2 142643\n2 2 23848\n0 2 29845\n0 -2 80462\n-1 1 91852\n-1 2 26526\n0 -1 136272\n1 1 108999",
"output": "3.0000000"
},
{
"input": "2 1\n1 0 1\n0 1 999999",
"output": "1.0000000"
},
{
"input": "2 999997\n1 1 1\n1 2 1",
"output": "-1"
}
] | 1,580,708,006
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 6
| 155
| 0
|
n,pop = list(map(int,input().split()))
arr = []
cnt = 0
for i in range(n):
x,y,p = list(map(int,input().split()))
cnt+=p
arr.append([abs(x),abs(y),p])
arr.sort()
if pop+cnt<1000000:
print(-1)
else:
cnt = 0
for i in range(n):
cnt+=arr[i][2]
if pop+cnt>=1000000:
u,v = arr[i][0],arr[i][1]
break
r = u**2+v**2
print(r**(0.5))
|
Title: Megacity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.
The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with *n* other locations, the *i*-th one has coordinates (*x**i*, *y**i*) with the population of *k**i* people. You can widen the city boundaries to a circle of radius *r*. In such case all locations inside the circle and on its border are included into the city.
Your goal is to write a program that will determine the minimum radius *r*, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.
Input Specification:
The first line of the input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=103; 1<=≤<=*s*<=<<=106) — the number of locatons around Tomsk city and the population of the city. Then *n* lines follow. The *i*-th line contains three integers — the *x**i* and *y**i* coordinate values of the *i*-th location and the number *k**i* of people in it (1<=≤<=*k**i*<=<<=106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.
It is guaranteed that no two locations are at the same point and no location is at point (0; 0).
Output Specification:
In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.
The answer is considered correct if the absolute or relative error don't exceed 10<=-<=6.
Demo Input:
['4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1\n', '4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1\n', '2 1\n1 1 999997\n2 2 1\n']
Demo Output:
['2.8284271\n', '1.4142136\n', '-1']
Note:
none
|
```python
n,pop = list(map(int,input().split()))
arr = []
cnt = 0
for i in range(n):
x,y,p = list(map(int,input().split()))
cnt+=p
arr.append([abs(x),abs(y),p])
arr.sort()
if pop+cnt<1000000:
print(-1)
else:
cnt = 0
for i in range(n):
cnt+=arr[i][2]
if pop+cnt>=1000000:
u,v = arr[i][0],arr[i][1]
break
r = u**2+v**2
print(r**(0.5))
```
| 0
|
|
362
|
B
|
Petya and Staircases
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
|
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
|
[
"10 5\n2 4 8 3 6\n",
"10 5\n2 4 5 7 9\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "10 5\n2 4 8 3 6",
"output": "NO"
},
{
"input": "10 5\n2 4 5 7 9",
"output": "YES"
},
{
"input": "10 9\n2 3 4 5 6 7 8 9 10",
"output": "NO"
},
{
"input": "5 2\n4 5",
"output": "NO"
},
{
"input": "123 13\n36 73 111 2 92 5 47 55 48 113 7 78 37",
"output": "YES"
},
{
"input": "10 10\n7 6 4 2 5 10 8 3 9 1",
"output": "NO"
},
{
"input": "12312 0",
"output": "YES"
},
{
"input": "9817239 1\n6323187",
"output": "YES"
},
{
"input": "1 1\n1",
"output": "NO"
},
{
"input": "5 4\n4 2 5 1",
"output": "NO"
},
{
"input": "5 3\n4 3 5",
"output": "NO"
},
{
"input": "500 3\n18 62 445",
"output": "YES"
},
{
"input": "500 50\n72 474 467 241 442 437 336 234 410 120 438 164 405 177 142 114 27 20 445 235 46 176 88 488 242 391 28 414 145 92 206 334 152 343 367 254 100 243 155 348 148 450 461 483 97 34 471 69 416 362",
"output": "NO"
},
{
"input": "500 8\n365 313 338 410 482 417 325 384",
"output": "YES"
},
{
"input": "1000000000 10\n2 3 5 6 8 9 123 874 1230 1000000000",
"output": "NO"
},
{
"input": "1000000000 10\n1 2 3 5 6 8 9 123 874 1230",
"output": "NO"
},
{
"input": "10 1\n1",
"output": "NO"
},
{
"input": "10 4\n1 2 4 5",
"output": "NO"
},
{
"input": "50 20\n22 33 17 23 27 5 26 31 41 20 8 24 6 3 4 29 40 25 13 16",
"output": "NO"
},
{
"input": "50 40\n14 27 19 30 31 20 28 11 37 29 23 33 7 26 22 16 1 6 18 3 47 36 38 2 48 9 41 8 5 50 4 45 44 25 39 12 43 42 40 46",
"output": "NO"
},
{
"input": "123 12\n35 95 47 99 79 122 58 94 31 57 18 10",
"output": "YES"
},
{
"input": "10 5\n1 3 5 7 9",
"output": "NO"
},
{
"input": "100 7\n2 3 5 6 8 9 100",
"output": "NO"
},
{
"input": "100 3\n98 99 100",
"output": "NO"
},
{
"input": "100 3\n97 98 99",
"output": "NO"
},
{
"input": "100 3\n96 98 99",
"output": "YES"
},
{
"input": "10 6\n2 3 5 6 8 9",
"output": "YES"
},
{
"input": "1000000000 10\n2 4 10 18 40 42 49 58 59 60",
"output": "NO"
},
{
"input": "10 3\n1 4 6",
"output": "NO"
},
{
"input": "8 3\n2 3 4",
"output": "NO"
},
{
"input": "100 3\n4 5 6",
"output": "NO"
},
{
"input": "10 2\n10 1",
"output": "NO"
},
{
"input": "10 1\n10",
"output": "NO"
},
{
"input": "4 2\n2 3",
"output": "YES"
},
{
"input": "2 1\n1",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "4 3\n2 3 4",
"output": "NO"
},
{
"input": "5 3\n4 2 3",
"output": "NO"
}
] | 1,607,051,751
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 307,200
|
n,m = map(int, input().split())
dirty_stairs = list(map(int, input().split()))
dirty_stairs.sort()
identifier = 1
if dirty_stairs[0] == 1 or dirty_stairs[-1] == n:
print("NO")
else:
for x in range(0, m-1):
if dirty_stairs[x] == dirty_stairs[x+1]-1 and dirty_stairs[x+1] == dirty_stairs[x+2]-1:
print("NO")
identifier = 0
break
if identifier == 1:
print("YES")
|
Title: Petya and Staircases
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
Output Specification:
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
Demo Input:
['10 5\n2 4 8 3 6\n', '10 5\n2 4 5 7 9\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n,m = map(int, input().split())
dirty_stairs = list(map(int, input().split()))
dirty_stairs.sort()
identifier = 1
if dirty_stairs[0] == 1 or dirty_stairs[-1] == n:
print("NO")
else:
for x in range(0, m-1):
if dirty_stairs[x] == dirty_stairs[x+1]-1 and dirty_stairs[x+1] == dirty_stairs[x+2]-1:
print("NO")
identifier = 0
break
if identifier == 1:
print("YES")
```
| 0
|
|
331
|
C1
|
The Great Julya Calendar
|
PROGRAMMING
| 1,100
|
[
"dp"
] | null | null |
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
|
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
|
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
|
[
"24\n"
] |
[
"5"
] |
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions:
| 20
|
[
{
"input": "24",
"output": "5"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "31",
"output": "6"
},
{
"input": "701",
"output": "116"
},
{
"input": "222",
"output": "39"
},
{
"input": "156",
"output": "28"
},
{
"input": "12343",
"output": "1778"
},
{
"input": "7984",
"output": "1178"
},
{
"input": "1000000",
"output": "128207"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "11",
"output": "3"
},
{
"input": "77",
"output": "14"
},
{
"input": "216",
"output": "37"
},
{
"input": "1468",
"output": "233"
},
{
"input": "1995",
"output": "305"
},
{
"input": "11010",
"output": "1568"
},
{
"input": "47320",
"output": "6746"
},
{
"input": "258634",
"output": "35024"
}
] | 1,679,671,924
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS1
| 6
| 92
| 0
|
n = int(input())
d = n//10
print(2*d + int(bool(n%10)))
|
Title: The Great Julya Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
Input Specification:
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
Output Specification:
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
Demo Input:
['24\n']
Demo Output:
['5']
Note:
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions:
|
```python
n = int(input())
d = n//10
print(2*d + int(bool(n%10)))
```
| 0
|
|
900
|
B
|
Position in Fraction
|
PROGRAMMING
| 1,300
|
[
"math",
"number theory"
] | null | null |
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
|
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
|
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
|
[
"1 2 0\n",
"2 3 7\n"
] |
[
"2",
"-1"
] |
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction.
| 1,000
|
[
{
"input": "1 2 0",
"output": "2"
},
{
"input": "2 3 7",
"output": "-1"
},
{
"input": "1 100000 1",
"output": "5"
},
{
"input": "1 7 7",
"output": "6"
},
{
"input": "99999 100000 8",
"output": "-1"
},
{
"input": "44102 73848 2",
"output": "132"
},
{
"input": "7 31 3",
"output": "15"
},
{
"input": "8880 81608 9",
"output": "161"
},
{
"input": "4942 62768 5",
"output": "122"
},
{
"input": "69168 84860 4",
"output": "107"
},
{
"input": "971 1883 3",
"output": "130"
},
{
"input": "1636 3269 6",
"output": "150"
},
{
"input": "6873 7769 3",
"output": "163"
},
{
"input": "13805 15538 3",
"output": "164"
},
{
"input": "10958 21926 3",
"output": "117"
},
{
"input": "8 51 0",
"output": "10"
},
{
"input": "1 10 1",
"output": "1"
},
{
"input": "1 9 0",
"output": "-1"
},
{
"input": "53 101 6",
"output": "-1"
},
{
"input": "1 10001 9",
"output": "5"
},
{
"input": "25102 31579 2",
"output": "174"
},
{
"input": "38790 39359 0",
"output": "212"
},
{
"input": "47117 78718 0",
"output": "213"
},
{
"input": "1 57 0",
"output": "1"
},
{
"input": "1 3 0",
"output": "-1"
},
{
"input": "1 100 0",
"output": "1"
},
{
"input": "2 3 0",
"output": "-1"
},
{
"input": "99971 99989 0",
"output": "9"
},
{
"input": "567 1580 0",
"output": "6"
},
{
"input": "45 97 0",
"output": "39"
},
{
"input": "35 111 4",
"output": "-1"
},
{
"input": "1 2 5",
"output": "1"
},
{
"input": "1 7 0",
"output": "-1"
}
] | 1,572,700,064
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 124
| 0
|
a,b,c=map(int,input().split())
a%=b
k=0
for i in range(1,b+1):
a*=10
if int(a/b)==c:
print(i)
k+=1
break
a%=b
if k==0:
print(-1)
|
Title: Position in Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
Input Specification:
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
Output Specification:
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Demo Input:
['1 2 0\n', '2 3 7\n']
Demo Output:
['2', '-1']
Note:
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction.
|
```python
a,b,c=map(int,input().split())
a%=b
k=0
for i in range(1,b+1):
a*=10
if int(a/b)==c:
print(i)
k+=1
break
a%=b
if k==0:
print(-1)
```
| 3
|
|
618
|
C
|
Constellation
|
PROGRAMMING
| 1,600
|
[
"geometry",
"implementation"
] | null | null |
Cat Noku has obtained a map of the night sky. On this map, he found a constellation with *n* stars numbered from 1 to *n*. For each *i*, the *i*-th star is located at coordinates (*x**i*,<=*y**i*). No two stars are located at the same position.
In the evening Noku is going to take a look at the night sky. He would like to find three distinct stars and form a triangle. The triangle must have positive area. In addition, all other stars must lie strictly outside of this triangle. He is having trouble finding the answer and would like your help. Your job is to find the indices of three stars that would form a triangle that satisfies all the conditions.
It is guaranteed that there is no line such that all stars lie on that line. It can be proven that if the previous condition is satisfied, there exists a solution to this problem.
|
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=100<=000).
Each of the next *n* lines contains two integers *x**i* and *y**i* (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109).
It is guaranteed that no two stars lie at the same point, and there does not exist a line such that all stars lie on that line.
|
Print three distinct integers on a single line — the indices of the three points that form a triangle that satisfies the conditions stated in the problem.
If there are multiple possible answers, you may print any of them.
|
[
"3\n0 1\n1 0\n1 1\n",
"5\n0 0\n0 2\n2 0\n2 2\n1 1\n"
] |
[
"1 2 3\n",
"1 3 5\n"
] |
In the first sample, we can print the three indices in any order.
In the second sample, we have the following picture.
Note that the triangle formed by starts 1, 4 and 3 doesn't satisfy the conditions stated in the problem, as point 5 is not strictly outside of this triangle (it lies on it's border).
| 1,500
|
[
{
"input": "3\n0 1\n1 0\n1 1",
"output": "1 2 3"
},
{
"input": "5\n0 0\n0 2\n2 0\n2 2\n1 1",
"output": "1 3 5"
},
{
"input": "3\n819934317 939682125\n487662889 8614219\n-557136619 382982369",
"output": "1 3 2"
},
{
"input": "10\n25280705 121178189\n219147240 -570920213\n-829849659 923854124\n18428128 -781819137\n-876779400 528386329\n-780997681 387686853\n-101900553 749998368\n58277314 355353788\n732128908 336416193\n840698381 600685123",
"output": "1 3 2"
},
{
"input": "10\n404775998 670757742\n30131431 723806809\n25599613 633170449\n13303280 387243789\n-33017802 -539177851\n1425218 149682549\n-47620079 -831223391\n-25996011 -398742031\n38471092 890600029\n-3745401 46270169",
"output": "1 2 3"
},
{
"input": "10\n13303280 387243789\n30131431 723806809\n404775998 670757742\n-25996011 -398742031\n25599613 633170449\n38471092 890600029\n-33017802 -539177851\n-47620079 -831223391\n1425218 149682549\n-3745401 46270169",
"output": "1 3 5"
},
{
"input": "10\n999999999 1\n999999998 1\n999999997 1\n1000000000 1\n999999996 1\n999999995 1\n999999994 1\n999999992 1\n999999993 1\n0 0",
"output": "1 2 10"
},
{
"input": "4\n0 1\n0 2\n0 3\n7 7",
"output": "1 4 2"
},
{
"input": "3\n0 0\n999999999 1\n999999998 1",
"output": "1 2 3"
},
{
"input": "10\n0 999999999\n0 1000000000\n-1 1000000000\n1 1000000000\n-2 1000000000\n2 1000000000\n-3 1000000000\n3 1000000000\n-4 1000000000\n4 1000000000",
"output": "1 2 3"
},
{
"input": "12\n1000000000 0\n1000000000 1\n1000000000 2\n1000000000 3\n1000000000 4\n1000000000 5\n1000000000 6\n1000000000 7\n1000000000 8\n1000000000 9\n1000000000 10\n999999999 5",
"output": "1 2 12"
},
{
"input": "12\n1000000000 0\n1000000000 1\n1000000000 2\n1000000000 3\n1000000000 4\n1000000000 5\n1000000000 6\n1000000000 7\n1000000000 8\n1000000000 9\n1000000000 10\n999999999 -1",
"output": "1 2 12"
},
{
"input": "12\n1000000000 0\n1000000000 1\n1000000000 2\n1000000000 3\n1000000000 4\n1000000000 5\n1000000000 6\n1000000000 7\n1000000000 8\n1000000000 9\n1000000000 10\n999999999 10",
"output": "1 2 12"
},
{
"input": "12\n1000000000 0\n1000000000 1\n1000000000 2\n1000000000 3\n1000000000 4\n1000000000 5\n1000000000 6\n1000000000 7\n1000000000 8\n1000000000 9\n1000000000 10\n999999999 1",
"output": "1 2 12"
},
{
"input": "11\n-1000000000 1\n-1000000000 2\n-1000000000 3\n-1000000000 4\n-1000000000 5\n-1000000000 6\n-1000000000 7\n-1000000000 8\n-1000000000 9\n-1000000000 10\n-999999999 5",
"output": "1 11 2"
},
{
"input": "11\n-1000000000 1\n-1000000000 2\n-1000000000 3\n-1000000000 4\n-1000000000 5\n-1000000000 6\n-1000000000 7\n-1000000000 8\n-1000000000 9\n-1000000000 10\n-999999999 7",
"output": "1 11 2"
},
{
"input": "11\n-1000000000 1\n-1000000000 2\n-1000000000 3\n-1000000000 4\n-1000000000 5\n-1000000000 6\n-1000000000 7\n-1000000000 8\n-1000000000 9\n-1000000000 10\n-999999999 8",
"output": "1 11 2"
},
{
"input": "11\n-1000000000 1\n-1000000000 2\n-1000000000 3\n-1000000000 4\n-1000000000 5\n-1000000000 6\n-1000000000 7\n-1000000000 8\n-1000000000 9\n-1000000000 10\n-999999999 10",
"output": "1 11 2"
},
{
"input": "11\n-1000000000 -1\n-1000000000 -2\n-1000000000 -3\n-1000000000 -4\n-1000000000 -5\n-1000000000 -6\n-1000000000 -7\n-1000000000 -8\n-1000000000 -9\n-1000000000 -10\n-999999999 -5",
"output": "1 2 11"
},
{
"input": "11\n-1000000000 -1\n-1000000000 -2\n-1000000000 -3\n-1000000000 -4\n-1000000000 -5\n-1000000000 -6\n-1000000000 -7\n-1000000000 -8\n-1000000000 -9\n-1000000000 -10\n-999999999 -1",
"output": "1 2 11"
},
{
"input": "11\n-1000000000 -1\n-1000000000 -2\n-1000000000 -3\n-1000000000 -4\n-1000000000 -5\n-1000000000 -6\n-1000000000 -7\n-1000000000 -8\n-1000000000 -9\n-1000000000 -10\n-999999999 -2",
"output": "1 2 11"
},
{
"input": "11\n-1000000000 -1\n-1000000000 -2\n-1000000000 -3\n-1000000000 -4\n-1000000000 -5\n-1000000000 -6\n-1000000000 -7\n-1000000000 -8\n-1000000000 -9\n-1000000000 -10\n-999999999 -4",
"output": "1 2 11"
},
{
"input": "11\n-1000000000 -1\n-1000000000 -2\n-1000000000 -3\n-1000000000 -4\n-1000000000 -5\n-1000000000 -6\n-1000000000 -7\n-1000000000 -8\n-1000000000 -9\n-1000000000 -10\n-999999999 -8",
"output": "1 2 11"
},
{
"input": "10\n2 1000000000\n8 1000000000\n9 1000000000\n3 1000000000\n4 1000000000\n5 1000000000\n6 1000000000\n1 1000000000\n7 1000000000\n0 0",
"output": "1 10 4"
},
{
"input": "10\n1000000000 1\n999999999 1\n999999998 1\n999999997 1\n999999996 1\n999999995 1\n999999994 1\n999999993 1\n999999992 1\n0 0",
"output": "1 2 10"
},
{
"input": "10\n999999999 1\n999999998 1\n999999997 1\n999999996 1\n999999995 1\n999999994 1\n999999993 1\n1000000000 1\n999999992 1\n0 0",
"output": "1 2 10"
},
{
"input": "4\n0 0\n1 0\n2 0\n1 100",
"output": "1 2 4"
},
{
"input": "4\n0 0\n3 0\n2 0\n1 1",
"output": "3 2 4"
},
{
"input": "4\n0 0\n1 1\n2 2\n3 4",
"output": "1 2 4"
},
{
"input": "4\n0 0\n0 1\n0 2\n1 1",
"output": "1 4 2"
},
{
"input": "4\n0 0\n2 0\n1 0\n1 1",
"output": "3 2 4"
},
{
"input": "4\n0 0\n1 1\n2 2\n5 -1",
"output": "1 4 2"
},
{
"input": "5\n0 1\n0 2\n0 3\n0 4\n10 10",
"output": "1 5 2"
},
{
"input": "4\n0 1\n0 2\n0 3\n1 1",
"output": "1 4 2"
},
{
"input": "4\n0 0\n1 0\n2 0\n2 1",
"output": "1 2 4"
},
{
"input": "4\n0 0\n-1 -1\n1 1\n100 0",
"output": "1 2 4"
},
{
"input": "4\n0 0\n2 0\n1 1\n1 0",
"output": "4 2 3"
},
{
"input": "4\n0 0\n1 0\n2 0\n3 1",
"output": "1 2 4"
},
{
"input": "3\n0 0\n12345691 12336918\n19349510 19335760",
"output": "1 3 2"
},
{
"input": "21\n0 19\n0 0\n0 8\n0 2\n0 18\n0 17\n0 1\n0 5\n0 16\n0 11\n0 10\n0 13\n0 12\n0 14\n0 6\n0 7\n0 3\n0 15\n0 4\n0 9\n1 1",
"output": "7 2 21"
},
{
"input": "10\n0 0\n1 -100\n1 100\n1 50\n1 0\n1 -50\n1 10\n1 -10\n1 5\n1 -5",
"output": "1 2 6"
},
{
"input": "3\n1 2\n2 1\n2 3",
"output": "1 2 3"
},
{
"input": "3\n-1000000000 -1000000000\n1000000000 -1000000000\n-1000000000 1000000000",
"output": "1 2 3"
},
{
"input": "10\n0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 1\n9 0",
"output": "1 2 9"
},
{
"input": "4\n1 1\n2 2\n3 3\n10 11",
"output": "1 2 4"
},
{
"input": "4\n0 0\n0 2\n0 1\n3 3",
"output": "1 4 3"
},
{
"input": "4\n0 0\n2 2\n1 1\n2 0",
"output": "1 4 3"
},
{
"input": "4\n0 1\n0 0\n0 5\n1 1",
"output": "1 2 4"
},
{
"input": "4\n1 0\n2 0\n3 0\n-7 -7",
"output": "1 4 2"
},
{
"input": "4\n0 0\n0 2\n0 1\n10 10",
"output": "1 4 3"
},
{
"input": "4\n-50000000 204926\n0 0\n8192 50000000\n16384 100000000",
"output": "1 2 3"
},
{
"input": "4\n65537 536870912\n0 536805376\n1 536870912\n-8191 0",
"output": "1 3 2"
},
{
"input": "4\n0 0\n131072 0\n131072 131072\n200000 0",
"output": "1 2 3"
},
{
"input": "3\n-536870912 10\n536870912 11\n-536870912 6",
"output": "1 3 2"
},
{
"input": "4\n3 7\n2 4\n1 2\n0 0",
"output": "1 3 2"
},
{
"input": "4\n0 0\n0 1\n0 2\n3 3",
"output": "1 4 2"
}
] | 1,454,133,412
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 98
| 748
| 13,004,800
|
n = int(input())
points = [list(map(int, input().split(' ')))+[i+1] for i in range(n)]
points.sort()
i, j, k = 0, 1, 2
while True:
if points[i][0] == points[j][0] and points[j][0] == points[k][0]:
k += 1
elif points[i][0] != points[j][0] and points[i][0] != points[k][0] and\
(points[j][1]-points[i][1])/(points[j][0]-points[i][0]) == (points[k][1]-points[i][1])/(points[k][0]-points[i][0]):
k += 1
else:
break
print(points[i][2], points[j][2], points[k][2])
|
Title: Constellation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Cat Noku has obtained a map of the night sky. On this map, he found a constellation with *n* stars numbered from 1 to *n*. For each *i*, the *i*-th star is located at coordinates (*x**i*,<=*y**i*). No two stars are located at the same position.
In the evening Noku is going to take a look at the night sky. He would like to find three distinct stars and form a triangle. The triangle must have positive area. In addition, all other stars must lie strictly outside of this triangle. He is having trouble finding the answer and would like your help. Your job is to find the indices of three stars that would form a triangle that satisfies all the conditions.
It is guaranteed that there is no line such that all stars lie on that line. It can be proven that if the previous condition is satisfied, there exists a solution to this problem.
Input Specification:
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=100<=000).
Each of the next *n* lines contains two integers *x**i* and *y**i* (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109).
It is guaranteed that no two stars lie at the same point, and there does not exist a line such that all stars lie on that line.
Output Specification:
Print three distinct integers on a single line — the indices of the three points that form a triangle that satisfies the conditions stated in the problem.
If there are multiple possible answers, you may print any of them.
Demo Input:
['3\n0 1\n1 0\n1 1\n', '5\n0 0\n0 2\n2 0\n2 2\n1 1\n']
Demo Output:
['1 2 3\n', '1 3 5\n']
Note:
In the first sample, we can print the three indices in any order.
In the second sample, we have the following picture.
Note that the triangle formed by starts 1, 4 and 3 doesn't satisfy the conditions stated in the problem, as point 5 is not strictly outside of this triangle (it lies on it's border).
|
```python
n = int(input())
points = [list(map(int, input().split(' ')))+[i+1] for i in range(n)]
points.sort()
i, j, k = 0, 1, 2
while True:
if points[i][0] == points[j][0] and points[j][0] == points[k][0]:
k += 1
elif points[i][0] != points[j][0] and points[i][0] != points[k][0] and\
(points[j][1]-points[i][1])/(points[j][0]-points[i][0]) == (points[k][1]-points[i][1])/(points[k][0]-points[i][0]):
k += 1
else:
break
print(points[i][2], points[j][2], points[k][2])
```
| 3
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,674,391,184
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
def findSquare(n:int, m:int, a:int):
counting = 0
sideN = n
sideM = m
while (n > 0 and m > 0):
if n >= a and m >= a:
n -= a
m -= a
counting += 1
elif n >= a:
n -= a
counting += 1
elif m >= a:
m -= a
counting += 1
else:
if n != 0 and sideM >= a:
sideM -= a
counting += 1
if m != 0 and sideN >= a:
sideN -= a
counting += 1
if sideM < a and sideN < a:
n = 0
m = 0
counting += 1
return counting
if __name__ == "__main__":
print(findSquare(6,6,4))
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
def findSquare(n:int, m:int, a:int):
counting = 0
sideN = n
sideM = m
while (n > 0 and m > 0):
if n >= a and m >= a:
n -= a
m -= a
counting += 1
elif n >= a:
n -= a
counting += 1
elif m >= a:
m -= a
counting += 1
else:
if n != 0 and sideM >= a:
sideM -= a
counting += 1
if m != 0 and sideN >= a:
sideN -= a
counting += 1
if sideM < a and sideN < a:
n = 0
m = 0
counting += 1
return counting
if __name__ == "__main__":
print(findSquare(6,6,4))
```
| 0
|
688
|
B
|
Lovely Palindromes
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
|
Print the *n*-th even-length palindrome number.
|
[
"1\n",
"10\n"
] |
[
"11\n",
"1001\n"
] |
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
| 1,000
|
[
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "2",
"output": "22"
},
{
"input": "3",
"output": "33"
},
{
"input": "4",
"output": "44"
},
{
"input": "5",
"output": "55"
},
{
"input": "6",
"output": "66"
},
{
"input": "7",
"output": "77"
},
{
"input": "8",
"output": "88"
},
{
"input": "9",
"output": "99"
},
{
"input": "13",
"output": "1331"
},
{
"input": "14",
"output": "1441"
},
{
"input": "15",
"output": "1551"
},
{
"input": "16",
"output": "1661"
},
{
"input": "17",
"output": "1771"
},
{
"input": "18",
"output": "1881"
},
{
"input": "19",
"output": "1991"
},
{
"input": "20",
"output": "2002"
},
{
"input": "26550",
"output": "2655005562"
},
{
"input": "16137",
"output": "1613773161"
},
{
"input": "91471",
"output": "9147117419"
},
{
"input": "41242",
"output": "4124224214"
},
{
"input": "30866",
"output": "3086666803"
},
{
"input": "4442231232741003009964183199713851566466079882929007253443848371978048610682180431295671905105980116",
"output": "44422312327410030099641831997138515664660798829290072534438483719780486106821804312956719051059801166110895015091765921340812860168408791738483443527009292889706646651583179913814699003001472321322444"
},
{
"input": "7806526084732010742850470902979267046151204705563201965818390803445852115032119297648697933316807252",
"output": "78065260847320107428504709029792670461512047055632019658183908034458521150321192976486979333168072522527086133397968467929112305112585443080938185691023655074021516407629792090740582470102374806256087"
},
{
"input": "2244399823612183124978103773977739948197709395321563148611877999503840818596183803323258317041875913",
"output": "22443998236121831249781037739777399481977093953215631486118779995038408185961838033232583170418759133195781407138523233083816958180483059997781168413651235939077918499377793773018794213812163289934422"
},
{
"input": "123451",
"output": "123451154321"
}
] | 1,610,494,577
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 92
| 921,600
|
n = str(input())
res = n
for j in range(len(n) - 1, -1, -1):
res += n[j]
print(res)
|
Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
|
```python
n = str(input())
res = n
for j in range(len(n) - 1, -1, -1):
res += n[j]
print(res)
```
| 3
|
|
585
|
E
|
Present for Vitalik the Philatelist
|
PROGRAMMING
| 2,900
|
[
"combinatorics",
"math",
"number theory"
] | null | null |
Vitalik the philatelist has a birthday today!
As he is a regular customer in a stamp store called 'Robin Bobin', the store management decided to make him a gift.
Vitalik wants to buy one stamp and the store will give him a non-empty set of the remaining stamps, such that the greatest common divisor (GCD) of the price of the stamps they give to him is more than one. If the GCD of prices of the purchased stamp and prices of present stamps set will be equal to 1, then Vitalik will leave the store completely happy.
The store management asks you to count the number of different situations in which Vitalik will leave the store completely happy. Since the required number of situations can be very large, you need to find the remainder of this number modulo 109<=+<=7. The situations are different if the stamps purchased by Vitalik are different, or if one of the present sets contains a stamp that the other present does not contain.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=5·105) — the number of distinct stamps, available for sale in the 'Robin Bobin' store.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (2<=≤<=*a**i*<=≤<=107), where *a**i* is the price of the *i*-th stamp.
|
Print a single integer — the remainder of the sought number of situations modulo 109<=+<=7.
|
[
"3\n2 3 2\n",
"2\n9 6\n"
] |
[
"5\n",
"0\n"
] |
In the first sample the following situations are possible:
- Vitalik buys the 1-st stamp, the store gives him the 2-nd stamp as a present; - Vitalik buys the 3-rd stamp, the store gives him the 2-nd stamp as a present; - Vitalik buys the 2-nd stamp, the store gives him the 1-st stamp as a present; - Vitalik buys the 2-nd stamp, the store gives him the 3-rd stamp as a present; - Vitalik buys the 2-nd stamp, the store gives him the 1-st and 3-rd stamps as a present.
| 2,250
|
[] | 1,692,727,875
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1692727875.6566317")# 1692727875.6566486
|
Title: Present for Vitalik the Philatelist
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitalik the philatelist has a birthday today!
As he is a regular customer in a stamp store called 'Robin Bobin', the store management decided to make him a gift.
Vitalik wants to buy one stamp and the store will give him a non-empty set of the remaining stamps, such that the greatest common divisor (GCD) of the price of the stamps they give to him is more than one. If the GCD of prices of the purchased stamp and prices of present stamps set will be equal to 1, then Vitalik will leave the store completely happy.
The store management asks you to count the number of different situations in which Vitalik will leave the store completely happy. Since the required number of situations can be very large, you need to find the remainder of this number modulo 109<=+<=7. The situations are different if the stamps purchased by Vitalik are different, or if one of the present sets contains a stamp that the other present does not contain.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=5·105) — the number of distinct stamps, available for sale in the 'Robin Bobin' store.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (2<=≤<=*a**i*<=≤<=107), where *a**i* is the price of the *i*-th stamp.
Output Specification:
Print a single integer — the remainder of the sought number of situations modulo 109<=+<=7.
Demo Input:
['3\n2 3 2\n', '2\n9 6\n']
Demo Output:
['5\n', '0\n']
Note:
In the first sample the following situations are possible:
- Vitalik buys the 1-st stamp, the store gives him the 2-nd stamp as a present; - Vitalik buys the 3-rd stamp, the store gives him the 2-nd stamp as a present; - Vitalik buys the 2-nd stamp, the store gives him the 1-st stamp as a present; - Vitalik buys the 2-nd stamp, the store gives him the 3-rd stamp as a present; - Vitalik buys the 2-nd stamp, the store gives him the 1-st and 3-rd stamps as a present.
|
```python
print("_RANDOM_GUESS_1692727875.6566317")# 1692727875.6566486
```
| 0
|
|
124
|
A
|
The number of positions
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
|
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
|
Print the single number — the number of the sought positions.
|
[
"3 1 1\n",
"5 2 3\n"
] |
[
"2\n",
"3\n"
] |
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
| 500
|
[
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 7",
"output": "5"
},
{
"input": "14 5 5",
"output": "6"
},
{
"input": "16 6 9",
"output": "10"
},
{
"input": "20 13 17",
"output": "7"
},
{
"input": "22 4 8",
"output": "9"
},
{
"input": "23 8 14",
"output": "15"
},
{
"input": "26 18 22",
"output": "8"
},
{
"input": "28 6 1",
"output": "2"
},
{
"input": "29 5 23",
"output": "24"
},
{
"input": "32 27 15",
"output": "5"
},
{
"input": "33 11 5",
"output": "6"
},
{
"input": "37 21 15",
"output": "16"
},
{
"input": "39 34 33",
"output": "5"
},
{
"input": "41 27 11",
"output": "12"
},
{
"input": "42 25 16",
"output": "17"
},
{
"input": "45 7 43",
"output": "38"
},
{
"input": "47 16 17",
"output": "18"
},
{
"input": "49 11 37",
"output": "38"
},
{
"input": "51 38 39",
"output": "13"
},
{
"input": "52 29 7",
"output": "8"
},
{
"input": "56 43 12",
"output": "13"
},
{
"input": "58 57 28",
"output": "1"
},
{
"input": "59 12 39",
"output": "40"
},
{
"input": "62 9 52",
"output": "53"
},
{
"input": "63 29 44",
"output": "34"
},
{
"input": "65 30 22",
"output": "23"
},
{
"input": "66 27 38",
"output": "39"
},
{
"input": "71 33 53",
"output": "38"
},
{
"input": "73 14 12",
"output": "13"
},
{
"input": "73 37 35",
"output": "36"
},
{
"input": "76 69 44",
"output": "7"
},
{
"input": "79 25 20",
"output": "21"
},
{
"input": "81 60 20",
"output": "21"
},
{
"input": "81 79 14",
"output": "2"
},
{
"input": "84 0 42",
"output": "43"
},
{
"input": "88 79 8",
"output": "9"
},
{
"input": "90 76 59",
"output": "14"
},
{
"input": "92 2 22",
"output": "23"
},
{
"input": "94 5 88",
"output": "89"
},
{
"input": "94 62 48",
"output": "32"
},
{
"input": "96 22 72",
"output": "73"
},
{
"input": "100 11 88",
"output": "89"
},
{
"input": "100 81 91",
"output": "19"
},
{
"input": "1 0 0",
"output": "1"
}
] | 1,671,709,527
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
a,b,c=map(int,input().split())
print(a-min(b,c))
|
Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
|
```python
a,b,c=map(int,input().split())
print(a-min(b,c))
```
| 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"
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{
"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"
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{
"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"
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{
"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"
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{
"input": "4 5\n3 5 5 5",
"output": "0"
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{
"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"
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{
"input": "2 1\n1 1",
"output": "0"
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{
"input": "4 10\n9 10 10 10",
"output": "0"
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{
"input": "5 2\n2 2 2 2 2",
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{
"input": "2 5\n4 5",
"output": "0"
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{
"input": "5 10\n10 10 10 10 10",
"output": "0"
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{
"input": "2 6\n6 6",
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{
"input": "2 9\n9 9",
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},
{
"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",
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},
{
"input": "1 5\n5",
"output": "0"
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{
"input": "3 1\n1 1 1",
"output": "0"
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{
"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",
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},
{
"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",
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{
"input": "2 2\n2 1",
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{
"input": "5 5\n5 5 5 4 5",
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{
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{
"input": "2 5\n5 5",
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{
"input": "1 7\n1",
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},
{
"input": "6 7\n1 1 1 1 1 1",
"output": "66"
},
{
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{
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{
"input": "3 2\n2 2 2",
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{
"input": "6 5\n5 5 5 5 5 5",
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{
"input": "10 5\n5 5 5 5 5 5 5 4 1 1",
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},
{
"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"
}
] | 1,499,705,628
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 11
| 62
| 5,529,600
|
n,k = map(int,input().split())
lists = list(map(int,input().split()))
sums = sum(lists)
l,r = 0,10**1000
m = (l+r)//2
while r-l != 1:
if 2*(sums+k*m) >= (2*k-1)*(len(lists)+m):
r = m
m = (l+r)//2
else:
l = m
m = (l+r)//2
print(r)
|
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
n,k = map(int,input().split())
lists = list(map(int,input().split()))
sums = sum(lists)
l,r = 0,10**1000
m = (l+r)//2
while r-l != 1:
if 2*(sums+k*m) >= (2*k-1)*(len(lists)+m):
r = m
m = (l+r)//2
else:
l = m
m = (l+r)//2
print(r)
```
| 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"
}
] | 1,603,055,334
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 218
| 0
|
n=int(input())
sol="NO"
# n(n+1)/2=tn
for i in range(n+1):
if int(i*(i+1)/2)==n:
sol="YES"
break
print(sol)
|
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
n=int(input())
sol="NO"
# n(n+1)/2=tn
for i in range(n+1):
if int(i*(i+1)/2)==n:
sol="YES"
break
print(sol)
```
| 3.9455
|
965
|
A
|
Paper Airplanes
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
|
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
|
Print a single integer — the minimum number of packs they should buy.
|
[
"5 3 2 3\n",
"5 3 100 1\n"
] |
[
"4\n",
"5\n"
] |
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
| 500
|
[
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51",
"output": "6"
},
{
"input": "2 400 23 57",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10000 10000 3 2",
"output": "16670000"
},
{
"input": "1 2 1 2",
"output": "1"
},
{
"input": "1 1 2 2",
"output": "1"
},
{
"input": "1 1 10 10",
"output": "1"
},
{
"input": "5324 5439 32 13",
"output": "69622"
},
{
"input": "9000 1 2432 1",
"output": "9000"
},
{
"input": "230 1234 9124 23",
"output": "10"
},
{
"input": "11 1 1 1",
"output": "11"
},
{
"input": "6246 8489 1227 9",
"output": "4858"
},
{
"input": "9 20 5 7",
"output": "6"
}
] | 1,668,142,673
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 18
| 62
| 0
|
from math import ceil
k,n,s,p = map(float,input().split())
print(ceil((ceil(n/s)*k)/p))
|
Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
|
```python
from math import ceil
k,n,s,p = map(float,input().split())
print(ceil((ceil(n/s)*k)/p))
```
| 3
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,605,528,890
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 186
| 0
|
# n=int(input())
# s1=0
# s2=0
# s3=0
# for _ in range(n):
# x,y,z=(map(int,input().split()))
# s1=s1+x
# s2=s2+y
# s3=s3+z
#
# if s1==0 and s2==0 and s3==0:
# print("YES")
# else:
# print("NO")
# l=[]
# for i in range(5):
# arr=list(map(int,input().split()))
# l.append(arr)
# # print(l)
# x,y=3,3
# for i in range(1,6):
# for j in range(1,6):
# if l[i-1][j-1]==1:
# print(abs(x-i)+abs(y-j))
# n,t=map(int,input().split())
# s=input()
#
# for i in range(t):
#
#
# s=s.replace("BG",'GB')
#
#
# print(s)
s=input()
for i in range(len(s)):
s = s.replace("--", '2')
s = s.replace("-.", "1")
s=s.replace(".",'0')
print(s)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
# n=int(input())
# s1=0
# s2=0
# s3=0
# for _ in range(n):
# x,y,z=(map(int,input().split()))
# s1=s1+x
# s2=s2+y
# s3=s3+z
#
# if s1==0 and s2==0 and s3==0:
# print("YES")
# else:
# print("NO")
# l=[]
# for i in range(5):
# arr=list(map(int,input().split()))
# l.append(arr)
# # print(l)
# x,y=3,3
# for i in range(1,6):
# for j in range(1,6):
# if l[i-1][j-1]==1:
# print(abs(x-i)+abs(y-j))
# n,t=map(int,input().split())
# s=input()
#
# for i in range(t):
#
#
# s=s.replace("BG",'GB')
#
#
# print(s)
s=input()
for i in range(len(s)):
s = s.replace("--", '2')
s = s.replace("-.", "1")
s=s.replace(".",'0')
print(s)
```
| 3.9535
|
450
|
B
|
Jzzhu and Sequences
|
PROGRAMMING
| 1,300
|
[
"implementation",
"math"
] | null | null |
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
|
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
|
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
|
[
"2 3\n3\n",
"0 -1\n2\n"
] |
[
"1\n",
"1000000006\n"
] |
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
| 1,000
|
[
{
"input": "2 3\n3",
"output": "1"
},
{
"input": "0 -1\n2",
"output": "1000000006"
},
{
"input": "-9 -11\n12345",
"output": "1000000005"
},
{
"input": "0 0\n1000000000",
"output": "0"
},
{
"input": "-1000000000 1000000000\n2000000000",
"output": "1000000000"
},
{
"input": "-12345678 12345678\n1912345678",
"output": "12345678"
},
{
"input": "728374857 678374857\n1928374839",
"output": "950000007"
},
{
"input": "278374837 992837483\n1000000000",
"output": "721625170"
},
{
"input": "-693849384 502938493\n982838498",
"output": "502938493"
},
{
"input": "-783928374 983738273\n992837483",
"output": "16261734"
},
{
"input": "-872837483 -682738473\n999999999",
"output": "190099010"
},
{
"input": "-892837483 -998273847\n999283948",
"output": "892837483"
},
{
"input": "-283938494 738473848\n1999999999",
"output": "716061513"
},
{
"input": "-278374857 819283838\n1",
"output": "721625150"
},
{
"input": "-1000000000 123456789\n1",
"output": "7"
},
{
"input": "-529529529 -524524524\n2",
"output": "475475483"
},
{
"input": "1 2\n2000000000",
"output": "2"
},
{
"input": "-1 -2\n2000000000",
"output": "1000000005"
},
{
"input": "1 2\n1999999999",
"output": "1"
},
{
"input": "1 2\n1999999998",
"output": "1000000006"
},
{
"input": "1 2\n1999999997",
"output": "1000000005"
},
{
"input": "1 2\n1999999996",
"output": "1000000006"
},
{
"input": "69975122 366233206\n1189460676",
"output": "703741923"
},
{
"input": "812229413 904420051\n806905621",
"output": "812229413"
},
{
"input": "872099024 962697902\n1505821695",
"output": "90598878"
},
{
"input": "887387283 909670917\n754835014",
"output": "112612724"
},
{
"input": "37759824 131342932\n854621399",
"output": "868657075"
},
{
"input": "-246822123 800496170\n626323615",
"output": "753177884"
},
{
"input": "-861439463 974126967\n349411083",
"output": "835566423"
},
{
"input": "-69811049 258093841\n1412447",
"output": "741906166"
},
{
"input": "844509330 -887335829\n123329059",
"output": "844509330"
},
{
"input": "83712471 -876177148\n1213284777",
"output": "40110388"
},
{
"input": "598730524 -718984219\n1282749880",
"output": "401269483"
},
{
"input": "-474244697 -745885656\n1517883612",
"output": "271640959"
},
{
"input": "-502583588 -894906953\n1154189557",
"output": "497416419"
},
{
"input": "-636523651 -873305815\n154879215",
"output": "763217843"
},
{
"input": "721765550 594845720\n78862386",
"output": "126919830"
},
{
"input": "364141461 158854993\n1337196589",
"output": "364141461"
},
{
"input": "878985260 677031952\n394707801",
"output": "798046699"
},
{
"input": "439527072 -24854079\n1129147002",
"output": "464381151"
},
{
"input": "840435009 -612103127\n565968986",
"output": "387896880"
},
{
"input": "875035447 -826471373\n561914518",
"output": "124964560"
},
{
"input": "-342526698 305357084\n70776744",
"output": "352116225"
},
{
"input": "-903244186 899202229\n1527859274",
"output": "899202229"
},
{
"input": "-839482546 815166320\n1127472130",
"output": "839482546"
},
{
"input": "-976992569 -958313041\n1686580818",
"output": "981320479"
},
{
"input": "-497338894 -51069176\n737081851",
"output": "502661113"
},
{
"input": "-697962643 -143148799\n1287886520",
"output": "856851208"
},
{
"input": "-982572938 -482658433\n1259858332",
"output": "982572938"
},
{
"input": "123123 78817\n2000000000",
"output": "78817"
},
{
"input": "1000000000 -1000000000\n3",
"output": "14"
},
{
"input": "-1000000000 1000000000\n6",
"output": "14"
},
{
"input": "2 3\n6",
"output": "1000000006"
},
{
"input": "0 -1\n6",
"output": "1"
},
{
"input": "500000000 -1000000000\n600000003",
"output": "500000014"
},
{
"input": "-1000000000 1000000000\n3",
"output": "999999993"
},
{
"input": "1 3\n6",
"output": "1000000005"
},
{
"input": "1 2\n12",
"output": "1000000006"
},
{
"input": "7 -1000000000\n3",
"output": "0"
},
{
"input": "-999999997 999999997\n6",
"output": "20"
},
{
"input": "3 4\n6",
"output": "1000000006"
},
{
"input": "-1 2\n6",
"output": "1000000004"
},
{
"input": "2 3\n12",
"output": "1000000006"
},
{
"input": "4 18\n6",
"output": "999999993"
},
{
"input": "1 2\n6",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n6",
"output": "999999993"
},
{
"input": "999999999 -999999999\n3",
"output": "16"
},
{
"input": "-1 0\n1",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n9",
"output": "14"
},
{
"input": "999999999 -1000000000\n12",
"output": "999999992"
},
{
"input": "1000000000 -7\n3",
"output": "0"
},
{
"input": "-5 5\n6",
"output": "999999997"
},
{
"input": "5 9\n6",
"output": "1000000003"
},
{
"input": "-15 -10\n1",
"output": "999999992"
}
] | 1,636,009,231
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 74
| 62
| 4,505,600
|
a, b = map(int, input().split())
n = int(input())
if(n%6==1):
val = a
elif(n%6==2):
val = b
elif(n%6==3):
val = b-a
elif(n%6==4):
val = -a
elif(n%6==5):
val = -b
else:
val = a-b
if(val<0):
val += 1000000007
print(val%1000000007)
|
Title: Jzzhu and Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
Output Specification:
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\n3\n', '0 -1\n2\n']
Demo Output:
['1\n', '1000000006\n']
Note:
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
|
```python
a, b = map(int, input().split())
n = int(input())
if(n%6==1):
val = a
elif(n%6==2):
val = b
elif(n%6==3):
val = b-a
elif(n%6==4):
val = -a
elif(n%6==5):
val = -b
else:
val = a-b
if(val<0):
val += 1000000007
print(val%1000000007)
```
| 3
|
|
5
|
B
|
Center Alignment
|
PROGRAMMING
| 1,200
|
[
"implementation",
"strings"
] |
B. Center Alignment
|
1
|
64
|
Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.
You are to implement the alignment in the shortest possible time. Good luck!
|
The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000.
|
Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better.
|
[
"This is\n\nCodeforces\nBeta\nRound\n5\n",
"welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck\n"
] |
[
"************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************\n",
"****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************\n"
] |
none
| 0
|
[
{
"input": "This is\n\nCodeforces\nBeta\nRound\n5",
"output": "************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************"
},
{
"input": "welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck",
"output": "****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************"
},
{
"input": "0\n2",
"output": "***\n*0*\n*2*\n***"
},
{
"input": "O\no\nd",
"output": "***\n*O*\n*o*\n*d*\n***"
},
{
"input": "0v uO M6Sy",
"output": "************\n*0v uO M6Sy*\n************"
},
{
"input": "fm v\nOL U W",
"output": "**********\n* fm v *\n*OL U W*\n**********"
},
{
"input": "vb\nJ\nyU\nZ",
"output": "****\n*vb*\n*J *\n*yU*\n* Z*\n****"
},
{
"input": "N\nSV\nEh\n6f\nX6\n9e",
"output": "****\n*N *\n*SV*\n*Eh*\n*6f*\n*X6*\n*9e*\n****"
},
{
"input": "Pj\nA\nFA\nP\nVJ\nU\nEb\nW",
"output": "****\n*Pj*\n*A *\n*FA*\n* P*\n*VJ*\n*U *\n*Eb*\n* W*\n****"
},
{
"input": "T\n7j\nS\nb\nq8\nVZ\nn\n4T\niZ\npA",
"output": "****\n*T *\n*7j*\n* S*\n*b *\n*q8*\n*VZ*\n* n*\n*4T*\n*iZ*\n*pA*\n****"
},
{
"input": "8\n\n\n\ny\nW\n\n\n\n3B\n\nw\nV\n\n\n\nL\nSr\n\n\nV\n\n5\n\nAq\n\n\n\nJ\nR\n\n04\nJ\nv\nhU\n\n\n\nY\nG\n4\n\nG\nb\n\n\n9\n\n6\nd\n\n2\n\n\nE\n7\n\nr\n\n\n\n\nKC\ns\nE\n\nab\n4\nx\n\n\n\n\n\nEe\n4\n\nl\n\np\n\nG\nM\n\n\nn\n\n\nm0\n\nw\n\n\nP\n\n\n\n0",
"output": "****\n*8 *\n* *\n* *\n* *\n* y*\n*W *\n* *\n* *\n* *\n*3B*\n* *\n* w*\n*V *\n* *\n* *\n* *\n* L*\n*Sr*\n* *\n* *\n*V *\n* *\n* 5*\n* *\n*Aq*\n* *\n* *\n* *\n*J *\n* R*\n* *\n*04*\n*J *\n* v*\n*hU*\n* *\n* *\n* *\n*Y *\n* G*\n*4 *\n* *\n* G*\n*b *\n* *\n* *\n* 9*\n* *\n*6 *\n* d*\n* *\n*2 *\n* *\n* *\n* E*\n*7 *\n* *\n* r*\n* *\n* *\n* *\n* *\n*KC*\n*s *\n* E*\n* *\n*ab*\n*4 *\n* x*\n* *\n* *\n* *\n* *\n* *\n*Ee*\n*4 *\n* *\n* l*\n* *\n*p *\n* *\n* G*\n*M *\n* *\n*..."
},
{
"input": "U",
"output": "***\n*U*\n***"
}
] | 1,634,323,181
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 122
| 0
|
import sys
linhas = []
for line in sys.stdin:
linhas.append(input())
print(linhas)
|
Title: Center Alignment
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.
You are to implement the alignment in the shortest possible time. Good luck!
Input Specification:
The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000.
Output Specification:
Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better.
Demo Input:
['This is\n\nCodeforces\nBeta\nRound\n5\n', 'welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck\n']
Demo Output:
['************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************\n', '****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************\n']
Note:
none
|
```python
import sys
linhas = []
for line in sys.stdin:
linhas.append(input())
print(linhas)
```
| 0
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,636,989,744
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 122
| 0
|
n=input()
c=0
d=0
for x in n:
if x.isupper():
c+=1
else:
d+=1
if c>d:
print(n.upper())
else:
print(n.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
n=input()
c=0
d=0
for x in n:
if x.isupper():
c+=1
else:
d+=1
if c>d:
print(n.upper())
else:
print(n.lower())
```
| 3.9695
|
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"
}
] | 1,572,090,734
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,c;
cin>>a>>b>>c;
int count=a;
int ans=0;
for(int i =1;i<a;i++)
{
int d=min(int(b/i),int(c/(a-i)));
//cout<<d<<endl;
ans=max(d,ans);
}
cout<<ans;
}
|
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
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,c;
cin>>a>>b>>c;
int count=a;
int ans=0;
for(int i =1;i<a;i++)
{
int d=min(int(b/i),int(c/(a-i)));
//cout<<d<<endl;
ans=max(d,ans);
}
cout<<ans;
}
```
| -1
|
|
266
|
B
|
Queue at the School
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"graph matchings",
"implementation",
"shortest paths"
] | null | null |
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
|
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
|
[
"5 1\nBGGBG\n",
"5 2\nBGGBG\n",
"4 1\nGGGB\n"
] |
[
"GBGGB\n",
"GGBGB\n",
"GGGB\n"
] |
none
| 500
|
[
{
"input": "5 1\nBGGBG",
"output": "GBGGB"
},
{
"input": "5 2\nBGGBG",
"output": "GGBGB"
},
{
"input": "4 1\nGGGB",
"output": "GGGB"
},
{
"input": "2 1\nBB",
"output": "BB"
},
{
"input": "2 1\nBG",
"output": "GB"
},
{
"input": "6 2\nBBGBBG",
"output": "GBBGBB"
},
{
"input": "8 3\nBBGBGBGB",
"output": "GGBGBBBB"
},
{
"input": "10 3\nBBGBBBBBBG",
"output": "GBBBBBGBBB"
},
{
"input": "22 7\nGBGGBGGGGGBBBGGBGBGBBB",
"output": "GGGGGGGGBGGBGGBBBBBBBB"
},
{
"input": "50 4\nGBBGBBBGGGGGBBGGBBBBGGGBBBGBBBGGBGGBGBBBGGBGGBGGBG",
"output": "GGBGBGBGBGBGGGBBGBGBGBGBBBGBGBGBGBGBGBGBGBGBGGBGBB"
},
{
"input": "50 8\nGGGGBGGBGGGBGBBBGGGGGGGGBBGBGBGBBGGBGGBGGGGGGGGBBG",
"output": "GGGGGGGGGGGGBGGBGBGBGBGBGGGGGGBGBGBGBGBGBGGBGGBGBB"
},
{
"input": "50 30\nBGGGGGGBGGBGBGGGGBGBBGBBBGGBBBGBGBGGGGGBGBBGBGBGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "27 6\nGBGBGBGGGGGGBGGBGGBBGBBBGBB",
"output": "GGGGGGGBGBGBGGGGGBGBBBBBBBB"
},
{
"input": "46 11\nBGGGGGBGBGGBGGGBBGBBGBBGGBBGBBGBGGGGGGGBGBGBGB",
"output": "GGGGGGGGGGGBGGGGGBBGBGBGBGBGBGBGBGBGBGBGBBBBBB"
},
{
"input": "50 6\nBGGBBBBGGBBBBBBGGBGBGBBBBGBBBBBBGBBBBBBBBBBBBBBBBB",
"output": "GGGGBBBBBGBGBGBGBBBGBBBBBBGBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 8\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGBBBBBBBBBBB"
},
{
"input": "50 13\nGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 1\nB",
"output": "B"
},
{
"input": "1 1\nG",
"output": "G"
},
{
"input": "1 50\nB",
"output": "B"
},
{
"input": "1 50\nG",
"output": "G"
},
{
"input": "50 50\nBBBBBBBBGGBBBBBBGBBBBBBBBBBBGBBBBBBBBBBBBBBGBBBBBB",
"output": "GGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nGGBBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBGGGGGGBG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBB"
},
{
"input": "6 3\nGGBBBG",
"output": "GGGBBB"
},
{
"input": "26 3\nGBBGBBBBBGGGBGBGGGBGBGGBBG",
"output": "GGBBBBGBGBGBGGGBGBGGGBGBBB"
},
{
"input": "46 3\nGGBBGGGGBBGBGBBBBBGGGBGGGBBGGGBBBGGBGGBBBGBGBB",
"output": "GGGGBGBGGGBBBBBGBGBGBGGGBGGBGBGBGBGBGBGBGBBBBB"
},
{
"input": "44 8\nBGBBBBBBBBBGGBBGBGBGGBBBBBGBBGBBBBBBBBBGBBGB",
"output": "GBBGBGBGBGBGBGBBBBGBBGBBBBBBBBBGBBGBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "30 25\nBGGBBGBGGBGBGBBGBGGGGBGBGGBBBB",
"output": "GGGGGGGGGGGGGGGBBBBBBBBBBBBBBB"
},
{
"input": "17 42\nBBGBGBGGGGGGBBGGG",
"output": "GGGGGGGGGGGBBBBBB"
},
{
"input": "30 50\nBGGBBGGGGGGGGBBGGGBBGGBBBGBBGG",
"output": "GGGGGGGGGGGGGGGGGGBBBBBBBBBBBB"
},
{
"input": "31 33\nBBGGBBGGBGBBBBGGBGBBBGBGGBBGBBB",
"output": "GGGGGGGGGGGGBBBBBBBBBBBBBBBBBBB"
},
{
"input": "2 2\nBG",
"output": "GB"
},
{
"input": "8 8\nGGGBGGBB",
"output": "GGGGGBBB"
},
{
"input": "34 34\nBGGBBBBGGBBGBGGGGGGBBGGGGGBGGBGGGB",
"output": "GGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBB"
},
{
"input": "34 20\nBBBBGBGGGBGGGBGBGGBBBBGGBBGGGBGBBG",
"output": "GGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nBBGBBBBBBBBBBBBGBBBGBBBBBBBGBBBBBBGBBBGBBGBBGBBBGG",
"output": "GGGGGGGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 10\nGGGGGGGGGG",
"output": "GGGGGGGGGG"
},
{
"input": "10 10\nBBBBBBBBBB",
"output": "BBBBBBBBBB"
},
{
"input": "10 10\nBGBGBGBGBG",
"output": "GGGGGBBBBB"
},
{
"input": "1 1\nB",
"output": "B"
}
] | 1,695,130,781
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 62
| 0
|
s=input().split()
n=int(s[0])
t=int(s[1])
q=input()
for i in range(t):
q=q.replace('BG','GB',n)
print(q)
|
Title: Queue at the School
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
Output Specification:
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
Demo Input:
['5 1\nBGGBG\n', '5 2\nBGGBG\n', '4 1\nGGGB\n']
Demo Output:
['GBGGB\n', 'GGBGB\n', 'GGGB\n']
Note:
none
|
```python
s=input().split()
n=int(s[0])
t=int(s[1])
q=input()
for i in range(t):
q=q.replace('BG','GB',n)
print(q)
```
| 3
|
|
237
|
C
|
Primes on Interval
|
PROGRAMMING
| 1,600
|
[
"binary search",
"number theory",
"two pointers"
] | null | null |
You've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.
Consider positive integers *a*, *a*<=+<=1, ..., *b* (*a*<=≤<=*b*). You want to find the minimum integer *l* (1<=≤<=*l*<=≤<=*b*<=-<=*a*<=+<=1) such that for any integer *x* (*a*<=≤<=*x*<=≤<=*b*<=-<=*l*<=+<=1) among *l* integers *x*, *x*<=+<=1, ..., *x*<=+<=*l*<=-<=1 there are at least *k* prime numbers.
Find and print the required minimum *l*. If no value *l* meets the described limitations, print -1.
|
A single line contains three space-separated integers *a*,<=*b*,<=*k* (1<=≤<=*a*,<=*b*,<=*k*<=≤<=106; *a*<=≤<=*b*).
|
In a single line print a single integer — the required minimum *l*. If there's no solution, print -1.
|
[
"2 4 2\n",
"6 13 1\n",
"1 4 3\n"
] |
[
"3\n",
"4\n",
"-1\n"
] |
none
| 1,500
|
[
{
"input": "2 4 2",
"output": "3"
},
{
"input": "6 13 1",
"output": "4"
},
{
"input": "1 4 3",
"output": "-1"
},
{
"input": "5 8 2",
"output": "4"
},
{
"input": "8 10 3",
"output": "-1"
},
{
"input": "1 5 2",
"output": "3"
},
{
"input": "6 8 3",
"output": "-1"
},
{
"input": "21 29 2",
"output": "9"
},
{
"input": "17 27 3",
"output": "11"
},
{
"input": "1 1000000 10000",
"output": "137970"
},
{
"input": "690059 708971 10000",
"output": "-1"
},
{
"input": "12357 534133 2",
"output": "138"
},
{
"input": "838069 936843 3",
"output": "142"
},
{
"input": "339554 696485 4",
"output": "168"
},
{
"input": "225912 522197 5",
"output": "190"
},
{
"input": "404430 864261 6",
"output": "236"
},
{
"input": "689973 807140 7",
"output": "236"
},
{
"input": "177146 548389 8",
"output": "240"
},
{
"input": "579857 857749 9",
"output": "300"
},
{
"input": "35648 527231 10",
"output": "280"
},
{
"input": "2 1000000 10000",
"output": "137970"
},
{
"input": "1 999999 9999",
"output": "137958"
},
{
"input": "5 5 10",
"output": "-1"
},
{
"input": "11 11 6",
"output": "-1"
},
{
"input": "4 4 95",
"output": "-1"
},
{
"input": "1 1000000 1000000",
"output": "-1"
},
{
"input": "1 1000000 78498",
"output": "999999"
},
{
"input": "1 1000000 78499",
"output": "-1"
},
{
"input": "3459 94738 1",
"output": "72"
},
{
"input": "1 1000000 1",
"output": "114"
},
{
"input": "1 1000000 78498",
"output": "999999"
},
{
"input": "1 1000000 78497",
"output": "999998"
},
{
"input": "1 1000000 78490",
"output": "999978"
},
{
"input": "1000 10000 13",
"output": "168"
},
{
"input": "100000 1000000 7821",
"output": "108426"
},
{
"input": "20 1000000 40000",
"output": "539580"
},
{
"input": "1000 900000 50000",
"output": "659334"
},
{
"input": "10000 1000000 60000",
"output": "793662"
},
{
"input": "9999 99999 8000",
"output": "86572"
},
{
"input": "50 150 20",
"output": "100"
},
{
"input": "999953 999953 1",
"output": "1"
},
{
"input": "999953 999953 2",
"output": "-1"
},
{
"input": "999931 999953 2",
"output": "23"
},
{
"input": "999906 999984 4",
"output": "52"
},
{
"input": "999940 999983 3",
"output": "26"
},
{
"input": "1 1 1",
"output": "-1"
},
{
"input": "1 1 1000000",
"output": "-1"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "1 3 1",
"output": "2"
},
{
"input": "1 3 2",
"output": "3"
},
{
"input": "1 4 2",
"output": "3"
},
{
"input": "1 5 2",
"output": "3"
},
{
"input": "1 5 3",
"output": "5"
},
{
"input": "2 5 2",
"output": "3"
},
{
"input": "3 5 1",
"output": "2"
}
] | 1,613,461,554
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 55
| 171
| 9,523,200
|
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
N=10**6+6
primes=[1]*N
p=2
while p*p<=N:
if primes[p]:
for j in range(p*p,N,p):
primes[j]=0
p+=1
primes[1]=primes[0]=0
def fn(x):
for i in range(n-x+1):
ones=aa[i+x-1]-(aa[i-1] if i>=1 else 0)
if ones<k:return False
return True
for _ in range(1):#nmbr()):
a,b,k=lst()
n=b-a+1
l=1;r=b-a+1
aa=[0]*(b-a+1)
c=0
for i in range(a,b+1):
aa[c]=primes[i]+aa[max(0,c-1)]
c+=1
# print(aa)
while l<=r:
mid=(l+r)>>1
# print(mid,fn(mid))
if fn(mid)==False:l=mid+1
else:r=mid-1
print(l if l<=n else -1)
|
Title: Primes on Interval
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.
Consider positive integers *a*, *a*<=+<=1, ..., *b* (*a*<=≤<=*b*). You want to find the minimum integer *l* (1<=≤<=*l*<=≤<=*b*<=-<=*a*<=+<=1) such that for any integer *x* (*a*<=≤<=*x*<=≤<=*b*<=-<=*l*<=+<=1) among *l* integers *x*, *x*<=+<=1, ..., *x*<=+<=*l*<=-<=1 there are at least *k* prime numbers.
Find and print the required minimum *l*. If no value *l* meets the described limitations, print -1.
Input Specification:
A single line contains three space-separated integers *a*,<=*b*,<=*k* (1<=≤<=*a*,<=*b*,<=*k*<=≤<=106; *a*<=≤<=*b*).
Output Specification:
In a single line print a single integer — the required minimum *l*. If there's no solution, print -1.
Demo Input:
['2 4 2\n', '6 13 1\n', '1 4 3\n']
Demo Output:
['3\n', '4\n', '-1\n']
Note:
none
|
```python
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
N=10**6+6
primes=[1]*N
p=2
while p*p<=N:
if primes[p]:
for j in range(p*p,N,p):
primes[j]=0
p+=1
primes[1]=primes[0]=0
def fn(x):
for i in range(n-x+1):
ones=aa[i+x-1]-(aa[i-1] if i>=1 else 0)
if ones<k:return False
return True
for _ in range(1):#nmbr()):
a,b,k=lst()
n=b-a+1
l=1;r=b-a+1
aa=[0]*(b-a+1)
c=0
for i in range(a,b+1):
aa[c]=primes[i]+aa[max(0,c-1)]
c+=1
# print(aa)
while l<=r:
mid=(l+r)>>1
# print(mid,fn(mid))
if fn(mid)==False:l=mid+1
else:r=mid-1
print(l if l<=n else -1)
```
| 3
|
|
835
|
B
|
The number on the board
|
PROGRAMMING
| 1,100
|
[
"greedy"
] | null | null |
Some natural number was written on the board. Its sum of digits was not less than *k*. But you were distracted a bit, and someone changed this number to *n*, replacing some digits with others. It's known that the length of the number didn't change.
You have to find the minimum number of digits in which these two numbers can differ.
|
The first line contains integer *k* (1<=≤<=*k*<=≤<=109).
The second line contains integer *n* (1<=≤<=*n*<=<<=10100000).
There are no leading zeros in *n*. It's guaranteed that this situation is possible.
|
Print the minimum number of digits in which the initial number and *n* can differ.
|
[
"3\n11\n",
"3\n99\n"
] |
[
"1\n",
"0\n"
] |
In the first example, the initial number could be 12.
In the second example the sum of the digits of *n* is not less than *k*. The initial number could be equal to *n*.
| 750
|
[
{
"input": "3\n11",
"output": "1"
},
{
"input": "3\n99",
"output": "0"
},
{
"input": "10\n5205602270",
"output": "0"
},
{
"input": "70\n3326631213",
"output": "6"
},
{
"input": "200\n1000000010000000000000000000010000000000000001000001000000000000000000000000000000000000000000000000",
"output": "22"
},
{
"input": "500\n1899337170458531693764539600958943248270674811247191310452938511077656066239840703432499357537079035",
"output": "6"
},
{
"input": "700\n9307216756404590162143344901558545760612901767837570518638460182990196397856220673189163417019781185",
"output": "32"
},
{
"input": "900\n7570423817272967027553082464863962024635217372307919506594193055572300657732661146354209508997483330",
"output": "91"
},
{
"input": "18\n900",
"output": "1"
},
{
"input": "23\n12138",
"output": "1"
},
{
"input": "16\n333",
"output": "2"
},
{
"input": "3\n12",
"output": "0"
},
{
"input": "3\n111",
"output": "0"
},
{
"input": "1\n100",
"output": "0"
},
{
"input": "17\n89",
"output": "0"
},
{
"input": "18\n99",
"output": "0"
},
{
"input": "42\n97779",
"output": "2"
},
{
"input": "2\n11",
"output": "0"
},
{
"input": "6\n33",
"output": "0"
},
{
"input": "45\n23456",
"output": "5"
},
{
"input": "3\n21",
"output": "0"
},
{
"input": "2\n2",
"output": "0"
},
{
"input": "108\n199999899899",
"output": "3"
},
{
"input": "6\n222",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "9\n9",
"output": "0"
},
{
"input": "5\n5",
"output": "0"
},
{
"input": "27\n888",
"output": "3"
}
] | 1,601,023,283
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 48
| 171
| 4,710,400
|
k = int(input())
n = list(input())
n.sort()
sum=0
for i in n:
sum+=int(i)
m=0
if k>sum:
j=0
while sum<k:
sum+=9-int(n[j])
j+=1
m+=1
print(m)
|
Title: The number on the board
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some natural number was written on the board. Its sum of digits was not less than *k*. But you were distracted a bit, and someone changed this number to *n*, replacing some digits with others. It's known that the length of the number didn't change.
You have to find the minimum number of digits in which these two numbers can differ.
Input Specification:
The first line contains integer *k* (1<=≤<=*k*<=≤<=109).
The second line contains integer *n* (1<=≤<=*n*<=<<=10100000).
There are no leading zeros in *n*. It's guaranteed that this situation is possible.
Output Specification:
Print the minimum number of digits in which the initial number and *n* can differ.
Demo Input:
['3\n11\n', '3\n99\n']
Demo Output:
['1\n', '0\n']
Note:
In the first example, the initial number could be 12.
In the second example the sum of the digits of *n* is not less than *k*. The initial number could be equal to *n*.
|
```python
k = int(input())
n = list(input())
n.sort()
sum=0
for i in n:
sum+=int(i)
m=0
if k>sum:
j=0
while sum<k:
sum+=9-int(n[j])
j+=1
m+=1
print(m)
```
| 3
|
|
379
|
A
|
New Year Candles
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
|
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
|
Print a single integer — the number of hours Vasily can light up the room for.
|
[
"4 2\n",
"6 3\n"
] |
[
"7\n",
"8\n"
] |
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
| 500
|
[
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,666,593,226
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
a,b=map(int,input().split())
h=a
c=a//b+a%b
while c>=b:
h+=c//b*b
c=c//b+c%b
print(h)
|
Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
|
```python
a,b=map(int,input().split())
h=a
c=a//b+a%b
while c>=b:
h+=c//b*b
c=c//b+c%b
print(h)
```
| 0
|
|
798
|
A
|
Mike and palindrome
|
PROGRAMMING
| 1,000
|
[
"brute force",
"constructive algorithms",
"strings"
] | null | null |
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
|
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
|
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
|
[
"abccaa\n",
"abbcca\n",
"abcda\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "abccaa",
"output": "YES"
},
{
"input": "abbcca",
"output": "NO"
},
{
"input": "abcda",
"output": "YES"
},
{
"input": "kyw",
"output": "YES"
},
{
"input": "fccf",
"output": "NO"
},
{
"input": "mnlm",
"output": "YES"
},
{
"input": "gqrk",
"output": "NO"
},
{
"input": "glxlg",
"output": "YES"
},
{
"input": "czhfc",
"output": "YES"
},
{
"input": "broon",
"output": "NO"
},
{
"input": "rmggmr",
"output": "NO"
},
{
"input": "wvxxzw",
"output": "YES"
},
{
"input": "ukvciu",
"output": "NO"
},
{
"input": "vrnwnrv",
"output": "YES"
},
{
"input": "vlkjkav",
"output": "YES"
},
{
"input": "guayhmg",
"output": "NO"
},
{
"input": "lkvhhvkl",
"output": "NO"
},
{
"input": "ffdsslff",
"output": "YES"
},
{
"input": "galjjtyw",
"output": "NO"
},
{
"input": "uosgwgsou",
"output": "YES"
},
{
"input": "qjwmjmljq",
"output": "YES"
},
{
"input": "ustrvrodf",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "qjfyjjyfjq",
"output": "NO"
},
{
"input": "ysxibbixsq",
"output": "YES"
},
{
"input": "howfslfwmh",
"output": "NO"
},
{
"input": "ekhajrjahke",
"output": "YES"
},
{
"input": "ucnolsloncw",
"output": "YES"
},
{
"input": "jrzsfrrkrtj",
"output": "NO"
},
{
"input": "typayzzyapyt",
"output": "NO"
},
{
"input": "uwdhkzokhdwu",
"output": "YES"
},
{
"input": "xokxpyyuafij",
"output": "NO"
},
{
"input": "eusneioiensue",
"output": "YES"
},
{
"input": "fuxpuajabpxuf",
"output": "YES"
},
{
"input": "guvggtfhlgruy",
"output": "NO"
},
{
"input": "cojhkhxxhkhjoc",
"output": "NO"
},
{
"input": "mhifbmmmmbmihm",
"output": "YES"
},
{
"input": "kxfqqncnebpami",
"output": "NO"
},
{
"input": "scfwrjevejrwfcs",
"output": "YES"
},
{
"input": "thdaonpepdoadht",
"output": "YES"
},
{
"input": "jsfzcbnhsccuqsj",
"output": "NO"
},
{
"input": "nn",
"output": "NO"
},
{
"input": "nm",
"output": "YES"
},
{
"input": "jdj",
"output": "YES"
},
{
"input": "bbcaa",
"output": "NO"
},
{
"input": "abcde",
"output": "NO"
},
{
"input": "abcdf",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "abecd",
"output": "NO"
},
{
"input": "abccacb",
"output": "NO"
},
{
"input": "aabc",
"output": "NO"
},
{
"input": "anpqb",
"output": "NO"
},
{
"input": "c",
"output": "YES"
},
{
"input": "abcdefg",
"output": "NO"
},
{
"input": "aanbb",
"output": "NO"
},
{
"input": "aabbb",
"output": "NO"
},
{
"input": "aaabbab",
"output": "NO"
},
{
"input": "ab",
"output": "YES"
},
{
"input": "aabbc",
"output": "NO"
},
{
"input": "ecabd",
"output": "NO"
},
{
"input": "abcdrty",
"output": "NO"
},
{
"input": "abcdmnp",
"output": "NO"
},
{
"input": "bbbbbb",
"output": "NO"
},
{
"input": "abcxuio",
"output": "NO"
},
{
"input": "abcdabcde",
"output": "NO"
},
{
"input": "abcxpoi",
"output": "NO"
},
{
"input": "aba",
"output": "YES"
},
{
"input": "aacbb",
"output": "NO"
},
{
"input": "abcedca",
"output": "NO"
},
{
"input": "abcdd",
"output": "NO"
},
{
"input": "abbcs",
"output": "NO"
},
{
"input": "aaabccc",
"output": "NO"
},
{
"input": "paxkxbq",
"output": "NO"
},
{
"input": "z",
"output": "YES"
},
{
"input": "b",
"output": "YES"
},
{
"input": "abcdefghi",
"output": "NO"
},
{
"input": "abcqr",
"output": "NO"
},
{
"input": "abcdc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aabcd",
"output": "NO"
},
{
"input": "abbba",
"output": "YES"
},
{
"input": "aaabbb",
"output": "NO"
},
{
"input": "bb",
"output": "NO"
},
{
"input": "aaacbbb",
"output": "NO"
},
{
"input": "abbzcca",
"output": "NO"
},
{
"input": "abxab",
"output": "NO"
},
{
"input": "bbb",
"output": "YES"
},
{
"input": "abcrtyu",
"output": "NO"
},
{
"input": "cbacb",
"output": "NO"
},
{
"input": "acbb",
"output": "NO"
},
{
"input": "ww",
"output": "NO"
},
{
"input": "aaaaaa",
"output": "NO"
},
{
"input": "jizzz",
"output": "NO"
},
{
"input": "aaakcba",
"output": "NO"
},
{
"input": "acbak",
"output": "NO"
},
{
"input": "bddeffd",
"output": "NO"
},
{
"input": "aaa",
"output": "YES"
},
{
"input": "afghqwe",
"output": "NO"
},
{
"input": "abcdfga",
"output": "NO"
}
] | 1,637,520,259
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
s = input()
count = 0
for i in range(0,len(s)//2):
if s[i] == s[len(s)-i-1]:
continue
else :
count = count + 1
if count <= 1 :
print("YES")
else :
print("NO")
|
Title: Mike and palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
Input Specification:
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
Output Specification:
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
Demo Input:
['abccaa\n', 'abbcca\n', 'abcda\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
s = input()
count = 0
for i in range(0,len(s)//2):
if s[i] == s[len(s)-i-1]:
continue
else :
count = count + 1
if count <= 1 :
print("YES")
else :
print("NO")
```
| 0
|
|
375
|
D
|
Tree and Queries
|
PROGRAMMING
| 2,400
|
[
"data structures",
"dfs and similar",
"trees"
] | null | null |
You have a rooted tree consisting of *n* vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to *n*. Then we represent the color of vertex *v* as *c**v*. The tree root is a vertex with number 1.
In this problem you need to answer to *m* queries. Each query is described by two integers *v**j*,<=*k**j*. The answer to query *v**j*,<=*k**j* is the number of such colors of vertices *x*, that the subtree of vertex *v**j* contains at least *k**j* vertices of color *x*.
You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105; 1<=≤<=*m*<=≤<=105). The next line contains a sequence of integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105). The next *n*<=-<=1 lines contain the edges of the tree. The *i*-th line contains the numbers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the vertices connected by an edge of the tree.
Next *m* lines contain the queries. The *j*-th line contains two integers *v**j*,<=*k**j* (1<=≤<=*v**j*<=≤<=*n*; 1<=≤<=*k**j*<=≤<=105).
|
Print *m* integers — the answers to the queries in the order the queries appear in the input.
|
[
"8 5\n1 2 2 3 3 2 3 3\n1 2\n1 5\n2 3\n2 4\n5 6\n5 7\n5 8\n1 2\n1 3\n1 4\n2 3\n5 3\n",
"4 1\n1 2 3 4\n1 2\n2 3\n3 4\n1 1\n"
] |
[
"2\n2\n1\n0\n1\n",
"4\n"
] |
A subtree of vertex *v* in a rooted tree with root *r* is a set of vertices {*u* : *dist*(*r*, *v*) + *dist*(*v*, *u*) = *dist*(*r*, *u*)}. Where *dist*(*x*, *y*) is the length (in edges) of the shortest path between vertices *x* and *y*.
| 2,000
|
[
{
"input": "8 5\n1 2 2 3 3 2 3 3\n1 2\n1 5\n2 3\n2 4\n5 6\n5 7\n5 8\n1 2\n1 3\n1 4\n2 3\n5 3",
"output": "2\n2\n1\n0\n1"
},
{
"input": "4 1\n1 2 3 4\n1 2\n2 3\n3 4\n1 1",
"output": "4"
}
] | 1,463,539,456
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 1,000
| 28,262,400
|
n, m = map(int, input().split())
c = list(map(int, input().split()))
g = [[] for i in range(n)]
for i in range(n-1):
a, b = map(int, input().split())
a, b = a-1, b-1
g[a].append(b)
g[b].append(a)
ans = [0 for i in range(m)]
q = [[] for i in range(n)]
u = [0 for i in range(n)]
for i in range(m):
v, k = map(int, input().split())
q[v-1].append((k, i))
def add(a, x, d):
i = x
while i <= n:
if not i in a: a[i] = 0
a[i] += d
i += i & -i
def query(a, x):
i = x
s = 0
while i > 0:
if i in a: s += a[i]
i -= i & -i
return s
def merge(a, b):
for c, y in b[0].items():
y0 = (a[0][c] if c in a[0] else 0)
a[0][c] = y0 + y
if y0:
add(a[1], y0, -1)
add(a[1], y0 + y, 1)
def dfs(x):
u[x] = 1
tmp = ({c[x]: 1}, {})
l = [tmp]
for y in g[x]:
if not u[y]:
l.append(dfs(y))
if len(l) == 1:
add(tmp[1], 1, 1)
else:
maxc = max(map(lambda x: len(x[0]), l[1:]))
ind = list(filter(lambda x: len(l[x][0]) == maxc, range(1, len(l))))[0]
l[0], l[ind] = l[ind], l[0]
for i in range(1, len(l)):
merge(l[0], l[i])
l = l[0]
for k, i in q[x]:
ans[i] = len(l[0]) - query(l[1], min(k-1, n))
return l
dfs(0)
print('\n'.join(map(str, ans)))
|
Title: Tree and Queries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a rooted tree consisting of *n* vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to *n*. Then we represent the color of vertex *v* as *c**v*. The tree root is a vertex with number 1.
In this problem you need to answer to *m* queries. Each query is described by two integers *v**j*,<=*k**j*. The answer to query *v**j*,<=*k**j* is the number of such colors of vertices *x*, that the subtree of vertex *v**j* contains at least *k**j* vertices of color *x*.
You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105; 1<=≤<=*m*<=≤<=105). The next line contains a sequence of integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105). The next *n*<=-<=1 lines contain the edges of the tree. The *i*-th line contains the numbers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the vertices connected by an edge of the tree.
Next *m* lines contain the queries. The *j*-th line contains two integers *v**j*,<=*k**j* (1<=≤<=*v**j*<=≤<=*n*; 1<=≤<=*k**j*<=≤<=105).
Output Specification:
Print *m* integers — the answers to the queries in the order the queries appear in the input.
Demo Input:
['8 5\n1 2 2 3 3 2 3 3\n1 2\n1 5\n2 3\n2 4\n5 6\n5 7\n5 8\n1 2\n1 3\n1 4\n2 3\n5 3\n', '4 1\n1 2 3 4\n1 2\n2 3\n3 4\n1 1\n']
Demo Output:
['2\n2\n1\n0\n1\n', '4\n']
Note:
A subtree of vertex *v* in a rooted tree with root *r* is a set of vertices {*u* : *dist*(*r*, *v*) + *dist*(*v*, *u*) = *dist*(*r*, *u*)}. Where *dist*(*x*, *y*) is the length (in edges) of the shortest path between vertices *x* and *y*.
|
```python
n, m = map(int, input().split())
c = list(map(int, input().split()))
g = [[] for i in range(n)]
for i in range(n-1):
a, b = map(int, input().split())
a, b = a-1, b-1
g[a].append(b)
g[b].append(a)
ans = [0 for i in range(m)]
q = [[] for i in range(n)]
u = [0 for i in range(n)]
for i in range(m):
v, k = map(int, input().split())
q[v-1].append((k, i))
def add(a, x, d):
i = x
while i <= n:
if not i in a: a[i] = 0
a[i] += d
i += i & -i
def query(a, x):
i = x
s = 0
while i > 0:
if i in a: s += a[i]
i -= i & -i
return s
def merge(a, b):
for c, y in b[0].items():
y0 = (a[0][c] if c in a[0] else 0)
a[0][c] = y0 + y
if y0:
add(a[1], y0, -1)
add(a[1], y0 + y, 1)
def dfs(x):
u[x] = 1
tmp = ({c[x]: 1}, {})
l = [tmp]
for y in g[x]:
if not u[y]:
l.append(dfs(y))
if len(l) == 1:
add(tmp[1], 1, 1)
else:
maxc = max(map(lambda x: len(x[0]), l[1:]))
ind = list(filter(lambda x: len(l[x][0]) == maxc, range(1, len(l))))[0]
l[0], l[ind] = l[ind], l[0]
for i in range(1, len(l)):
merge(l[0], l[i])
l = l[0]
for k, i in q[x]:
ans[i] = len(l[0]) - query(l[1], min(k-1, n))
return l
dfs(0)
print('\n'.join(map(str, ans)))
```
| 0
|
|
47
|
B
|
Coins
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
B. Coins
|
2
|
256
|
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
|
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
|
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
|
[
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] |
[
"CBA",
"ACB"
] |
none
| 1,000
|
[
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "A>C\nA>B\nB<C",
"output": "BCA"
},
{
"input": "C<B\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nA>B\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nA<C",
"output": "ABC"
},
{
"input": "C<B\nB<A\nC>A",
"output": "Impossible"
},
{
"input": "B<C\nC<A\nA>B",
"output": "BCA"
},
{
"input": "A>B\nC<B\nC<A",
"output": "CBA"
},
{
"input": "B>A\nC>B\nA>C",
"output": "Impossible"
},
{
"input": "B<A\nC>B\nC>A",
"output": "BAC"
},
{
"input": "A<B\nC>B\nA<C",
"output": "ABC"
},
{
"input": "A<B\nC<A\nB<C",
"output": "Impossible"
},
{
"input": "A>C\nC<B\nB>A",
"output": "CAB"
},
{
"input": "C>A\nA<B\nB>C",
"output": "ACB"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "B>C\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA<B\nB>C",
"output": "ACB"
},
{
"input": "B>A\nA>C\nB>C",
"output": "CAB"
},
{
"input": "B<A\nA<C\nC<B",
"output": "Impossible"
},
{
"input": "A<C\nB>C\nA>B",
"output": "Impossible"
},
{
"input": "B>A\nC<A\nC>B",
"output": "Impossible"
},
{
"input": "A>C\nC>B\nB<A",
"output": "BCA"
},
{
"input": "B<C\nB<A\nA>C",
"output": "BCA"
},
{
"input": "A>B\nC>B\nA<C",
"output": "BAC"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA>B\nB>C",
"output": "Impossible"
},
{
"input": "B>A\nB>C\nA<C",
"output": "ACB"
},
{
"input": "B>C\nC<A\nB<A",
"output": "CBA"
},
{
"input": "C>A\nB>A\nB>C",
"output": "ACB"
},
{
"input": "B<A\nB>C\nA<C",
"output": "Impossible"
},
{
"input": "B<C\nA<B\nC>A",
"output": "ABC"
},
{
"input": "C<B\nB>A\nA>C",
"output": "CAB"
},
{
"input": "A>B\nC>A\nB<C",
"output": "BAC"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "B>C\nC>A\nA>B",
"output": "Impossible"
},
{
"input": "B<A\nB>C\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nC>A",
"output": "ABC"
},
{
"input": "A>C\nA>B\nB>C",
"output": "CBA"
},
{
"input": "B<C\nA>B\nA<C",
"output": "BAC"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "C>B\nA<B\nC<A",
"output": "Impossible"
},
{
"input": "A<C\nC<B\nA>B",
"output": "Impossible"
},
{
"input": "B>C\nA<B\nA<C",
"output": "ACB"
},
{
"input": "C>A\nA<B\nC>B",
"output": "ABC"
}
] | 1,587,801,545
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 218
| 307,200
|
d = {"A": 1 , "B" : 1 , "C" : 1}
for _ in range(3):
q = input()
a = q[0]
e= q[1]
b = q[2]
if(q[1]=="<"):
d[a] -=1
d[b]+= 1
else:
d[a] +=1
d[b] -= 1
s = ""
for i in d:
if d[i] == -1:
s+=i
break
for i in d:
if d[i] == 1:
s+=i
break
for i in d:
if d[i] == 3:
s+=i
break
if(len(s)!=3):
print("Impossible")
else:
print(s)
|
Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none
|
```python
d = {"A": 1 , "B" : 1 , "C" : 1}
for _ in range(3):
q = input()
a = q[0]
e= q[1]
b = q[2]
if(q[1]=="<"):
d[a] -=1
d[b]+= 1
else:
d[a] +=1
d[b] -= 1
s = ""
for i in d:
if d[i] == -1:
s+=i
break
for i in d:
if d[i] == 1:
s+=i
break
for i in d:
if d[i] == 3:
s+=i
break
if(len(s)!=3):
print("Impossible")
else:
print(s)
```
| 3.944928
|
870
|
A
|
Search for Pretty Integers
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list.
|
Print the smallest pretty integer.
|
[
"2 3\n4 2\n5 7 6\n",
"8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n"
] |
[
"25\n",
"1\n"
] |
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.
| 500
|
[
{
"input": "2 3\n4 2\n5 7 6",
"output": "25"
},
{
"input": "8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "1 1\n9\n1",
"output": "19"
},
{
"input": "9 1\n5 4 2 3 6 1 7 9 8\n9",
"output": "9"
},
{
"input": "5 3\n7 2 5 8 6\n3 1 9",
"output": "12"
},
{
"input": "4 5\n5 2 6 4\n8 9 1 3 7",
"output": "12"
},
{
"input": "5 9\n4 2 1 6 7\n2 3 4 5 6 7 8 9 1",
"output": "1"
},
{
"input": "9 9\n5 4 3 2 1 6 7 8 9\n3 2 1 5 4 7 8 9 6",
"output": "1"
},
{
"input": "9 5\n2 3 4 5 6 7 8 9 1\n4 2 1 6 7",
"output": "1"
},
{
"input": "9 9\n1 2 3 4 5 6 7 8 9\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "9 9\n1 2 3 4 5 6 7 8 9\n9 8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "9 9\n9 8 7 6 5 4 3 2 1\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "9 9\n9 8 7 6 5 4 3 2 1\n9 8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "1 1\n8\n9",
"output": "89"
},
{
"input": "1 1\n9\n8",
"output": "89"
},
{
"input": "1 1\n1\n2",
"output": "12"
},
{
"input": "1 1\n2\n1",
"output": "12"
},
{
"input": "1 1\n9\n9",
"output": "9"
},
{
"input": "1 1\n1\n1",
"output": "1"
},
{
"input": "4 5\n3 2 4 5\n1 6 5 9 8",
"output": "5"
},
{
"input": "3 2\n4 5 6\n1 5",
"output": "5"
},
{
"input": "5 4\n1 3 5 6 7\n2 4 3 9",
"output": "3"
},
{
"input": "5 5\n1 3 5 7 9\n2 4 6 8 9",
"output": "9"
},
{
"input": "2 2\n1 8\n2 8",
"output": "8"
},
{
"input": "5 5\n5 6 7 8 9\n1 2 3 4 5",
"output": "5"
},
{
"input": "5 5\n1 2 3 4 5\n1 2 3 4 5",
"output": "1"
},
{
"input": "5 5\n1 2 3 4 5\n2 3 4 5 6",
"output": "2"
},
{
"input": "2 2\n1 5\n2 5",
"output": "5"
},
{
"input": "4 4\n1 3 5 8\n2 4 6 8",
"output": "8"
},
{
"input": "3 3\n1 5 3\n2 5 7",
"output": "5"
},
{
"input": "3 3\n3 6 8\n2 6 9",
"output": "6"
},
{
"input": "2 2\n1 4\n2 4",
"output": "4"
},
{
"input": "5 3\n3 4 5 6 7\n1 5 9",
"output": "5"
},
{
"input": "4 4\n1 2 3 4\n2 5 6 7",
"output": "2"
},
{
"input": "5 5\n1 2 3 4 5\n9 2 1 7 5",
"output": "1"
},
{
"input": "2 2\n1 3\n2 3",
"output": "3"
},
{
"input": "3 3\n3 2 1\n3 2 1",
"output": "1"
},
{
"input": "3 3\n1 3 5\n2 3 6",
"output": "3"
},
{
"input": "3 3\n5 6 7\n5 6 7",
"output": "5"
},
{
"input": "1 2\n5\n2 5",
"output": "5"
},
{
"input": "3 3\n2 4 9\n7 8 9",
"output": "9"
},
{
"input": "3 3\n1 2 4\n3 4 5",
"output": "4"
},
{
"input": "3 2\n1 4 9\n2 4",
"output": "4"
},
{
"input": "3 3\n3 5 6\n1 5 9",
"output": "5"
},
{
"input": "3 2\n1 2 4\n3 4",
"output": "4"
},
{
"input": "2 4\n8 9\n1 2 3 9",
"output": "9"
},
{
"input": "1 2\n9\n8 9",
"output": "9"
},
{
"input": "3 2\n1 2 4\n4 2",
"output": "2"
},
{
"input": "2 3\n4 5\n1 3 5",
"output": "5"
},
{
"input": "3 2\n1 2 3\n2 3",
"output": "2"
},
{
"input": "4 3\n1 3 5 9\n2 8 9",
"output": "9"
},
{
"input": "2 2\n1 9\n9 2",
"output": "9"
}
] | 1,605,831,828
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 109
| 0
|
# https://codeforces.com/problemset/problem/870/A
# 900
n, m = map(int, input().split())
nl = min(map(int, input().split()))
ml = min(map(int, input().split()))
if nl == ml:
print(nl)
elif nl < ml:
print(f"{nl}{ml}")
else:
print(f"{ml}{nl}")
|
Title: Search for Pretty Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list.
Output Specification:
Print the smallest pretty integer.
Demo Input:
['2 3\n4 2\n5 7 6\n', '8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n']
Demo Output:
['25\n', '1\n']
Note:
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.
|
```python
# https://codeforces.com/problemset/problem/870/A
# 900
n, m = map(int, input().split())
nl = min(map(int, input().split()))
ml = min(map(int, input().split()))
if nl == ml:
print(nl)
elif nl < ml:
print(f"{nl}{ml}")
else:
print(f"{ml}{nl}")
```
| 0
|
|
682
|
A
|
Alyona and Numbers
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"math",
"number theory"
] | null | null |
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
|
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
|
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
|
[
"6 12\n",
"11 14\n",
"1 5\n",
"3 8\n",
"5 7\n",
"21 21\n"
] |
[
"14\n",
"31\n",
"1\n",
"5\n",
"7\n",
"88\n"
] |
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
| 500
|
[
{
"input": "6 12",
"output": "14"
},
{
"input": "11 14",
"output": "31"
},
{
"input": "1 5",
"output": "1"
},
{
"input": "3 8",
"output": "5"
},
{
"input": "5 7",
"output": "7"
},
{
"input": "21 21",
"output": "88"
},
{
"input": "10 15",
"output": "30"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 1000000",
"output": "200000"
},
{
"input": "1000000 1",
"output": "200000"
},
{
"input": "1000000 1000000",
"output": "200000000000"
},
{
"input": "944 844",
"output": "159348"
},
{
"input": "368 984",
"output": "72423"
},
{
"input": "792 828",
"output": "131155"
},
{
"input": "920 969",
"output": "178296"
},
{
"input": "640 325",
"output": "41600"
},
{
"input": "768 170",
"output": "26112"
},
{
"input": "896 310",
"output": "55552"
},
{
"input": "320 154",
"output": "9856"
},
{
"input": "744 999",
"output": "148652"
},
{
"input": "630 843",
"output": "106218"
},
{
"input": "54 688",
"output": "7431"
},
{
"input": "478 828",
"output": "79157"
},
{
"input": "902 184",
"output": "33194"
},
{
"input": "31 29",
"output": "180"
},
{
"input": "751 169",
"output": "25384"
},
{
"input": "879 14",
"output": "2462"
},
{
"input": "7 858",
"output": "1201"
},
{
"input": "431 702",
"output": "60512"
},
{
"input": "855 355",
"output": "60705"
},
{
"input": "553 29",
"output": "3208"
},
{
"input": "721767 525996",
"output": "75929310986"
},
{
"input": "805191 74841",
"output": "12052259926"
},
{
"input": "888615 590981",
"output": "105030916263"
},
{
"input": "4743 139826",
"output": "132638943"
},
{
"input": "88167 721374",
"output": "12720276292"
},
{
"input": "171591 13322",
"output": "457187060"
},
{
"input": "287719 562167",
"output": "32349225415"
},
{
"input": "371143 78307",
"output": "5812618980"
},
{
"input": "487271 627151",
"output": "61118498984"
},
{
"input": "261436 930642",
"output": "48660664382"
},
{
"input": "377564 446782",
"output": "33737759810"
},
{
"input": "460988 28330",
"output": "2611958008"
},
{
"input": "544412 352983",
"output": "38433636199"
},
{
"input": "660540 869123",
"output": "114818101284"
},
{
"input": "743964 417967",
"output": "62190480238"
},
{
"input": "827388 966812",
"output": "159985729411"
},
{
"input": "910812 515656",
"output": "93933134534"
},
{
"input": "26940 64501",
"output": "347531388"
},
{
"input": "110364 356449",
"output": "7867827488"
},
{
"input": "636358 355531",
"output": "45248999219"
},
{
"input": "752486 871672",
"output": "131184195318"
},
{
"input": "803206 420516",
"output": "67552194859"
},
{
"input": "919334 969361",
"output": "178233305115"
},
{
"input": "35462 261309",
"output": "1853307952"
},
{
"input": "118887 842857",
"output": "20040948031"
},
{
"input": "202311 358998",
"output": "14525848875"
},
{
"input": "285735 907842",
"output": "51880446774"
},
{
"input": "401863 456686",
"output": "36705041203"
},
{
"input": "452583 972827",
"output": "88056992428"
},
{
"input": "235473 715013",
"output": "33673251230"
},
{
"input": "318897 263858",
"output": "16828704925"
},
{
"input": "402321 812702",
"output": "65393416268"
},
{
"input": "518449 361546",
"output": "37488632431"
},
{
"input": "634577 910391",
"output": "115542637921"
},
{
"input": "685297 235043",
"output": "32214852554"
},
{
"input": "801425 751183",
"output": "120403367155"
},
{
"input": "884849 300028",
"output": "53095895155"
},
{
"input": "977 848872",
"output": "165869588"
},
{
"input": "51697 397716",
"output": "4112144810"
},
{
"input": "834588 107199",
"output": "17893399803"
},
{
"input": "918012 688747",
"output": "126455602192"
},
{
"input": "1436 237592",
"output": "68236422"
},
{
"input": "117564 753732",
"output": "17722349770"
},
{
"input": "200988 302576",
"output": "12162829017"
},
{
"input": "284412 818717",
"output": "46570587880"
},
{
"input": "400540 176073",
"output": "14104855884"
},
{
"input": "483964 724917",
"output": "70166746198"
},
{
"input": "567388 241058",
"output": "27354683301"
},
{
"input": "650812 789902",
"output": "102815540084"
},
{
"input": "400999 756281",
"output": "60653584944"
},
{
"input": "100 101",
"output": "2020"
},
{
"input": "100 102",
"output": "2040"
},
{
"input": "103 100",
"output": "2060"
},
{
"input": "100 104",
"output": "2080"
},
{
"input": "3 4",
"output": "3"
},
{
"input": "11 23",
"output": "50"
},
{
"input": "8 14",
"output": "23"
},
{
"input": "23423 34234",
"output": "160372597"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "999999 999999",
"output": "199999600001"
},
{
"input": "82 99",
"output": "1624"
},
{
"input": "21 18",
"output": "75"
},
{
"input": "234 234",
"output": "10952"
},
{
"input": "4 4",
"output": "4"
},
{
"input": "6 13",
"output": "15"
},
{
"input": "3 9",
"output": "6"
},
{
"input": "99999 99999",
"output": "1999960001"
},
{
"input": "34 33",
"output": "225"
},
{
"input": "2 2",
"output": "0"
},
{
"input": "333 1",
"output": "66"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "8 2",
"output": "3"
},
{
"input": "2179 2218",
"output": "966605"
},
{
"input": "1000000 999999",
"output": "199999800000"
},
{
"input": "873828 774207",
"output": "135304750879"
},
{
"input": "13 19",
"output": "50"
},
{
"input": "1648 576469",
"output": "190004183"
},
{
"input": "11 13",
"output": "28"
},
{
"input": "5 8",
"output": "8"
},
{
"input": "650074 943659",
"output": "122689636154"
},
{
"input": "1 3",
"output": "0"
},
{
"input": "54 43",
"output": "465"
},
{
"input": "14 9",
"output": "26"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "543 534",
"output": "57993"
},
{
"input": "321 123",
"output": "7896"
},
{
"input": "21 3",
"output": "12"
},
{
"input": "2 1",
"output": "0"
},
{
"input": "4 3",
"output": "3"
},
{
"input": "47474 74747",
"output": "709707816"
},
{
"input": "4 9",
"output": "8"
},
{
"input": "7 4",
"output": "6"
},
{
"input": "9 4",
"output": "8"
},
{
"input": "12414 4214",
"output": "10462520"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "253 821",
"output": "41542"
},
{
"input": "2 4",
"output": "2"
}
] | 1,612,118,404
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 128
| 733
| 0
|
n,m = map(int,input().split())
gr = n if n>m else m
sm = n+m-gr
save=0
st=0
for i in range(1,sm+1):
gr+=1
temp=gr//5
if i%5==0:
st+=1
temp-=1*st
save+=temp
print(save)
|
Title: Alyona and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
Output Specification:
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
Demo Input:
['6 12\n', '11 14\n', '1 5\n', '3 8\n', '5 7\n', '21 21\n']
Demo Output:
['14\n', '31\n', '1\n', '5\n', '7\n', '88\n']
Note:
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
|
```python
n,m = map(int,input().split())
gr = n if n>m else m
sm = n+m-gr
save=0
st=0
for i in range(1,sm+1):
gr+=1
temp=gr//5
if i%5==0:
st+=1
temp-=1*st
save+=temp
print(save)
```
| 3
|
Subsets and Splits
Successful Python Submissions
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.