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3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*Β·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=β€<=<=*k*,<=*w*<=<=β€<=<=1000, 0<=β€<=*n*<=β€<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer β the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,696,402,131
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 46
| 0
|
k, n, w = map(int, input().split())
total = (w * (w + 1)) // 2
total *= k
if total - n > 0:
print(total - n)
else:
print(0)
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*Β·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=β€<=<=*k*,<=*w*<=<=β€<=<=1000, 0<=β€<=*n*<=β€<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer β the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
k, n, w = map(int, input().split())
total = (w * (w + 1)) // 2
total *= k
if total - n > 0:
print(total - n)
else:
print(0)
```
| 3
|
|
414
|
B
|
Mashmokh and ACM
|
PROGRAMMING
| 1,400
|
[
"combinatorics",
"dp",
"number theory"
] | null | null |
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of *l* integers *b*1,<=*b*2,<=...,<=*b**l* (1<=β€<=*b*1<=β€<=*b*2<=β€<=...<=β€<=*b**l*<=β€<=*n*) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all *i* (1<=β€<=*i*<=β€<=*l*<=-<=1).
Given *n* and *k* find the number of good sequences of length *k*. As the answer can be rather large print it modulo 1000000007 (109<=+<=7).
|
The first line of input contains two space-separated integers *n*,<=*k*Β (1<=β€<=*n*,<=*k*<=β€<=2000).
|
Output a single integer β the number of good sequences of length *k* modulo 1000000007 (109<=+<=7).
|
[
"3 2\n",
"6 4\n",
"2 1\n"
] |
[
"5\n",
"39\n",
"2\n"
] |
In the first sample the good sequences are: [1,β1],β[2,β2],β[3,β3],β[1,β2],β[1,β3].
| 1,000
|
[
{
"input": "3 2",
"output": "5"
},
{
"input": "6 4",
"output": "39"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "1478 194",
"output": "312087753"
},
{
"input": "1415 562",
"output": "953558593"
},
{
"input": "1266 844",
"output": "735042656"
},
{
"input": "680 1091",
"output": "351905328"
},
{
"input": "1229 1315",
"output": "100240813"
},
{
"input": "1766 1038",
"output": "435768250"
},
{
"input": "1000 1",
"output": "1000"
},
{
"input": "2000 100",
"output": "983281065"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2000 1000",
"output": "228299266"
},
{
"input": "1928 1504",
"output": "81660104"
},
{
"input": "2000 2000",
"output": "585712681"
},
{
"input": "29 99",
"output": "23125873"
},
{
"input": "56 48",
"output": "20742237"
},
{
"input": "209 370",
"output": "804680894"
},
{
"input": "83 37",
"output": "22793555"
},
{
"input": "49 110",
"output": "956247348"
},
{
"input": "217 3",
"output": "4131"
},
{
"input": "162 161",
"output": "591739753"
},
{
"input": "273 871",
"output": "151578252"
},
{
"input": "43 1640",
"output": "173064407"
},
{
"input": "1472 854",
"output": "748682383"
},
{
"input": "1639 1056",
"output": "467464129"
},
{
"input": "359 896",
"output": "770361185"
},
{
"input": "1544 648",
"output": "9278889"
},
{
"input": "436 1302",
"output": "874366220"
},
{
"input": "1858 743",
"output": "785912917"
},
{
"input": "991 1094",
"output": "483493131"
},
{
"input": "1013 1550",
"output": "613533467"
},
{
"input": "675 741",
"output": "474968598"
},
{
"input": "1420 1223",
"output": "922677437"
},
{
"input": "1544 1794",
"output": "933285446"
},
{
"input": "1903 1612",
"output": "620810276"
},
{
"input": "500 1304",
"output": "706176027"
},
{
"input": "525 314",
"output": "245394744"
},
{
"input": "39 1930",
"output": "992125404"
},
{
"input": "1895 753",
"output": "180474828"
},
{
"input": "1722 1474",
"output": "742424590"
},
{
"input": "1153 1823",
"output": "791493066"
},
{
"input": "1409 734",
"output": "627413973"
},
{
"input": "478 1301",
"output": "476483030"
},
{
"input": "1887 1729",
"output": "730033374"
},
{
"input": "1610 774",
"output": "50897314"
},
{
"input": "1770 679",
"output": "235295539"
},
{
"input": "987 1292",
"output": "560110556"
},
{
"input": "1707 1117",
"output": "237674323"
},
{
"input": "1424 1431",
"output": "184145444"
},
{
"input": "86 1078",
"output": "252515343"
},
{
"input": "1066 995",
"output": "180753612"
},
{
"input": "1024 133",
"output": "392603027"
},
{
"input": "659 974",
"output": "397026719"
},
{
"input": "1349 1606",
"output": "522392901"
},
{
"input": "473 211",
"output": "809550224"
},
{
"input": "634 1825",
"output": "438513382"
},
{
"input": "22 373",
"output": "907321755"
},
{
"input": "531 147",
"output": "242883376"
},
{
"input": "1307 1247",
"output": "21512331"
},
{
"input": "415 735",
"output": "393705804"
},
{
"input": "1659 1501",
"output": "225266660"
},
{
"input": "1454 296",
"output": "750032659"
},
{
"input": "158 772",
"output": "850911301"
},
{
"input": "2000 1",
"output": "2000"
}
] | 1,535,295,445
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 65
| 639
| 0
|
def C(n,k):
w=1
for i in range(n-k+1,n+1):
w*=i
for i in range (1,k+1):
w//=i
return w
n,k=[int(x) for x in input().split()]
ans=0
for i in range(1,n+1):
pp=1;
i1=i;
for j in range(2,i+1):
sch=0;
while(i1%j==0):
sch+=1
i1//=j
if (sch):
pp*=C(sch+k-1,sch)
pp%=1000000007
ans+=pp
ans%=1000000007
print(ans)
# Made By Mostafa_Khaled
|
Title: Mashmokh and ACM
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of *l* integers *b*1,<=*b*2,<=...,<=*b**l* (1<=β€<=*b*1<=β€<=*b*2<=β€<=...<=β€<=*b**l*<=β€<=*n*) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all *i* (1<=β€<=*i*<=β€<=*l*<=-<=1).
Given *n* and *k* find the number of good sequences of length *k*. As the answer can be rather large print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line of input contains two space-separated integers *n*,<=*k*Β (1<=β€<=*n*,<=*k*<=β€<=2000).
Output Specification:
Output a single integer β the number of good sequences of length *k* modulo 1000000007 (109<=+<=7).
Demo Input:
['3 2\n', '6 4\n', '2 1\n']
Demo Output:
['5\n', '39\n', '2\n']
Note:
In the first sample the good sequences are: [1,β1],β[2,β2],β[3,β3],β[1,β2],β[1,β3].
|
```python
def C(n,k):
w=1
for i in range(n-k+1,n+1):
w*=i
for i in range (1,k+1):
w//=i
return w
n,k=[int(x) for x in input().split()]
ans=0
for i in range(1,n+1):
pp=1;
i1=i;
for j in range(2,i+1):
sch=0;
while(i1%j==0):
sch+=1
i1//=j
if (sch):
pp*=C(sch+k-1,sch)
pp%=1000000007
ans+=pp
ans%=1000000007
print(ans)
# Made By Mostafa_Khaled
```
| 3
|
|
987
|
A
|
Infinity Gauntlet
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:
- the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color.
Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.
|
In the first line of input there is one integer $n$ ($0 \le n \le 6$)Β β the number of Gems in Infinity Gauntlet.
In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.
|
In the first line output one integer $m$ ($0 \le m \le 6$)Β β the number of absent Gems.
Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.
|
[
"4\nred\npurple\nyellow\norange\n",
"0\n"
] |
[
"2\nSpace\nTime\n",
"6\nTime\nMind\nSoul\nPower\nReality\nSpace\n"
] |
In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.
In the second sample Thanos doesn't have any Gems, so he needs all six.
| 500
|
[
{
"input": "4\nred\npurple\nyellow\norange",
"output": "2\nSpace\nTime"
},
{
"input": "0",
"output": "6\nMind\nSpace\nPower\nTime\nReality\nSoul"
},
{
"input": "6\npurple\nblue\nyellow\nred\ngreen\norange",
"output": "0"
},
{
"input": "1\npurple",
"output": "5\nTime\nReality\nSoul\nSpace\nMind"
},
{
"input": "3\nblue\norange\npurple",
"output": "3\nTime\nReality\nMind"
},
{
"input": "2\nyellow\nred",
"output": "4\nPower\nSoul\nSpace\nTime"
},
{
"input": "1\ngreen",
"output": "5\nReality\nSpace\nPower\nSoul\nMind"
},
{
"input": "2\npurple\ngreen",
"output": "4\nReality\nMind\nSpace\nSoul"
},
{
"input": "1\nblue",
"output": "5\nPower\nReality\nSoul\nTime\nMind"
},
{
"input": "2\npurple\nblue",
"output": "4\nMind\nSoul\nTime\nReality"
},
{
"input": "2\ngreen\nblue",
"output": "4\nReality\nMind\nPower\nSoul"
},
{
"input": "3\npurple\ngreen\nblue",
"output": "3\nMind\nReality\nSoul"
},
{
"input": "1\norange",
"output": "5\nReality\nTime\nPower\nSpace\nMind"
},
{
"input": "2\npurple\norange",
"output": "4\nReality\nMind\nTime\nSpace"
},
{
"input": "2\norange\ngreen",
"output": "4\nSpace\nMind\nReality\nPower"
},
{
"input": "3\norange\npurple\ngreen",
"output": "3\nReality\nSpace\nMind"
},
{
"input": "2\norange\nblue",
"output": "4\nTime\nMind\nReality\nPower"
},
{
"input": "3\nblue\ngreen\norange",
"output": "3\nPower\nMind\nReality"
},
{
"input": "4\nblue\norange\ngreen\npurple",
"output": "2\nMind\nReality"
},
{
"input": "1\nred",
"output": "5\nTime\nSoul\nMind\nPower\nSpace"
},
{
"input": "2\nred\npurple",
"output": "4\nMind\nSpace\nTime\nSoul"
},
{
"input": "2\nred\ngreen",
"output": "4\nMind\nSpace\nPower\nSoul"
},
{
"input": "3\nred\npurple\ngreen",
"output": "3\nSoul\nSpace\nMind"
},
{
"input": "2\nblue\nred",
"output": "4\nMind\nTime\nPower\nSoul"
},
{
"input": "3\nred\nblue\npurple",
"output": "3\nTime\nMind\nSoul"
},
{
"input": "3\nred\nblue\ngreen",
"output": "3\nSoul\nPower\nMind"
},
{
"input": "4\npurple\nblue\ngreen\nred",
"output": "2\nMind\nSoul"
},
{
"input": "2\norange\nred",
"output": "4\nPower\nMind\nTime\nSpace"
},
{
"input": "3\nred\norange\npurple",
"output": "3\nMind\nSpace\nTime"
},
{
"input": "3\nred\norange\ngreen",
"output": "3\nMind\nSpace\nPower"
},
{
"input": "4\nred\norange\ngreen\npurple",
"output": "2\nSpace\nMind"
},
{
"input": "3\nblue\norange\nred",
"output": "3\nPower\nMind\nTime"
},
{
"input": "4\norange\nblue\npurple\nred",
"output": "2\nTime\nMind"
},
{
"input": "4\ngreen\norange\nred\nblue",
"output": "2\nMind\nPower"
},
{
"input": "5\npurple\norange\nblue\nred\ngreen",
"output": "1\nMind"
},
{
"input": "1\nyellow",
"output": "5\nPower\nSoul\nReality\nSpace\nTime"
},
{
"input": "2\npurple\nyellow",
"output": "4\nTime\nReality\nSpace\nSoul"
},
{
"input": "2\ngreen\nyellow",
"output": "4\nSpace\nReality\nPower\nSoul"
},
{
"input": "3\npurple\nyellow\ngreen",
"output": "3\nSoul\nReality\nSpace"
},
{
"input": "2\nblue\nyellow",
"output": "4\nTime\nReality\nPower\nSoul"
},
{
"input": "3\nyellow\nblue\npurple",
"output": "3\nSoul\nReality\nTime"
},
{
"input": "3\ngreen\nyellow\nblue",
"output": "3\nSoul\nReality\nPower"
},
{
"input": "4\nyellow\nblue\ngreen\npurple",
"output": "2\nReality\nSoul"
},
{
"input": "2\nyellow\norange",
"output": "4\nTime\nSpace\nReality\nPower"
},
{
"input": "3\nyellow\npurple\norange",
"output": "3\nSpace\nReality\nTime"
},
{
"input": "3\norange\nyellow\ngreen",
"output": "3\nSpace\nReality\nPower"
},
{
"input": "4\ngreen\nyellow\norange\npurple",
"output": "2\nSpace\nReality"
},
{
"input": "3\nyellow\nblue\norange",
"output": "3\nTime\nReality\nPower"
},
{
"input": "4\norange\npurple\nblue\nyellow",
"output": "2\nReality\nTime"
},
{
"input": "4\nblue\norange\nyellow\ngreen",
"output": "2\nReality\nPower"
},
{
"input": "5\ngreen\nyellow\norange\nblue\npurple",
"output": "1\nReality"
},
{
"input": "3\nyellow\npurple\nred",
"output": "3\nTime\nSoul\nSpace"
},
{
"input": "3\nred\ngreen\nyellow",
"output": "3\nPower\nSoul\nSpace"
},
{
"input": "4\nred\npurple\ngreen\nyellow",
"output": "2\nSpace\nSoul"
},
{
"input": "3\nred\nyellow\nblue",
"output": "3\nPower\nSoul\nTime"
},
{
"input": "4\nblue\nyellow\nred\npurple",
"output": "2\nTime\nSoul"
},
{
"input": "4\nblue\nyellow\nred\ngreen",
"output": "2\nSoul\nPower"
},
{
"input": "5\nred\nyellow\ngreen\nblue\npurple",
"output": "1\nSoul"
},
{
"input": "3\nred\nyellow\norange",
"output": "3\nPower\nSpace\nTime"
},
{
"input": "4\norange\ngreen\nyellow\nred",
"output": "2\nPower\nSpace"
},
{
"input": "5\norange\nred\ngreen\nyellow\npurple",
"output": "1\nSpace"
},
{
"input": "4\nyellow\nred\norange\nblue",
"output": "2\nTime\nPower"
},
{
"input": "5\npurple\nblue\norange\nyellow\nred",
"output": "1\nTime"
},
{
"input": "5\norange\nblue\nyellow\nred\ngreen",
"output": "1\nPower"
}
] | 1,607,150,355
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 64
| 140
| 0
|
lst = []
ans = {'purple' : 'Power', 'green' : 'Time', 'blue': 'Space', "orange" : "Soul", "red" : "Reality", "yellow" : "Mind"}
for i in range(int(input())):
lst.append(input())
a = []
for i in ans.keys():
if i not in lst:
a.append(ans[i])
print(len(a))
for i in a:
print(i)
|
Title: Infinity Gauntlet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:
- the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color.
Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.
Input Specification:
In the first line of input there is one integer $n$ ($0 \le n \le 6$)Β β the number of Gems in Infinity Gauntlet.
In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.
Output Specification:
In the first line output one integer $m$ ($0 \le m \le 6$)Β β the number of absent Gems.
Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.
Demo Input:
['4\nred\npurple\nyellow\norange\n', '0\n']
Demo Output:
['2\nSpace\nTime\n', '6\nTime\nMind\nSoul\nPower\nReality\nSpace\n']
Note:
In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.
In the second sample Thanos doesn't have any Gems, so he needs all six.
|
```python
lst = []
ans = {'purple' : 'Power', 'green' : 'Time', 'blue': 'Space', "orange" : "Soul", "red" : "Reality", "yellow" : "Mind"}
for i in range(int(input())):
lst.append(input())
a = []
for i in ans.keys():
if i not in lst:
a.append(ans[i])
print(len(a))
for i in a:
print(i)
```
| 3
|
|
618
|
A
|
Slime Combining
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1.
You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right.
|
The first line of the input will contain a single integer, *n* (1<=β€<=*n*<=β€<=100<=000).
|
Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left.
|
[
"1\n",
"2\n",
"3\n",
"8\n"
] |
[
"1\n",
"2\n",
"2 1\n",
"4\n"
] |
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4
| 500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "2 1"
},
{
"input": "8",
"output": "4"
},
{
"input": "100000",
"output": "17 16 11 10 8 6"
},
{
"input": "12345",
"output": "14 13 6 5 4 1"
},
{
"input": "32",
"output": "6"
},
{
"input": "70958",
"output": "17 13 11 9 6 4 3 2"
},
{
"input": "97593",
"output": "17 15 14 13 12 11 9 6 5 4 1"
},
{
"input": "91706",
"output": "17 15 14 11 10 6 5 4 2"
},
{
"input": "85371",
"output": "17 15 12 11 9 7 6 5 4 2 1"
},
{
"input": "97205",
"output": "17 15 14 13 12 10 9 8 6 5 3 1"
},
{
"input": "34768",
"output": "16 11 10 9 8 7 5"
},
{
"input": "12705",
"output": "14 13 9 8 6 1"
},
{
"input": "30151",
"output": "15 14 13 11 9 8 7 3 2 1"
},
{
"input": "4974",
"output": "13 10 9 7 6 4 3 2"
},
{
"input": "32728",
"output": "15 14 13 12 11 10 9 8 7 5 4"
},
{
"input": "8192",
"output": "14"
},
{
"input": "65536",
"output": "17"
},
{
"input": "32",
"output": "6"
},
{
"input": "256",
"output": "9"
},
{
"input": "4096",
"output": "13"
},
{
"input": "33301",
"output": "16 10 5 3 1"
},
{
"input": "16725",
"output": "15 9 7 5 3 1"
},
{
"input": "149",
"output": "8 5 3 1"
},
{
"input": "16277",
"output": "14 13 12 11 10 9 8 5 3 1"
},
{
"input": "99701",
"output": "17 16 11 9 7 6 5 3 1"
}
] | 1,454,088,757
| 1,357
|
Python 3
|
OK
|
TESTS
| 27
| 61
| 0
|
n = int(input())
import math
s = ''
a = 1
while n > 0 :
a = int(math.log2(n)) + 1
s += str(a) + ' '
n -= 2 ** (a - 1)
print(s)
|
Title: Slime Combining
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1.
You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right.
Input Specification:
The first line of the input will contain a single integer, *n* (1<=β€<=*n*<=β€<=100<=000).
Output Specification:
Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left.
Demo Input:
['1\n', '2\n', '3\n', '8\n']
Demo Output:
['1\n', '2\n', '2 1\n', '4\n']
Note:
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4
|
```python
n = int(input())
import math
s = ''
a = 1
while n > 0 :
a = int(math.log2(n)) + 1
s += str(a) + ' '
n -= 2 ** (a - 1)
print(s)
```
| 3
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=β€<=*n*<=β€<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit β Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991βββ19βββ10βββ1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,640,304,801
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 154
| 0
|
if __name__ == '__main__':
number = input()
spell_num = 0
while len(number) >= 2:
n = 0
for i in number:
n += int(i)
number = str(n)
spell_num += 1
print(spell_num)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=β€<=*n*<=β€<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit β Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991βββ19βββ10βββ1. After three transformations the number becomes one-digit.
|
```python
if __name__ == '__main__':
number = input()
spell_num = 0
while len(number) >= 2:
n = 0
for i in number:
n += int(i)
number = str(n)
spell_num += 1
print(spell_num)
```
| 3.9615
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=β€<=|*s*|<=β€<=50). The second line contains the string *t* (1<=β€<=|*t*|<=β€<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,587,320,078
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 19
| 280
| 0
|
a=str(input())
b=str(input())
x=int(0)
y=int(0)
while x<len(b):
if b[x]==a[y]:
y=y+1
x=x+1
print(y+1)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=β€<=|*s*|<=β€<=50). The second line contains the string *t* (1<=β€<=|*t*|<=β€<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
a=str(input())
b=str(input())
x=int(0)
y=int(0)
while x<len(b):
if b[x]==a[y]:
y=y+1
x=x+1
print(y+1)
```
| 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 β 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* β 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"
}
] | 1,668,252,812
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 124
| 0
|
n=input()
l=[i for i in n]
upper=0
lower=0
for i in l:
if i.isupper():
upper=upper+1
else:
lower=lower+1
if upper>lower:
for i in l:
print(i.upper(),end="")
elif lower>upper:
for i in l:
print(i.lower(),end="")
else:
for i in l:
print(i.lower(),end="")
|
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 β 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* β 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()
l=[i for i in n]
upper=0
lower=0
for i in l:
if i.isupper():
upper=upper+1
else:
lower=lower+1
if upper>lower:
for i in l:
print(i.upper(),end="")
elif lower>upper:
for i in l:
print(i.lower(),end="")
else:
for i in l:
print(i.lower(),end="")
```
| 3.969
|
114
|
A
|
Cifera
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
|
The first input line contains integer number *k*, the second line contains integer number *l* (2<=β€<=*k*,<=*l*<=β€<=231<=-<=1).
|
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number β the importance of number *l*.
|
[
"5\n25\n",
"3\n8\n"
] |
[
"YES\n1\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "5\n25",
"output": "YES\n1"
},
{
"input": "3\n8",
"output": "NO"
},
{
"input": "123\n123",
"output": "YES\n0"
},
{
"input": "99\n970300",
"output": "NO"
},
{
"input": "1000\n6666666",
"output": "NO"
},
{
"input": "59\n3571",
"output": "NO"
},
{
"input": "256\n16777217",
"output": "NO"
},
{
"input": "4638\n21511044",
"output": "YES\n1"
},
{
"input": "24\n191102976",
"output": "YES\n5"
},
{
"input": "52010\n557556453",
"output": "NO"
},
{
"input": "61703211\n1750753082",
"output": "NO"
},
{
"input": "137\n2571353",
"output": "YES\n2"
},
{
"input": "8758\n1746157336",
"output": "NO"
},
{
"input": "2\n64",
"output": "YES\n5"
},
{
"input": "96\n884736",
"output": "YES\n2"
},
{
"input": "1094841453\n1656354409",
"output": "NO"
},
{
"input": "1154413\n1229512809",
"output": "NO"
},
{
"input": "2442144\n505226241",
"output": "NO"
},
{
"input": "11548057\n1033418098",
"output": "NO"
},
{
"input": "581\n196122941",
"output": "YES\n2"
},
{
"input": "146\n1913781536",
"output": "NO"
},
{
"input": "945916\n1403881488",
"output": "NO"
},
{
"input": "68269\n365689065",
"output": "NO"
},
{
"input": "30\n900",
"output": "YES\n1"
},
{
"input": "6\n1296",
"output": "YES\n3"
},
{
"input": "1470193122\n1420950405",
"output": "NO"
},
{
"input": "90750\n1793111557",
"output": "NO"
},
{
"input": "1950054\n1664545956",
"output": "NO"
},
{
"input": "6767692\n123762320",
"output": "NO"
},
{
"input": "1437134\n1622348229",
"output": "NO"
},
{
"input": "444103\n1806462642",
"output": "NO"
},
{
"input": "2592\n6718464",
"output": "YES\n1"
},
{
"input": "50141\n366636234",
"output": "NO"
},
{
"input": "835\n582182875",
"output": "YES\n2"
},
{
"input": "156604\n902492689",
"output": "NO"
},
{
"input": "27385965\n1742270058",
"output": "NO"
},
{
"input": "3\n9",
"output": "YES\n1"
},
{
"input": "35\n1838265625",
"output": "YES\n5"
},
{
"input": "8\n4096",
"output": "YES\n3"
},
{
"input": "85955\n945811082",
"output": "NO"
},
{
"input": "54958832\n956670209",
"output": "NO"
},
{
"input": "1475381\n1348159738",
"output": "NO"
},
{
"input": "7313241\n413670642",
"output": "NO"
},
{
"input": "582470\n2116368165",
"output": "NO"
},
{
"input": "26859739\n595086170",
"output": "NO"
},
{
"input": "249766393\n1582130",
"output": "NO"
},
{
"input": "11734\n137686756",
"output": "YES\n1"
},
{
"input": "925093\n1098566745",
"output": "NO"
},
{
"input": "40\n1600",
"output": "YES\n1"
},
{
"input": "2147483647\n2147483647",
"output": "YES\n0"
},
{
"input": "2147483646\n2147483647",
"output": "NO"
},
{
"input": "2147483647\n2147483646",
"output": "NO"
},
{
"input": "2\n2147483647",
"output": "NO"
},
{
"input": "2\n1073741825",
"output": "NO"
},
{
"input": "2\n1073741824",
"output": "YES\n29"
},
{
"input": "10000\n10",
"output": "NO"
},
{
"input": "10\n10000",
"output": "YES\n3"
},
{
"input": "10\n2000000000",
"output": "NO"
},
{
"input": "10\n1000000000",
"output": "YES\n8"
},
{
"input": "5\n1808548329",
"output": "NO"
},
{
"input": "2\n2147483646",
"output": "NO"
},
{
"input": "25\n125",
"output": "NO"
},
{
"input": "6\n18",
"output": "NO"
},
{
"input": "5\n30",
"output": "NO"
}
] | 1,588,874,869
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 64
| 312
| 20,172,800
|
n =int(input())
m = int(input())
i=0
val =n
while(val<=m):
if(val==m):
print("YES")
print(i)
exit()
val *=n
i+=1
print("NO")
|
Title: Cifera
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
Input Specification:
The first input line contains integer number *k*, the second line contains integer number *l* (2<=β€<=*k*,<=*l*<=β€<=231<=-<=1).
Output Specification:
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number β the importance of number *l*.
Demo Input:
['5\n25\n', '3\n8\n']
Demo Output:
['YES\n1\n', 'NO\n']
Note:
none
|
```python
n =int(input())
m = int(input())
i=0
val =n
while(val<=m):
if(val==m):
print("YES")
print(i)
exit()
val *=n
i+=1
print("NO")
```
| 3
|
|
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<=β€<=*n*<=β€<=105) β the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=*n*) β the elements of array.
The third line contains integer *m* (1<=β€<=*m*<=β€<=105) β the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=β€<=*b**i*<=β€<=*n*) β 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 Π‘++. 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"
}
] | 1,688,375,778
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 372
| 24,883,200
|
size = int(input())
main = list(map(int, input().split()))
twoDarr = [[0] for _ in range(size+1)]
counter = 1
for x in main:
if twoDarr[x] == [0]:
twoDarr[x][0] = counter
else:
twoDarr[x].append(counter)
counter+=1
qurries = int(input())
list_qur = list(map(int, input().split()))
petya = 0
vasya = 0
for y in list_qur:
petya+= min(twoDarr[y])
vasya+= size - max(twoDarr[y]) + 1
print(f'{petya} {vasya}')
|
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<=β€<=*n*<=β€<=105) β the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=*n*) β the elements of array.
The third line contains integer *m* (1<=β€<=*m*<=β€<=105) β the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=β€<=*b**i*<=β€<=*n*) β 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 Π‘++. 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
size = int(input())
main = list(map(int, input().split()))
twoDarr = [[0] for _ in range(size+1)]
counter = 1
for x in main:
if twoDarr[x] == [0]:
twoDarr[x][0] = counter
else:
twoDarr[x].append(counter)
counter+=1
qurries = int(input())
list_qur = list(map(int, input().split()))
petya = 0
vasya = 0
for y in list_qur:
petya+= min(twoDarr[y])
vasya+= size - max(twoDarr[y]) + 1
print(f'{petya} {vasya}')
```
| 3
|
|
447
|
A
|
DZY Loves Hash
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x*Β *mod*Β *p*. Operation *a*Β *mod*Β *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.
|
The first line contains two integers, *p* and *n* (2<=β€<=*p*,<=*n*<=β€<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=β€<=*x**i*<=β€<=109).
|
Output a single integer β the answer to the problem.
|
[
"10 5\n0\n21\n53\n41\n53\n",
"5 5\n0\n1\n2\n3\n4\n"
] |
[
"4\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "10 5\n0\n21\n53\n41\n53",
"output": "4"
},
{
"input": "5 5\n0\n1\n2\n3\n4",
"output": "-1"
},
{
"input": "10 6\n811966798\n734823552\n790326404\n929189974\n414343256\n560346537",
"output": "4"
},
{
"input": "2 2\n788371161\n801743052",
"output": "-1"
},
{
"input": "10 6\n812796223\n122860157\n199259103\n597650585\n447742024\n521549402",
"output": "3"
},
{
"input": "300 2\n822454942\n119374431",
"output": "-1"
},
{
"input": "300 2\n823284367\n507345500",
"output": "-1"
},
{
"input": "2 5\n791579811\n35613889\n997079893\n878677665\n693717467",
"output": "2"
},
{
"input": "20 5\n793926268\n28931770\n842870287\n974950617\n859404206",
"output": "-1"
},
{
"input": "100 15\n805069249\n778178198\n633897389\n844316223\n146759898\n870527016\n82668754\n42060733\n943602929\n979451110\n746979598\n47406033\n607284062\n850870259\n229415316",
"output": "5"
},
{
"input": "100 15\n806204335\n189490323\n718805086\n716787474\n262315718\n822030596\n894644222\n724054623\n141384399\n579354205\n192622443\n672556242\n97417563\n243354557\n208957882",
"output": "8"
},
{
"input": "100 15\n807033760\n577461392\n275221433\n532633429\n295714486\n783298996\n255799943\n99107143\n729119412\n59302896\n37640015\n313610861\n630550567\n534283052\n681062462",
"output": "8"
},
{
"input": "100 15\n808103310\n136224397\n360129131\n405104681\n263786657\n734802577\n67808179\n928584682\n926900882\n511722343\n483348395\n938695534\n120684068\n74152694\n808088675",
"output": "9"
},
{
"input": "2 2\n2\n2",
"output": "2"
},
{
"input": "300 2\n0\n300",
"output": "2"
},
{
"input": "2 2\n0\n0",
"output": "2"
}
] | 1,405,770,245
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 93
| 0
|
p,n = map(int, input().split())
hash_table = []
for i in range(0, n):
x = int(input())
if x % p in hash_table:
print(i + 1)
quit()
else:
hash_table.append(x % p)
print(-1)
|
Title: DZY Loves Hash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x*Β *mod*Β *p*. Operation *a*Β *mod*Β *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.
Input Specification:
The first line contains two integers, *p* and *n* (2<=β€<=*p*,<=*n*<=β€<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=β€<=*x**i*<=β€<=109).
Output Specification:
Output a single integer β the answer to the problem.
Demo Input:
['10 5\n0\n21\n53\n41\n53\n', '5 5\n0\n1\n2\n3\n4\n']
Demo Output:
['4\n', '-1\n']
Note:
none
|
```python
p,n = map(int, input().split())
hash_table = []
for i in range(0, n):
x = int(input())
if x % p in hash_table:
print(i + 1)
quit()
else:
hash_table.append(x % p)
print(-1)
```
| 3
|
|
632
|
A
|
Grandma Laura and Apples
|
PROGRAMMING
| 1,200
|
[] | null | null |
Grandma Laura came to the market to sell some apples. During the day she sold all the apples she had. But grandma is old, so she forgot how many apples she had brought to the market.
She precisely remembers she had *n* buyers and each of them bought exactly half of the apples she had at the moment of the purchase and also she gave a half of an apple to some of them as a gift (if the number of apples at the moment of purchase was odd), until she sold all the apples she had.
So each buyer took some integral positive number of apples, but maybe he didn't pay for a half of an apple (if the number of apples at the moment of the purchase was odd).
For each buyer grandma remembers if she gave a half of an apple as a gift or not. The cost of an apple is *p* (the number *p* is even).
Print the total money grandma should have at the end of the day to check if some buyers cheated her.
|
The first line contains two integers *n* and *p* (1<=β€<=*n*<=β€<=40,<=2<=β€<=*p*<=β€<=1000) β the number of the buyers and the cost of one apple. It is guaranteed that the number *p* is even.
The next *n* lines contains the description of buyers. Each buyer is described with the string half if he simply bought half of the apples and with the string halfplus if grandma also gave him a half of an apple as a gift.
It is guaranteed that grandma has at least one apple at the start of the day and she has no apples at the end of the day.
|
Print the only integer *a* β the total money grandma should have at the end of the day.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
|
[
"2 10\nhalf\nhalfplus\n",
"3 10\nhalfplus\nhalfplus\nhalfplus\n"
] |
[
"15\n",
"55\n"
] |
In the first sample at the start of the day the grandma had two apples. First she sold one apple and then she sold a half of the second apple and gave a half of the second apple as a present to the second buyer.
| 0
|
[
{
"input": "2 10\nhalf\nhalfplus",
"output": "15"
},
{
"input": "3 10\nhalfplus\nhalfplus\nhalfplus",
"output": "55"
},
{
"input": "10 328\nhalf\nhalfplus\nhalfplus\nhalf\nhalfplus\nhalf\nhalf\nhalf\nhalfplus\nhalfplus",
"output": "258300"
},
{
"input": "1 2\nhalfplus",
"output": "1"
},
{
"input": "5 6\nhalf\nhalf\nhalfplus\nhalf\nhalfplus",
"output": "114"
},
{
"input": "10 2\nhalfplus\nhalfplus\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalfplus",
"output": "1703"
},
{
"input": "20 6\nhalf\nhalfplus\nhalfplus\nhalfplus\nhalf\nhalf\nhalfplus\nhalf\nhalfplus\nhalfplus\nhalfplus\nhalf\nhalf\nhalfplus\nhalfplus\nhalf\nhalfplus\nhalfplus\nhalfplus\nhalfplus",
"output": "6056877"
},
{
"input": "30 68\nhalf\nhalf\nhalf\nhalfplus\nhalfplus\nhalf\nhalf\nhalf\nhalf\nhalf\nhalf\nhalf\nhalf\nhalfplus\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalfplus\nhalfplus\nhalf\nhalf\nhalf\nhalfplus\nhalf\nhalfplus\nhalfplus",
"output": "59847312584"
},
{
"input": "40 562\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalf\nhalf\nhalf\nhalfplus\nhalf\nhalf\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalf\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalfplus\nhalf\nhalf\nhalfplus\nhalf\nhalfplus\nhalfplus\nhalf\nhalf\nhalfplus\nhalf\nhalf\nhalfplus\nhalfplus",
"output": "486750560068982"
}
] | 1,649,766,476
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 9
| 46
| 0
|
n, p = (int(x) for x in input().split())
b = [input() for _ in range(n)]
a = 0
ans = 0
for i in reversed(b):
a *= 2
if i == 'halfplus':
a += 1
ans += p//2
ans += p * (a//2)
print(ans)
|
Title: Grandma Laura and Apples
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grandma Laura came to the market to sell some apples. During the day she sold all the apples she had. But grandma is old, so she forgot how many apples she had brought to the market.
She precisely remembers she had *n* buyers and each of them bought exactly half of the apples she had at the moment of the purchase and also she gave a half of an apple to some of them as a gift (if the number of apples at the moment of purchase was odd), until she sold all the apples she had.
So each buyer took some integral positive number of apples, but maybe he didn't pay for a half of an apple (if the number of apples at the moment of the purchase was odd).
For each buyer grandma remembers if she gave a half of an apple as a gift or not. The cost of an apple is *p* (the number *p* is even).
Print the total money grandma should have at the end of the day to check if some buyers cheated her.
Input Specification:
The first line contains two integers *n* and *p* (1<=β€<=*n*<=β€<=40,<=2<=β€<=*p*<=β€<=1000) β the number of the buyers and the cost of one apple. It is guaranteed that the number *p* is even.
The next *n* lines contains the description of buyers. Each buyer is described with the string half if he simply bought half of the apples and with the string halfplus if grandma also gave him a half of an apple as a gift.
It is guaranteed that grandma has at least one apple at the start of the day and she has no apples at the end of the day.
Output Specification:
Print the only integer *a* β the total money grandma should have at the end of the day.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['2 10\nhalf\nhalfplus\n', '3 10\nhalfplus\nhalfplus\nhalfplus\n']
Demo Output:
['15\n', '55\n']
Note:
In the first sample at the start of the day the grandma had two apples. First she sold one apple and then she sold a half of the second apple and gave a half of the second apple as a present to the second buyer.
|
```python
n, p = (int(x) for x in input().split())
b = [input() for _ in range(n)]
a = 0
ans = 0
for i in reversed(b):
a *= 2
if i == 'halfplus':
a += 1
ans += p//2
ans += p * (a//2)
print(ans)
```
| 3
|
|
750
|
A
|
New Year and Hurry
|
PROGRAMMING
| 800
|
[
"binary search",
"brute force",
"implementation",
"math"
] | null | null |
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5Β·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
|
The only line of the input contains two integers *n* and *k* (1<=β€<=*n*<=β€<=10, 1<=β€<=*k*<=β€<=240)Β β the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
|
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
|
[
"3 222\n",
"4 190\n",
"7 1\n"
] |
[
"2\n",
"4\n",
"7\n"
] |
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5β+β10β=β15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5β+β10β+β15β+β20β=β50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
| 500
|
[
{
"input": "3 222",
"output": "2"
},
{
"input": "4 190",
"output": "4"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "10 135",
"output": "6"
},
{
"input": "10 136",
"output": "5"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "9 240",
"output": "0"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "9 235",
"output": "1"
},
{
"input": "9 236",
"output": "0"
},
{
"input": "5 225",
"output": "2"
},
{
"input": "5 226",
"output": "1"
},
{
"input": "4 210",
"output": "3"
},
{
"input": "4 211",
"output": "2"
},
{
"input": "4 191",
"output": "3"
},
{
"input": "10 165",
"output": "5"
},
{
"input": "10 166",
"output": "4"
},
{
"input": "8 100",
"output": "7"
},
{
"input": "8 101",
"output": "6"
},
{
"input": "8 60",
"output": "8"
},
{
"input": "8 61",
"output": "7"
},
{
"input": "10 15",
"output": "9"
},
{
"input": "10 16",
"output": "8"
},
{
"input": "4 100",
"output": "4"
},
{
"input": "4 101",
"output": "4"
},
{
"input": "7 167",
"output": "4"
},
{
"input": "10 164",
"output": "5"
},
{
"input": "9 170",
"output": "4"
},
{
"input": "8 160",
"output": "5"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "8 123",
"output": "6"
},
{
"input": "2 99",
"output": "2"
},
{
"input": "10 88",
"output": "7"
},
{
"input": "1 235",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 240",
"output": "0"
},
{
"input": "1 55",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "3 240",
"output": "0"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "2 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "2 236",
"output": "0"
},
{
"input": "10 2",
"output": "9"
},
{
"input": "3 239",
"output": "0"
},
{
"input": "1 237",
"output": "0"
},
{
"input": "7 8",
"output": "7"
},
{
"input": "10 235",
"output": "1"
}
] | 1,686,063,242
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 46
| 0
|
a = input().split(" ")
n = int(a[0])
time = 240 - int(a[1])
for i in range(1,n+1):
if time >= 5*i:
time -= 5*i
i += 1
elif time < 5*i:
break
print(i-1)
|
Title: New Year and Hurry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5Β·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
Input Specification:
The only line of the input contains two integers *n* and *k* (1<=β€<=*n*<=β€<=10, 1<=β€<=*k*<=β€<=240)Β β the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
Output Specification:
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
Demo Input:
['3 222\n', '4 190\n', '7 1\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5β+β10β=β15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5β+β10β+β15β+β20β=β50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
|
```python
a = input().split(" ")
n = int(a[0])
time = 240 - int(a[1])
for i in range(1,n+1):
if time >= 5*i:
time -= 5*i
i += 1
elif time < 5*i:
break
print(i-1)
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,622,039,947
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 0
|
print('YES' if ''.join([_ for _ in reversed(input())]) == input() else 'NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
print('YES' if ''.join([_ for _ in reversed(input())]) == input() else 'NO')
```
| 3.969
|
32
|
A
|
Reconnaissance
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Reconnaissance
|
2
|
256
|
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different.
|
The first line contains two integers *n* and *d* (1<=β€<=*n*<=β€<=1000,<=1<=β€<=*d*<=β€<=109) β amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers β heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
|
Output one number β amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*.
|
[
"5 10\n10 20 50 60 65\n",
"5 1\n55 30 29 31 55\n"
] |
[
"6\n",
"6\n"
] |
none
| 500
|
[
{
"input": "5 10\n10 20 50 60 65",
"output": "6"
},
{
"input": "5 1\n55 30 29 31 55",
"output": "6"
},
{
"input": "6 10\n4 6 4 1 9 3",
"output": "30"
},
{
"input": "7 100\n19 1694 261 162 1 234 513",
"output": "8"
},
{
"input": "8 42\n37 53 74 187 568 22 5 65",
"output": "20"
},
{
"input": "10 4\n11 6 76 49 28 20 57 152 5 32",
"output": "4"
},
{
"input": "100 100\n51 93 101 960 2 477 213 129 663 925 254 78 1486 274 160 481 132 156 412 372 5 57 152 298 1771 7 359 468 254 406 202 929 221 366 552 97 555 29 822 118 539 140 992 854 7 163 134 103 940 30 409 1003 398 43 555 79 107 40 23 103 643 171 310 382 770 337 18 189 570 177 29 54 855 171 205 291 299 935 620 180 114 358 88 292 118 400 218 537 369 60 683 192 13 537 59 824 264 191 3 300",
"output": "2404"
}
] | 1,567,926,188
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 528
| 307,200
|
n, d = map(int, input().split())
data = sorted(map(int, input().split()))
res = 0
for i in range(n):
for j in range(i + 1, n):
if data[j] - data[i] > d:
break
res += 2
print(res)
|
Title: Reconnaissance
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different.
Input Specification:
The first line contains two integers *n* and *d* (1<=β€<=*n*<=β€<=1000,<=1<=β€<=*d*<=β€<=109) β amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers β heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
Output Specification:
Output one number β amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*.
Demo Input:
['5 10\n10 20 50 60 65\n', '5 1\n55 30 29 31 55\n']
Demo Output:
['6\n', '6\n']
Note:
none
|
```python
n, d = map(int, input().split())
data = sorted(map(int, input().split()))
res = 0
for i in range(n):
for j in range(i + 1, n):
if data[j] - data[i] > d:
break
res += 2
print(res)
```
| 3.867428
|
219
|
A
|
k-String
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
|
The first input line contains integer *k* (1<=β€<=*k*<=β€<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=β€<=|*s*|<=β€<=1000, where |*s*| is the length of string *s*.
|
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
|
[
"2\naazz\n",
"3\nabcabcabz\n"
] |
[
"azaz\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
},
{
"input": "5\naaaaa",
"output": "aaaaa"
},
{
"input": "3\naabaaaaabb",
"output": "-1"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "2\nbabac",
"output": "-1"
},
{
"input": "3\nbbbccc",
"output": "bcbcbc"
},
{
"input": "2\naa",
"output": "aa"
},
{
"input": "250\ncececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece",
"output": "cececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece"
},
{
"input": "15\nabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaa",
"output": "aaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbc"
},
{
"input": "1\naaa",
"output": "aaa"
},
{
"input": "1\naabaab",
"output": "aaaabb"
},
{
"input": "2\naabbbbccccccdddddddd",
"output": "abbcccddddabbcccdddd"
},
{
"input": "3\naaaaaaaaacccdddddd",
"output": "aaacddaaacddaaacdd"
},
{
"input": "3\naaaaaaaaacccbbbbbb",
"output": "aaabbcaaabbcaaabbc"
},
{
"input": "2\naazzzz",
"output": "azzazz"
},
{
"input": "2\naaaa",
"output": "aaaa"
},
{
"input": "2\naaaazz",
"output": "aazaaz"
},
{
"input": "2\naaaabb",
"output": "aabaab"
},
{
"input": "2\naabbbb",
"output": "abbabb"
},
{
"input": "2\naaaazzzz",
"output": "aazzaazz"
},
{
"input": "2\naaazz",
"output": "-1"
},
{
"input": "2\nbbbbaa",
"output": "abbabb"
},
{
"input": "2\naaaaaabbbb",
"output": "aaabbaaabb"
},
{
"input": "2\naaaaaazz",
"output": "aaazaaaz"
},
{
"input": "2\naaaabbbb",
"output": "aabbaabb"
},
{
"input": "2\naabaab",
"output": "aabaab"
},
{
"input": "1\naaaaa",
"output": "aaaaa"
},
{
"input": "2\nbbaaaa",
"output": "aabaab"
},
{
"input": "2\nacaccc",
"output": "accacc"
}
] | 1,686,755,031
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 42
| 124
| 0
|
k = int(input())
word = [*input()]
flag = 0
letters = list(set(word))
numbers = []
for i in range(len(letters)):
if word.count(letters[i]) % k != 0:
flag = 1
else:
numbers.append(word.count(letters[i]) // k)
if flag == 1:
print(-1)
else:
copy = []
for t in range(len(letters)):
for l in range(numbers[t]):
copy.append(letters[t])
q = "".join(copy)
last = []
for z in range(k):
last.append(q)
print("".join(last))
|
Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=β€<=*k*<=β€<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=β€<=|*s*|<=β€<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none
|
```python
k = int(input())
word = [*input()]
flag = 0
letters = list(set(word))
numbers = []
for i in range(len(letters)):
if word.count(letters[i]) % k != 0:
flag = 1
else:
numbers.append(word.count(letters[i]) // k)
if flag == 1:
print(-1)
else:
copy = []
for t in range(len(letters)):
for l in range(numbers[t]):
copy.append(letters[t])
q = "".join(copy)
last = []
for z in range(k):
last.append(q)
print("".join(last))
```
| 3
|
|
119
|
A
|
Epic Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
|
The only string contains space-separated integers *a*, *b* and *n* (1<=β€<=*a*,<=*b*,<=*n*<=β€<=100) β the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
|
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
|
[
"3 5 9\n",
"1 1 100\n"
] |
[
"0",
"1"
] |
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*,β*b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*,β0)β=β*gcd*(0,β*x*)β=β*x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3,β9)β=β3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5,β6)β=β1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3,β5)β=β1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5,β4)β=β1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3,β3)β=β3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5,β0)β=β5 stones from the heap. As 0β<β5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
| 500
|
[
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,550,000,537
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 216
| 0
|
a, b, n = map(int, input().split())
def gcd(a, b):
if a == 0:
return b
elif b == 0:
return a
m = max(a, b)
n = min(a, b)
r = m % n
while r > 0:
m = n
n = r
r = m % n
return n
while True:
x = gcd(a, n)
if n < x:
print(1)
break
n -= x
y = gcd(b, n)
if n < y:
print(0)
break
n -= y
|
Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=β€<=*a*,<=*b*,<=*n*<=β€<=100) β the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*,β*b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*,β0)β=β*gcd*(0,β*x*)β=β*x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3,β9)β=β3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5,β6)β=β1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3,β5)β=β1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5,β4)β=β1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3,β3)β=β3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5,β0)β=β5 stones from the heap. As 0β<β5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
|
```python
a, b, n = map(int, input().split())
def gcd(a, b):
if a == 0:
return b
elif b == 0:
return a
m = max(a, b)
n = min(a, b)
r = m % n
while r > 0:
m = n
n = r
r = m % n
return n
while True:
x = gcd(a, n)
if n < x:
print(1)
break
n -= x
y = gcd(b, n)
if n < y:
print(0)
break
n -= y
```
| 3
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer β the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,697,840,189
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 92
| 0
|
n = int(input())
p = 0
for i in range(n):
s = input()
if s.count("1")>=2:
p+=1
print(p)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer β the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
p = 0
for i in range(n):
s = input()
if s.count("1")>=2:
p+=1
print(p)
```
| 3
|
|
286
|
A
|
Lucky Permutation
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"math"
] | null | null |
A permutation *p* of size *n* is the sequence *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct integers, each of them is from 1 to *n* (1<=β€<=*p**i*<=β€<=*n*).
A lucky permutation is such permutation *p*, that any integer *i* (1<=β€<=*i*<=β€<=*n*) meets this condition *p**p**i*<==<=*n*<=-<=*i*<=+<=1.
You have integer *n*. Find some lucky permutation *p* of size *n*.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=105) β the required permutation size.
|
Print "-1" (without the quotes) if the lucky permutation *p* of size *n* doesn't exist.
Otherwise, print *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=β€<=*p**i*<=β€<=*n*) after a space β the required permutation.
If there are multiple answers, you can print any of them.
|
[
"1\n",
"2\n",
"4\n",
"5\n"
] |
[
"1 \n",
"-1\n",
"2 4 1 3 \n",
"2 5 3 1 4 \n"
] |
none
| 500
|
[
{
"input": "1",
"output": "1 "
},
{
"input": "2",
"output": "-1"
},
{
"input": "4",
"output": "2 4 1 3 "
},
{
"input": "5",
"output": "2 5 3 1 4 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "100000",
"output": "2 100000 4 99998 6 99996 8 99994 10 99992 12 99990 14 99988 16 99986 18 99984 20 99982 22 99980 24 99978 26 99976 28 99974 30 99972 32 99970 34 99968 36 99966 38 99964 40 99962 42 99960 44 99958 46 99956 48 99954 50 99952 52 99950 54 99948 56 99946 58 99944 60 99942 62 99940 64 99938 66 99936 68 99934 70 99932 72 99930 74 99928 76 99926 78 99924 80 99922 82 99920 84 99918 86 99916 88 99914 90 99912 92 99910 94 99908 96 99906 98 99904 100 99902 102 99900 104 99898 106 99896 108 99894 110 99892 112 99890 114..."
},
{
"input": "6",
"output": "-1"
},
{
"input": "7",
"output": "-1"
},
{
"input": "8",
"output": "2 8 4 6 3 5 1 7 "
},
{
"input": "9",
"output": "2 9 4 7 5 3 6 1 8 "
},
{
"input": "99997",
"output": "2 99997 4 99995 6 99993 8 99991 10 99989 12 99987 14 99985 16 99983 18 99981 20 99979 22 99977 24 99975 26 99973 28 99971 30 99969 32 99967 34 99965 36 99963 38 99961 40 99959 42 99957 44 99955 46 99953 48 99951 50 99949 52 99947 54 99945 56 99943 58 99941 60 99939 62 99937 64 99935 66 99933 68 99931 70 99929 72 99927 74 99925 76 99923 78 99921 80 99919 82 99917 84 99915 86 99913 88 99911 90 99909 92 99907 94 99905 96 99903 98 99901 100 99899 102 99897 104 99895 106 99893 108 99891 110 99889 112 99887 114 ..."
},
{
"input": "10000",
"output": "2 10000 4 9998 6 9996 8 9994 10 9992 12 9990 14 9988 16 9986 18 9984 20 9982 22 9980 24 9978 26 9976 28 9974 30 9972 32 9970 34 9968 36 9966 38 9964 40 9962 42 9960 44 9958 46 9956 48 9954 50 9952 52 9950 54 9948 56 9946 58 9944 60 9942 62 9940 64 9938 66 9936 68 9934 70 9932 72 9930 74 9928 76 9926 78 9924 80 9922 82 9920 84 9918 86 9916 88 9914 90 9912 92 9910 94 9908 96 9906 98 9904 100 9902 102 9900 104 9898 106 9896 108 9894 110 9892 112 9890 114 9888 116 9886 118 9884 120 9882 122 9880 124 9878 126 9..."
},
{
"input": "10001",
"output": "2 10001 4 9999 6 9997 8 9995 10 9993 12 9991 14 9989 16 9987 18 9985 20 9983 22 9981 24 9979 26 9977 28 9975 30 9973 32 9971 34 9969 36 9967 38 9965 40 9963 42 9961 44 9959 46 9957 48 9955 50 9953 52 9951 54 9949 56 9947 58 9945 60 9943 62 9941 64 9939 66 9937 68 9935 70 9933 72 9931 74 9929 76 9927 78 9925 80 9923 82 9921 84 9919 86 9917 88 9915 90 9913 92 9911 94 9909 96 9907 98 9905 100 9903 102 9901 104 9899 106 9897 108 9895 110 9893 112 9891 114 9889 116 9887 118 9885 120 9883 122 9881 124 9879 126 9..."
},
{
"input": "10002",
"output": "-1"
},
{
"input": "10003",
"output": "-1"
},
{
"input": "10004",
"output": "2 10004 4 10002 6 10000 8 9998 10 9996 12 9994 14 9992 16 9990 18 9988 20 9986 22 9984 24 9982 26 9980 28 9978 30 9976 32 9974 34 9972 36 9970 38 9968 40 9966 42 9964 44 9962 46 9960 48 9958 50 9956 52 9954 54 9952 56 9950 58 9948 60 9946 62 9944 64 9942 66 9940 68 9938 70 9936 72 9934 74 9932 76 9930 78 9928 80 9926 82 9924 84 9922 86 9920 88 9918 90 9916 92 9914 94 9912 96 9910 98 9908 100 9906 102 9904 104 9902 106 9900 108 9898 110 9896 112 9894 114 9892 116 9890 118 9888 120 9886 122 9884 124 9882 126..."
},
{
"input": "25",
"output": "2 25 4 23 6 21 8 19 10 17 12 15 13 11 14 9 16 7 18 5 20 3 22 1 24 "
},
{
"input": "29",
"output": "2 29 4 27 6 25 8 23 10 21 12 19 14 17 15 13 16 11 18 9 20 7 22 5 24 3 26 1 28 "
},
{
"input": "33",
"output": "2 33 4 31 6 29 8 27 10 25 12 23 14 21 16 19 17 15 18 13 20 11 22 9 24 7 26 5 28 3 30 1 32 "
},
{
"input": "9",
"output": "2 9 4 7 5 3 6 1 8 "
},
{
"input": "13",
"output": "2 13 4 11 6 9 7 5 8 3 10 1 12 "
},
{
"input": "17",
"output": "2 17 4 15 6 13 8 11 9 7 10 5 12 3 14 1 16 "
},
{
"input": "50001",
"output": "2 50001 4 49999 6 49997 8 49995 10 49993 12 49991 14 49989 16 49987 18 49985 20 49983 22 49981 24 49979 26 49977 28 49975 30 49973 32 49971 34 49969 36 49967 38 49965 40 49963 42 49961 44 49959 46 49957 48 49955 50 49953 52 49951 54 49949 56 49947 58 49945 60 49943 62 49941 64 49939 66 49937 68 49935 70 49933 72 49931 74 49929 76 49927 78 49925 80 49923 82 49921 84 49919 86 49917 88 49915 90 49913 92 49911 94 49909 96 49907 98 49905 100 49903 102 49901 104 49899 106 49897 108 49895 110 49893 112 49891 114 ..."
},
{
"input": "20001",
"output": "2 20001 4 19999 6 19997 8 19995 10 19993 12 19991 14 19989 16 19987 18 19985 20 19983 22 19981 24 19979 26 19977 28 19975 30 19973 32 19971 34 19969 36 19967 38 19965 40 19963 42 19961 44 19959 46 19957 48 19955 50 19953 52 19951 54 19949 56 19947 58 19945 60 19943 62 19941 64 19939 66 19937 68 19935 70 19933 72 19931 74 19929 76 19927 78 19925 80 19923 82 19921 84 19919 86 19917 88 19915 90 19913 92 19911 94 19909 96 19907 98 19905 100 19903 102 19901 104 19899 106 19897 108 19895 110 19893 112 19891 114 ..."
},
{
"input": "30001",
"output": "2 30001 4 29999 6 29997 8 29995 10 29993 12 29991 14 29989 16 29987 18 29985 20 29983 22 29981 24 29979 26 29977 28 29975 30 29973 32 29971 34 29969 36 29967 38 29965 40 29963 42 29961 44 29959 46 29957 48 29955 50 29953 52 29951 54 29949 56 29947 58 29945 60 29943 62 29941 64 29939 66 29937 68 29935 70 29933 72 29931 74 29929 76 29927 78 29925 80 29923 82 29921 84 29919 86 29917 88 29915 90 29913 92 29911 94 29909 96 29907 98 29905 100 29903 102 29901 104 29899 106 29897 108 29895 110 29893 112 29891 114 ..."
},
{
"input": "99996",
"output": "2 99996 4 99994 6 99992 8 99990 10 99988 12 99986 14 99984 16 99982 18 99980 20 99978 22 99976 24 99974 26 99972 28 99970 30 99968 32 99966 34 99964 36 99962 38 99960 40 99958 42 99956 44 99954 46 99952 48 99950 50 99948 52 99946 54 99944 56 99942 58 99940 60 99938 62 99936 64 99934 66 99932 68 99930 70 99928 72 99926 74 99924 76 99922 78 99920 80 99918 82 99916 84 99914 86 99912 88 99910 90 99908 92 99906 94 99904 96 99902 98 99900 100 99898 102 99896 104 99894 106 99892 108 99890 110 99888 112 99886 114 ..."
},
{
"input": "99999",
"output": "-1"
},
{
"input": "101",
"output": "2 101 4 99 6 97 8 95 10 93 12 91 14 89 16 87 18 85 20 83 22 81 24 79 26 77 28 75 30 73 32 71 34 69 36 67 38 65 40 63 42 61 44 59 46 57 48 55 50 53 51 49 52 47 54 45 56 43 58 41 60 39 62 37 64 35 66 33 68 31 70 29 72 27 74 25 76 23 78 21 80 19 82 17 84 15 86 13 88 11 90 9 92 7 94 5 96 3 98 1 100 "
},
{
"input": "505",
"output": "2 505 4 503 6 501 8 499 10 497 12 495 14 493 16 491 18 489 20 487 22 485 24 483 26 481 28 479 30 477 32 475 34 473 36 471 38 469 40 467 42 465 44 463 46 461 48 459 50 457 52 455 54 453 56 451 58 449 60 447 62 445 64 443 66 441 68 439 70 437 72 435 74 433 76 431 78 429 80 427 82 425 84 423 86 421 88 419 90 417 92 415 94 413 96 411 98 409 100 407 102 405 104 403 106 401 108 399 110 397 112 395 114 393 116 391 118 389 120 387 122 385 124 383 126 381 128 379 130 377 132 375 134 373 136 371 138 369 140 367 142 ..."
},
{
"input": "565",
"output": "2 565 4 563 6 561 8 559 10 557 12 555 14 553 16 551 18 549 20 547 22 545 24 543 26 541 28 539 30 537 32 535 34 533 36 531 38 529 40 527 42 525 44 523 46 521 48 519 50 517 52 515 54 513 56 511 58 509 60 507 62 505 64 503 66 501 68 499 70 497 72 495 74 493 76 491 78 489 80 487 82 485 84 483 86 481 88 479 90 477 92 475 94 473 96 471 98 469 100 467 102 465 104 463 106 461 108 459 110 457 112 455 114 453 116 451 118 449 120 447 122 445 124 443 126 441 128 439 130 437 132 435 134 433 136 431 138 429 140 427 142 ..."
}
] | 1,379,097,997
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 780
| 3,276,800
|
n=int(input())
L=[0]*(n+1)
X=[False]*(n+1)
if(n%4!=0 and n%4!=1):
print(-1)
else:
for i in range(1,n+1):
if(X[i]):
continue
X[i]=True
X[n-i+1]=True
for j in range(i+1,n+1):
if(X[j]):
continue
X[j]=True
X[n-j+1]=True
L[i]=j
L[n-i+1]=n-j+1
L[j]=n-i+1
L[n-j+1]=i
break
if(n%4==1):
L[n//2+1]=n//2+1
for i in range(1,n):
print(L[i],end=" ")
print(L[n])
|
Title: Lucky Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation *p* of size *n* is the sequence *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct integers, each of them is from 1 to *n* (1<=β€<=*p**i*<=β€<=*n*).
A lucky permutation is such permutation *p*, that any integer *i* (1<=β€<=*i*<=β€<=*n*) meets this condition *p**p**i*<==<=*n*<=-<=*i*<=+<=1.
You have integer *n*. Find some lucky permutation *p* of size *n*.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=105) β the required permutation size.
Output Specification:
Print "-1" (without the quotes) if the lucky permutation *p* of size *n* doesn't exist.
Otherwise, print *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=β€<=*p**i*<=β€<=*n*) after a space β the required permutation.
If there are multiple answers, you can print any of them.
Demo Input:
['1\n', '2\n', '4\n', '5\n']
Demo Output:
['1 \n', '-1\n', '2 4 1 3 \n', '2 5 3 1 4 \n']
Note:
none
|
```python
n=int(input())
L=[0]*(n+1)
X=[False]*(n+1)
if(n%4!=0 and n%4!=1):
print(-1)
else:
for i in range(1,n+1):
if(X[i]):
continue
X[i]=True
X[n-i+1]=True
for j in range(i+1,n+1):
if(X[j]):
continue
X[j]=True
X[n-j+1]=True
L[i]=j
L[n-i+1]=n-j+1
L[j]=n-i+1
L[n-j+1]=i
break
if(n%4==1):
L[n//2+1]=n//2+1
for i in range(1,n):
print(L[i],end=" ")
print(L[n])
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
|
Output one number β 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"
}
] | 1,622,093,606
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 124
| 0
|
n=input()
nlist=[int(a) for a in n.split()]
width=nlist[0]
length=nlist[1]
if length%2==0:
print(int(length*width/2))
else:
print(int(length//2*width+width//2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
Output Specification:
Output one number β 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
n=input()
nlist=[int(a) for a in n.split()]
width=nlist[0]
length=nlist[1]
if length%2==0:
print(int(length*width/2))
else:
print(int(length//2*width+width//2))
```
| 3.969
|
505
|
A
|
Mr. Kitayuta's Gift
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Mr. Kitayuta has kindly given you a string *s* consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter into *s* to make it a palindrome. A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not.
You can choose any lowercase English letter, and insert it to any position of *s*, possibly to the beginning or the end of *s*. You have to insert a letter even if the given string is already a palindrome.
If it is possible to insert one lowercase English letter into *s* so that the resulting string will be a palindrome, print the string after the insertion. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one palindrome that can be obtained, you are allowed to print any of them.
|
The only line of the input contains a string *s* (1<=β€<=|*s*|<=β€<=10). Each character in *s* is a lowercase English letter.
|
If it is possible to turn *s* into a palindrome by inserting one lowercase English letter, print the resulting string in a single line. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one solution, any of them will be accepted.
|
[
"revive\n",
"ee\n",
"kitayuta\n"
] |
[
"reviver\n",
"eye",
"NA\n"
] |
For the first sample, insert 'r' to the end of "revive" to obtain a palindrome "reviver".
For the second sample, there is more than one solution. For example, "eve" will also be accepted.
For the third sample, it is not possible to turn "kitayuta" into a palindrome by just inserting one letter.
| 500
|
[
{
"input": "revive",
"output": "reviver"
},
{
"input": "ee",
"output": "eee"
},
{
"input": "kitayuta",
"output": "NA"
},
{
"input": "evima",
"output": "NA"
},
{
"input": "a",
"output": "aa"
},
{
"input": "yutampo",
"output": "NA"
},
{
"input": "fft",
"output": "tfft"
},
{
"input": "shuseki",
"output": "NA"
},
{
"input": "lsdijfjisl",
"output": "lsdijfjidsl"
},
{
"input": "ewcdcwerp",
"output": "NA"
},
{
"input": "noon",
"output": "nooon"
},
{
"input": "add",
"output": "adda"
},
{
"input": "iq",
"output": "qiq"
},
{
"input": "hogloid",
"output": "NA"
},
{
"input": "yosupo",
"output": "NA"
},
{
"input": "stat",
"output": "stats"
},
{
"input": "level",
"output": "levvel"
},
{
"input": "iwiwi",
"output": "iwiiwi"
},
{
"input": "racecar",
"output": "raceecar"
},
{
"input": "dog",
"output": "NA"
},
{
"input": "squirrel",
"output": "NA"
},
{
"input": "slime",
"output": "NA"
},
{
"input": "palindrome",
"output": "NA"
},
{
"input": "nqwcvcwqn",
"output": "nqwcvvcwqn"
},
{
"input": "lsdijfjisl",
"output": "lsdijfjidsl"
},
{
"input": "jewiiwefj",
"output": "jfewiiwefj"
},
{
"input": "vnwmvvmonv",
"output": "NA"
},
{
"input": "zennyuu",
"output": "NA"
},
{
"input": "mrtoudai",
"output": "NA"
},
{
"input": "z",
"output": "zz"
},
{
"input": "babccba",
"output": "babccbab"
},
{
"input": "battaba",
"output": "abattaba"
},
{
"input": "bbb",
"output": "bbbb"
},
{
"input": "zaa",
"output": "zaaz"
},
{
"input": "abbabab",
"output": "NA"
},
{
"input": "racecar",
"output": "raceecar"
},
{
"input": "zzz",
"output": "zzzz"
},
{
"input": "eviver",
"output": "reviver"
},
{
"input": "aaaaa",
"output": "aaaaaa"
},
{
"input": "eee",
"output": "eeee"
}
] | 1,459,292,943
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 62
| 4,608,000
|
a=input()
for i in range(26):
for j in range(len(a)+1):
b=a[:j]+chr(97+i)+a[j:]
if b==b[::-1]:print(b);exit()
print('NA')
|
Title: Mr. Kitayuta's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Kitayuta has kindly given you a string *s* consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter into *s* to make it a palindrome. A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not.
You can choose any lowercase English letter, and insert it to any position of *s*, possibly to the beginning or the end of *s*. You have to insert a letter even if the given string is already a palindrome.
If it is possible to insert one lowercase English letter into *s* so that the resulting string will be a palindrome, print the string after the insertion. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one palindrome that can be obtained, you are allowed to print any of them.
Input Specification:
The only line of the input contains a string *s* (1<=β€<=|*s*|<=β€<=10). Each character in *s* is a lowercase English letter.
Output Specification:
If it is possible to turn *s* into a palindrome by inserting one lowercase English letter, print the resulting string in a single line. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one solution, any of them will be accepted.
Demo Input:
['revive\n', 'ee\n', 'kitayuta\n']
Demo Output:
['reviver\n', 'eye', 'NA\n']
Note:
For the first sample, insert 'r' to the end of "revive" to obtain a palindrome "reviver".
For the second sample, there is more than one solution. For example, "eve" will also be accepted.
For the third sample, it is not possible to turn "kitayuta" into a palindrome by just inserting one letter.
|
```python
a=input()
for i in range(26):
for j in range(len(a)+1):
b=a[:j]+chr(97+i)+a[j:]
if b==b[::-1]:print(b);exit()
print('NA')
```
| 3
|
|
552
|
B
|
Vanya and Books
|
PROGRAMMING
| 1,200
|
[
"implementation",
"math"
] | null | null |
Vanya got an important task β he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=109) β the number of books in the library.
|
Print the number of digits needed to number all the books.
|
[
"13\n",
"4\n"
] |
[
"17\n",
"4\n"
] |
Note to the first test. The books get numbers 1,β2,β3,β4,β5,β6,β7,β8,β9,β10,β11,β12,β13, which totals to 17 digits.
Note to the second sample. The books get numbers 1,β2,β3,β4, which totals to 4 digits.
| 1,000
|
[
{
"input": "13",
"output": "17"
},
{
"input": "4",
"output": "4"
},
{
"input": "100",
"output": "192"
},
{
"input": "99",
"output": "189"
},
{
"input": "1000000000",
"output": "8888888899"
},
{
"input": "1000000",
"output": "5888896"
},
{
"input": "999",
"output": "2889"
},
{
"input": "55",
"output": "101"
},
{
"input": "222222222",
"output": "1888888896"
},
{
"input": "8",
"output": "8"
},
{
"input": "13",
"output": "17"
},
{
"input": "313",
"output": "831"
},
{
"input": "1342",
"output": "4261"
},
{
"input": "30140",
"output": "139594"
},
{
"input": "290092",
"output": "1629447"
},
{
"input": "2156660",
"output": "13985516"
},
{
"input": "96482216",
"output": "760746625"
},
{
"input": "943006819",
"output": "8375950269"
},
{
"input": "1",
"output": "1"
},
{
"input": "7",
"output": "7"
},
{
"input": "35",
"output": "61"
},
{
"input": "996",
"output": "2880"
},
{
"input": "6120",
"output": "23373"
},
{
"input": "30660",
"output": "142194"
},
{
"input": "349463",
"output": "1985673"
},
{
"input": "8171970",
"output": "56092686"
},
{
"input": "36123011",
"output": "277872985"
},
{
"input": "986747865",
"output": "8769619683"
},
{
"input": "9",
"output": "9"
},
{
"input": "10",
"output": "11"
},
{
"input": "11",
"output": "13"
},
{
"input": "101",
"output": "195"
},
{
"input": "1000",
"output": "2893"
},
{
"input": "1001",
"output": "2897"
},
{
"input": "9999",
"output": "38889"
},
{
"input": "10000",
"output": "38894"
},
{
"input": "10001",
"output": "38899"
},
{
"input": "99999",
"output": "488889"
},
{
"input": "100000",
"output": "488895"
},
{
"input": "100001",
"output": "488901"
},
{
"input": "999999",
"output": "5888889"
},
{
"input": "1000001",
"output": "5888903"
},
{
"input": "9999999",
"output": "68888889"
},
{
"input": "10000000",
"output": "68888897"
},
{
"input": "10000001",
"output": "68888905"
},
{
"input": "99999999",
"output": "788888889"
},
{
"input": "100000000",
"output": "788888898"
},
{
"input": "100000001",
"output": "788888907"
},
{
"input": "999999999",
"output": "8888888889"
},
{
"input": "95",
"output": "181"
},
{
"input": "995",
"output": "2877"
},
{
"input": "9995",
"output": "38873"
},
{
"input": "99995",
"output": "488869"
},
{
"input": "999999995",
"output": "8888888853"
},
{
"input": "999999998",
"output": "8888888880"
}
] | 1,606,545,148
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 55
| 156
| 0
|
# RawCoder : https://clck.ru/S8wGG
n = int(input())
if(n <= 9):print(n)
else:
le = len(str(n))
s = ""
req = 0
for k in range(1,le):
s += str(k)
print((n * le) - (int(s)*9))
|
Title: Vanya and Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got an important task β he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=109) β the number of books in the library.
Output Specification:
Print the number of digits needed to number all the books.
Demo Input:
['13\n', '4\n']
Demo Output:
['17\n', '4\n']
Note:
Note to the first test. The books get numbers 1,β2,β3,β4,β5,β6,β7,β8,β9,β10,β11,β12,β13, which totals to 17 digits.
Note to the second sample. The books get numbers 1,β2,β3,β4, which totals to 4 digits.
|
```python
# RawCoder : https://clck.ru/S8wGG
n = int(input())
if(n <= 9):print(n)
else:
le = len(str(n))
s = ""
req = 0
for k in range(1,le):
s += str(k)
print((n * le) - (int(s)*9))
```
| 3
|
|
735
|
B
|
Urbanization
|
PROGRAMMING
| 1,100
|
[
"greedy",
"number theory",
"sortings"
] | null | null |
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
|
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=β€<=*n*,<=*n*1,<=*n*2<=β€<=100<=000, *n*1<=+<=*n*2<=β€<=*n*)Β β the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
|
Print one real valueΒ β the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"2 1 1\n1 5\n",
"4 2 1\n1 4 2 3\n"
] |
[
"6.00000000\n",
"6.50000000\n"
] |
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub>β+β*a*<sub class="lower-index">4</sub>)β/β2β+β*a*<sub class="lower-index">2</sub>β=β(3β+β2)β/β2β+β4β=β6.5
| 1,000
|
[
{
"input": "2 1 1\n1 5",
"output": "6.00000000"
},
{
"input": "4 2 1\n1 4 2 3",
"output": "6.50000000"
},
{
"input": "3 1 2\n1 2 3",
"output": "4.50000000"
},
{
"input": "10 4 6\n3 5 7 9 12 25 67 69 83 96",
"output": "88.91666667"
},
{
"input": "19 7 12\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 100000 100000",
"output": "47052.10714286"
},
{
"input": "100 9 6\n109 711 40 95 935 48 228 253 308 726 816 534 252 8 966 363 162 508 84 83 807 506 748 178 45 30 106 108 764 698 825 198 336 353 158 790 64 262 403 334 577 571 742 541 946 602 279 621 910 776 421 886 29 133 114 394 762 965 339 263 750 530 49 80 124 31 322 292 27 590 960 278 111 932 849 491 561 744 469 511 106 271 156 160 836 363 149 473 457 543 976 809 490 29 85 626 265 88 995 946",
"output": "1849.66666667"
},
{
"input": "69 6 63\n53475 22876 79144 6335 33763 79104 65441 45527 65847 94406 74670 43529 75330 19403 67629 56187 57949 23071 64910 54409 55348 18056 855 24961 50565 6622 26467 33989 22660 79469 41246 13965 79706 14422 16075 93378 81313 48173 13470 97348 2346 27452 59427 29925 29847 73823 32021 10988 24609 98855 90919 45939 17203 8439 43007 40138 55693 30314 71734 33458 66850 4011 20089 20546 92090 50842 78859 62756 40177",
"output": "135712.88888889"
},
{
"input": "69 6 9\n2612 17461 69001 33130 10662 85485 88195 45974 16712 81365 67119 87797 15559 20197 74716 92979 97268 49466 68603 48351 99905 35606 54242 98603 68232 54398 82637 49647 38979 46171 54680 23334 15892 92186 69670 29711 67999 2220 32317 717 70667 68262 86760 55720 97158 61122 7251 138 21022 27197 12691 59331 13576 66999 38332 13574 83484 66646 17704 33065 98583 80259 64631 16745 69431 40747 82089 82788 32739",
"output": "183129.44444444"
}
] | 1,515,217,097
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 187
| 30,924,800
|
def solve():
n,n1,n2 = map(int, input().split())
lo = min([n1, n2])
hi = max([n1, n2])
l = list(map(int,input().split()))
l.sort(reverse=True)
#print(l)
s1 = sum(l[:lo])
s2 = sum(l[lo:lo+hi])
#print(s1)
#print(s2)
ans = (s1/lo) + (s2/hi)
print(ans)
solve()
|
Title: Urbanization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
Input Specification:
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=β€<=*n*,<=*n*1,<=*n*2<=β€<=100<=000, *n*1<=+<=*n*2<=β€<=*n*)Β β the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
Output Specification:
Print one real valueΒ β the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['2 1 1\n1 5\n', '4 2 1\n1 4 2 3\n']
Demo Output:
['6.00000000\n', '6.50000000\n']
Note:
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub>β+β*a*<sub class="lower-index">4</sub>)β/β2β+β*a*<sub class="lower-index">2</sub>β=β(3β+β2)β/β2β+β4β=β6.5
|
```python
def solve():
n,n1,n2 = map(int, input().split())
lo = min([n1, n2])
hi = max([n1, n2])
l = list(map(int,input().split()))
l.sort(reverse=True)
#print(l)
s1 = sum(l[:lo])
s2 = sum(l[lo:lo+hi])
#print(s1)
#print(s2)
ans = (s1/lo) + (s2/hi)
print(ans)
solve()
```
| 3
|
|
770
|
A
|
New Password
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
|
The first line contains two positive integers *n* and *k* (2<=β€<=*n*<=β€<=100, 2<=β€<=*k*<=β€<=*min*(*n*,<=26)) β the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
|
Print any password which satisfies all conditions given by Innokentiy.
|
[
"4 3\n",
"6 6\n",
"5 2\n"
] |
[
"java\n",
"python\n",
"phphp\n"
] |
In the first test there is one of the appropriate new passwords β java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords β python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords β phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
| 500
|
[
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,658,781,925
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 46
| 0
|
n, k =map(int,input().split())
wordlist = []
c = k
for i in range(n):
if len(wordlist)==n:
break
wordlist.append(chr(97+(i%k)))
if len(wordlist)==n:
break
print(''.join(wordlist))
|
Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=β€<=*n*<=β€<=100, 2<=β€<=*k*<=β€<=*min*(*n*,<=26)) β the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords β java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords β python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords β phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
|
```python
n, k =map(int,input().split())
wordlist = []
c = k
for i in range(n):
if len(wordlist)==n:
break
wordlist.append(chr(97+(i%k)))
if len(wordlist)==n:
break
print(''.join(wordlist))
```
| 3
|
|
716
|
B
|
Complete the Word
|
PROGRAMMING
| 1,300
|
[
"greedy",
"two pointers"
] | null | null |
ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?
|
The first and only line of the input contains a single string *s* (1<=β€<=|*s*|<=β€<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.
|
If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them.
|
[
"ABC??FGHIJK???OPQR?TUVWXY?\n",
"WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n",
"??????????????????????????\n",
"AABCDEFGHIJKLMNOPQRSTUVW??M\n"
] |
[
"ABCDEFGHIJKLMNOPQRZTUVWXYS",
"-1",
"MNBVCXZLKJHGFDSAQPWOEIRUYT",
"-1"
] |
In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is β-β1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer.
| 1,000
|
[
{
"input": "ABC??FGHIJK???OPQR?TUVWXY?",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO",
"output": "-1"
},
{
"input": "??????????????????????????",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "AABCDEFGHIJKLMNOPQRSTUVW??M",
"output": "-1"
},
{
"input": "QWERTYUIOPASDFGHJKL???????",
"output": "QWERTYUIOPASDFGHJKLBCMNVXZ"
},
{
"input": "ABABABBAB????????????ABABABABA???????????ABABABABA?????????KLCSJB?????????Z",
"output": "ABABABBABAAAAAAAAAAAAABABABABAAAAAAAAAAAAABABABABADEFGHIMNOKLCSJBPQRTUVWXYZ"
},
{
"input": "Q?E?T?U?O?A?D?G?J?L?X?V?MMQ?E?T?U?O?A?D?G?J?L?X?V?N",
"output": "QAEATAUAOAAADAGAJALAXAVAMMQBECTFUHOIAKDPGRJSLWXYVZN"
},
{
"input": "???????????????????????????",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZA"
},
{
"input": "EJMGJAXCHXYIKZSQKUGRCLSTWDLNCVZIGXGWILAVFBEIGOHWGVEPRJTHWEDQRPOVZUQOSRVTIHFFHJMCLOWGHCIGJBCAAVBJFMJEFTEGFXZFVRZOXAFOFVXRAIZEWIKILFLYDZVDADYWYWYJXAGDFGNZBQKKKTGWPINLCDBZVULROGAKEKXXTWNYKQBMLQMQRUYOWUTWMNTJVGUXENHXWMFWMSBKVNGXSNFFTRTTGEGBBHMFZTKNJQDYUQOXVDWTDHZCCQNYYIOFPMKYQIGEEYBCKBAYVCTWARVMHIENKXKFXNXEFUHUNRQPEDFUBMKNQOYCQHGTLRHLWUAVZJDRBRTSVQHBKRDJFKKYEZAJWJKATRFZLNELPYGFUIWBXLIWVTHUILJHTQKDGRNCFTFELCOQPJDBYSPYJOUDKIFRCKEMJPUXTTAMHVENEVMNTZLUYSUALQOUPPRLZHCYICXAQFFRQZAAJNFKVRJDMDXFTBRJSAAHTSVG",
"output": "-1"
},
{
"input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
"output": "-1"
},
{
"input": "DMWSBHPGSJJD?EEV?CYAXQCCGNNQWNN?OMEDD?VC?CTKNQQPYXKKJFAYMJ?FMPXXCLKOL?OTRCE",
"output": "-1"
},
{
"input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
"output": "-1"
},
{
"input": "E?BIVQUPQQEJNMINFD?NKV?IROHPTGUIPMEVYPII?LZJMRI?FTKKKBHPOVQZZSAPDDWVSPVHOBT",
"output": "-1"
},
{
"input": "FDQHJSNDDXHJLWVZVXJZUGKVHWCZVRWVZTIURLMJNGAMCUBDGVSIDEYRJZOLDISDNTOEKLSNLBSOQZLJVPAMLEBAVUNBXNKMLZBGJJQCGCSKBFSEEDXEVSWGZHFJIZJESPZIKIONJWTFFYYZKIDBSDNPJVAUHQMRFKIJWCEGTBVZHWZEKLPHGZVKZFAFAQRNKHGACNRTSXQKKCYBMEMKNKKSURKHOSMEVUXNGOCVCLVVSKULGBKFPCEKVRAJMBWCFFFSCCNDOSEKXEFFZETTUZHMQETWCVZASTTULYOPBNMOMXMVUEEEYZHSMRPAEIHUKNPNJTARJKQKIOXDJASSQPQQHEQIQJQLVPIJRCFVOVECHBOCRYWQEDXZLJXUDZUBFTRWEWNYTSKGDBEBWFFLMUYWELNVAAXSMKYEZXQFKKHJTZKMKMYOBTVXAOVBRMAMHTBDDYMDGQYEEBYZUBMUCKLKXCZGTWVZAYJOXZVGUYNXOVAPXQVE",
"output": "-1"
},
{
"input": "KMNTIOJTLEKZW?JALAZYWYMKWRXTLAKNMDJLICZMETAKHVPTDOLAPCGHOEYSNIUJZVLPBTZ?YSR",
"output": "-1"
},
{
"input": "?MNURVAKIVSOGITVJZEZCAOZEFVNZERAHVNCVCYKTJVEHK?ZMDL?CROLIDFSG?EIFHYKELMQRBVLE?CERELHDVFODJ?LBGJVFPO?CVMPBW?DPGZMVA?BKPXQQCRMKHJWDNAJSGOTGLBNSWMXMKAQ?MWMXCNRSGHTL?LGLAHSDHAGZRGTNDFI?KJ?GSAWOEPOENXTJCVJGMYOFIQKKDWOCIKPGCMFEKNEUPFGBCBYQCM?EQSAX?HZ?MFKAUHOHRKZZSIVZCAKYIKBDJYOCZJRYNLSOKGAEGQRQ?TBURXXLHAFCNVGAUVWBXZILMHWSBYJTIMWPNEGATPURPTJYFWKHRL?QPYUQ?HKDDHWAHOWUSONQKSZFIYFMFUJAMIYAMPNBGVPJSDFDFSAHDWWGEAKXLHBURNTIMCUZIAFAOCVNKPJRNLNGSJVMGKQ?IFQSRHTZGKHGXFJBDGPLCUUMEWNOSCONIVCLAOAPPSFFLCPRIXTKNBSSOVM",
"output": "-1"
},
{
"input": "MRHKVVRBFEIFWIZGWCATJPBSZWNYANEWSSEVFQUUVNJKQOKVIGYBPFSZFTBUCNQEJEYVOWSPYER",
"output": "-1"
},
{
"input": "CNRFBWKRTQTDFOMIGPPGDBHPRNRXFASDDBCZXHORGXDRSIORLJEROJBLLEHLNBILBPX?KHQLCOUPTKUADCDNHNBWMVNUUVUFPIRXSPNUCCRLJTDSUIUDLBKNKMXSAVBJDUGWIMNBIUWJX?TCBDEBNDYUGPS?MQSSEIIUGEE?XXKW?CMFQMWUAEXTSZNNOCPHBAEAKWALYBBMUMQZXUKTQPWNMZKIDECWIZFHKQIUJZRSBZPQFUQNVKQZMYJDHXZWXFHIZ?HWPIPIWV?JMIYKEJDNPMKTTOY?NTOMZZXTNMWQENYRWFYM?WLJJFCIJSETZSJORBZZHAFWYKGQJAPYQQXUWROOZUDOJJLNCDRSGUKYAZLLENGUICGOYPLJQ?POSKHPMOFJMAOXCITWWL?LOEDKHZPQFZZCTB?JYZNXZSDREAMGGXHMCFTQNOUALEYHULSDQVOXZIWFHNNHHG?FYUOCQNKBLFGGZ?YNFNVLRMENYBDWMDSP",
"output": "-1"
},
{
"input": "KSRVTPFVRJWNPYUZMXBRLKVXIQPPBYVSYKRQPNGKTKRPFMKLIYFACFKBIQGPAXLEUESVGPBBXLY",
"output": "-1"
},
{
"input": "LLVYUOXHBHUZSAPUMQEKWSQAFRKSMEENXDQYOPQFXNNFXSRBGXFUIRBFJDSDKQIDMCPPTWRJOZCRHZYZPBVUJPQXHNALAOCJDTTBDZWYDBVPMNSQNVMLHHUJAOIWFSEJEJSRBYREOZKHEXTBAXPTISPGIPOYBFFEJNAKKXAEPNGKWYGEJTNEZIXAWRSCEIRTKNEWSKSGKNIKDEOVXGYVEVFRGTNDFNWIFDRZQEJQZYIWNZXCONVZAKKKETPTPPXZMIVDWPGXOFODRNJZBATKGXAPXYHTUUFFASCHOLSMVSWBIJBAENEGNQTWKKOJUYQNXWDCDXBXBJOOWETWLQMGKHAJEMGXMYNVEHRAEGZOJJQPZGYRHXRNKMSWFYDIZLIBUTSKIKGQJZLGZQFJVIMNOHNZJKWVVPFMFACVXKJKTBZRXRZDJKSWSXBBKWIKEICSZEIPTOJCKJQYYPNUPRNPQNNCVITNXPLAKQBYAIQGNAHXDUQWQLYN",
"output": "-1"
},
{
"input": "PVCKCT?KLTFPIBBIHODCAABEQLJKQECRUJUSHSXPMBEVBKHQTIKQLBLTIRQZPOGPWMMNWWCUKAD",
"output": "-1"
},
{
"input": "BRTYNUVBBWMFDSRXAMLNSBIN???WDDQVPCSWGJTHLRAKTPFKGVLHAKNRIEYIDDRDZLLTBRKXRVRSPBSLXIZRRBEVMHJSAFPLZAIHFVTTEKDO?DYWKILEYRM?VHSEQCBYZZRZMICVZRYA?ONCSZOPGZUMIHJQJPIFX?YJMIERCMKTSFTDZIKEZPLDEOOCJLQIZ?RPHUEQHPNNSBRQRTDGLWNSCZ?WQVIZPTOETEXYI?DRQUOMREPUTOAJKFNBGYNWMGCAOELXEPLLZEYHTVLT?ETJJXLHJMAUDQESNQ?ZCGNDGI?JSGUXQV?QAWQIYKXBKCCSWNRTGHPZF?CSWDQSAZIWQNHOWHYAEZNXRMPAZEQQPPIBQQJEDHJEDHVXNEDETEN?ZHEPJJ?VVDYGPJUWGCBMB?ANFJHJXQVAJWCAZEZXZX?BACPPXORNENMCRMQPIYKNPHX?NSKGEABWWVLHQ?ESWLJUPQJSFIUEGMGHEYVLYEDWJG?L",
"output": "-1"
},
{
"input": "TESTEIGHTYFOUR",
"output": "-1"
},
{
"input": "ABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "-1"
},
{
"input": "?????????????????????????",
"output": "-1"
},
{
"input": "Q?RYJPGLNQ",
"output": "-1"
},
{
"input": "ABCDEFGHIJKLMNOPQRZTUVWXYS",
"output": "ABCDEFGHIJKLMNOPQRZTUVWXYS"
},
{
"input": "AACDEFGHIJKLMNOPQRZTUVWXYS",
"output": "-1"
},
{
"input": "ZA?ABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "ZAZABCDEFGHIJKLMNOPQRSTUVWXY"
},
{
"input": "AABBCCDDEEFFGGHHIIJJKKLLMMNNOOPPQQRRSSTTUUVVWWXXYYZZ",
"output": "-1"
},
{
"input": "ABCDEFGHIJKLMNOPQRSTUVWXYYYZABC",
"output": "-1"
},
{
"input": "????",
"output": "-1"
},
{
"input": "ABCDEFGHIJKLMNOPQRZTUVWXYS??",
"output": "ABCDEFGHIJKLMNOPQRZTUVWXYSAA"
},
{
"input": "A",
"output": "-1"
},
{
"input": "NKBDABACEFGGGIJLLLLMMMOMPQWZSSRHHTTUWUWVXYY",
"output": "-1"
},
{
"input": "AA",
"output": "-1"
},
{
"input": "BAAAAAAAAAAAAAAAAAAAAAAAAAAAXA?CDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "BAAAAAAAAAAAAAAAAAAAAAAAAAAAXABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "???DEFGHIJKL??L?PQRSTUVW???",
"output": "-1"
},
{
"input": "?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A?A",
"output": "-1"
},
{
"input": "A?AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA?????????????AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA???????????AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA??????????",
"output": "-1"
},
{
"input": "ZXXCVBNMLKJHGFDSAQWWERTYUIOP",
"output": "-1"
},
{
"input": "ABCA???????????????????????",
"output": "ABCADEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "AB",
"output": "-1"
},
{
"input": "ABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "BCDEFGHIAJKLMNOPQARSTUVWXYZ",
"output": "-1"
},
{
"input": "AABBCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "-1"
},
{
"input": "???AA????????????????????BB???",
"output": "-1"
},
{
"input": "ABC??FGAHIJ??MNOPQRST??WXYZ",
"output": "ABCDEFGAHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBN",
"output": "-1"
},
{
"input": "??????AAAAAAAAAAAAAAAAAAAABBBBB??????????AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBB????????????AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA??????????AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAASSSSSSSSFFFFFFF?????????DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD??????WWWWWWWWWWEEEEEEEEEEERRRRRRRRRRRRRRRRRRRRRRRRRHHHHHHHHHH?????",
"output": "-1"
}
] | 1,583,428,246
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 115
| 764
| 819,200
|
import math
#q=int(input())
q=1
for _ in range(q):
s=list(input())
n=len(s)
flag=False
file=[]
for i in range(n-25):
l=s[i:i+26]
k=[0]*26
ques=0
for j in range(26):
if l[j]!='?':
k[ord(l[j])-65]+=1
else:
ques+=1
if max(k)==1 or ques==26:
flag=True
check=i
file=k
break
else:
continue
if flag:
k=0
for i in range(check,check+26):
for j in range(k,26):
if s[i]=='?':
if file[j]==0:
s[i]=chr(65+j)
k=j+1
for i in range(n):
if s[i]=='?':
s[i]='A'
print(''.join(s))
else:
print(-1)
|
Title: Complete the Word
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?
Input Specification:
The first and only line of the input contains a single string *s* (1<=β€<=|*s*|<=β€<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.
Output Specification:
If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them.
Demo Input:
['ABC??FGHIJK???OPQR?TUVWXY?\n', 'WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n', '??????????????????????????\n', 'AABCDEFGHIJKLMNOPQRSTUVW??M\n']
Demo Output:
['ABCDEFGHIJKLMNOPQRZTUVWXYS', '-1', 'MNBVCXZLKJHGFDSAQPWOEIRUYT', '-1']
Note:
In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is β-β1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer.
|
```python
import math
#q=int(input())
q=1
for _ in range(q):
s=list(input())
n=len(s)
flag=False
file=[]
for i in range(n-25):
l=s[i:i+26]
k=[0]*26
ques=0
for j in range(26):
if l[j]!='?':
k[ord(l[j])-65]+=1
else:
ques+=1
if max(k)==1 or ques==26:
flag=True
check=i
file=k
break
else:
continue
if flag:
k=0
for i in range(check,check+26):
for j in range(k,26):
if s[i]=='?':
if file[j]==0:
s[i]=chr(65+j)
k=j+1
for i in range(n):
if s[i]=='?':
s[i]='A'
print(''.join(s))
else:
print(-1)
```
| 3
|
|
466
|
A
|
Cheap Travel
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
|
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=β€<=*n*,<=*m*,<=*a*,<=*b*<=β€<=1000) β the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
|
Print a single integer β the minimum sum in rubles that Ann will need to spend.
|
[
"6 2 1 2\n",
"5 2 2 3\n"
] |
[
"6\n",
"8\n"
] |
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets.
| 500
|
[
{
"input": "6 2 1 2",
"output": "6"
},
{
"input": "5 2 2 3",
"output": "8"
},
{
"input": "10 3 5 1",
"output": "4"
},
{
"input": "1000 1 1000 1000",
"output": "1000000"
},
{
"input": "1000 3 1000 1000",
"output": "334000"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10 2 1 1",
"output": "5"
},
{
"input": "1 1000 1 2",
"output": "1"
},
{
"input": "1 1000 3 2",
"output": "2"
},
{
"input": "10 3 1 2",
"output": "7"
},
{
"input": "995 1 2 1",
"output": "995"
},
{
"input": "556 2 16 15",
"output": "4170"
},
{
"input": "477 2 16 14",
"output": "3346"
},
{
"input": "101 110 1 100",
"output": "100"
},
{
"input": "9 3 3 10",
"output": "27"
},
{
"input": "100 8 10 1",
"output": "13"
},
{
"input": "6 4 1 3",
"output": "5"
},
{
"input": "8 5 2 8",
"output": "14"
},
{
"input": "1000 2 1 1000",
"output": "1000"
}
] | 1,697,909,715
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 19
| 62
| 0
|
n, m, a, b=map(int,input().split())
if b/m>=a:
print(n*a)
else:
s=(n//m)*b
x= n%m
s+=min(x*a,b)
print(s)
|
Title: Cheap Travel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
Input Specification:
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=β€<=*n*,<=*m*,<=*a*,<=*b*<=β€<=1000) β the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
Output Specification:
Print a single integer β the minimum sum in rubles that Ann will need to spend.
Demo Input:
['6 2 1 2\n', '5 2 2 3\n']
Demo Output:
['6\n', '8\n']
Note:
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets.
|
```python
n, m, a, b=map(int,input().split())
if b/m>=a:
print(n*a)
else:
s=(n//m)*b
x= n%m
s+=min(x*a,b)
print(s)
```
| 3
|
|
714
|
A
|
Meeting of Old Friends
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] | null | null |
Today an outstanding event is going to happen in the forestΒ β hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive.
Calculate the number of minutes they will be able to spend together.
|
The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=β€<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=β€<=1018, *l*1<=β€<=*r*1, *l*2<=β€<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
|
Print one integerΒ β the number of minutes Sonya and Filya will be able to spend together.
|
[
"1 10 9 20 1\n",
"1 100 50 200 75\n"
] |
[
"2\n",
"50\n"
] |
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
| 500
|
[
{
"input": "1 10 9 20 1",
"output": "2"
},
{
"input": "1 100 50 200 75",
"output": "50"
},
{
"input": "6 6 5 8 9",
"output": "1"
},
{
"input": "1 1000000000 1 1000000000 1",
"output": "999999999"
},
{
"input": "5 100 8 8 8",
"output": "0"
},
{
"input": "1 1000000000000000000 2 99999999999999999 1000000000",
"output": "99999999999999997"
},
{
"input": "1 1 1 1 1",
"output": "0"
},
{
"input": "1 2 3 4 5",
"output": "0"
},
{
"input": "1 1000000000 2 999999999 3141592",
"output": "999999997"
},
{
"input": "24648817341102 41165114064236 88046848035 13602161452932 10000831349205",
"output": "0"
},
{
"input": "1080184299348 34666828555290 6878390132365 39891656267344 15395310291636",
"output": "27788438422925"
},
{
"input": "11814 27385 22309 28354 23595",
"output": "5076"
},
{
"input": "4722316546398 36672578279675 796716437180 33840047334985 13411035401708",
"output": "29117730788587"
},
{
"input": "14300093617438 14381698008501 6957847034861 32510754974307 66056597033082",
"output": "81604391064"
},
{
"input": "700062402405871919 762322967106512617 297732773882447821 747309903322652819 805776739998108178",
"output": "47247500916780901"
},
{
"input": "59861796371397621 194872039092923459 668110259718450585 841148673332698972 928360292123223779",
"output": "0"
},
{
"input": "298248781360904821 346420922793050061 237084570581741798 726877079564549183 389611850470532358",
"output": "48172141432145241"
},
{
"input": "420745791717606818 864206437350900994 764928840030524015 966634105370748487 793326512080703489",
"output": "99277597320376979"
},
{
"input": "519325240668210886 776112702001665034 360568516809443669 875594219634943179 994594983925273138",
"output": "256787461333454149"
},
{
"input": "170331212821058551 891149660635282032 125964175621755330 208256491683509799 526532153531983174",
"output": "37925278862451249"
},
{
"input": "1 3 3 5 3",
"output": "0"
},
{
"input": "1 5 8 10 9",
"output": "0"
},
{
"input": "1 2 4 5 10",
"output": "0"
},
{
"input": "1 2 2 3 5",
"output": "1"
},
{
"input": "2 4 3 7 3",
"output": "1"
},
{
"input": "1 2 9 10 1",
"output": "0"
},
{
"input": "5 15 1 10 5",
"output": "5"
},
{
"input": "1 4 9 20 25",
"output": "0"
},
{
"input": "2 4 1 2 5",
"output": "1"
},
{
"input": "10 1000 1 100 2",
"output": "91"
},
{
"input": "1 3 3 8 10",
"output": "1"
},
{
"input": "4 6 6 8 9",
"output": "1"
},
{
"input": "2 3 1 4 3",
"output": "1"
},
{
"input": "1 2 2 3 100",
"output": "1"
},
{
"input": "1 2 100 120 2",
"output": "0"
},
{
"input": "1 3 5 7 4",
"output": "0"
},
{
"input": "1 3 5 7 5",
"output": "0"
},
{
"input": "1 4 8 10 6",
"output": "0"
},
{
"input": "1 2 5 6 100",
"output": "0"
},
{
"input": "1 2 5 10 20",
"output": "0"
},
{
"input": "1 2 5 6 7",
"output": "0"
},
{
"input": "2 5 7 12 6",
"output": "0"
},
{
"input": "10 20 50 100 80",
"output": "0"
},
{
"input": "1 2 5 10 2",
"output": "0"
},
{
"input": "1 2 5 6 4",
"output": "0"
},
{
"input": "5 9 1 2 3",
"output": "0"
},
{
"input": "50 100 1 20 3",
"output": "0"
},
{
"input": "10 20 3 7 30",
"output": "0"
},
{
"input": "1 5 10 10 100",
"output": "0"
},
{
"input": "100 101 1 2 3",
"output": "0"
},
{
"input": "1 5 10 20 6",
"output": "0"
},
{
"input": "1 10 15 25 5",
"output": "0"
},
{
"input": "1 2 5 10 3",
"output": "0"
},
{
"input": "2 3 5 6 100",
"output": "0"
},
{
"input": "1 2 4 5 6",
"output": "0"
},
{
"input": "6 10 1 2 40",
"output": "0"
},
{
"input": "20 30 1 5 1",
"output": "0"
},
{
"input": "20 40 50 100 50",
"output": "0"
},
{
"input": "1 1 4 9 2",
"output": "0"
},
{
"input": "1 2 5 6 1",
"output": "0"
},
{
"input": "1 100 400 500 450",
"output": "0"
},
{
"input": "5 6 1 2 5",
"output": "0"
},
{
"input": "1 10 21 30 50",
"output": "0"
},
{
"input": "100 200 300 400 101",
"output": "0"
},
{
"input": "2 8 12 16 9",
"output": "0"
},
{
"input": "1 5 7 9 6",
"output": "0"
},
{
"input": "300 400 100 200 101",
"output": "0"
},
{
"input": "1 2 2 3 10",
"output": "1"
},
{
"input": "1 10 100 200 5",
"output": "0"
},
{
"input": "1 3 3 4 4",
"output": "1"
},
{
"input": "10 20 30 40 25",
"output": "0"
},
{
"input": "1 2 5 10 1",
"output": "0"
},
{
"input": "2 4 8 10 1",
"output": "0"
},
{
"input": "2 5 10 15 7",
"output": "0"
},
{
"input": "100 200 5 10 1",
"output": "0"
},
{
"input": "1 2 100 200 300",
"output": "0"
},
{
"input": "30 100 10 20 25",
"output": "0"
},
{
"input": "10 20 1 5 6",
"output": "0"
},
{
"input": "4 5 1 2 4",
"output": "0"
},
{
"input": "11 100 1 9 1000",
"output": "0"
},
{
"input": "1 1 10 10 228",
"output": "0"
},
{
"input": "5 7 10 20 15",
"output": "0"
},
{
"input": "1 3 8 9 7",
"output": "0"
},
{
"input": "1 10 2 8 8",
"output": "6"
},
{
"input": "1 5 9 15 1",
"output": "0"
},
{
"input": "1 3 5 6 12",
"output": "0"
},
{
"input": "1 100 500 1000 3",
"output": "0"
},
{
"input": "1 1 1 1 2",
"output": "1"
},
{
"input": "1 1000 100 1000 200",
"output": "900"
},
{
"input": "4 5 1 4 1",
"output": "1"
},
{
"input": "1 5 5 7 3",
"output": "1"
},
{
"input": "1 4 4 10 11",
"output": "1"
},
{
"input": "1 1 3 4 100",
"output": "0"
},
{
"input": "1 4 3 5 6",
"output": "2"
},
{
"input": "10 100 20 30 40",
"output": "11"
},
{
"input": "5 9 1 11 7",
"output": "4"
}
] | 1,603,271,179
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 96
| 155
| 0
|
l1,r1,l2,r2,k=map(int,input().split())
ans=0
# if l2>=l1 and r1>=r2:
# ans=r2-l2+1
# if l2<=k<=r2:
# ans-=1
# else:
# pass
# elif l1>=l2 and r2<=r1 and r2>=l1:
# ans=r2-l1+1
# if l1<=k<=r2:
# ans-=1
# else:
# pass
# elif l2>=l1 and r2>=r1 and r1>=l2:
# ans=r1-l2+1
# if l2<=k<=r2:
# ans-=1
# else:
# pass
# elif l2<=l1 and r2>=r1:
# ans=r1-l1+1
# if l1<=k<=r1:
# ans-=1
# else:
# pass
# else:
# ans=0
# print(ans)
l=max(l1,l2)
r=min(r1,r2)
if l>r:
ans=0
else:
if l<=k<=r:
ans=r-l
else:
ans=r-l+1
print(ans)
|
Title: Meeting of Old Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today an outstanding event is going to happen in the forestΒ β hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input Specification:
The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=β€<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=β€<=1018, *l*1<=β€<=*r*1, *l*2<=β€<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output Specification:
Print one integerΒ β the number of minutes Sonya and Filya will be able to spend together.
Demo Input:
['1 10 9 20 1\n', '1 100 50 200 75\n']
Demo Output:
['2\n', '50\n']
Note:
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
|
```python
l1,r1,l2,r2,k=map(int,input().split())
ans=0
# if l2>=l1 and r1>=r2:
# ans=r2-l2+1
# if l2<=k<=r2:
# ans-=1
# else:
# pass
# elif l1>=l2 and r2<=r1 and r2>=l1:
# ans=r2-l1+1
# if l1<=k<=r2:
# ans-=1
# else:
# pass
# elif l2>=l1 and r2>=r1 and r1>=l2:
# ans=r1-l2+1
# if l2<=k<=r2:
# ans-=1
# else:
# pass
# elif l2<=l1 and r2>=r1:
# ans=r1-l1+1
# if l1<=k<=r1:
# ans-=1
# else:
# pass
# else:
# ans=0
# print(ans)
l=max(l1,l2)
r=min(r1,r2)
if l>r:
ans=0
else:
if l<=k<=r:
ans=r-l
else:
ans=r-l+1
print(ans)
```
| 3
|
|
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<=β€<=*n*<=β€<=100).
Then follow *n* lines. Each describes a laptop as *speed* *ram* *hdd* *cost*. Besides,
- *speed*, *ram*, *hdd* and *cost* are integers - 1000<=β€<=*speed*<=β€<=4200 is the processor's speed in megahertz - 256<=β€<=*ram*<=β€<=4096 the RAM volume in megabytes - 1<=β€<=*hdd*<=β€<=500 is the HDD in gigabytes - 100<=β€<=*cost*<=β€<=1000 is price in tugriks
All laptops have different prices.
|
Print a single number β 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"
}
] | 1,620,832,010
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 124
| 6,963,200
|
x = []
n = int(input())
for i in range(n):
a, b, c, d = map(int, input().split())
x.append((a,b,c,d))
y = 1001
for i in range(n):
f = 0
for j in range(n):
if x[i][0] < x[j][0] and x[i][1] < x[j][1] and x[i][2] < x[j][2]:
f = 1
break
if f == 0:
if x[i][3] < y:
y = x[i][3]
ans = i
print(ans + 1)
|
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<=β€<=*n*<=β€<=100).
Then follow *n* lines. Each describes a laptop as *speed* *ram* *hdd* *cost*. Besides,
- *speed*, *ram*, *hdd* and *cost* are integers - 1000<=β€<=*speed*<=β€<=4200 is the processor's speed in megahertz - 256<=β€<=*ram*<=β€<=4096 the RAM volume in megabytes - 1<=β€<=*hdd*<=β€<=500 is the HDD in gigabytes - 100<=β€<=*cost*<=β€<=1000 is price in tugriks
All laptops have different prices.
Output Specification:
Print a single number β 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
x = []
n = int(input())
for i in range(n):
a, b, c, d = map(int, input().split())
x.append((a,b,c,d))
y = 1001
for i in range(n):
f = 0
for j in range(n):
if x[i][0] < x[j][0] and x[i][1] < x[j][1] and x[i][2] < x[j][2]:
f = 1
break
if f == 0:
if x[i][3] < y:
y = x[i][3]
ans = i
print(ans + 1)
```
| 3.95603
|
935
|
A
|
Fafa and his Company
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
|
The input consists of a single line containing a positive integer *n* (2<=β€<=*n*<=β€<=105) β the number of employees in Fafa's company.
|
Print a single integer representing the answer to the problem.
|
[
"2\n",
"10\n"
] |
[
"1\n",
"3\n"
] |
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
| 500
|
[
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
},
{
"input": "1024",
"output": "10"
},
{
"input": "99999",
"output": "11"
},
{
"input": "10007",
"output": "1"
},
{
"input": "4096",
"output": "12"
},
{
"input": "65536",
"output": "16"
},
{
"input": "40320",
"output": "95"
},
{
"input": "30030",
"output": "63"
},
{
"input": "161",
"output": "3"
},
{
"input": "1000",
"output": "15"
},
{
"input": "10000",
"output": "24"
},
{
"input": "777",
"output": "7"
},
{
"input": "121",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "40000",
"output": "34"
},
{
"input": "99990",
"output": "47"
},
{
"input": "98765",
"output": "3"
},
{
"input": "56789",
"output": "3"
},
{
"input": "13579",
"output": "3"
},
{
"input": "97531",
"output": "3"
},
{
"input": "12345",
"output": "7"
},
{
"input": "54321",
"output": "7"
},
{
"input": "83160",
"output": "127"
},
{
"input": "9",
"output": "2"
},
{
"input": "21",
"output": "3"
},
{
"input": "11",
"output": "1"
},
{
"input": "15",
"output": "3"
}
] | 1,689,489,538
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 46
| 0
|
s = int(input())
t = 0
for i in range(1,s):
s -= 1
if i%s==0:
t += 1
print(t)
|
Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=β€<=*n*<=β€<=105) β the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
|
```python
s = int(input())
t = 0
for i in range(1,s):
s -= 1
if i%s==0:
t += 1
print(t)
```
| 3
|
|
369
|
C
|
Valera and Elections
|
PROGRAMMING
| 1,600
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
The city Valera lives in is going to hold elections to the city Parliament.
The city has *n* districts and *n*<=-<=1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to *n*, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are *n* candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to *n*, inclusive. If the candidate number *i* will be elected in the city Parliament, he will perform exactly one promise β to repair all problem roads on the way from the *i*-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
|
The first line contains a single integer *n* (2<=β€<=*n*<=β€<=105) β the number of districts in the city.
Then *n*<=-<=1 lines follow. Each line contains the description of a city road as three positive integers *x**i*, *y**i*, *t**i* (1<=β€<=*x**i*,<=*y**i*<=β€<=*n*, 1<=β€<=*t**i*<=β€<=2) β the districts connected by the *i*-th bidirectional road and the road type. If *t**i* equals to one, then the *i*-th road isn't the problem road; if *t**i* equals to two, then the *i*-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
|
In the first line print a single non-negative number *k* β the minimum size of the required subset of candidates. Then on the second line print *k* space-separated integers *a*1,<=*a*2,<=... *a**k* β the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
|
[
"5\n1 2 2\n2 3 2\n3 4 2\n4 5 2\n",
"5\n1 2 1\n2 3 2\n2 4 1\n4 5 1\n",
"5\n1 2 2\n1 3 2\n1 4 2\n1 5 2\n"
] |
[
"1\n5 \n",
"1\n3 \n",
"4\n5 4 3 2 \n"
] |
none
| 1,500
|
[
{
"input": "5\n1 2 2\n2 3 2\n3 4 2\n4 5 2",
"output": "1\n5 "
},
{
"input": "5\n1 2 1\n2 3 2\n2 4 1\n4 5 1",
"output": "1\n3 "
},
{
"input": "5\n1 2 2\n1 3 2\n1 4 2\n1 5 2",
"output": "4\n5 4 3 2 "
},
{
"input": "5\n1 5 1\n5 4 2\n4 3 1\n3 2 2",
"output": "1\n2 "
},
{
"input": "2\n1 2 1",
"output": "0"
},
{
"input": "10\n7 5 1\n2 1 2\n8 7 2\n2 4 1\n4 5 2\n9 5 1\n3 2 2\n2 10 1\n6 5 2",
"output": "3\n8 6 3 "
},
{
"input": "2\n2 1 1",
"output": "0"
},
{
"input": "2\n1 2 2",
"output": "1\n2 "
},
{
"input": "5\n3 1 1\n4 5 1\n1 4 1\n1 2 1",
"output": "0"
},
{
"input": "5\n1 3 2\n5 4 2\n2 1 2\n4 3 2",
"output": "2\n5 2 "
},
{
"input": "10\n1 9 1\n3 2 2\n1 2 2\n4 7 2\n3 5 2\n4 3 2\n10 3 2\n7 8 2\n3 6 1",
"output": "3\n8 10 5 "
},
{
"input": "10\n7 9 2\n2 6 2\n7 4 1\n5 4 2\n3 2 1\n8 5 2\n4 3 2\n7 10 1\n1 2 2",
"output": "3\n9 8 6 "
},
{
"input": "10\n3 9 1\n2 10 2\n1 7 1\n3 4 1\n7 8 2\n1 2 1\n5 3 1\n5 6 2\n2 3 2",
"output": "3\n6 10 8 "
},
{
"input": "10\n1 10 2\n10 9 2\n10 8 2\n9 7 2\n8 6 1\n7 5 1\n6 4 1\n5 3 1\n4 2 1",
"output": "2\n7 8 "
},
{
"input": "10\n1 10 2\n10 9 2\n10 8 2\n9 7 2\n8 6 2\n7 5 2\n6 4 2\n5 3 2\n4 2 2",
"output": "2\n3 2 "
},
{
"input": "4\n1 2 2\n2 3 1\n2 4 2",
"output": "1\n4 "
}
] | 1,679,031,180
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 74
| 483
| 54,476,800
|
import sys
input = sys.stdin.readline
from collections import defaultdict
n = int(input())
d = [defaultdict(int) for i in range(n+1)]
for i in range(n-1):
a, b, c = map(int, input().split())
d[a][b] = c
d[b][a] = c
q = [(1, -1)]
x = [0]*(n+1)
ew = []
while q:
a, b = q.pop()
if a > 0:
q.append((~a, b))
for i in d[a]:
if i != b:
q.append((i, a))
else:
a = ~a
if a != 1:
if x[a] == 0:
if d[a][b] == 2:
ew.append(a)
x[b] = 1
else:
if x[b] != 1:
x[b] = 1
print(len(ew))
print(' '.join(map(str, ew)))
|
Title: Valera and Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The city Valera lives in is going to hold elections to the city Parliament.
The city has *n* districts and *n*<=-<=1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to *n*, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are *n* candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to *n*, inclusive. If the candidate number *i* will be elected in the city Parliament, he will perform exactly one promise β to repair all problem roads on the way from the *i*-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input Specification:
The first line contains a single integer *n* (2<=β€<=*n*<=β€<=105) β the number of districts in the city.
Then *n*<=-<=1 lines follow. Each line contains the description of a city road as three positive integers *x**i*, *y**i*, *t**i* (1<=β€<=*x**i*,<=*y**i*<=β€<=*n*, 1<=β€<=*t**i*<=β€<=2) β the districts connected by the *i*-th bidirectional road and the road type. If *t**i* equals to one, then the *i*-th road isn't the problem road; if *t**i* equals to two, then the *i*-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output Specification:
In the first line print a single non-negative number *k* β the minimum size of the required subset of candidates. Then on the second line print *k* space-separated integers *a*1,<=*a*2,<=... *a**k* β the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Demo Input:
['5\n1 2 2\n2 3 2\n3 4 2\n4 5 2\n', '5\n1 2 1\n2 3 2\n2 4 1\n4 5 1\n', '5\n1 2 2\n1 3 2\n1 4 2\n1 5 2\n']
Demo Output:
['1\n5 \n', '1\n3 \n', '4\n5 4 3 2 \n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
from collections import defaultdict
n = int(input())
d = [defaultdict(int) for i in range(n+1)]
for i in range(n-1):
a, b, c = map(int, input().split())
d[a][b] = c
d[b][a] = c
q = [(1, -1)]
x = [0]*(n+1)
ew = []
while q:
a, b = q.pop()
if a > 0:
q.append((~a, b))
for i in d[a]:
if i != b:
q.append((i, a))
else:
a = ~a
if a != 1:
if x[a] == 0:
if d[a][b] == 2:
ew.append(a)
x[b] = 1
else:
if x[b] != 1:
x[b] = 1
print(len(ew))
print(' '.join(map(str, ew)))
```
| 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"
}
] | 1,698,429,095
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 124
| 0
|
n=input()
item=[x for x in n]
s=item[0]
s=s.upper()
item[0]=s
s1=''.join(item)
print(s1)
|
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
n=input()
item=[x for x in n]
s=item[0]
s=s.upper()
item[0]=s
s1=''.join(item)
print(s1)
```
| 3
|
|
306
|
A
|
Candies
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarpus has got *n* candies and *m* friends (*n*<=β₯<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
|
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=β€<=*n*,<=*m*<=β€<=100;*n*<=β₯<=*m*) β the number of candies and the number of Polycarpus's friends.
|
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
|
[
"12 3\n",
"15 4\n",
"18 7\n"
] |
[
"4 4 4 ",
"3 4 4 4 ",
"2 2 2 3 3 3 3 "
] |
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
| 500
|
[
{
"input": "12 4",
"output": "3 3 3 3 "
},
{
"input": "15 6",
"output": "2 2 2 3 3 3 "
},
{
"input": "18 8",
"output": "2 2 2 2 2 2 3 3 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "100 1",
"output": "100 "
},
{
"input": "100 100",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "100 10",
"output": "10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "64 8",
"output": "8 8 8 8 8 8 8 8 "
},
{
"input": "81 27",
"output": "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 "
},
{
"input": "100 99",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 "
},
{
"input": "100 51",
"output": "1 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 "
},
{
"input": "100 49",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 "
},
{
"input": "100 13",
"output": "7 7 7 7 8 8 8 8 8 8 8 8 8 "
},
{
"input": "97 11",
"output": "8 8 9 9 9 9 9 9 9 9 9 "
},
{
"input": "19 3",
"output": "6 6 7 "
},
{
"input": "99 10",
"output": "9 10 10 10 10 10 10 10 10 10 "
},
{
"input": "88 5",
"output": "17 17 18 18 18 "
},
{
"input": "87 34",
"output": "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 "
},
{
"input": "97 55",
"output": "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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
}
] | 1,609,756,060
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 216
| 0
|
n,m=map(int,input().split())
l=[]
if n%m==0:
for i in range(m):
l.append(n//m)
print(*l)
else:
for i in range(n%m):
l.append((n//m)+1)
for i in range(m-n%m):
l.append(n//m)
print(*l)
|
Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has got *n* candies and *m* friends (*n*<=β₯<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
Input Specification:
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=β€<=*n*,<=*m*<=β€<=100;*n*<=β₯<=*m*) β the number of candies and the number of Polycarpus's friends.
Output Specification:
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
Demo Input:
['12 3\n', '15 4\n', '18 7\n']
Demo Output:
['4 4 4 ', '3 4 4 4 ', '2 2 2 3 3 3 3 ']
Note:
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
|
```python
n,m=map(int,input().split())
l=[]
if n%m==0:
for i in range(m):
l.append(n//m)
print(*l)
else:
for i in range(n%m):
l.append((n//m)+1)
for i in range(m-n%m):
l.append(n//m)
print(*l)
```
| 3
|
|
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<=β€<=*x*<=β€<=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"
}
] | 1,683,471,121
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 62
| 0
|
i=0
ls=[0]
c=0
n = int(input())
while 2**i <= n:
i+=1
i-=1
for j in range(i,-1,-1):
c+=n//(2**j)
n=n%(2**j)
print(c)
|
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<=β€<=*x*<=β€<=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
i=0
ls=[0]
c=0
n = int(input())
while 2**i <= n:
i+=1
i-=1
for j in range(i,-1,-1):
c+=n//(2**j)
n=n%(2**j)
print(c)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,627,894,187
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 154
| 6,963,200
|
n = int(input())
x = 0
y = 0
z = 0
for i in range(n):
lst = list(map(int, input().strip().split()))[:3]
x += lst[0]
y += lst[1]
z += lst[2]
if (x == 0 and y == 0 and z == 0):
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
x = 0
y = 0
z = 0
for i in range(n):
lst = list(map(int, input().strip().split()))[:3]
x += lst[0]
y += lst[1]
z += lst[2]
if (x == 0 and y == 0 and z == 0):
print("YES")
else:
print("NO")
```
| 3.94853
|
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*<=β€<=10000,<=1<=<<=*m*<=β€<=10).
|
Print a single integer β 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"
}
] | 1,697,791,583
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
import math
stairs, m = map(int, input().split())
found = False
if m > stairs:
print(-1)
exit()
for i in range(math.ceil(stairs / 2), stairs + 1):
if i % m == 0 and stairs >= math.ceil(i / m) * m:
print(i)
found = True
break
if not found:
print(-1)
|
Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=β€<=10000,<=1<=<<=*m*<=β€<=10).
Output Specification:
Print a single integer β 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
import math
stairs, m = map(int, input().split())
found = False
if m > stairs:
print(-1)
exit()
for i in range(math.ceil(stairs / 2), stairs + 1):
if i % m == 0 and stairs >= math.ceil(i / m) * m:
print(i)
found = True
break
if not found:
print(-1)
```
| 3
|
|
80
|
A
|
Panoramix's Prediction
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Panoramix's Prediction
|
2
|
256
|
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 isΒ not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
|
The first and only input line contains two positive integers β *n* and *m* (2<=β€<=*n*<=<<=*m*<=β€<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=β€<=*n*<=<<=*m*<=β€<=4.
|
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
|
[
"3 5\n",
"7 11\n",
"7 9\n"
] |
[
"YES",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,673,202,482
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 45
| 124
| 0
|
n,m = map(int,input().split())
if m > n:
f = 1
for i in range(1,m//2+1):
if m % i == 0:
f = f + 1
if f == 2:
li = []
for j in range(n+1,m+1):
c = 1
for k in range(1,n+1//2+1):
if j % k == 0:
c = c + 1
if c == 2:
li.append(j)
if len(li) == 1:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
|
Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 isΒ not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers β *n* and *m* (2<=β€<=*n*<=<<=*m*<=β€<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=β€<=*n*<=<<=*m*<=β€<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none
|
```python
n,m = map(int,input().split())
if m > n:
f = 1
for i in range(1,m//2+1):
if m % i == 0:
f = f + 1
if f == 2:
li = []
for j in range(n+1,m+1):
c = 1
for k in range(1,n+1//2+1):
if j % k == 0:
c = c + 1
if c == 2:
li.append(j)
if len(li) == 1:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
```
| 3.969
|
355
|
A
|
Vasya and Digital Root
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"implementation"
] | null | null |
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to:
1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=β₯<=10.
For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=β€<=101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist.
|
The first line contains two integers *k* and *d* (1<=β€<=*k*<=β€<=1000;β0<=β€<=*d*<=β€<=9).
|
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes.
|
[
"4 4\n",
"5 1\n",
"1 0\n"
] |
[
"5881\n",
"36172\n",
"0\n"
] |
For the first test sample *dr*(5881)ββ=ββ*dr*(22)ββ=ββ4.
For the second test sample *dr*(36172)ββ=ββ*dr*(19)ββ=ββ*dr*(10)ββ=ββ1.
| 500
|
[
{
"input": "4 4",
"output": "5881"
},
{
"input": "5 1",
"output": "36172"
},
{
"input": "1 0",
"output": "0"
},
{
"input": "8 7",
"output": "49722154"
},
{
"input": "487 0",
"output": "No solution"
},
{
"input": "1000 5",
"output": "8541939554067890866522280268745476436249986028349767396372181155840878549622667946850256234534972693110974918858266403731194206972478044933297639886527448596769215803533001453375065914421371731616055420973164037664278812596299678416020519508892847037891229851414508562230407367486468987019052183250172396304562086008837592345867873765321840214188417303688776985319268802181355472294386101622570417737061113209187893810568585166094583478900129912239498334853726870963804475563182775380744565964067602555515611220..."
},
{
"input": "22 9",
"output": "1583569962049529809017"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "13 5",
"output": "1381199538344"
},
{
"input": "100 4",
"output": "6334594910586850938286642284598905674550356974741186703111536643493065423553455569335256292313330478"
},
{
"input": "123 6",
"output": "928024873067884441426263446866614165147002631091527531801777528825238463822318502518751375671158771476735217071878592158343"
},
{
"input": "1000 1",
"output": "8286301124628812353504240076754144327937426329149605334362213339655339076564408659154706137278060590992944494591503606137350736487608756923833530346502466262820452589925067370165968733865814927433418675056573256434073937686361155637721866942352171450747045834987797118866710087297111065178077368748085213082452303815796793489599773148508108295035303578345492871662297456131736137780231762177312635688688714815857818196180724774924848693916003108422682889382923194020205691379066085156078824413573001257245677878..."
},
{
"input": "2 0",
"output": "No solution"
},
{
"input": "734 9",
"output": "5509849803670339733829077693143634799621955270111335907079347964026719040571586127009915057683769302171314977999063915868539391500563742827163274052101515706840652002966522709635011152141196057419086708927225560622675363856445980167733179728663010064912099615416068178748694469047950713834326493597331720572208847439692450327661109751421257198843242305082523510866664350537162158359215265173356615680034808012842300294492281197211603826994471586252822908597603049772690875861970190564793056757768783375525854981..."
},
{
"input": "678 8",
"output": "3301967993506605598118564082793505826927835671912383741219911930496842130418974223636865915672261642456247377827650506657877850580145623499927271391838907804651235401527392426584047219626357010023552497909436550723659221336486898100975437974320483591226280567200180225706948265372905918038750624429412331582504280650041845010449084641487447573160867860208332424835101416924485616494780952529083292227777966546236453553361466209621076748915774965082618181512654546592160909206650552581723190500273752213154329310..."
},
{
"input": "955 7",
"output": "4875434946733568640983465009954221247849488705968833681097920555785434899849497268074436910608289709905212840964404347113134616236366794383005890642796609027376389191650656756216171636192669456464756898600086886269167613161503734300581107122411830728903919402846291350458047685924037685489537178939190129043010338580479169957795695942333133962326316127076129681213167918954090336000635320714955444899171270809399782177230616239894234246885245402806465700760528496316658100834632585364274381823984214942419830421..."
},
{
"input": "893 3",
"output": "3154491812688062338683413382839715419754844054478504300541293341098785797116419835470049101334759365561276155814822131363018164033585874216523127145546903121862283071300185033613164338905028463571111541628115658108609505120357131336651371062955497690723492519748325195227665653129911625684144804656937323976632567108677478936761775342496303735237936919652618323430255701996987753367609559178855599470625167628439986055634187527493497208780060336400261449926469512996188738133678473883670714775784527941804249702..."
},
{
"input": "998 2",
"output": "8948712698877635315965401396781625629035528026219922557326466734622505808493494219669540192792500692387387200924494124219975316993592377253517258369463882172533672158172302951620486954085880606055358057621166471042557383036617719864238933843342304818076672889894622975857106353903546493307325157763617269195570831067239463586554245706859061059010215520785892192165179414199200952765077228454366556358805840526959104496983177402562569401945586771345953166346316987259989961516385311376707950154520512125143059966..."
},
{
"input": "960 6",
"output": "7291446744949293530598651243840704118065364362420848463900543089429494124955982767375712583398715647208330285855809398453361266463042342917037983463277320070057956978767965467358862479678812136400444143489366786562672928727263605336304125661306952421127807059398289873947797478996261161224877129724944902005212399176751167053423457968483336961277157597291131065544350665072172392437597673561840137077580044817979332136630042866681837917711758227272499659805765131669208911408670581308412686469802437930679571593..."
}
] | 1,591,623,170
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 108
| 0
|
k,d=input().split();
k=int(k);
d=int(d);
if(d==0):
if(k==1):
print("0");
else:
print("No solution");
else:
r=d*(pow(10,k-1));
print(r);
|
Title: Vasya and Digital Root
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to:
1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=β₯<=10.
For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=β€<=101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist.
Input Specification:
The first line contains two integers *k* and *d* (1<=β€<=*k*<=β€<=1000;β0<=β€<=*d*<=β€<=9).
Output Specification:
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes.
Demo Input:
['4 4\n', '5 1\n', '1 0\n']
Demo Output:
['5881\n', '36172\n', '0\n']
Note:
For the first test sample *dr*(5881)ββ=ββ*dr*(22)ββ=ββ4.
For the second test sample *dr*(36172)ββ=ββ*dr*(19)ββ=ββ*dr*(10)ββ=ββ1.
|
```python
k,d=input().split();
k=int(k);
d=int(d);
if(d==0):
if(k==1):
print("0");
else:
print("No solution");
else:
r=d*(pow(10,k-1));
print(r);
```
| 3
|
|
679
|
A
|
Bear and Prime 100
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"interactive",
"math"
] | null | null |
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden numberΒ β an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
|
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
|
Up to 20 times you can ask a queryΒ β print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden numberΒ β one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
|
[
"yes\nno\nyes\n",
"no\nyes\nno\nno\nno\n"
] |
[
"2\n80\n5\ncomposite\n",
"58\n59\n78\n78\n2\nprime\n"
] |
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2,β100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
| 750
|
[
{
"input": "30",
"output": "composite 4"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "2",
"output": "prime 16"
},
{
"input": "7",
"output": "prime 16"
},
{
"input": "9",
"output": "composite 3"
},
{
"input": "13",
"output": "prime 15"
},
{
"input": "55",
"output": "composite 6"
},
{
"input": "89",
"output": "prime 15"
},
{
"input": "3",
"output": "prime 16"
},
{
"input": "4",
"output": "composite 2"
},
{
"input": "6",
"output": "composite 4"
},
{
"input": "8",
"output": "composite 2"
},
{
"input": "11",
"output": "prime 15"
},
{
"input": "12",
"output": "composite 2"
},
{
"input": "25",
"output": "composite 4"
},
{
"input": "36",
"output": "composite 2"
},
{
"input": "46",
"output": "composite 10"
},
{
"input": "47",
"output": "prime 15"
},
{
"input": "49",
"output": "composite 5"
},
{
"input": "51",
"output": "composite 8"
},
{
"input": "53",
"output": "prime 15"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "64",
"output": "composite 2"
},
{
"input": "81",
"output": "composite 3"
},
{
"input": "91",
"output": "composite 7"
},
{
"input": "93",
"output": "composite 12"
},
{
"input": "94",
"output": "composite 16"
},
{
"input": "95",
"output": "composite 9"
},
{
"input": "96",
"output": "composite 2"
},
{
"input": "97",
"output": "prime 15"
},
{
"input": "99",
"output": "composite 3"
},
{
"input": "100",
"output": "composite 2"
},
{
"input": "26",
"output": "composite 7"
},
{
"input": "94",
"output": "composite 16"
},
{
"input": "25",
"output": "composite 4"
},
{
"input": "46",
"output": "composite 10"
},
{
"input": "49",
"output": "composite 5"
},
{
"input": "82",
"output": "composite 14"
},
{
"input": "87",
"output": "composite 11"
},
{
"input": "85",
"output": "composite 8"
},
{
"input": "74",
"output": "composite 13"
},
{
"input": "86",
"output": "composite 15"
},
{
"input": "93",
"output": "composite 12"
},
{
"input": "58",
"output": "composite 11"
},
{
"input": "47",
"output": "prime 15"
},
{
"input": "22",
"output": "composite 6"
},
{
"input": "23",
"output": "prime 15"
},
{
"input": "62",
"output": "composite 12"
},
{
"input": "38",
"output": "composite 9"
},
{
"input": "91",
"output": "composite 7"
},
{
"input": "34",
"output": "composite 8"
},
{
"input": "95",
"output": "composite 9"
},
{
"input": "4",
"output": "composite 2"
},
{
"input": "16",
"output": "composite 2"
},
{
"input": "96",
"output": "composite 2"
},
{
"input": "69",
"output": "composite 10"
},
{
"input": "65",
"output": "composite 7"
},
{
"input": "6",
"output": "composite 4"
},
{
"input": "18",
"output": "composite 4"
}
] | 1,622,581,869
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 59
| 155
| 0
|
prime=[1 for i in range(101)]
prime[0]=0
prime[1]=0
for i in range(2,101):
if prime[i]==1:
for j in range(i*i,101,i):
prime[j]=0
ans=[]
for i in range(2,101):
if prime[i]:
ans.append(i)
ans=ans[:-9]
ans.append(4)
ans.append(9)
ans.append(25)
ans.append(49)
#print(len(ans))
#print(ans)
count=0
for i in ans:
print(i,flush=True)
s=input()
if s=="yes":
count+=1
if count<=1:
print("prime",flush=True)
else:
print("composite",flush=True)
|
Title: Bear and Prime 100
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden numberΒ β an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input Specification:
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output Specification:
Up to 20 times you can ask a queryΒ β print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden numberΒ β one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
Demo Input:
['yes\nno\nyes\n', 'no\nyes\nno\nno\nno\n']
Demo Output:
['2\n80\n5\ncomposite\n', '58\n59\n78\n78\n2\nprime\n']
Note:
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2,β100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
|
```python
prime=[1 for i in range(101)]
prime[0]=0
prime[1]=0
for i in range(2,101):
if prime[i]==1:
for j in range(i*i,101,i):
prime[j]=0
ans=[]
for i in range(2,101):
if prime[i]:
ans.append(i)
ans=ans[:-9]
ans.append(4)
ans.append(9)
ans.append(25)
ans.append(49)
#print(len(ans))
#print(ans)
count=0
for i in ans:
print(i,flush=True)
s=input()
if s=="yes":
count+=1
if count<=1:
print("prime",flush=True)
else:
print("composite",flush=True)
```
| 3
|
|
678
|
A
|
Johny Likes Numbers
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
|
The only line contains two integers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=109).
|
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
|
[
"5 3\n",
"25 13\n",
"26 13\n"
] |
[
"6\n",
"26\n",
"39\n"
] |
none
| 0
|
[
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,609,962,847
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 171
| 0
|
def numbers(n, k):
return n + k - n%k
n, k = map(int, input().split())
print(numbers(n, k))
|
Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none
|
```python
def numbers(n, k):
return n + k - n%k
n, k = map(int, input().split())
print(numbers(n, k))
```
| 3
|
|
567
|
A
|
Lineland Mail
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* β a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values ββ*min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
|
The first line of the input contains integer *n* (2<=β€<=*n*<=β€<=105) β the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=β€<=*x**i*<=β€<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
|
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
|
[
"4\n-5 -2 2 7\n",
"2\n-1 1\n"
] |
[
"3 12\n3 9\n4 7\n5 12\n",
"2 2\n2 2\n"
] |
none
| 500
|
[
{
"input": "4\n-5 -2 2 7",
"output": "3 12\n3 9\n4 7\n5 12"
},
{
"input": "2\n-1 1",
"output": "2 2\n2 2"
},
{
"input": "3\n-1 0 1",
"output": "1 2\n1 1\n1 2"
},
{
"input": "4\n-1 0 1 3",
"output": "1 4\n1 3\n1 2\n2 4"
},
{
"input": "3\n-1000000000 0 1000000000",
"output": "1000000000 2000000000\n1000000000 1000000000\n1000000000 2000000000"
},
{
"input": "2\n-1000000000 1000000000",
"output": "2000000000 2000000000\n2000000000 2000000000"
},
{
"input": "10\n1 10 12 15 59 68 130 912 1239 9123",
"output": "9 9122\n2 9113\n2 9111\n3 9108\n9 9064\n9 9055\n62 8993\n327 8211\n327 7884\n7884 9122"
},
{
"input": "5\n-2 -1 0 1 2",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "5\n-2 -1 0 1 3",
"output": "1 5\n1 4\n1 3\n1 3\n2 5"
},
{
"input": "3\n-10000 1 10000",
"output": "10001 20000\n9999 10001\n9999 20000"
},
{
"input": "5\n-1000000000 -999999999 -999999998 -999999997 -999999996",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "10\n-857422304 -529223472 82412729 145077145 188538640 265299215 527377039 588634631 592896147 702473706",
"output": "328198832 1559896010\n328198832 1231697178\n62664416 939835033\n43461495 1002499449\n43461495 1045960944\n76760575 1122721519\n61257592 1384799343\n4261516 1446056935\n4261516 1450318451\n109577559 1559896010"
},
{
"input": "10\n-876779400 -829849659 -781819137 -570920213 18428128 25280705 121178189 219147240 528386329 923854124",
"output": "46929741 1800633524\n46929741 1753703783\n48030522 1705673261\n210898924 1494774337\n6852577 905425996\n6852577 902060105\n95897484 997957589\n97969051 1095926640\n309239089 1405165729\n395467795 1800633524"
},
{
"input": "30\n-15 1 21 25 30 40 59 60 77 81 97 100 103 123 139 141 157 158 173 183 200 215 226 231 244 256 267 279 289 292",
"output": "16 307\n16 291\n4 271\n4 267\n5 262\n10 252\n1 233\n1 232\n4 215\n4 211\n3 195\n3 192\n3 189\n16 169\n2 154\n2 156\n1 172\n1 173\n10 188\n10 198\n15 215\n11 230\n5 241\n5 246\n12 259\n11 271\n11 282\n10 294\n3 304\n3 307"
},
{
"input": "10\n-1000000000 -999999999 -999999997 -999999996 -999999995 -999999994 -999999992 -999999990 -999999988 -999999986",
"output": "1 14\n1 13\n1 11\n1 10\n1 9\n1 8\n2 8\n2 10\n2 12\n2 14"
},
{
"input": "50\n-50000 -49459 -48875 -48456 -48411 -48096 -47901 -47500 -47150 -46808 -46687 -46679 -46337 -45747 -45604 -45194 -44752 -44242 -44231 -44122 -43636 -43274 -42916 -42881 -42386 -42095 -41830 -41618 -41145 -40897 -40534 -40007 -39761 -39389 -39104 -38909 -38630 -38561 -38364 -38255 -38214 -38084 -37959 -37607 -37202 -36890 -36681 -36136 -36123 -35886",
"output": "541 14114\n541 13573\n419 12989\n45 12570\n45 12525\n195 12210\n195 12015\n350 11614\n342 11264\n121 10922\n8 10801\n8 10793\n342 10451\n143 9861\n143 9718\n410 9308\n442 8866\n11 8356\n11 8345\n109 8236\n362 7750\n358 7388\n35 7084\n35 7119\n291 7614\n265 7905\n212 8170\n212 8382\n248 8855\n248 9103\n363 9466\n246 9993\n246 10239\n285 10611\n195 10896\n195 11091\n69 11370\n69 11439\n109 11636\n41 11745\n41 11786\n125 11916\n125 12041\n352 12393\n312 12798\n209 13110\n209 13319\n13 13864\n13 13877\n237 141..."
},
{
"input": "3\n-1000000000 999999999 1000000000",
"output": "1999999999 2000000000\n1 1999999999\n1 2000000000"
}
] | 1,683,139,273
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 171
| 16,281,600
|
n=int(input())
lst=list(map(int,input().split()))
print(lst[1]-lst[0],lst[-1]-lst[0])
x,y=0,0
for i in range(1,n-1):
ch=""
x,y=lst[i]-lst[i-1],lst[i+1]-lst[i]
if x<y:
ch+=str(x)+" "
else:
ch+=str(y)+" "
x,y=lst[i]-lst[0],lst[n-1]-lst[i]
if x<y:
ch+=str(y)
else:
ch+=str(x)
print(ch)
print(lst[n-1]-lst[n-2],lst[n-1]-lst[0])
|
Title: Lineland Mail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* β a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values ββ*min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
Input Specification:
The first line of the input contains integer *n* (2<=β€<=*n*<=β€<=105) β the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=β€<=*x**i*<=β€<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
Output Specification:
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
Demo Input:
['4\n-5 -2 2 7\n', '2\n-1 1\n']
Demo Output:
['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n']
Note:
none
|
```python
n=int(input())
lst=list(map(int,input().split()))
print(lst[1]-lst[0],lst[-1]-lst[0])
x,y=0,0
for i in range(1,n-1):
ch=""
x,y=lst[i]-lst[i-1],lst[i+1]-lst[i]
if x<y:
ch+=str(x)+" "
else:
ch+=str(y)+" "
x,y=lst[i]-lst[0],lst[n-1]-lst[i]
if x<y:
ch+=str(y)
else:
ch+=str(x)
print(ch)
print(lst[n-1]-lst[n-2],lst[n-1]-lst[0])
```
| 3
|
|
1,005
|
A
|
Tanya and Stairways
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
|
The first line contains $n$ ($1 \le n \le 1000$) β the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) β all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
|
In the first line, output $t$ β the number of stairways that Tanya climbed. In the second line, output $t$ numbers β the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
|
[
"7\n1 2 3 1 2 3 4\n",
"4\n1 1 1 1\n",
"5\n1 2 3 4 5\n",
"5\n1 2 1 2 1\n"
] |
[
"2\n3 4 ",
"4\n1 1 1 1 ",
"1\n5 ",
"3\n2 2 1 "
] |
none
| 0
|
[
{
"input": "7\n1 2 3 1 2 3 4",
"output": "2\n3 4 "
},
{
"input": "4\n1 1 1 1",
"output": "4\n1 1 1 1 "
},
{
"input": "5\n1 2 3 4 5",
"output": "1\n5 "
},
{
"input": "5\n1 2 1 2 1",
"output": "3\n2 2 1 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "48\n1 2 3 4 1 2 3 1 1 2 3 1 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 1 2 1 2 1 1 2 1 2 1 2 3 1 2 1 2 1",
"output": "20\n4 3 1 3 4 1 4 4 4 1 2 2 2 1 2 2 3 2 2 1 "
},
{
"input": "2\n1 2",
"output": "1\n2 "
},
{
"input": "3\n1 1 2",
"output": "2\n1 2 "
},
{
"input": "4\n1 1 2 3",
"output": "2\n1 3 "
},
{
"input": "8\n1 2 3 1 2 3 4 5",
"output": "2\n3 5 "
},
{
"input": "5\n1 1 1 2 3",
"output": "3\n1 1 3 "
}
] | 1,633,034,716
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 16
| 77
| 6,963,200
|
n = int(input())
l = list(map(int, input().split()))
cur_stair = 1
ladders_length = []
ladders_cnt = 1
for i in range(1, len(l)):
if l[i] <= l[i-1]:
ladders_length.append(cur_stair)
ladders_cnt += 1
cur_stair = 0
cur_stair += 1
ladders_length.append(cur_stair)
print(ladders_cnt)
for x in ladders_length:
print(x, end=' ')
# Thu Sep 30 2021 23:45:16 GMT+0300 (ΠΠΎΡΠΊΠ²Π°, ΡΡΠ°Π½Π΄Π°ΡΡΠ½ΠΎΠ΅ Π²ΡΠ΅ΠΌΡ)
|
Title: Tanya and Stairways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Input Specification:
The first line contains $n$ ($1 \le n \le 1000$) β the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) β all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Output Specification:
In the first line, output $t$ β the number of stairways that Tanya climbed. In the second line, output $t$ numbers β the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
Demo Input:
['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n']
Demo Output:
['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 ']
Note:
none
|
```python
n = int(input())
l = list(map(int, input().split()))
cur_stair = 1
ladders_length = []
ladders_cnt = 1
for i in range(1, len(l)):
if l[i] <= l[i-1]:
ladders_length.append(cur_stair)
ladders_cnt += 1
cur_stair = 0
cur_stair += 1
ladders_length.append(cur_stair)
print(ladders_cnt)
for x in ladders_length:
print(x, end=' ')
# Thu Sep 30 2021 23:45:16 GMT+0300 (ΠΠΎΡΠΊΠ²Π°, ΡΡΠ°Π½Π΄Π°ΡΡΠ½ΠΎΠ΅ Π²ΡΠ΅ΠΌΡ)
```
| 3
|
|
710
|
A
|
King Moves
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
|
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
|
Print the only integer *x* β the number of moves permitted for the king.
|
[
"e4\n"
] |
[
"8\n"
] |
none
| 0
|
[
{
"input": "e4",
"output": "8"
},
{
"input": "a1",
"output": "3"
},
{
"input": "h8",
"output": "3"
},
{
"input": "a4",
"output": "5"
},
{
"input": "g7",
"output": "8"
},
{
"input": "e1",
"output": "5"
},
{
"input": "b2",
"output": "8"
},
{
"input": "c7",
"output": "8"
},
{
"input": "h6",
"output": "5"
},
{
"input": "c8",
"output": "5"
},
{
"input": "h2",
"output": "5"
},
{
"input": "h5",
"output": "5"
},
{
"input": "a8",
"output": "3"
},
{
"input": "f8",
"output": "5"
},
{
"input": "h1",
"output": "3"
},
{
"input": "f2",
"output": "8"
},
{
"input": "e8",
"output": "5"
},
{
"input": "h3",
"output": "5"
},
{
"input": "b8",
"output": "5"
},
{
"input": "g8",
"output": "5"
},
{
"input": "d8",
"output": "5"
},
{
"input": "h4",
"output": "5"
},
{
"input": "b1",
"output": "5"
},
{
"input": "a2",
"output": "5"
}
] | 1,549,122,017
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 93
| 0
|
move = input()
c = move[0]
d = int(move[1])
moves = 0
if c=='a' or c=='h':
moves = 3
if 1<d<8:
moves += 2
else:
moves = 8
if d==1 or d==8:
moves -=3
print(moves)
|
Title: King Moves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
Input Specification:
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Output Specification:
Print the only integer *x* β the number of moves permitted for the king.
Demo Input:
['e4\n']
Demo Output:
['8\n']
Note:
none
|
```python
move = input()
c = move[0]
d = int(move[1])
moves = 0
if c=='a' or c=='h':
moves = 3
if 1<d<8:
moves += 2
else:
moves = 8
if d==1 or d==8:
moves -=3
print(moves)
```
| 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<=β€<=*x*<=β€<=1<=000<=000)Β β 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"
}
] | 1,697,627,559
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 46
| 0
|
n = int(input())
print(n % 5 and n // 5 + 1 or 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<=β€<=*x*<=β€<=1<=000<=000)Β β 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
n = int(input())
print(n % 5 and n // 5 + 1 or n // 5)
```
| 3
|
|
219
|
A
|
k-String
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
|
The first input line contains integer *k* (1<=β€<=*k*<=β€<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=β€<=|*s*|<=β€<=1000, where |*s*| is the length of string *s*.
|
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
|
[
"2\naazz\n",
"3\nabcabcabz\n"
] |
[
"azaz\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
},
{
"input": "5\naaaaa",
"output": "aaaaa"
},
{
"input": "3\naabaaaaabb",
"output": "-1"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "2\nbabac",
"output": "-1"
},
{
"input": "3\nbbbccc",
"output": "bcbcbc"
},
{
"input": "2\naa",
"output": "aa"
},
{
"input": "250\ncececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece",
"output": "cececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece"
},
{
"input": "15\nabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaa",
"output": "aaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbc"
},
{
"input": "1\naaa",
"output": "aaa"
},
{
"input": "1\naabaab",
"output": "aaaabb"
},
{
"input": "2\naabbbbccccccdddddddd",
"output": "abbcccddddabbcccdddd"
},
{
"input": "3\naaaaaaaaacccdddddd",
"output": "aaacddaaacddaaacdd"
},
{
"input": "3\naaaaaaaaacccbbbbbb",
"output": "aaabbcaaabbcaaabbc"
},
{
"input": "2\naazzzz",
"output": "azzazz"
},
{
"input": "2\naaaa",
"output": "aaaa"
},
{
"input": "2\naaaazz",
"output": "aazaaz"
},
{
"input": "2\naaaabb",
"output": "aabaab"
},
{
"input": "2\naabbbb",
"output": "abbabb"
},
{
"input": "2\naaaazzzz",
"output": "aazzaazz"
},
{
"input": "2\naaazz",
"output": "-1"
},
{
"input": "2\nbbbbaa",
"output": "abbabb"
},
{
"input": "2\naaaaaabbbb",
"output": "aaabbaaabb"
},
{
"input": "2\naaaaaazz",
"output": "aaazaaaz"
},
{
"input": "2\naaaabbbb",
"output": "aabbaabb"
},
{
"input": "2\naabaab",
"output": "aabaab"
},
{
"input": "1\naaaaa",
"output": "aaaaa"
},
{
"input": "2\nbbaaaa",
"output": "aabaab"
},
{
"input": "2\nacaccc",
"output": "accacc"
}
] | 1,662,273,164
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 42
| 124
| 0
|
#ζεεε
₯εγ―γγγͺοΌοΌ
#carpe diem
'''
βββ βββ ββββ ββββ βββ βββββββββ
βββ βββ βββββ βββββ βββ βββββββββ
ββββββββββββ βββββββββββ ββββββββββββββ ββββββ βββββββββ ββββββββββ
βββ βββ βββββββββββ βββ βββ
ββββββββ βββ βββ βββ βββ βββ βββ
ββββββββ βββ βββ βββ βββ βββ
'''
#ζεεε
₯εγ―γγγͺοΌοΌ
#carpe diem
k=int(input())
s=input()
ans=''
for i in set(s):
ans+=i*(s.count(i)//k)
if len(ans*k)==len(s):
print(ans*k)
else:
print(-1)
#carpe diem
#carpe diem
|
Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=β€<=*k*<=β€<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=β€<=|*s*|<=β€<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none
|
```python
#ζεεε
₯εγ―γγγͺοΌοΌ
#carpe diem
'''
βββ βββ ββββ ββββ βββ βββββββββ
βββ βββ βββββ βββββ βββ βββββββββ
ββββββββββββ βββββββββββ ββββββββββββββ ββββββ βββββββββ ββββββββββ
βββ βββ βββββββββββ βββ βββ
ββββββββ βββ βββ βββ βββ βββ βββ
ββββββββ βββ βββ βββ βββ βββ
'''
#ζεεε
₯εγ―γγγͺοΌοΌ
#carpe diem
k=int(input())
s=input()
ans=''
for i in set(s):
ans+=i*(s.count(i)//k)
if len(ans*k)==len(s):
print(ans*k)
else:
print(-1)
#carpe diem
#carpe diem
```
| 3
|
|
753
|
A
|
Santa Claus and Candies
|
PROGRAMMING
| 1,000
|
[
"dp",
"greedy",
"math"
] | null | null |
Santa Claus has *n* candies, he dreams to give them as gifts to children.
What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.
|
The only line contains positive integer number *n* (1<=β€<=*n*<=β€<=1000) β number of candies Santa Claus has.
|
Print to the first line integer number *k* β maximal number of kids which can get candies.
Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*.
If there are many solutions, print any of them.
|
[
"5\n",
"9\n",
"2\n"
] |
[
"2\n2 3\n",
"3\n3 5 1\n",
"1\n2 \n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n1 4 "
},
{
"input": "9",
"output": "3\n1 2 6 "
},
{
"input": "2",
"output": "1\n2 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "3",
"output": "2\n1 2 "
},
{
"input": "1000",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 54 "
},
{
"input": "4",
"output": "2\n1 3 "
},
{
"input": "6",
"output": "3\n1 2 3 "
},
{
"input": "7",
"output": "3\n1 2 4 "
},
{
"input": "8",
"output": "3\n1 2 5 "
},
{
"input": "10",
"output": "4\n1 2 3 4 "
},
{
"input": "11",
"output": "4\n1 2 3 5 "
},
{
"input": "12",
"output": "4\n1 2 3 6 "
},
{
"input": "13",
"output": "4\n1 2 3 7 "
},
{
"input": "14",
"output": "4\n1 2 3 8 "
},
{
"input": "15",
"output": "5\n1 2 3 4 5 "
},
{
"input": "16",
"output": "5\n1 2 3 4 6 "
},
{
"input": "20",
"output": "5\n1 2 3 4 10 "
},
{
"input": "21",
"output": "6\n1 2 3 4 5 6 "
},
{
"input": "22",
"output": "6\n1 2 3 4 5 7 "
},
{
"input": "27",
"output": "6\n1 2 3 4 5 12 "
},
{
"input": "28",
"output": "7\n1 2 3 4 5 6 7 "
},
{
"input": "29",
"output": "7\n1 2 3 4 5 6 8 "
},
{
"input": "35",
"output": "7\n1 2 3 4 5 6 14 "
},
{
"input": "36",
"output": "8\n1 2 3 4 5 6 7 8 "
},
{
"input": "37",
"output": "8\n1 2 3 4 5 6 7 9 "
},
{
"input": "44",
"output": "8\n1 2 3 4 5 6 7 16 "
},
{
"input": "45",
"output": "9\n1 2 3 4 5 6 7 8 9 "
},
{
"input": "46",
"output": "9\n1 2 3 4 5 6 7 8 10 "
},
{
"input": "230",
"output": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 40 "
},
{
"input": "231",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 "
},
{
"input": "232",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 "
},
{
"input": "239",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 29 "
},
{
"input": "629",
"output": "34\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 68 "
},
{
"input": "630",
"output": "35\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 "
},
{
"input": "631",
"output": "35\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 "
},
{
"input": "945",
"output": "42\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 84 "
},
{
"input": "946",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 "
},
{
"input": "947",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 44 "
},
{
"input": "989",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 86 "
},
{
"input": "990",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 "
},
{
"input": "991",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 "
},
{
"input": "956",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 53 "
},
{
"input": "981",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 78 "
},
{
"input": "867",
"output": "41\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 47 "
},
{
"input": "906",
"output": "42\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 45 "
},
{
"input": "999",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 53 "
},
{
"input": "100",
"output": "13\n1 2 3 4 5 6 7 8 9 10 11 12 22 "
},
{
"input": "126",
"output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 "
}
] | 1,574,756,497
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 49
| 140
| 0
|
a=int(input())
# n2 + n -2*a
r=int((-1+(1+8*a)**0.5)//2)
print(r)
extra=a-(r*(r+1))//2
ans=[i for i in range(1,r+1)]
ans[-1]+=extra
print(*ans)
|
Title: Santa Claus and Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus has *n* candies, he dreams to give them as gifts to children.
What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.
Input Specification:
The only line contains positive integer number *n* (1<=β€<=*n*<=β€<=1000) β number of candies Santa Claus has.
Output Specification:
Print to the first line integer number *k* β maximal number of kids which can get candies.
Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*.
If there are many solutions, print any of them.
Demo Input:
['5\n', '9\n', '2\n']
Demo Output:
['2\n2 3\n', '3\n3 5 1\n', '1\n2 \n']
Note:
none
|
```python
a=int(input())
# n2 + n -2*a
r=int((-1+(1+8*a)**0.5)//2)
print(r)
extra=a-(r*(r+1))//2
ans=[i for i in range(1,r+1)]
ans[-1]+=extra
print(*ans)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,663,238,979
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
num=int(input())
m,n,k=0,0,0
for i in range(num):
a,b,c=map(int,input().split())
m+=a;n+=b;k+=c
if m==0 and n==0 and k==0:
print('YES')
else:
print('NO')
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
num=int(input())
m,n,k=0,0,0
for i in range(num):
a,b,c=map(int,input().split())
m+=a;n+=b;k+=c
if m==0 and n==0 and k==0:
print('YES')
else:
print('NO')
```
| 3.977
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=Γ<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=β€<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=β€<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer β the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,695,208,726
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
raw = 0
column = 0
for i in range(5):
curr = input()
if curr.count("1") > 0:
raw = i
curr = list(map(int, curr.split(' ')))
for j in range(5):
if curr[j] == 1:
column = j
print(abs(raw - 2) + abs(column - 2))
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=Γ<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=β€<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=β€<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer β the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
raw = 0
column = 0
for i in range(5):
curr = input()
if curr.count("1") > 0:
raw = i
curr = list(map(int, curr.split(' ')))
for j in range(5):
if curr[j] == 1:
column = j
print(abs(raw - 2) + abs(column - 2))
```
| 3
|
|
224
|
A
|
Parallelepiped
|
PROGRAMMING
| 1,100
|
[
"brute force",
"geometry",
"math"
] | null | null |
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
|
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
|
Print a single number β the sum of all edges of the parallelepiped.
|
[
"1 1 1\n",
"4 6 6\n"
] |
[
"12\n",
"28\n"
] |
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
| 500
|
[
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{
"input": "1022 584 112",
"output": "380"
},
{
"input": "66 174 319",
"output": "184"
},
{
"input": "912 276 1748",
"output": "444"
},
{
"input": "65 156 60",
"output": "120"
},
{
"input": "1 10000 10000",
"output": "40008"
},
{
"input": "1485 55 27",
"output": "332"
},
{
"input": "152 108 4104",
"output": "528"
},
{
"input": "1656 6900 1350",
"output": "740"
},
{
"input": "12 14 42",
"output": "60"
},
{
"input": "615 18 1230",
"output": "856"
},
{
"input": "680 60 408",
"output": "336"
},
{
"input": "644 966 6",
"output": "1308"
},
{
"input": "1 432 432",
"output": "1736"
},
{
"input": "2239 2239 1",
"output": "8964"
},
{
"input": "4106 8212 2",
"output": "16436"
},
{
"input": "10000 10000 10000",
"output": "1200"
},
{
"input": "3623 3623 1",
"output": "14500"
},
{
"input": "9801 9801 9801",
"output": "1188"
},
{
"input": "10000 1 10000",
"output": "40008"
},
{
"input": "9 9 9",
"output": "36"
},
{
"input": "9801 9702 9702",
"output": "1184"
}
] | 1,601,894,767
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 27
| 278
| 0
|
import math
a,b,c=map(int,input().split())
a1=(a*c/b)**0.5
a2=(a*b/c)**0.5
a3=(b*c/a)**0.5
print(int(4*(a1+a2+a3)))
|
Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number β the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
|
```python
import math
a,b,c=map(int,input().split())
a1=(a*c/b)**0.5
a2=(a*b/c)**0.5
a3=(b*c/a)**0.5
print(int(4*(a1+a2+a3)))
```
| 3
|
|
918
|
A
|
Eleven
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
|
The first and only line of input contains an integer *n* (1<=β€<=*n*<=β€<=1000).
|
Print Eleven's new name on the first and only line of output.
|
[
"8\n",
"15\n"
] |
[
"OOOoOooO\n",
"OOOoOooOooooOoo\n"
] |
none
| 500
|
[
{
"input": "8",
"output": "OOOoOooO"
},
{
"input": "15",
"output": "OOOoOooOooooOoo"
},
{
"input": "85",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooo"
},
{
"input": "381",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooo"
},
{
"input": "805",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1000",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1",
"output": "O"
},
{
"input": "2",
"output": "OO"
},
{
"input": "3",
"output": "OOO"
},
{
"input": "5",
"output": "OOOoO"
},
{
"input": "17",
"output": "OOOoOooOooooOoooo"
},
{
"input": "49",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooo"
},
{
"input": "256",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooo"
},
{
"input": "512",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "933",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "61",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooo"
},
{
"input": "781",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "999",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
}
] | 1,595,848,697
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 18
| 109
| 6,656,000
|
a=int(input())
b=1
c=2
x=[]
s=""
while(b<=a):
d=b+c
x. append(b)
b=c
c=d
for i in range(1,a+1):
if i in x:
s=s+'O'
else:
s=s+'o'
print(s)
|
Title: Eleven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
Input Specification:
The first and only line of input contains an integer *n* (1<=β€<=*n*<=β€<=1000).
Output Specification:
Print Eleven's new name on the first and only line of output.
Demo Input:
['8\n', '15\n']
Demo Output:
['OOOoOooO\n', 'OOOoOooOooooOoo\n']
Note:
none
|
```python
a=int(input())
b=1
c=2
x=[]
s=""
while(b<=a):
d=b+c
x. append(b)
b=c
c=d
for i in range(1,a+1):
if i in x:
s=s+'O'
else:
s=s+'o'
print(s)
```
| 3
|
|
766
|
A
|
Mahmoud and Longest Uncommon Subsequence
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"strings"
] | null | null |
While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.
Given two strings *a* and *b*, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.
A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don't have to be consecutive, for example, strings "ac", "bc", "abc" and "a" are subsequences of string "abc" while strings "abbc" and "acb" are not. The empty string is a subsequence of any string. Any string is a subsequence of itself.
|
The first line contains string *a*, and the second lineΒ β string *b*. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters.
|
If there's no uncommon subsequence, print "-1". Otherwise print the length of the longest uncommon subsequence of *a* and *b*.
|
[
"abcd\ndefgh\n",
"a\na\n"
] |
[
"5\n",
"-1\n"
] |
In the first example: you can choose "defgh" from string *b* as it is the longest subsequence of string *b* that doesn't appear as a subsequence of string *a*.
| 500
|
[
{
"input": "abcd\ndefgh",
"output": "5"
},
{
"input": "a\na",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaacccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaadddddddddddddddddddddddddddddddddddddddddddddddddd",
"output": "100"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "199"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nbbbbbbbbbbbbbbbbbbb",
"output": "99"
},
{
"input": "abcde\nfghij",
"output": "5"
},
{
"input": "abcde\nabcdf",
"output": "5"
},
{
"input": "abcde\nbbcde",
"output": "5"
},
{
"input": "abcde\neabcd",
"output": "5"
},
{
"input": "abcdefgh\nabdcefgh",
"output": "8"
},
{
"input": "mmmmm\nmnmmm",
"output": "5"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaa",
"output": "34"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\nzabcdefghijklmnopqrstuvwxy",
"output": "26"
},
{
"input": "a\nab",
"output": "2"
},
{
"input": "b\nab",
"output": "2"
},
{
"input": "ab\nb",
"output": "2"
},
{
"input": "ab\nc",
"output": "2"
},
{
"input": "aaaaaa\naaaaaa",
"output": "-1"
},
{
"input": "abacaba\nabacaba",
"output": "-1"
},
{
"input": "aabb\nbbaa",
"output": "4"
},
{
"input": "ab\nba",
"output": "2"
},
{
"input": "abcd\nabc",
"output": "4"
},
{
"input": "abaa\nabaa",
"output": "-1"
},
{
"input": "ab\nab",
"output": "-1"
},
{
"input": "ab\nabcd",
"output": "4"
},
{
"input": "abc\nabcd",
"output": "4"
},
{
"input": "mo\nmomo",
"output": "4"
},
{
"input": "koooooooooooooooo\nloooooooooooooooo",
"output": "17"
},
{
"input": "aaa\naa",
"output": "3"
},
{
"input": "abc\nabc",
"output": "-1"
},
{
"input": "abcd\nabcd",
"output": "-1"
},
{
"input": "abc\ncba",
"output": "3"
},
{
"input": "ahc\nahc",
"output": "-1"
},
{
"input": "abc\nbac",
"output": "3"
},
{
"input": "aa\naaa",
"output": "3"
},
{
"input": "aaa\naaa",
"output": "-1"
},
{
"input": "abc\nacb",
"output": "3"
},
{
"input": "abc\nab",
"output": "3"
},
{
"input": "abb\nabb",
"output": "-1"
},
{
"input": "abc\ncab",
"output": "3"
},
{
"input": "aaaaaa\naaaaa",
"output": "6"
},
{
"input": "aa\naab",
"output": "3"
},
{
"input": "len\nlena",
"output": "4"
},
{
"input": "aaaaa\naa",
"output": "5"
},
{
"input": "aaa\naaaa",
"output": "4"
},
{
"input": "bcd\nabcd",
"output": "4"
},
{
"input": "aaabbc\naaaccc",
"output": "6"
},
{
"input": "abcd\nzycd",
"output": "4"
},
{
"input": "baa\nzaa",
"output": "3"
},
{
"input": "asdf\nadfs",
"output": "4"
},
{
"input": "abcdefgh\nabcdefgh",
"output": "-1"
},
{
"input": "aba\naab",
"output": "3"
},
{
"input": "aba\naba",
"output": "-1"
},
{
"input": "abcccc\naccccc",
"output": "6"
},
{
"input": "acb\nabc",
"output": "3"
},
{
"input": "a\naa",
"output": "2"
},
{
"input": "abababab\nab",
"output": "8"
},
{
"input": "ababab\nab",
"output": "6"
}
] | 1,673,968,077
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 73
| 46
| 102,400
|
a=input()
b=input()
if a==b:
res=-1
else:
res=max(len(a),len(b))
print(res)
|
Title: Mahmoud and Longest Uncommon Subsequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.
Given two strings *a* and *b*, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.
A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don't have to be consecutive, for example, strings "ac", "bc", "abc" and "a" are subsequences of string "abc" while strings "abbc" and "acb" are not. The empty string is a subsequence of any string. Any string is a subsequence of itself.
Input Specification:
The first line contains string *a*, and the second lineΒ β string *b*. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters.
Output Specification:
If there's no uncommon subsequence, print "-1". Otherwise print the length of the longest uncommon subsequence of *a* and *b*.
Demo Input:
['abcd\ndefgh\n', 'a\na\n']
Demo Output:
['5\n', '-1\n']
Note:
In the first example: you can choose "defgh" from string *b* as it is the longest subsequence of string *b* that doesn't appear as a subsequence of string *a*.
|
```python
a=input()
b=input()
if a==b:
res=-1
else:
res=max(len(a),len(b))
print(res)
```
| 3
|
|
707
|
C
|
Pythagorean Triples
|
PROGRAMMING
| 1,500
|
[
"math",
"number theory"
] | null | null |
Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.
For example, triples (3,<=4,<=5), (5,<=12,<=13) and (6,<=8,<=10) are Pythagorean triples.
Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.
Katya had no problems with completing this task. Will you do the same?
|
The only line of the input contains single integer *n* (1<=β€<=*n*<=β€<=109)Β β the length of some side of a right triangle.
|
Print two integers *m* and *k* (1<=β€<=*m*,<=*k*<=β€<=1018), such that *n*, *m* and *k* form a Pythagorean triple, in the only line.
In case if there is no any Pythagorean triple containing integer *n*, print <=-<=1 in the only line. If there are many answers, print any of them.
|
[
"3\n",
"6\n",
"1\n",
"17\n",
"67\n"
] |
[
"4 5",
"8 10",
"-1",
"144 145",
"2244 2245"
] |
Illustration for the first sample.
| 1,500
|
[
{
"input": "3",
"output": "4 5"
},
{
"input": "6",
"output": "8 10"
},
{
"input": "1",
"output": "-1"
},
{
"input": "17",
"output": "144 145"
},
{
"input": "67",
"output": "2244 2245"
},
{
"input": "10",
"output": "24 26"
},
{
"input": "14",
"output": "48 50"
},
{
"input": "22",
"output": "120 122"
},
{
"input": "23",
"output": "264 265"
},
{
"input": "246",
"output": "15128 15130"
},
{
"input": "902",
"output": "203400 203402"
},
{
"input": "1000000000",
"output": "1250000000 750000000"
},
{
"input": "1998",
"output": "998000 998002"
},
{
"input": "2222222",
"output": "1234567654320 1234567654322"
},
{
"input": "2222226",
"output": "1234572098768 1234572098770"
},
{
"input": "1111110",
"output": "308641358024 308641358026"
},
{
"input": "9999998",
"output": "24999990000000 24999990000002"
},
{
"input": "1024",
"output": "1280 768"
},
{
"input": "8388608",
"output": "10485760 6291456"
},
{
"input": "4",
"output": "5 3"
},
{
"input": "8",
"output": "10 6"
},
{
"input": "16",
"output": "20 12"
},
{
"input": "492",
"output": "615 369"
},
{
"input": "493824",
"output": "617280 370368"
},
{
"input": "493804",
"output": "617255 370353"
},
{
"input": "493800",
"output": "617250 370350"
},
{
"input": "2048",
"output": "2560 1536"
},
{
"input": "8388612",
"output": "10485765 6291459"
},
{
"input": "44",
"output": "55 33"
},
{
"input": "444",
"output": "555 333"
},
{
"input": "4444",
"output": "5555 3333"
},
{
"input": "44444",
"output": "55555 33333"
},
{
"input": "444444",
"output": "555555 333333"
},
{
"input": "4444444",
"output": "5555555 3333333"
},
{
"input": "100000000",
"output": "125000000 75000000"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "4 5"
},
{
"input": "5",
"output": "12 13"
},
{
"input": "7",
"output": "24 25"
},
{
"input": "9",
"output": "40 41"
},
{
"input": "11",
"output": "60 61"
},
{
"input": "13",
"output": "84 85"
},
{
"input": "15",
"output": "112 113"
},
{
"input": "19",
"output": "180 181"
},
{
"input": "111",
"output": "6160 6161"
},
{
"input": "113",
"output": "6384 6385"
},
{
"input": "115",
"output": "6612 6613"
},
{
"input": "117",
"output": "6844 6845"
},
{
"input": "119",
"output": "7080 7081"
},
{
"input": "111111",
"output": "6172827160 6172827161"
},
{
"input": "111113",
"output": "6173049384 6173049385"
},
{
"input": "111115",
"output": "6173271612 6173271613"
},
{
"input": "111117",
"output": "6173493844 6173493845"
},
{
"input": "111119",
"output": "6173716080 6173716081"
},
{
"input": "9999993",
"output": "49999930000024 49999930000025"
},
{
"input": "9999979",
"output": "49999790000220 49999790000221"
},
{
"input": "9999990",
"output": "24999950000024 24999950000026"
},
{
"input": "9999991",
"output": "49999910000040 49999910000041"
},
{
"input": "9999992",
"output": "12499990 7499994"
},
{
"input": "9999973",
"output": "49999730000364 49999730000365"
},
{
"input": "9999994",
"output": "24999970000008 24999970000010"
},
{
"input": "9999995",
"output": "49999950000012 49999950000013"
},
{
"input": "9999996",
"output": "12499995 7499997"
},
{
"input": "9999997",
"output": "49999970000004 49999970000005"
},
{
"input": "9999978",
"output": "24999890000120 24999890000122"
},
{
"input": "99999993",
"output": "4999999300000024 4999999300000025"
},
{
"input": "99999979",
"output": "4999997900000220 4999997900000221"
},
{
"input": "99999990",
"output": "2499999500000024 2499999500000026"
},
{
"input": "99999991",
"output": "4999999100000040 4999999100000041"
},
{
"input": "99999992",
"output": "124999990 74999994"
},
{
"input": "99999973",
"output": "4999997300000364 4999997300000365"
},
{
"input": "99999994",
"output": "2499999700000008 2499999700000010"
},
{
"input": "99999995",
"output": "4999999500000012 4999999500000013"
},
{
"input": "99999996",
"output": "124999995 74999997"
},
{
"input": "99999997",
"output": "4999999700000004 4999999700000005"
},
{
"input": "99999978",
"output": "2499998900000120 2499998900000122"
},
{
"input": "987654323",
"output": "487730530870294164 487730530870294165"
},
{
"input": "2",
"output": "-1"
},
{
"input": "4",
"output": "5 3"
},
{
"input": "8",
"output": "10 6"
},
{
"input": "64",
"output": "80 48"
},
{
"input": "999999999",
"output": "499999999000000000 499999999000000001"
},
{
"input": "16",
"output": "20 12"
},
{
"input": "999999937",
"output": "499999937000001984 499999937000001985"
},
{
"input": "999999998",
"output": "249999999000000000 249999999000000002"
},
{
"input": "433494437",
"output": "93958713454973484 93958713454973485"
},
{
"input": "484916147",
"output": "117571834810662804 117571834810662805"
},
{
"input": "999999929",
"output": "499999929000002520 499999929000002521"
},
{
"input": "982451653",
"output": "482605625241216204 482605625241216205"
},
{
"input": "2048",
"output": "2560 1536"
}
] | 1,656,248,682
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 90
| 62
| 28,364,800
|
n = int(input())
if n % 2 == 0:
x = (n // 2) ** 2
n1 = x - 1
n2 = x + 1
else:
x = (n ** 2) // 2
n1 = x
n2 = x + 1
if n1 != n != n2 and n1 > 0 and n2 > 0:
print(n1, n2)
else:
print(-1)
|
Title: Pythagorean Triples
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.
For example, triples (3,<=4,<=5), (5,<=12,<=13) and (6,<=8,<=10) are Pythagorean triples.
Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.
Katya had no problems with completing this task. Will you do the same?
Input Specification:
The only line of the input contains single integer *n* (1<=β€<=*n*<=β€<=109)Β β the length of some side of a right triangle.
Output Specification:
Print two integers *m* and *k* (1<=β€<=*m*,<=*k*<=β€<=1018), such that *n*, *m* and *k* form a Pythagorean triple, in the only line.
In case if there is no any Pythagorean triple containing integer *n*, print <=-<=1 in the only line. If there are many answers, print any of them.
Demo Input:
['3\n', '6\n', '1\n', '17\n', '67\n']
Demo Output:
['4 5', '8 10', '-1', '144 145', '2244 2245']
Note:
Illustration for the first sample.
|
```python
n = int(input())
if n % 2 == 0:
x = (n // 2) ** 2
n1 = x - 1
n2 = x + 1
else:
x = (n ** 2) // 2
n1 = x
n2 = x + 1
if n1 != n != n2 and n1 > 0 and n2 > 0:
print(n1, n2)
else:
print(-1)
```
| 3
|
|
343
|
A
|
Rational Resistance
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance *R*0<==<=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
1. one resistor; 1. an element and one resistor plugged in sequence; 1. an element and one resistor plugged in parallel.
With the consecutive connection the resistance of the new element equals *R*<==<=*R**e*<=+<=*R*0. With the parallel connection the resistance of the new element equals . In this case *R**e* equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.
|
The single input line contains two space-separated integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists.
|
Print a single number β the answer to the problem.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
|
[
"1 1\n",
"3 2\n",
"199 200\n"
] |
[
"1\n",
"3\n",
"200\n"
] |
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5305da389756aab6423d918a08ced468f05604df.png" style="max-width: 100.0%;max-height: 100.0%;"/>. We cannot make this element using two resistors.
| 500
|
[
{
"input": "1 1",
"output": "1"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "199 200",
"output": "200"
},
{
"input": "1 1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "21 8",
"output": "7"
},
{
"input": "18 55",
"output": "21"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "2 5",
"output": "4"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "3 5",
"output": "4"
},
{
"input": "13 4",
"output": "7"
},
{
"input": "21 17",
"output": "9"
},
{
"input": "5 8",
"output": "5"
},
{
"input": "13 21",
"output": "7"
},
{
"input": "74 99",
"output": "28"
},
{
"input": "2377 1055",
"output": "33"
},
{
"input": "645597 134285",
"output": "87"
},
{
"input": "29906716 35911991",
"output": "92"
},
{
"input": "3052460231 856218974",
"output": "82"
},
{
"input": "288565475053 662099878640",
"output": "88"
},
{
"input": "11504415412768 12754036168327",
"output": "163"
},
{
"input": "9958408561221547 4644682781404278",
"output": "196"
},
{
"input": "60236007668635342 110624799949034113",
"output": "179"
},
{
"input": "4 43470202936783249",
"output": "10867550734195816"
},
{
"input": "16 310139055712567491",
"output": "19383690982035476"
},
{
"input": "15 110897893734203629",
"output": "7393192915613582"
},
{
"input": "439910263967866789 38",
"output": "11576585893891241"
},
{
"input": "36 316049483082136289",
"output": "8779152307837131"
},
{
"input": "752278442523506295 52",
"output": "14466893125452056"
},
{
"input": "4052739537881 6557470319842",
"output": "62"
},
{
"input": "44945570212853 72723460248141",
"output": "67"
},
{
"input": "498454011879264 806515533049393",
"output": "72"
},
{
"input": "8944394323791464 5527939700884757",
"output": "77"
},
{
"input": "679891637638612258 420196140727489673",
"output": "86"
},
{
"input": "1 923438",
"output": "923438"
},
{
"input": "3945894354376 1",
"output": "3945894354376"
},
{
"input": "999999999999999999 5",
"output": "200000000000000004"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "999999999999999991 1000000000000000000",
"output": "111111111111111120"
},
{
"input": "999999999999999993 999999999999999991",
"output": "499999999999999998"
},
{
"input": "3 1000000000000000000",
"output": "333333333333333336"
},
{
"input": "1000000000000000000 3",
"output": "333333333333333336"
},
{
"input": "10000000000 1000000001",
"output": "100000019"
},
{
"input": "2 999999999999999999",
"output": "500000000000000001"
},
{
"input": "999999999999999999 2",
"output": "500000000000000001"
},
{
"input": "2 1000000001",
"output": "500000002"
},
{
"input": "123 1000000000000000000",
"output": "8130081300813023"
}
] | 1,671,956,136
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 53
| 154
| 0
|
import sys
input = sys.stdin.readline
def gcd(a, b):
if(b == 0):
return 0
return a // b + gcd(b, a % b)
a, b = map(int, input().split())
print(gcd(a, b))
|
Title: Rational Resistance
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance *R*0<==<=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
1. one resistor; 1. an element and one resistor plugged in sequence; 1. an element and one resistor plugged in parallel.
With the consecutive connection the resistance of the new element equals *R*<==<=*R**e*<=+<=*R*0. With the parallel connection the resistance of the new element equals . In this case *R**e* equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.
Input Specification:
The single input line contains two space-separated integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists.
Output Specification:
Print a single number β the answer to the problem.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Demo Input:
['1 1\n', '3 2\n', '199 200\n']
Demo Output:
['1\n', '3\n', '200\n']
Note:
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5305da389756aab6423d918a08ced468f05604df.png" style="max-width: 100.0%;max-height: 100.0%;"/>. We cannot make this element using two resistors.
|
```python
import sys
input = sys.stdin.readline
def gcd(a, b):
if(b == 0):
return 0
return a // b + gcd(b, a % b)
a, b = map(int, input().split())
print(gcd(a, b))
```
| 3
|
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 β those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number β the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,659,241,652
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 92
| 0
|
n = int(input())
lst = list(map(int, input().split()))
day=0
flag=0
while True:
day=0
for i in range(len(lst)):
if n<=lst[i]:
print(day+1)
flag=1
break
else:
n-=lst[i]
day+=1
if flag==1:
break
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 β those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number β the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
n = int(input())
lst = list(map(int, input().split()))
day=0
flag=0
while True:
day=0
for i in range(len(lst)):
if n<=lst[i]:
print(day+1)
flag=1
break
else:
n-=lst[i]
day+=1
if flag==1:
break
```
| 3
|
|
432
|
A
|
Choosing Teams
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
|
The first line contains two integers, *n* and *k* (1<=β€<=*n*<=β€<=2000;Β 1<=β€<=*k*<=β€<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=β€<=*y**i*<=β€<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
|
Print a single number β the answer to the problem.
|
[
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
| 500
|
[
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,686,679,701
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 46
| 0
|
n, m = map(int, input().split())
ans = 0
temp = 5 - m
l = list(map(int, input().split()))
for i in range(n):
temp2 = l[i]
if(temp2 <= temp):
ans += 1
print(ans // 3)
|
Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=β€<=*n*<=β€<=2000;Β 1<=β€<=*k*<=β€<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=β€<=*y**i*<=β€<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number β the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
|
```python
n, m = map(int, input().split())
ans = 0
temp = 5 - m
l = list(map(int, input().split()))
for i in range(n):
temp2 = l[i]
if(temp2 <= temp):
ans += 1
print(ans // 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<=β€<=*m*<=β€<=*n*<=β€<=109) β 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* β 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"
}
] | 1,678,695,565
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 26
| 62
| 0
|
from math import comb
n,m=map(int,input().split())
maxi=comb(n-m+1,2)
if m%n==0:
mini=m*comb(m//n,2)
else:
mini=n%m*(comb(n//m+1,2))+(m-n%m)*(comb(n//m,2))
print(mini,maxi)
|
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<=β€<=*m*<=β€<=*n*<=β€<=109) β 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* β 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 comb
n,m=map(int,input().split())
maxi=comb(n-m+1,2)
if m%n==0:
mini=m*comb(m//n,2)
else:
mini=n%m*(comb(n//m+1,2))+(m-n%m)*(comb(n//m,2))
print(mini,maxi)
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalizationΒ» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,673,110,701
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
n = int(input())
b=[]
for i in range(n):
a = input()
b.append(a)
for i in b:
if len(i) <= 10:
print(i)
else:
f=[]
f.append(i[0])
f.append(len(i)-2)
f.append(i[len(i)-1])
for j in f:
print(j,end='')
print()
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalizationΒ» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n = int(input())
b=[]
for i in range(n):
a = input()
b.append(a)
for i in b:
if len(i) <= 10:
print(i)
else:
f=[]
f.append(i[0])
f.append(len(i)-2)
f.append(i[len(i)-1])
for j in f:
print(j,end='')
print()
```
| 3.977
|
499
|
B
|
Lecture
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
|
The first line contains two integers, *n* and *m* (1<=β€<=*n*<=β€<=3000, 1<=β€<=*m*<=β€<=3000) β the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* β the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
|
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
|
[
"4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n",
"5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n"
] |
[
"codeforces round letter round\n",
"hbnyiyc joll joll un joll\n"
] |
none
| 500
|
[
{
"input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest",
"output": "codeforces round letter round"
},
{
"input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll",
"output": "hbnyiyc joll joll un joll"
},
{
"input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b",
"output": "f jpggvr b hnxsd b"
},
{
"input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb",
"output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs"
},
{
"input": "1 1\namit am\namit",
"output": "am"
},
{
"input": "1 1\na c\na",
"output": "a"
}
] | 1,671,486,460
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 62
| 102,400
|
n, m = input().split()
d = dict()
for _ in range(int(m)):
a, b = input().split()
v = min((a, b), key=len)
d[a] = v
print(*[d[k] for k in input().split()])
|
Title: Lecture
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
Input Specification:
The first line contains two integers, *n* and *m* (1<=β€<=*n*<=β€<=3000, 1<=β€<=*m*<=β€<=3000) β the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* β the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output Specification:
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
Demo Input:
['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n']
Demo Output:
['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n']
Note:
none
|
```python
n, m = input().split()
d = dict()
for _ in range(int(m)):
a, b = input().split()
v = min((a, b), key=len)
d[a] = v
print(*[d[k] for k in input().split()])
```
| 3
|
|
825
|
A
|
Binary Protocol
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
|
The first line contains one integer number *n* (1<=β€<=*n*<=β€<=89) β length of the string *s*.
The second line contains string *s* β sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
|
Print the decoded number.
|
[
"3\n111\n",
"9\n110011101\n"
] |
[
"3\n",
"2031\n"
] |
none
| 0
|
[
{
"input": "3\n111",
"output": "3"
},
{
"input": "9\n110011101",
"output": "2031"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n100",
"output": "100"
},
{
"input": "5\n10001",
"output": "1001"
},
{
"input": "14\n11001100011000",
"output": "202002000"
},
{
"input": "31\n1000011111111100011110111111111",
"output": "100090049"
},
{
"input": "53\n10110111011110111110111111011111110111111110111111111",
"output": "123456789"
},
{
"input": "89\n11111111101111111110111111111011111111101111111110111111111011111111101111111110111111111",
"output": "999999999"
},
{
"input": "10\n1000000000",
"output": "1000000000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "4\n1110",
"output": "30"
},
{
"input": "8\n10101010",
"output": "11110"
}
] | 1,500,218,132
| 632
|
Python 3
|
OK
|
TESTS
| 13
| 62
| 4,608,000
|
def main():
N = int(input())
S = list(input())
count = 0
res = ""
for s in S:
if s == "1":
count += 1
else:
res += str(count)
count = 0
res += str(count)
print(res)
if __name__ == "__main__":
# global stime
# stime = time.clock()
main()
|
Title: Binary Protocol
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
Input Specification:
The first line contains one integer number *n* (1<=β€<=*n*<=β€<=89) β length of the string *s*.
The second line contains string *s* β sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
Output Specification:
Print the decoded number.
Demo Input:
['3\n111\n', '9\n110011101\n']
Demo Output:
['3\n', '2031\n']
Note:
none
|
```python
def main():
N = int(input())
S = list(input())
count = 0
res = ""
for s in S:
if s == "1":
count += 1
else:
res += str(count)
count = 0
res += str(count)
print(res)
if __name__ == "__main__":
# global stime
# stime = time.clock()
main()
```
| 3
|
|
437
|
C
|
The Child and Toy
|
PROGRAMMING
| 1,400
|
[
"graphs",
"greedy",
"sortings"
] | null | null |
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of *n* parts and *m* ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part *i* as *v**i*. The child spend *v**f*1<=+<=*v**f*2<=+<=...<=+<=*v**f**k* energy for removing part *i* where *f*1,<=*f*2,<=...,<=*f**k* are the parts that are directly connected to the *i*-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all *n* parts.
|
The first line contains two integers *n* and *m* (1<=β€<=*n*<=β€<=1000; 0<=β€<=*m*<=β€<=2000). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (0<=β€<=*v**i*<=β€<=105). Then followed *m* lines, each line contains two integers *x**i* and *y**i*, representing a rope from part *x**i* to part *y**i* (1<=β€<=*x**i*,<=*y**i*<=β€<=*n*;Β *x**i*<=β <=*y**i*).
Consider all the parts are numbered from 1 to *n*.
|
Output the minimum total energy the child should spend to remove all *n* parts of the toy.
|
[
"4 3\n10 20 30 40\n1 4\n1 2\n2 3\n",
"4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4\n",
"7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4\n"
] |
[
"40\n",
"400\n",
"160\n"
] |
One of the optimal sequence of actions in the first sample is:
- First, remove part 3, cost of the action is 20. - Then, remove part 2, cost of the action is 10. - Next, remove part 4, cost of the action is 10. - At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20β+β10β+β10β+β0β=β40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts.
| 1,500
|
[
{
"input": "4 3\n10 20 30 40\n1 4\n1 2\n2 3",
"output": "40"
},
{
"input": "4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4",
"output": "400"
},
{
"input": "7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4",
"output": "160"
},
{
"input": "1 0\n23333",
"output": "0"
},
{
"input": "5 4\n1 2 2 2 2\n1 2\n1 3\n1 4\n1 5",
"output": "4"
},
{
"input": "10 30\n3 6 17 15 13 15 6 12 9 1\n3 8\n1 10\n4 7\n1 7\n3 7\n2 9\n8 10\n3 1\n3 4\n8 6\n10 3\n3 9\n2 3\n10 4\n2 10\n5 8\n9 5\n6 1\n2 1\n7 2\n7 6\n7 10\n4 8\n5 6\n3 6\n4 1\n8 9\n7 9\n4 2\n5 10",
"output": "188"
},
{
"input": "3 3\n1 1 1\n1 2\n2 3\n3 1",
"output": "3"
}
] | 1,596,740,055
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 108
| 6,963,200
|
n, m = map(int, input().split())
cost = [0] + list(map(int, input().split()))
res = 0
for i in range(m):
x, y = map(int, input().split())
res += min(cost[x], cost[y])
print(res)
|
Title: The Child and Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of *n* parts and *m* ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part *i* as *v**i*. The child spend *v**f*1<=+<=*v**f*2<=+<=...<=+<=*v**f**k* energy for removing part *i* where *f*1,<=*f*2,<=...,<=*f**k* are the parts that are directly connected to the *i*-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all *n* parts.
Input Specification:
The first line contains two integers *n* and *m* (1<=β€<=*n*<=β€<=1000; 0<=β€<=*m*<=β€<=2000). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (0<=β€<=*v**i*<=β€<=105). Then followed *m* lines, each line contains two integers *x**i* and *y**i*, representing a rope from part *x**i* to part *y**i* (1<=β€<=*x**i*,<=*y**i*<=β€<=*n*;Β *x**i*<=β <=*y**i*).
Consider all the parts are numbered from 1 to *n*.
Output Specification:
Output the minimum total energy the child should spend to remove all *n* parts of the toy.
Demo Input:
['4 3\n10 20 30 40\n1 4\n1 2\n2 3\n', '4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4\n', '7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4\n']
Demo Output:
['40\n', '400\n', '160\n']
Note:
One of the optimal sequence of actions in the first sample is:
- First, remove part 3, cost of the action is 20. - Then, remove part 2, cost of the action is 10. - Next, remove part 4, cost of the action is 10. - At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20β+β10β+β10β+β0β=β40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts.
|
```python
n, m = map(int, input().split())
cost = [0] + list(map(int, input().split()))
res = 0
for i in range(m):
x, y = map(int, input().split())
res += min(cost[x], cost[y])
print(res)
```
| 3
|
|
897
|
A
|
Scarborough Fair
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
|
The first line contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=β€<=*l*<=β€<=*r*<=β€<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
|
Output string *s* after performing *m* operations described above.
|
[
"3 1\nioi\n1 1 i n\n",
"5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n"
] |
[
"noi",
"gaaak"
] |
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak.
| 500
|
[
{
"input": "3 1\nioi\n1 1 i n",
"output": "noi"
},
{
"input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g",
"output": "gaaak"
},
{
"input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n4 5 a e\n3 9 f a\n1 2 c h\n4 8 a c\n3 5 e d\n3 4 g f\n2 3 d h\n2 3 d e\n1 7 d g\n2 6 e g\n2 3 d g\n5 5 h h\n2 8 g d\n8 9 a f\n5 9 c e\n1 7 f d\n1 6 e e\n5 7 c a\n8 9 b b\n2 6 e b\n6 6 g h\n1 2 b b\n1 5 a f\n5 8 f h\n1 5 e g\n3 9 f h\n6 8 g a\n4 6 h g\n1 5 f a\n5 6 a c\n4 8 e d\n1 4 d g\n7 8 b f\n5 6 h b\n3 9 c e\n1 9 b a",
"output": "aahaddddh"
},
{
"input": "28 45\ndcbbaddjhbeefjadjchgkhgggfha\n10 25 c a\n13 19 a f\n12 28 e d\n12 27 e a\n9 20 b e\n7 17 g d\n22 26 j j\n8 16 c g\n14 16 a d\n3 10 f c\n10 26 d b\n8 17 i e\n10 19 d i\n6 21 c j\n7 22 b k\n17 19 a i\n4 18 j k\n8 25 a g\n10 27 j e\n9 18 g d\n16 23 h a\n17 26 k e\n8 16 h f\n1 15 d f\n22 28 k k\n11 20 c k\n6 11 b h\n17 17 e i\n15 22 g h\n8 18 c f\n4 16 e a\n8 25 b c\n6 24 d g\n5 9 f j\n12 19 i h\n4 25 e f\n15 25 c j\n15 27 e e\n11 20 b f\n19 27 e k\n2 21 d a\n9 27 k e\n14 24 b a\n3 6 i g\n2 26 k f",
"output": "fcbbajjfjaaefefehfahfagggfha"
},
{
"input": "87 5\nnfinedeojadjmgafnaogekfjkjfncnliagfchjfcmellgigjjcaaoeakdolchjcecljdeblmheimkibkgdkcdml\n47 56 a k\n51 81 o d\n5 11 j h\n48 62 j d\n16 30 k m",
"output": "nfinedeohadjmgafnaogemfjmjfncnliagfchjfcmellgigddckkdekkddlchdcecljdeblmheimkibkgdkcdml"
},
{
"input": "5 16\nacfbb\n1 2 e f\n2 5 a f\n2 3 b e\n4 4 f a\n2 3 f a\n1 2 b e\n4 5 c d\n2 4 e c\n1 4 e a\n1 3 d c\n3 5 e b\n3 5 e b\n2 2 e d\n1 3 e c\n3 3 a e\n1 5 a a",
"output": "acebb"
},
{
"input": "94 13\nbcaaaaaaccacddcdaacbdaabbcbaddbccbccbbbddbadddcccbddadddaadbdababadaacdcdbcdadabdcdcbcbcbcbbcd\n52 77 d d\n21 92 d b\n45 48 c b\n20 25 d a\n57 88 d b\n3 91 b d\n64 73 a a\n5 83 b d\n2 69 c c\n28 89 a b\n49 67 c b\n41 62 a c\n49 87 b c",
"output": "bcaaaaaaccacddcdaacddaaddcdbdddccdccddddddbdddddcdddcdddccdddcdcdcdcccdcddcdcdcddcdcdcdcdcdbcd"
},
{
"input": "67 39\nacbcbccccbabaabcabcaaaaaaccbcbbcbaaaacbbcccbcbabbcacccbbabbabbabaac\n4 36 a b\n25 38 a a\n3 44 b c\n35 57 b a\n4 8 a c\n20 67 c a\n30 66 b b\n27 40 a a\n2 56 a b\n10 47 c a\n22 65 c b\n29 42 a b\n1 46 c b\n57 64 b c\n20 29 b a\n14 51 c a\n12 55 b b\n20 20 a c\n2 57 c a\n22 60 c b\n16 51 c c\n31 64 a c\n17 30 c a\n23 36 c c\n28 67 a c\n37 40 a c\n37 50 b c\n29 48 c b\n2 34 b c\n21 53 b a\n26 63 a c\n23 28 c a\n51 56 c b\n32 61 b b\n64 67 b b\n21 67 b c\n8 53 c c\n40 62 b b\n32 38 c c",
"output": "accccccccaaaaaaaaaaaaaaaaaaaccccccccccccccccccccccccccccccccccccccc"
},
{
"input": "53 33\nhhcbhfafeececbhadfbdbehdfacfchbhdbfebdfeghebfcgdhehfh\n27 41 h g\n18 35 c b\n15 46 h f\n48 53 e g\n30 41 b c\n12 30 b f\n10 37 e f\n18 43 a h\n10 52 d a\n22 48 c e\n40 53 f d\n7 12 b h\n12 51 f a\n3 53 g a\n19 41 d h\n22 29 b h\n2 30 a b\n26 28 e h\n25 35 f a\n19 31 h h\n44 44 d e\n19 22 e c\n29 44 d h\n25 33 d h\n3 53 g c\n18 44 h b\n19 28 f e\n3 22 g h\n8 17 c a\n37 51 d d\n3 28 e h\n27 50 h h\n27 46 f b",
"output": "hhcbhfbfhfababbbbbbbbbbbbbbbbbeaaeaaeaaeabebdeaahahdh"
},
{
"input": "83 10\nfhbecdgadecabbbecedcgfdcefcbgechbedagecgdgfgdaahchdgchbeaedgafdefecdchceececfcdhcdh\n9 77 e e\n26 34 b g\n34 70 b a\n40 64 e g\n33 78 h f\n14 26 a a\n17 70 d g\n56 65 a c\n8 41 d c\n11 82 c b",
"output": "fhbecdgacebabbbebegbgfgbefbggebhgegagebgggfggaafbfggbfagbgggbfggfebgbfbeebebfbdhbdh"
},
{
"input": "1 4\ne\n1 1 c e\n1 1 e a\n1 1 e c\n1 1 d a",
"output": "a"
},
{
"input": "71 21\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n61 61 a a\n32 56 a a\n10 67 a a\n7 32 a a\n26 66 a a\n41 55 a a\n49 55 a a\n4 61 a a\n53 59 a a\n37 58 a a\n7 63 a a\n39 40 a a\n51 64 a a\n27 37 a a\n22 71 a a\n4 45 a a\n7 8 a a\n43 46 a a\n19 28 a a\n51 54 a a\n14 67 a a",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "30 4\neaaddabedcbbcccddbabdecadcecce\n2 17 c a\n16 29 e e\n16 21 c b\n7 11 b c",
"output": "eaaddacedacbaaaddbabdecadcecce"
},
{
"input": "48 30\naaaabaabbaababbbaabaabaababbabbbaabbbaabaaaaaaba\n3 45 a b\n1 14 a a\n15 32 a b\n37 47 a b\n9 35 a b\n36 39 b b\n6 26 a b\n36 44 a a\n28 44 b a\n29 31 b a\n20 39 a a\n45 45 a b\n21 32 b b\n7 43 a b\n14 48 a b\n14 33 a b\n39 44 a a\n9 36 b b\n4 23 b b\n9 42 b b\n41 41 b a\n30 47 a b\n8 42 b a\n14 38 b b\n3 15 a a\n35 47 b b\n14 34 a b\n38 43 a b\n1 35 b a\n16 28 b a",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbb"
},
{
"input": "89 29\nbabaabaaabaaaababbbbbbbabbbaaaaababbaababababbababaaabbababaaabbbbaaabaaaaaabaaabaabbabab\n39 70 b b\n3 56 b b\n5 22 b a\n4 39 a b\n41 87 b b\n34 41 a a\n10 86 a b\n29 75 a b\n2 68 a a\n27 28 b b\n42 51 b a\n18 61 a a\n6 67 b a\n47 63 a a\n8 68 a b\n4 74 b a\n19 65 a b\n8 55 a b\n5 30 a a\n3 65 a b\n16 57 a b\n34 56 b a\n1 70 a b\n59 68 b b\n29 57 b a\n47 49 b b\n49 73 a a\n32 61 b b\n29 42 a a",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaabbbbbbbbbbbbbab"
},
{
"input": "59 14\nfbebcfabdefbaaedcefdeecababcabebadfbccaaedaebfdaefdbbcbebbe\n5 32 e f\n8 46 e e\n31 43 e f\n3 10 e a\n53 54 f d\n55 59 d a\n39 58 e b\n54 56 f a\n9 40 b e\n28 37 d a\n7 35 e b\n7 56 c f\n23 26 e a\n15 44 e d",
"output": "fbabcfabdffbaafdfffdfffababfabfbaafdffaafdabbfdabfdbbfbbbbe"
},
{
"input": "7 17\nbbaabab\n3 5 a b\n5 7 a a\n5 5 a a\n4 4 b a\n7 7 a a\n5 6 b b\n1 3 b a\n6 7 a b\n4 6 a b\n6 6 a a\n2 4 b a\n1 7 b a\n4 6 b b\n2 5 b b\n2 5 a b\n1 4 a a\n4 4 b a",
"output": "abbabaa"
},
{
"input": "100 1\ndebaaagbfdgehagadabfgheegggfghghgeeeabgceffeffggcbcegfgebbdhebhfagcgadcbdbabddbcadgbgdebdfehceehcaef\n13 99 f c",
"output": "debaaagbfdgehagadabcgheegggcghghgeeeabgcecceccggcbcegcgebbdhebhcagcgadcbdbabddbcadgbgdebdcehceehcaef"
},
{
"input": "1 1\na\n1 1 a b",
"output": "b"
},
{
"input": "100 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n1 100 a b",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2 2\naa\n1 2 a b\n1 2 b c",
"output": "cc"
},
{
"input": "3 3\naaa\n1 3 a b\n1 3 b c\n1 3 c d",
"output": "ddd"
},
{
"input": "2 2\naa\n2 2 a b\n1 1 a b",
"output": "bb"
}
] | 1,612,445,465
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 62
| 0
|
n,m=map(int,input().split())
s=input()
for i in range(m):
l,r,c1,c2=input().split()
s=s[:int(l)-1]+s[int(l)-1:int(r)].replace(c1,c2)+s[int(r):]
print(s)
|
Title: Scarborough Fair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
Input Specification:
The first line contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=β€<=*l*<=β€<=*r*<=β€<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
Output Specification:
Output string *s* after performing *m* operations described above.
Demo Input:
['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n']
Demo Output:
['noi', 'gaaak']
Note:
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak.
|
```python
n,m=map(int,input().split())
s=input()
for i in range(m):
l,r,c1,c2=input().split()
s=s[:int(l)-1]+s[int(l)-1:int(r)].replace(c1,c2)+s[int(r):]
print(s)
```
| 3
|
|
217
|
A
|
Ice Skating
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null |
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
|
The first line of input contains a single integer *n* (1<=β€<=*n*<=β€<=100) β the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=β€<=*x**i*,<=*y**i*<=β€<=1000) β the coordinates of the *i*-th snow drift.
Note that the north direction coinΡides with the direction of *Oy* axis, so the east direction coinΡides with the direction of the *Ox* axis. All snow drift's locations are distinct.
|
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
|
[
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] |
[
"1\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182 184\n806 663\n468 459",
"output": "21"
},
{
"input": "17\n660 646\n440 442\n689 618\n441 415\n922 865\n950 972\n312 366\n203 229\n873 860\n219 199\n344 308\n169 176\n961 992\n153 84\n201 230\n987 938\n834 815",
"output": "16"
},
{
"input": "11\n798 845\n722 911\n374 270\n629 537\n748 856\n831 885\n486 641\n751 829\n609 492\n98 27\n654 663",
"output": "10"
},
{
"input": "1\n321 88",
"output": "0"
},
{
"input": "9\n811 859\n656 676\n76 141\n945 951\n497 455\n18 55\n335 294\n267 275\n656 689",
"output": "7"
},
{
"input": "7\n948 946\n130 130\n761 758\n941 938\n971 971\n387 385\n509 510",
"output": "6"
},
{
"input": "6\n535 699\n217 337\n508 780\n180 292\n393 112\n732 888",
"output": "5"
},
{
"input": "14\n25 23\n499 406\n193 266\n823 751\n219 227\n101 138\n978 992\n43 74\n997 932\n237 189\n634 538\n774 740\n842 767\n742 802",
"output": "13"
},
{
"input": "12\n548 506\n151 198\n370 380\n655 694\n654 690\n407 370\n518 497\n819 827\n765 751\n802 771\n741 752\n653 662",
"output": "11"
},
{
"input": "40\n685 711\n433 403\n703 710\n491 485\n616 619\n288 282\n884 871\n367 352\n500 511\n977 982\n51 31\n576 564\n508 519\n755 762\n22 20\n368 353\n232 225\n953 955\n452 436\n311 330\n967 988\n369 364\n791 803\n150 149\n651 661\n118 93\n398 387\n748 766\n852 852\n230 228\n555 545\n515 519\n667 678\n867 862\n134 146\n859 863\n96 99\n486 469\n303 296\n780 786",
"output": "38"
},
{
"input": "3\n175 201\n907 909\n388 360",
"output": "2"
},
{
"input": "7\n312 298\n86 78\n73 97\n619 594\n403 451\n538 528\n71 86",
"output": "6"
},
{
"input": "19\n802 820\n368 248\n758 794\n455 378\n876 888\n771 814\n245 177\n586 555\n844 842\n364 360\n820 856\n731 624\n982 975\n825 856\n122 121\n862 896\n42 4\n792 841\n828 820",
"output": "16"
},
{
"input": "32\n643 877\n842 614\n387 176\n99 338\n894 798\n652 728\n611 648\n622 694\n579 781\n243 46\n322 305\n198 438\n708 579\n246 325\n536 459\n874 593\n120 277\n989 907\n223 110\n35 130\n761 692\n690 661\n518 766\n226 93\n678 597\n725 617\n661 574\n775 496\n56 416\n14 189\n358 359\n898 901",
"output": "31"
},
{
"input": "32\n325 327\n20 22\n72 74\n935 933\n664 663\n726 729\n785 784\n170 171\n315 314\n577 580\n984 987\n313 317\n434 435\n962 961\n55 54\n46 44\n743 742\n434 433\n617 612\n332 332\n883 886\n940 936\n793 792\n645 644\n611 607\n418 418\n465 465\n219 218\n167 164\n56 54\n403 405\n210 210",
"output": "29"
},
{
"input": "32\n652 712\n260 241\n27 154\n188 16\n521 351\n518 356\n452 540\n790 827\n339 396\n336 551\n897 930\n828 627\n27 168\n180 113\n134 67\n794 671\n812 711\n100 241\n686 813\n138 289\n384 506\n884 932\n913 959\n470 508\n730 734\n373 478\n788 862\n392 426\n148 68\n113 49\n713 852\n924 894",
"output": "29"
},
{
"input": "14\n685 808\n542 677\n712 747\n832 852\n187 410\n399 338\n626 556\n530 635\n267 145\n215 209\n559 684\n944 949\n753 596\n601 823",
"output": "13"
},
{
"input": "5\n175 158\n16 2\n397 381\n668 686\n957 945",
"output": "4"
},
{
"input": "5\n312 284\n490 509\n730 747\n504 497\n782 793",
"output": "4"
},
{
"input": "2\n802 903\n476 348",
"output": "1"
},
{
"input": "4\n325 343\n425 442\n785 798\n275 270",
"output": "3"
},
{
"input": "28\n462 483\n411 401\n118 94\n111 127\n5 6\n70 52\n893 910\n73 63\n818 818\n182 201\n642 633\n900 886\n893 886\n684 700\n157 173\n953 953\n671 660\n224 225\n832 801\n152 157\n601 585\n115 101\n739 722\n611 606\n659 642\n461 469\n702 689\n649 653",
"output": "25"
},
{
"input": "36\n952 981\n885 900\n803 790\n107 129\n670 654\n143 132\n66 58\n813 819\n849 837\n165 198\n247 228\n15 39\n619 618\n105 138\n868 855\n965 957\n293 298\n613 599\n227 212\n745 754\n723 704\n877 858\n503 487\n678 697\n592 595\n155 135\n962 982\n93 89\n660 673\n225 212\n967 987\n690 680\n804 813\n489 518\n240 221\n111 124",
"output": "34"
},
{
"input": "30\n89 3\n167 156\n784 849\n943 937\n144 95\n24 159\n80 120\n657 683\n585 596\n43 147\n909 964\n131 84\n345 389\n333 321\n91 126\n274 325\n859 723\n866 922\n622 595\n690 752\n902 944\n127 170\n426 383\n905 925\n172 284\n793 810\n414 510\n890 884\n123 24\n267 255",
"output": "29"
},
{
"input": "5\n664 666\n951 941\n739 742\n844 842\n2 2",
"output": "4"
},
{
"input": "3\n939 867\n411 427\n757 708",
"output": "2"
},
{
"input": "36\n429 424\n885 972\n442 386\n512 511\n751 759\n4 115\n461 497\n496 408\n8 23\n542 562\n296 331\n448 492\n412 395\n109 166\n622 640\n379 355\n251 262\n564 586\n66 115\n275 291\n666 611\n629 534\n510 567\n635 666\n738 803\n420 369\n92 17\n101 144\n141 92\n258 258\n184 235\n492 456\n311 210\n394 357\n531 512\n634 636",
"output": "34"
},
{
"input": "29\n462 519\n871 825\n127 335\n156 93\n576 612\n885 830\n634 779\n340 105\n744 795\n716 474\n93 139\n563 805\n137 276\n177 101\n333 14\n391 437\n873 588\n817 518\n460 597\n572 670\n140 303\n392 441\n273 120\n862 578\n670 639\n410 161\n544 577\n193 116\n252 195",
"output": "28"
},
{
"input": "23\n952 907\n345 356\n812 807\n344 328\n242 268\n254 280\n1000 990\n80 78\n424 396\n595 608\n755 813\n383 380\n55 56\n598 633\n203 211\n508 476\n600 593\n206 192\n855 882\n517 462\n967 994\n642 657\n493 488",
"output": "22"
},
{
"input": "10\n579 816\n806 590\n830 787\n120 278\n677 800\n16 67\n188 251\n559 560\n87 67\n104 235",
"output": "8"
},
{
"input": "23\n420 424\n280 303\n515 511\n956 948\n799 803\n441 455\n362 369\n299 289\n823 813\n982 967\n876 878\n185 157\n529 551\n964 989\n655 656\n1 21\n114 112\n45 56\n935 937\n1000 997\n934 942\n360 366\n648 621",
"output": "22"
},
{
"input": "23\n102 84\n562 608\n200 127\n952 999\n465 496\n322 367\n728 690\n143 147\n855 867\n861 866\n26 59\n300 273\n255 351\n192 246\n70 111\n365 277\n32 104\n298 319\n330 354\n241 141\n56 125\n315 298\n412 461",
"output": "22"
},
{
"input": "7\n429 506\n346 307\n99 171\n853 916\n322 263\n115 157\n906 924",
"output": "6"
},
{
"input": "3\n1 1\n2 1\n2 2",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "5\n1 1\n1 2\n2 2\n3 1\n3 3",
"output": "0"
},
{
"input": "6\n1 1\n1 2\n2 2\n3 1\n3 2\n3 3",
"output": "0"
},
{
"input": "20\n1 1\n2 2\n3 3\n3 9\n4 4\n5 2\n5 5\n5 7\n5 8\n6 2\n6 6\n6 9\n7 7\n8 8\n9 4\n9 7\n9 9\n10 2\n10 9\n10 10",
"output": "1"
},
{
"input": "21\n1 1\n1 9\n2 1\n2 2\n2 5\n2 6\n2 9\n3 3\n3 8\n4 1\n4 4\n5 5\n5 8\n6 6\n7 7\n8 8\n9 9\n10 4\n10 10\n11 5\n11 11",
"output": "1"
},
{
"input": "22\n1 1\n1 3\n1 4\n1 8\n1 9\n1 11\n2 2\n3 3\n4 4\n4 5\n5 5\n6 6\n6 8\n7 7\n8 3\n8 4\n8 8\n9 9\n10 10\n11 4\n11 9\n11 11",
"output": "3"
},
{
"input": "50\n1 1\n2 2\n2 9\n3 3\n4 4\n4 9\n4 16\n4 24\n5 5\n6 6\n7 7\n8 8\n8 9\n8 20\n9 9\n10 10\n11 11\n12 12\n13 13\n14 7\n14 14\n14 16\n14 25\n15 4\n15 6\n15 15\n15 22\n16 6\n16 16\n17 17\n18 18\n19 6\n19 19\n20 20\n21 21\n22 6\n22 22\n23 23\n24 6\n24 7\n24 8\n24 9\n24 24\n25 1\n25 3\n25 5\n25 7\n25 23\n25 24\n25 25",
"output": "7"
},
{
"input": "55\n1 1\n1 14\n2 2\n2 19\n3 1\n3 3\n3 8\n3 14\n3 23\n4 1\n4 4\n5 5\n5 8\n5 15\n6 2\n6 3\n6 4\n6 6\n7 7\n8 8\n8 21\n9 9\n10 1\n10 10\n11 9\n11 11\n12 12\n13 13\n14 14\n15 15\n15 24\n16 5\n16 16\n17 5\n17 10\n17 17\n17 18\n17 22\n17 27\n18 18\n19 19\n20 20\n21 20\n21 21\n22 22\n23 23\n24 14\n24 24\n25 25\n26 8\n26 11\n26 26\n27 3\n27 27\n28 28",
"output": "5"
},
{
"input": "3\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "6\n4 4\n3 4\n5 4\n4 5\n4 3\n3 1",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n1 2",
"output": "0"
},
{
"input": "8\n1 3\n1 1\n4 1\n2 2\n2 5\n5 9\n5 1\n5 4",
"output": "1"
},
{
"input": "10\n1 1\n1 2\n1 3\n1 4\n5 5\n6 6\n7 7\n8 8\n9 9\n100 100",
"output": "6"
},
{
"input": "7\n1 1\n2 2\n3 3\n4 4\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "6\n1 1\n2 1\n2 2\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "4\n3 1\n2 1\n2 2\n1 2",
"output": "0"
},
{
"input": "6\n1 1\n2 2\n2 1\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "3\n1 2\n1 3\n1 4",
"output": "0"
},
{
"input": "4\n1 1\n2 2\n1 2\n2 1",
"output": "0"
},
{
"input": "4\n1 3\n2 1\n3 2\n3 1",
"output": "1"
},
{
"input": "7\n1 1\n1 2\n2 2\n3 3\n3 4\n4 4\n1 4",
"output": "0"
},
{
"input": "21\n12 12\n13 12\n12 11\n13 13\n10 10\n11 10\n11 11\n501 500\n501 501\n503 502\n500 500\n503 503\n502 501\n502 502\n700 700\n702 702\n703 702\n701 701\n702 701\n703 703\n701 700",
"output": "2"
},
{
"input": "6\n1 11\n6 8\n11 10\n1 10\n11 11\n6 9",
"output": "1"
},
{
"input": "4\n1 1\n2 2\n3 2\n3 1",
"output": "0"
},
{
"input": "3\n1 2\n3 4\n3 2",
"output": "0"
},
{
"input": "3\n1 1\n1 2\n2 2",
"output": "0"
},
{
"input": "4\n5 5\n5 4\n6 3\n6 4",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n2 1",
"output": "0"
}
] | 1,644,778,159
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 76
| 92
| 0
|
n = int(input())
coords = []
for _ in range(n):
nx, ny = map(int, input().split())
coords.append((nx, ny))
def search(x, y, v):
v.append((x, y))
for nx, ny in coords:
if (nx == x and ny != y) or (nx != x and ny == y):
if (nx, ny) not in v:
search(nx, ny, v)
return v
c = -1
while coords:
x, y = coords[0]
c += 1
visited = search(x, y, [])
for coord in visited:
coords.remove(coord)
print(c)
|
Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=β€<=*n*<=β€<=100) β the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=β€<=*x**i*,<=*y**i*<=β€<=1000) β the coordinates of the *i*-th snow drift.
Note that the north direction coinΡides with the direction of *Oy* axis, so the east direction coinΡides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
n = int(input())
coords = []
for _ in range(n):
nx, ny = map(int, input().split())
coords.append((nx, ny))
def search(x, y, v):
v.append((x, y))
for nx, ny in coords:
if (nx == x and ny != y) or (nx != x and ny == y):
if (nx, ny) not in v:
search(nx, ny, v)
return v
c = -1
while coords:
x, y = coords[0]
c += 1
visited = search(x, y, [])
for coord in visited:
coords.remove(coord)
print(c)
```
| 3
|
|
779
|
A
|
Pupils Redistribution
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
In Berland each high school student is characterized by academic performance β integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known β integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
|
The first line of the input contains integer number *n* (1<=β€<=*n*<=β€<=100) β number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=β€<=*b**i*<=β€<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
|
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
|
[
"4\n5 4 4 4\n5 5 4 5\n",
"6\n1 1 1 1 1 1\n5 5 5 5 5 5\n",
"1\n5\n3\n",
"9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n"
] |
[
"1\n",
"3\n",
"-1\n",
"4\n"
] |
none
| 500
|
[
{
"input": "4\n5 4 4 4\n5 5 4 5",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1\n5 5 5 5 5 5",
"output": "3"
},
{
"input": "1\n5\n3",
"output": "-1"
},
{
"input": "9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1",
"output": "4"
},
{
"input": "1\n1\n2",
"output": "-1"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "8\n1 1 2 2 3 3 4 4\n4 4 5 5 1 1 1 1",
"output": "2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1\n2 2 2 2 2 2 2 2 2 2",
"output": "5"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "2\n1 1\n1 1",
"output": "0"
},
{
"input": "2\n1 2\n1 1",
"output": "-1"
},
{
"input": "2\n2 2\n1 1",
"output": "1"
},
{
"input": "2\n1 2\n2 1",
"output": "0"
},
{
"input": "2\n1 1\n2 2",
"output": "1"
},
{
"input": "5\n5 5 5 5 5\n5 5 5 5 5",
"output": "0"
},
{
"input": "5\n5 5 5 3 5\n5 3 5 5 5",
"output": "0"
},
{
"input": "5\n2 3 2 3 3\n2 3 2 2 2",
"output": "1"
},
{
"input": "5\n4 4 1 4 2\n1 2 4 2 2",
"output": "1"
},
{
"input": "50\n4 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\n4 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",
"output": "0"
},
{
"input": "50\n1 3 1 3 3 3 1 3 3 3 3 1 1 1 3 3 3 1 3 1 1 1 3 1 3 1 3 3 3 1 3 1 1 3 3 3 1 1 1 1 3 3 1 1 1 3 3 1 1 1\n1 3 1 3 3 1 1 3 1 3 3 1 1 1 1 3 3 1 3 1 1 3 1 1 3 1 1 1 1 3 3 1 3 3 3 3 1 3 3 3 3 3 1 1 3 3 1 1 3 1",
"output": "0"
},
{
"input": "50\n1 1 1 4 1 1 4 1 4 1 1 4 1 1 4 1 1 4 1 1 4 1 4 4 4 1 1 4 1 4 4 4 4 4 4 4 1 4 1 1 1 1 4 1 4 4 1 1 1 4\n1 4 4 1 1 4 1 4 4 1 1 4 1 4 1 1 4 1 1 1 4 4 1 1 4 1 4 1 1 4 4 4 4 1 1 4 4 1 1 1 4 1 4 1 4 1 1 1 4 4",
"output": "0"
},
{
"input": "50\n3 5 1 3 3 4 3 4 2 5 2 1 2 2 5 5 4 5 4 2 1 3 4 2 3 3 3 2 4 3 5 5 5 5 5 5 2 5 2 2 5 4 4 1 5 3 4 2 1 3\n3 5 3 2 5 3 4 4 5 2 3 4 4 4 2 2 4 4 4 3 3 5 5 4 3 1 4 4 5 5 4 1 2 5 5 4 1 2 3 4 5 5 3 2 3 4 3 5 1 1",
"output": "3"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n1 1 3 1 3 1 1 3 1 1 3 1 3 1 1 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 1 1 1 3 1 1 1 3 1 1 3 3 1 3 3 1 3 1 3 3 3 3 1 1 3 3 3 1 1 3 1 3 3 3 1 3 3 3 3 3 1 3 3 3 3 1 3 1 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 1 1 3 1 1 1\n1 1 1 3 3 3 3 3 3 3 1 3 3 3 1 3 3 3 3 3 3 1 3 3 1 3 3 1 1 1 3 3 3 3 3 3 3 1 1 3 3 3 1 1 3 3 1 1 1 3 3 3 1 1 3 1 1 3 3 1 1 3 3 3 3 3 3 1 3 3 3 1 1 3 3 3 1 1 3 3 1 3 1 3 3 1 1 3 3 1 1 3 1 3 3 3 1 3 1 3",
"output": "0"
},
{
"input": "100\n2 4 5 2 5 5 4 4 5 4 4 5 2 5 5 4 5 2 5 2 2 4 5 4 4 4 2 4 2 2 4 2 4 2 2 2 4 5 5 5 4 2 4 5 4 4 2 5 4 2 5 4 5 4 5 4 5 5 5 4 2 2 4 5 2 5 5 2 5 2 4 4 4 5 5 2 2 2 4 4 2 2 2 5 5 2 2 4 5 4 2 4 4 2 5 2 4 4 4 4\n4 4 2 5 2 2 4 2 5 2 5 4 4 5 2 4 5 4 5 2 2 2 2 5 4 5 2 4 2 2 5 2 5 2 4 5 5 5 2 5 4 4 4 4 5 2 2 4 2 4 2 4 5 5 5 4 5 4 5 5 5 2 5 4 4 4 4 4 2 5 5 4 2 4 4 5 5 2 4 4 4 2 2 2 5 4 2 2 4 5 4 4 4 4 2 2 4 5 5 2",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 4 3 1 4 2 1 3 1 1 2 4 4 4 4 1 1 4 1 4 4 1 1 2 3 3 3 2 4 2 3 3 3 1 3 4 2 2 1 3 4 4 3 2 2 2 4 2 1 2 1 2 2 1 1 4 2 1 3 2 4 4 4 2 3 1 3 1 3 2 2 2 2 4 4 1 3 1 1 4 2 3 3 4 4 2 4 4 2 4 3 3 1 3 2 4\n3 1 4 4 2 1 1 1 1 1 1 3 1 1 3 4 3 2 2 4 2 1 4 4 4 4 1 2 3 4 2 3 3 4 3 3 2 4 2 2 2 1 2 4 4 4 2 1 3 4 3 3 4 2 4 4 3 2 4 2 4 2 4 4 1 4 3 1 4 3 3 3 3 1 2 2 2 2 4 1 2 1 3 4 3 1 3 3 4 2 3 3 2 1 3 4 2 1 1 2",
"output": "0"
},
{
"input": "100\n2 4 5 2 1 5 5 2 1 5 1 5 1 1 1 3 4 5 1 1 2 3 3 1 5 5 4 4 4 1 1 1 5 2 3 5 1 2 2 1 1 1 2 2 1 2 4 4 5 1 3 2 5 3 5 5 3 2 2 2 1 3 4 4 4 4 4 5 3 1 4 1 5 4 4 5 4 5 2 4 4 3 1 2 1 4 5 3 3 3 3 2 2 2 3 5 3 1 3 4\n3 2 5 1 5 4 4 3 5 5 5 2 1 4 4 3 2 3 3 5 5 4 5 5 2 1 2 4 4 3 5 1 1 5 1 3 2 5 2 4 4 2 4 2 4 2 3 2 5 1 4 4 1 1 1 5 3 5 1 1 4 5 1 1 2 2 5 3 5 1 1 1 2 3 3 2 3 2 4 4 5 4 2 1 3 4 1 1 2 4 1 5 3 1 2 1 3 4 1 3",
"output": "0"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n1 4 4 1 4 4 1 1 4 1 1 1 1 4 4 4 4 1 1 1 1 1 1 4 4 4 1 1 4 4 1 1 1 1 4 4 4 4 4 1 1 4 4 1 1 1 4 1 1 1 1 4 4 4 4 4 4 1 4 4 4 4 1 1 1 4 1 4 1 1 1 1 4 1 1 1 4 4 4 1 4 4 1 4 4 4 4 4 1 4 1 1 4 1 4 1 1 1 4 4\n4 1 1 4 4 4 1 4 4 4 1 1 4 1 1 4 1 4 4 4 1 1 4 1 4 1 1 1 4 4 1 4 1 4 1 4 4 1 1 4 1 4 1 1 1 4 1 4 4 4 1 4 1 4 4 4 4 1 4 1 1 4 1 1 4 4 4 1 4 1 4 1 4 4 4 1 1 4 1 4 4 4 4 1 1 1 1 1 4 4 1 4 1 4 1 1 1 4 4 1",
"output": "1"
},
{
"input": "100\n5 2 5 2 2 3 3 2 5 3 2 5 3 3 3 5 2 2 5 5 3 3 5 3 2 2 2 3 2 2 2 2 3 5 3 3 2 3 2 5 3 3 5 3 2 2 5 5 5 5 5 2 3 2 2 2 2 3 2 5 2 2 2 3 5 5 5 3 2 2 2 3 5 3 2 5 5 3 5 5 5 3 2 5 2 3 5 3 2 5 5 3 5 2 3 3 2 2 2 2\n5 3 5 3 3 5 2 5 3 2 3 3 5 2 5 2 2 5 2 5 2 5 3 3 5 3 2 2 2 3 5 3 2 2 3 2 2 5 5 2 3 2 3 3 5 3 2 5 2 2 2 3 3 5 3 3 5 2 2 2 3 3 2 2 3 5 3 5 5 3 3 2 5 3 5 2 3 2 5 5 3 2 5 5 2 2 2 2 3 2 2 5 2 5 2 2 3 3 2 5",
"output": "1"
},
{
"input": "100\n4 4 5 4 3 5 5 2 4 5 5 5 3 4 4 2 5 2 5 3 3 3 3 5 3 2 2 2 4 4 4 4 3 3 4 5 3 2 2 2 4 4 5 3 4 5 4 5 5 2 4 2 5 2 3 4 4 5 2 2 4 4 5 5 5 3 5 4 5 5 5 4 3 3 2 4 3 5 5 5 2 4 2 5 4 3 5 3 2 3 5 2 5 2 2 5 4 5 4 3\n5 4 2 4 3 5 2 5 5 3 4 5 4 5 3 3 5 5 2 3 4 2 3 5 2 2 2 4 2 5 2 4 4 5 2 2 4 4 5 5 2 3 4 2 4 5 2 5 2 2 4 5 5 3 5 5 5 4 3 4 4 3 5 5 3 4 5 3 2 3 4 3 4 4 2 5 3 4 5 5 3 5 3 3 4 3 5 3 2 2 4 5 4 5 5 2 3 4 3 5",
"output": "1"
},
{
"input": "100\n1 4 2 2 2 1 4 5 5 5 4 4 5 5 1 3 2 1 4 5 2 3 4 4 5 4 4 4 4 5 1 3 5 5 3 3 3 3 5 1 4 3 5 1 2 4 1 3 5 5 1 3 3 3 1 3 5 4 4 2 2 5 5 5 2 3 2 5 1 3 5 4 5 3 2 2 3 2 3 3 2 5 2 4 2 3 4 1 3 1 3 1 5 1 5 2 3 5 4 5\n1 2 5 3 2 3 4 2 5 1 2 5 3 4 3 3 4 1 5 5 1 3 3 1 1 4 1 4 2 5 4 1 3 4 5 3 2 2 1 4 5 5 2 3 3 5 5 4 2 3 3 5 3 3 5 4 4 5 3 5 1 1 4 4 4 1 3 5 5 5 4 2 4 5 3 2 2 2 5 5 5 1 4 3 1 3 1 2 2 4 5 1 3 2 4 5 1 5 2 5",
"output": "1"
},
{
"input": "100\n3 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 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 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 3 3 3 3 3 3 3 3 3\n3 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 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 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 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n5 2 2 2 5 2 5 5 5 2 5 2 5 5 5 5 5 5 2 2 2 5 5 2 5 2 2 5 2 5 5 2 5 2 5 2 5 5 5 5 5 2 2 2 2 5 5 2 5 5 5 2 5 5 5 2 5 5 5 2 2 2 5 2 2 2 5 5 2 5 5 5 2 5 2 2 5 2 2 2 5 5 5 5 2 5 2 5 2 2 5 2 5 2 2 2 2 5 5 2\n5 5 2 2 5 5 2 5 2 2 5 5 5 5 2 5 5 2 5 2 2 5 2 2 5 2 5 2 2 5 2 5 2 5 5 2 2 5 5 5 2 5 5 2 5 5 5 2 2 5 5 5 2 5 5 5 2 2 2 5 5 5 2 2 5 5 2 2 2 5 2 5 5 2 5 2 5 2 2 5 5 2 2 5 5 2 2 5 2 2 5 2 2 2 5 5 2 2 2 5",
"output": "1"
},
{
"input": "100\n3 3 2 2 1 2 3 3 2 2 1 1 3 3 1 1 1 2 1 2 3 2 3 3 3 1 2 3 1 2 1 2 3 3 2 1 1 1 1 1 2 2 3 2 1 1 3 3 1 3 3 1 3 1 3 3 3 2 1 2 3 1 3 2 2 2 2 2 2 3 1 3 1 2 2 1 2 3 2 3 3 1 2 1 1 3 1 1 1 2 1 2 2 2 3 2 3 2 1 1\n1 3 1 2 1 1 1 1 1 2 1 2 1 3 2 2 3 2 1 1 2 2 2 1 1 3 2 3 2 1 2 2 3 2 3 1 3 1 1 2 3 1 2 1 3 2 1 2 3 2 3 3 3 2 2 2 3 1 3 1 1 2 1 3 1 3 1 3 3 3 1 3 3 2 1 3 3 3 3 3 2 1 2 2 3 3 2 1 2 2 1 3 3 1 3 2 2 1 1 3",
"output": "1"
},
{
"input": "100\n5 3 3 2 5 3 2 4 2 3 3 5 3 4 5 4 3 3 4 3 2 3 3 4 5 4 2 4 2 4 5 3 3 4 5 3 5 3 5 3 3 2 5 3 4 5 2 5 2 2 4 2 2 2 2 5 4 5 4 3 5 4 2 5 5 3 4 5 2 3 2 2 2 5 3 2 2 2 3 3 5 2 3 2 4 5 3 3 3 5 2 3 3 3 5 4 5 5 5 2\n4 4 4 5 5 3 5 5 4 3 5 4 3 4 3 3 5 3 5 5 3 3 3 5 5 4 4 3 2 5 4 3 3 4 5 3 5 2 4 2 2 2 5 3 5 2 5 5 3 3 2 3 3 4 2 5 2 5 2 4 2 4 2 3 3 4 2 2 2 4 4 3 3 3 4 3 3 3 5 5 3 4 2 2 3 5 5 2 3 4 5 4 5 3 4 2 5 3 2 4",
"output": "3"
},
{
"input": "100\n5 3 4 4 2 5 1 1 4 4 3 5 5 1 4 4 2 5 3 2 1 1 3 2 4 4 4 2 5 2 2 3 1 4 1 4 4 5 3 5 1 4 1 4 1 5 5 3 5 5 1 5 3 5 1 3 3 4 5 3 2 2 4 5 2 5 4 2 4 4 1 1 4 2 4 1 2 2 4 3 4 1 1 1 4 3 5 1 2 1 4 5 4 4 2 1 4 1 3 2\n1 1 1 1 4 2 1 4 1 1 3 5 4 3 5 2 2 4 2 2 4 1 3 4 4 5 1 1 2 2 2 1 4 1 4 4 1 5 5 2 3 5 1 5 4 2 3 2 2 5 4 1 1 4 5 2 4 5 4 4 3 3 2 4 3 4 5 5 4 2 4 2 1 2 3 2 2 5 5 3 1 3 4 3 4 4 5 3 1 1 3 5 1 4 4 2 2 1 4 5",
"output": "2"
},
{
"input": "100\n3 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 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 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 3 3 3 3 3 3 3 3 3\n3 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 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 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 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 4 3 3 4 4 3 3 3 4 3 4 3 4 4 3 3 3 3 3 3 4 3 3 4 3 3 3 3 4 3 3 3 4 4 4 3 3 4 4 4 3 4 4 3 3 4 3 3 3 4 4 4 3 4 3 3 3 3 3 3 3 4 4 3 3 3 3 4 3 3 3 3 3 4 4 3 3 3 3 3 4 3 4 4 4 4 3 4 3 4 4 4 4 3 3\n4 3 3 3 3 4 4 3 4 4 4 3 3 4 4 3 4 4 4 4 3 4 3 3 3 4 4 4 3 4 3 4 4 3 3 4 3 3 3 3 3 4 3 3 3 3 4 4 4 3 3 4 3 4 4 4 4 3 4 4 3 3 4 3 3 4 3 4 3 4 4 4 4 3 3 4 3 4 4 4 3 3 4 4 4 4 4 3 3 3 4 3 3 4 3 3 3 3 3 3",
"output": "5"
},
{
"input": "100\n4 2 5 2 5 4 2 5 5 4 4 2 4 4 2 4 4 5 2 5 5 2 2 4 4 5 4 5 5 5 2 2 2 2 4 4 5 2 4 4 4 2 2 5 5 4 5 4 4 2 4 5 4 2 4 5 4 2 4 5 4 4 4 4 4 5 4 2 5 2 5 5 5 5 4 2 5 5 4 4 2 5 2 5 2 5 4 2 4 2 4 5 2 5 2 4 2 4 2 4\n5 4 5 4 5 2 2 4 5 2 5 5 5 5 5 4 4 4 4 5 4 5 5 2 4 4 4 4 5 2 4 4 5 5 2 5 2 5 5 4 4 5 2 5 2 5 2 5 4 5 2 5 2 5 2 4 4 5 4 2 5 5 4 2 2 2 5 4 2 2 4 4 4 5 5 2 5 2 2 4 4 4 2 5 4 5 2 2 5 4 4 5 5 4 5 5 4 5 2 5",
"output": "5"
},
{
"input": "100\n3 4 5 3 5 4 5 4 4 4 2 4 5 4 3 2 3 4 3 5 2 5 2 5 4 3 4 2 5 2 5 3 4 5 2 5 4 2 4 5 4 3 2 4 4 5 2 5 5 3 3 5 2 4 4 2 3 3 2 5 5 5 2 4 5 5 4 2 2 5 3 3 2 4 4 2 4 5 5 2 5 5 3 2 5 2 4 4 3 3 5 4 5 5 2 5 4 5 4 3\n4 3 5 5 2 4 2 4 5 5 5 2 3 3 3 3 5 5 5 5 3 5 2 3 5 2 3 2 2 5 5 3 5 3 4 2 2 5 3 3 3 3 5 2 4 5 3 5 3 4 4 4 5 5 3 4 4 2 2 4 4 5 3 2 4 5 5 4 5 2 2 3 5 4 5 5 2 5 4 3 2 3 2 5 4 5 3 4 5 5 3 5 2 2 4 4 3 2 5 2",
"output": "4"
},
{
"input": "100\n4 1 1 2 1 4 4 1 4 5 5 5 2 2 1 3 5 2 1 5 2 1 2 4 4 2 1 2 2 2 4 3 1 4 2 2 3 1 1 4 4 5 4 4 4 5 1 4 1 4 3 1 2 1 2 4 1 2 5 2 1 4 3 4 1 4 2 1 1 1 5 3 3 1 4 1 3 1 4 1 1 2 2 2 3 1 4 3 4 4 5 2 5 4 3 3 3 2 2 1\n5 1 4 4 3 4 4 5 2 3 3 4 4 2 3 2 3 1 3 1 1 4 1 5 4 3 2 4 3 3 3 2 3 4 1 5 4 2 4 2 2 2 5 3 1 2 5 3 2 2 1 1 2 2 3 5 1 2 5 3 2 1 1 2 1 2 4 3 5 4 5 3 2 4 1 3 4 1 4 4 5 4 4 5 4 2 5 3 4 1 4 2 4 2 4 5 4 5 4 2",
"output": "6"
},
{
"input": "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\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "100\n3 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 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 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 3 3 3 3 3 3 3 3 3\n3 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 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 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 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n4 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 1 4 4 4 4 4 4 4 4 4 4\n4 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 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",
"output": "0"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 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 1 2 2 2 2 2 2 2 2 2\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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",
"output": "1"
},
{
"input": "100\n3 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 1 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 1 3 1 3 3 3 3 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 4 3 3 3 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 1 3 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3",
"output": "1"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\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": "50"
},
{
"input": "100\n3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5\n3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1",
"output": "25"
},
{
"input": "100\n3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5\n2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4",
"output": "50"
},
{
"input": "100\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "40"
},
{
"input": "100\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5\n2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3",
"output": "30"
},
{
"input": "5\n4 4 4 4 5\n4 5 5 5 5",
"output": "-1"
},
{
"input": "4\n1 1 1 1\n3 3 3 3",
"output": "2"
},
{
"input": "6\n1 1 2 2 3 4\n1 2 3 3 4 4",
"output": "-1"
},
{
"input": "4\n1 1 1 2\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n2 2 2\n4 4 4",
"output": "-1"
},
{
"input": "2\n1 2\n3 4",
"output": "-1"
},
{
"input": "6\n1 1 1 3 3 3\n2 2 2 4 4 4",
"output": "-1"
},
{
"input": "5\n1 2 2 2 2\n1 1 1 1 3",
"output": "-1"
},
{
"input": "2\n1 3\n2 2",
"output": "-1"
},
{
"input": "2\n1 3\n4 5",
"output": "-1"
},
{
"input": "4\n1 2 3 4\n5 5 5 5",
"output": "-1"
},
{
"input": "2\n1 3\n2 4",
"output": "-1"
},
{
"input": "2\n1 2\n4 4",
"output": "-1"
},
{
"input": "2\n1 2\n3 3",
"output": "-1"
},
{
"input": "10\n4 4 4 4 2 3 3 3 3 1\n2 2 2 2 4 1 1 1 1 3",
"output": "-1"
},
{
"input": "6\n1 2 3 3 4 4\n1 1 2 2 3 4",
"output": "-1"
},
{
"input": "5\n3 3 3 3 1\n1 1 1 1 3",
"output": "-1"
},
{
"input": "2\n1 1\n2 3",
"output": "-1"
},
{
"input": "8\n1 1 2 2 3 3 3 3\n2 2 2 2 1 1 1 1",
"output": "2"
},
{
"input": "5\n1 1 1 3 3\n1 1 1 1 2",
"output": "-1"
},
{
"input": "6\n2 2 3 3 4 4\n2 3 4 5 5 5",
"output": "-1"
},
{
"input": "6\n1 1 2 2 3 4\n3 3 4 4 1 2",
"output": "-1"
},
{
"input": "4\n1 2 3 3\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n3 3 3",
"output": "-1"
},
{
"input": "5\n3 3 3 2 2\n2 2 2 3 3",
"output": "-1"
},
{
"input": "10\n1 2 3 4 1 2 3 4 1 2\n1 2 3 4 1 2 3 4 3 4",
"output": "-1"
},
{
"input": "2\n2 2\n1 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n1 1 4",
"output": "-1"
},
{
"input": "4\n3 4 4 4\n3 3 4 4",
"output": "-1"
}
] | 1,657,310,353
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 46
| 0
|
n = int(input())
A = map(int, input().split())
B = map(int, input().split())
dic_a = {1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
dic_b = dic_a.copy()
for x, y in zip(A, B):
dic_a[x] += 1
dic_b[y] += 1
c = 0
res = 0
for i in range(1, 6):
value = dic_a[i]-dic_b[i]
if abs(value) % 2 != 0:
c = 1
else:
res += abs(value)//2
if c == 1:
print(-1)
else:
print(res // 2)
|
Title: Pupils Redistribution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland each high school student is characterized by academic performance β integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known β integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
Input Specification:
The first line of the input contains integer number *n* (1<=β€<=*n*<=β€<=100) β number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=β€<=*b**i*<=β€<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
Output Specification:
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
Demo Input:
['4\n5 4 4 4\n5 5 4 5\n', '6\n1 1 1 1 1 1\n5 5 5 5 5 5\n', '1\n5\n3\n', '9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n']
Demo Output:
['1\n', '3\n', '-1\n', '4\n']
Note:
none
|
```python
n = int(input())
A = map(int, input().split())
B = map(int, input().split())
dic_a = {1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
dic_b = dic_a.copy()
for x, y in zip(A, B):
dic_a[x] += 1
dic_b[y] += 1
c = 0
res = 0
for i in range(1, 6):
value = dic_a[i]-dic_b[i]
if abs(value) % 2 != 0:
c = 1
else:
res += abs(value)//2
if c == 1:
print(-1)
else:
print(res // 2)
```
| 3
|
|
272
|
A
|
Dima and Friends
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
|
In a single line print the answer to the problem.
|
[
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] |
[
"3\n",
"2\n",
"3\n"
] |
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
| 500
|
[
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 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",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,567,686,855
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 248
| 0
|
n=int(input())
ans=0
l=[int(x) for x in input().split()]
x=n+1
y=sum(l)
for i in range(1,6):
if (y+i)%x!=1:
ans+=1
print(ans)
|
Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
|
```python
n=int(input())
ans=0
l=[int(x) for x in input().split()]
x=n+1
y=sum(l)
for i in range(1,6):
if (y+i)%x!=1:
ans+=1
print(ans)
```
| 3
|
|
401
|
C
|
Team
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"greedy",
"implementation"
] | null | null |
Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork.
For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that:
- there wouldn't be a pair of any side-adjacent cards with zeroes in a row; - there wouldn't be a group of three consecutive cards containing numbers one.
Today Vanya brought *n* cards with zeroes and *m* cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way.
|
The first line contains two integers: *n* (1<=β€<=*n*<=β€<=106) β the number of cards containing number 0; *m* (1<=β€<=*m*<=β€<=106) β the number of cards containing number 1.
|
In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1.
|
[
"1 2\n",
"4 8\n",
"4 10\n",
"1 5\n"
] |
[
"101\n",
"110110110101\n",
"11011011011011\n",
"-1\n"
] |
none
| 1,500
|
[
{
"input": "1 2",
"output": "101"
},
{
"input": "4 8",
"output": "110110110101"
},
{
"input": "4 10",
"output": "11011011011011"
},
{
"input": "1 5",
"output": "-1"
},
{
"input": "3 4",
"output": "1010101"
},
{
"input": "3 10",
"output": "-1"
},
{
"input": "74 99",
"output": "11011011011011011011011011011011011011011011011011011011011011011011011010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101"
},
{
"input": "19 30",
"output": "1101101101101101101101101101101010101010101010101"
},
{
"input": "33 77",
"output": "-1"
},
{
"input": "3830 6966",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "1000000 1000000",
"output": "1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101..."
},
{
"input": "1027 2030",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "4610 4609",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "3342 3339",
"output": "-1"
},
{
"input": "7757 7755",
"output": "-1"
},
{
"input": "10 8",
"output": "-1"
},
{
"input": "4247 8495",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "7101 14204",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "9801 19605",
"output": "-1"
},
{
"input": "4025 6858",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "7129 13245",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "8826 12432",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "6322 9256",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "8097 14682",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "6196 6197",
"output": "1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101..."
},
{
"input": "1709 2902",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "455 512",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101..."
},
{
"input": "1781 1272",
"output": "-1"
},
{
"input": "3383 5670",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "954 1788",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "9481 15554",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "9079 100096",
"output": "-1"
},
{
"input": "481533 676709",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "423472 564888",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "227774 373297",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "42346 51898",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "739107 1000000",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "455043 798612",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "801460 801459",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "303498 503791",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "518822 597833",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "32342 64687",
"output": "-1"
},
{
"input": "873192 873189",
"output": "-1"
},
{
"input": "384870 450227",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "201106 208474",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "775338 980888",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "263338 393171",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "241043 330384",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "307203 614408",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "379310 417986",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "661101 785111",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "284634 319008",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "500000 1000000",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "499999 1000000",
"output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..."
},
{
"input": "3 1",
"output": "-1"
},
{
"input": "14124 242112",
"output": "-1"
},
{
"input": "2 1",
"output": "010"
},
{
"input": "2 3",
"output": "10101"
},
{
"input": "1 4",
"output": "11011"
},
{
"input": "2 6",
"output": "11011011"
},
{
"input": "2 5",
"output": "1101101"
},
{
"input": "2 4",
"output": "110101"
},
{
"input": "2 3",
"output": "10101"
},
{
"input": "2 2",
"output": "1010"
},
{
"input": "2 1",
"output": "010"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1 3",
"output": "1101"
},
{
"input": "3 2",
"output": "01010"
},
{
"input": "5 4",
"output": "010101010"
},
{
"input": "4 3",
"output": "0101010"
},
{
"input": "11 10",
"output": "010101010101010101010"
},
{
"input": "8 7",
"output": "010101010101010"
},
{
"input": "4 2",
"output": "-1"
},
{
"input": "6 5",
"output": "01010101010"
},
{
"input": "3 7",
"output": "1101101101"
},
{
"input": "5 1",
"output": "-1"
},
{
"input": "10 1",
"output": "-1"
},
{
"input": "4 9",
"output": "1101101101101"
},
{
"input": "6 4",
"output": "-1"
},
{
"input": "12 10",
"output": "-1"
},
{
"input": "4 100",
"output": "-1"
}
] | 1,610,887,380
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 81
| 280
| 13,004,800
|
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
z,zz=input,lambda:list(map(int,z().split()))
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from re import *
from sys import *
from math import *
from heapq import *
from queue import *
from bisect import *
from string import *
from itertools import *
from collections import *
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from collections import Counter as cc
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def output(answer):stdout.write(str(answer))
###########################---Test-Case---#################################
"""
If you Know me , Then you probably don't know me !
"""
###########################---START-CODING---##############################
n,m=zzz()
if n>m+1 or 2*n+2<m:
print(-1)
elif n==m+1:
print('01'*m+'0')
elif n==m:
print('01'*n)
elif 2*n+2==m:
print('110'*n+'11')
else:
t=m-n-1
print('110'*t+'10'*(n-t)+'1')
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork.
For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that:
- there wouldn't be a pair of any side-adjacent cards with zeroes in a row; - there wouldn't be a group of three consecutive cards containing numbers one.
Today Vanya brought *n* cards with zeroes and *m* cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way.
Input Specification:
The first line contains two integers: *n* (1<=β€<=*n*<=β€<=106) β the number of cards containing number 0; *m* (1<=β€<=*m*<=β€<=106) β the number of cards containing number 1.
Output Specification:
In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1.
Demo Input:
['1 2\n', '4 8\n', '4 10\n', '1 5\n']
Demo Output:
['101\n', '110110110101\n', '11011011011011\n', '-1\n']
Note:
none
|
```python
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
z,zz=input,lambda:list(map(int,z().split()))
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from re import *
from sys import *
from math import *
from heapq import *
from queue import *
from bisect import *
from string import *
from itertools import *
from collections import *
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from collections import Counter as cc
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def output(answer):stdout.write(str(answer))
###########################---Test-Case---#################################
"""
If you Know me , Then you probably don't know me !
"""
###########################---START-CODING---##############################
n,m=zzz()
if n>m+1 or 2*n+2<m:
print(-1)
elif n==m+1:
print('01'*m+'0')
elif n==m:
print('01'*n)
elif 2*n+2==m:
print('110'*n+'11')
else:
t=m-n-1
print('110'*t+'10'*(n-t)+'1')
```
| 3
|
|
766
|
A
|
Mahmoud and Longest Uncommon Subsequence
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"strings"
] | null | null |
While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.
Given two strings *a* and *b*, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.
A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don't have to be consecutive, for example, strings "ac", "bc", "abc" and "a" are subsequences of string "abc" while strings "abbc" and "acb" are not. The empty string is a subsequence of any string. Any string is a subsequence of itself.
|
The first line contains string *a*, and the second lineΒ β string *b*. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters.
|
If there's no uncommon subsequence, print "-1". Otherwise print the length of the longest uncommon subsequence of *a* and *b*.
|
[
"abcd\ndefgh\n",
"a\na\n"
] |
[
"5\n",
"-1\n"
] |
In the first example: you can choose "defgh" from string *b* as it is the longest subsequence of string *b* that doesn't appear as a subsequence of string *a*.
| 500
|
[
{
"input": "abcd\ndefgh",
"output": "5"
},
{
"input": "a\na",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaacccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaadddddddddddddddddddddddddddddddddddddddddddddddddd",
"output": "100"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "199"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nbbbbbbbbbbbbbbbbbbb",
"output": "99"
},
{
"input": "abcde\nfghij",
"output": "5"
},
{
"input": "abcde\nabcdf",
"output": "5"
},
{
"input": "abcde\nbbcde",
"output": "5"
},
{
"input": "abcde\neabcd",
"output": "5"
},
{
"input": "abcdefgh\nabdcefgh",
"output": "8"
},
{
"input": "mmmmm\nmnmmm",
"output": "5"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaa",
"output": "34"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\nzabcdefghijklmnopqrstuvwxy",
"output": "26"
},
{
"input": "a\nab",
"output": "2"
},
{
"input": "b\nab",
"output": "2"
},
{
"input": "ab\nb",
"output": "2"
},
{
"input": "ab\nc",
"output": "2"
},
{
"input": "aaaaaa\naaaaaa",
"output": "-1"
},
{
"input": "abacaba\nabacaba",
"output": "-1"
},
{
"input": "aabb\nbbaa",
"output": "4"
},
{
"input": "ab\nba",
"output": "2"
},
{
"input": "abcd\nabc",
"output": "4"
},
{
"input": "abaa\nabaa",
"output": "-1"
},
{
"input": "ab\nab",
"output": "-1"
},
{
"input": "ab\nabcd",
"output": "4"
},
{
"input": "abc\nabcd",
"output": "4"
},
{
"input": "mo\nmomo",
"output": "4"
},
{
"input": "koooooooooooooooo\nloooooooooooooooo",
"output": "17"
},
{
"input": "aaa\naa",
"output": "3"
},
{
"input": "abc\nabc",
"output": "-1"
},
{
"input": "abcd\nabcd",
"output": "-1"
},
{
"input": "abc\ncba",
"output": "3"
},
{
"input": "ahc\nahc",
"output": "-1"
},
{
"input": "abc\nbac",
"output": "3"
},
{
"input": "aa\naaa",
"output": "3"
},
{
"input": "aaa\naaa",
"output": "-1"
},
{
"input": "abc\nacb",
"output": "3"
},
{
"input": "abc\nab",
"output": "3"
},
{
"input": "abb\nabb",
"output": "-1"
},
{
"input": "abc\ncab",
"output": "3"
},
{
"input": "aaaaaa\naaaaa",
"output": "6"
},
{
"input": "aa\naab",
"output": "3"
},
{
"input": "len\nlena",
"output": "4"
},
{
"input": "aaaaa\naa",
"output": "5"
},
{
"input": "aaa\naaaa",
"output": "4"
},
{
"input": "bcd\nabcd",
"output": "4"
},
{
"input": "aaabbc\naaaccc",
"output": "6"
},
{
"input": "abcd\nzycd",
"output": "4"
},
{
"input": "baa\nzaa",
"output": "3"
},
{
"input": "asdf\nadfs",
"output": "4"
},
{
"input": "abcdefgh\nabcdefgh",
"output": "-1"
},
{
"input": "aba\naab",
"output": "3"
},
{
"input": "aba\naba",
"output": "-1"
},
{
"input": "abcccc\naccccc",
"output": "6"
},
{
"input": "acb\nabc",
"output": "3"
},
{
"input": "a\naa",
"output": "2"
},
{
"input": "abababab\nab",
"output": "8"
},
{
"input": "ababab\nab",
"output": "6"
}
] | 1,645,781,406
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 73
| 46
| 0
|
a = str(input())
b = str(input())
if len(a) > len(b):
print(len(a))
elif len(a) < len(b):
print(len(b))
elif len(a) == len(b):
if a == b:
print(-1)
else:
print(len(a))
|
Title: Mahmoud and Longest Uncommon Subsequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.
Given two strings *a* and *b*, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.
A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don't have to be consecutive, for example, strings "ac", "bc", "abc" and "a" are subsequences of string "abc" while strings "abbc" and "acb" are not. The empty string is a subsequence of any string. Any string is a subsequence of itself.
Input Specification:
The first line contains string *a*, and the second lineΒ β string *b*. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters.
Output Specification:
If there's no uncommon subsequence, print "-1". Otherwise print the length of the longest uncommon subsequence of *a* and *b*.
Demo Input:
['abcd\ndefgh\n', 'a\na\n']
Demo Output:
['5\n', '-1\n']
Note:
In the first example: you can choose "defgh" from string *b* as it is the longest subsequence of string *b* that doesn't appear as a subsequence of string *a*.
|
```python
a = str(input())
b = str(input())
if len(a) > len(b):
print(len(a))
elif len(a) < len(b):
print(len(b))
elif len(a) == len(b):
if a == b:
print(-1)
else:
print(len(a))
```
| 3
|
|
510
|
A
|
Fox And Snake
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
|
The only line contains two integers: *n* and *m* (3<=β€<=*n*,<=*m*<=β€<=50).
*n* is an odd number.
|
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
|
[
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] |
[
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] |
none
| 500
|
[
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########"
},
{
"input": "3 5",
"output": "#####\n....#\n#####"
},
{
"input": "3 6",
"output": "######\n.....#\n######"
},
{
"input": "7 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "7 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####"
},
{
"input": "49 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 27",
"output": "###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n....................."
},
{
"input": "11 15",
"output": "###############\n..............#\n###############\n#..............\n###############\n..............#\n###############\n#..............\n###############\n..............#\n###############"
},
{
"input": "11 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "19 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "23 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "49 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "33 43",
"output": "###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#.................."
},
{
"input": "33 44",
"output": "############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#......."
},
{
"input": "45 45",
"output": "#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#########################################..."
},
{
"input": "45 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "45 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###"
},
{
"input": "45 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####"
},
{
"input": "5 4",
"output": "####\n...#\n####\n#...\n####"
},
{
"input": "5 5",
"output": "#####\n....#\n#####\n#....\n#####"
},
{
"input": "5 6",
"output": "######\n.....#\n######\n#.....\n######"
},
{
"input": "7 5",
"output": "#####\n....#\n#####\n#....\n#####\n....#\n#####"
},
{
"input": "7 6",
"output": "######\n.....#\n######\n#.....\n######\n.....#\n######"
},
{
"input": "7 7",
"output": "#######\n......#\n#######\n#......\n#######\n......#\n#######"
}
] | 1,688,188,673
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 0
|
i, j = map(int, input().split())
o = 0
for i in range(i):
if i%2 == 0:
print("#"*j)
elif i == (4*o)+1:
print("."*(j-1)+"#")
o = o+1
else:
print("#"+"."*(j-1))
|
Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=β€<=*n*,<=*m*<=β€<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none
|
```python
i, j = map(int, input().split())
o = 0
for i in range(i):
if i%2 == 0:
print("#"*j)
elif i == (4*o)+1:
print("."*(j-1)+"#")
o = o+1
else:
print("#"+"."*(j-1))
```
| 3
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=β€<=*x*1,<=*x*2,<=*x*3<=β€<=100)Β β the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integerΒ β the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,684,086,649
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 46
| 0
|
x1, x2, x3 = map(int, input().split())
# Sort the points
points = [x1, x2, x3]
points.sort()
# Calculate the middle point and the total distance
middle_point = points[1]
total_distance = abs(middle_point - x1) + abs(middle_point - x2) + abs(middle_point - x3)
print(total_distance)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=β€<=*x*1,<=*x*2,<=*x*3<=β€<=100)Β β the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integerΒ β the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
x1, x2, x3 = map(int, input().split())
# Sort the points
points = [x1, x2, x3]
points.sort()
# Calculate the middle point and the total distance
middle_point = points[1]
total_distance = abs(middle_point - x1) + abs(middle_point - x2) + abs(middle_point - x3)
print(total_distance)
```
| 3
|
|
702
|
A
|
Maximum Increase
|
PROGRAMMING
| 800
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
|
The first line contains single positive integer *n* (1<=β€<=*n*<=β€<=105) β the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=109).
|
Print the maximum length of an increasing subarray of the given array.
|
[
"5\n1 7 2 11 15\n",
"6\n100 100 100 100 100 100\n",
"3\n1 2 3\n"
] |
[
"3\n",
"1\n",
"3\n"
] |
none
| 0
|
[
{
"input": "5\n1 7 2 11 15",
"output": "3"
},
{
"input": "6\n100 100 100 100 100 100",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n1 2 3 3 4",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "3\n1 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "2"
},
{
"input": "11\n1 2 3 1 2 3 2 1 2 3 4",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "9"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n3 2 1",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "1\n1234394",
"output": "1"
}
] | 1,687,585,615
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 77
| 12,800,000
|
n=int(input())
ans = 1
j = 1
l=list(map(int,input().split()))
for i in range(1,n):
if l[i]>l[i-1]:
j+=1
else:
j=1
if j>ans:
ans=j
print(ans)
|
Title: Maximum Increase
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
Input Specification:
The first line contains single positive integer *n* (1<=β€<=*n*<=β€<=105) β the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=109).
Output Specification:
Print the maximum length of an increasing subarray of the given array.
Demo Input:
['5\n1 7 2 11 15\n', '6\n100 100 100 100 100 100\n', '3\n1 2 3\n']
Demo Output:
['3\n', '1\n', '3\n']
Note:
none
|
```python
n=int(input())
ans = 1
j = 1
l=list(map(int,input().split()))
for i in range(1,n):
if l[i]>l[i-1]:
j+=1
else:
j=1
if j>ans:
ans=j
print(ans)
```
| 3
|
|
612
|
B
|
HDD is Outdated Technology
|
PROGRAMMING
| 1,200
|
[
"implementation",
"math"
] | null | null |
HDD hard drives group data by sectors. All files are split to fragments and each of them are written in some sector of hard drive. Note the fragments can be written in sectors in arbitrary order.
One of the problems of HDD hard drives is the following: the magnetic head should move from one sector to another to read some file.
Find the time need to read file split to *n* fragments. The *i*-th sector contains the *f**i*-th fragment of the file (1<=β€<=*f**i*<=β€<=*n*). Note different sectors contains the different fragments. At the start the magnetic head is in the position that contains the first fragment. The file are reading in the following manner: at first the first fragment is read, then the magnetic head moves to the sector that contains the second fragment, then the second fragment is read and so on until the *n*-th fragment is read. The fragments are read in the order from the first to the *n*-th.
It takes |*a*<=-<=*b*| time units to move the magnetic head from the sector *a* to the sector *b*. Reading a fragment takes no time.
|
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=2Β·105) β the number of fragments.
The second line contains *n* different integers *f**i* (1<=β€<=*f**i*<=β€<=*n*) β the number of the fragment written in the *i*-th sector.
|
Print the only integer β the number of time units needed to read the file.
|
[
"3\n3 1 2\n",
"5\n1 3 5 4 2\n"
] |
[
"3\n",
"10\n"
] |
In the second example the head moves in the following way:
- 1->2 means movement from the sector 1 to the sector 5, i.e. it takes 4 time units - 2->3 means movement from the sector 5 to the sector 2, i.e. it takes 3 time units - 3->4 means movement from the sector 2 to the sector 4, i.e. it takes 2 time units - 4->5 means movement from the sector 4 to the sector 3, i.e. it takes 1 time units
So the answer to the second example is 4β+β3β+β2β+β1β=β10.
| 0
|
[
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "5\n1 3 5 4 2",
"output": "10"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5",
"output": "40"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "10\n1 3 5 7 9 10 8 6 4 2",
"output": "45"
},
{
"input": "100\n11 9 35 34 51 74 16 67 26 21 14 80 84 79 7 61 28 3 53 43 42 5 56 36 69 30 22 88 1 27 65 91 46 31 59 50 17 96 25 18 64 55 78 2 63 24 95 48 93 13 38 76 89 94 15 90 45 81 52 87 83 73 44 49 23 82 85 75 86 33 47 19 58 97 37 20 40 10 92 4 6 68 77 54 71 12 62 60 100 39 41 99 72 29 57 8 70 32 66 98",
"output": "3580"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "99"
},
{
"input": "100\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2",
"output": "4950"
}
] | 1,565,979,050
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 265
| 14,233,600
|
def hdd(n, lst):
res, answer = [0] * n, 0
for i in range(n):
res[lst[i] - 1] = i
for i in range(n - 1):
answer += abs(res[i] - res[i + 1])
return answer
m = int(input())
a = [int(j) for j in input().split()]
print(hdd(m, a))
|
Title: HDD is Outdated Technology
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
HDD hard drives group data by sectors. All files are split to fragments and each of them are written in some sector of hard drive. Note the fragments can be written in sectors in arbitrary order.
One of the problems of HDD hard drives is the following: the magnetic head should move from one sector to another to read some file.
Find the time need to read file split to *n* fragments. The *i*-th sector contains the *f**i*-th fragment of the file (1<=β€<=*f**i*<=β€<=*n*). Note different sectors contains the different fragments. At the start the magnetic head is in the position that contains the first fragment. The file are reading in the following manner: at first the first fragment is read, then the magnetic head moves to the sector that contains the second fragment, then the second fragment is read and so on until the *n*-th fragment is read. The fragments are read in the order from the first to the *n*-th.
It takes |*a*<=-<=*b*| time units to move the magnetic head from the sector *a* to the sector *b*. Reading a fragment takes no time.
Input Specification:
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=2Β·105) β the number of fragments.
The second line contains *n* different integers *f**i* (1<=β€<=*f**i*<=β€<=*n*) β the number of the fragment written in the *i*-th sector.
Output Specification:
Print the only integer β the number of time units needed to read the file.
Demo Input:
['3\n3 1 2\n', '5\n1 3 5 4 2\n']
Demo Output:
['3\n', '10\n']
Note:
In the second example the head moves in the following way:
- 1->2 means movement from the sector 1 to the sector 5, i.e. it takes 4 time units - 2->3 means movement from the sector 5 to the sector 2, i.e. it takes 3 time units - 3->4 means movement from the sector 2 to the sector 4, i.e. it takes 2 time units - 4->5 means movement from the sector 4 to the sector 3, i.e. it takes 1 time units
So the answer to the second example is 4β+β3β+β2β+β1β=β10.
|
```python
def hdd(n, lst):
res, answer = [0] * n, 0
for i in range(n):
res[lst[i] - 1] = i
for i in range(n - 1):
answer += abs(res[i] - res[i + 1])
return answer
m = int(input())
a = [int(j) for j in input().split()]
print(hdd(m, a))
```
| 3
|
|
431
|
A
|
Black Square
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
|
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=β€<=*a*1,<=*a*2,<=*a*3,<=*a*4<=β€<=104).
The second line contains string *s* (1<=β€<=|*s*|<=β€<=105), where the *Ρ*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
|
Print a single integer β the total number of calories that Jury wastes.
|
[
"1 2 3 4\n123214\n",
"1 5 3 2\n11221\n"
] |
[
"13\n",
"13\n"
] |
none
| 500
|
[
{
"input": "1 2 3 4\n123214",
"output": "13"
},
{
"input": "1 5 3 2\n11221",
"output": "13"
},
{
"input": "5 5 5 1\n3422",
"output": "16"
},
{
"input": "4 3 2 1\n2",
"output": "3"
},
{
"input": "5651 6882 6954 4733\n2442313421",
"output": "60055"
},
{
"input": "0 0 0 0\n4132",
"output": "0"
},
{
"input": "3163 5778 83 7640\n11141442444",
"output": "64270"
},
{
"input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442",
"output": "420780"
},
{
"input": "0 0 0 0\n1",
"output": "0"
},
{
"input": "1 2 3 4\n4",
"output": "4"
},
{
"input": "2343 7653 1242 5432\n1",
"output": "2343"
},
{
"input": "2343 7653 1242 5432\n2",
"output": "7653"
},
{
"input": "2343 7653 1242 5432\n3",
"output": "1242"
},
{
"input": "2343 7653 1242 5432\n4",
"output": "5432"
},
{
"input": "1 2 3 4\n123412",
"output": "13"
},
{
"input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111",
"output": "2650"
},
{
"input": "1 2 3 4\n11111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "1 2 3 4\n23123231321231231231231231221232123121312321",
"output": "87"
},
{
"input": "1 2 3 4\n1111111111111222222222233333333333444444444444444",
"output": "126"
},
{
"input": "2 3 1 4\n121321232412342112312313213123123412131231231232",
"output": "105"
}
] | 1,671,652,836
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 49
| 93
| 1,740,800
|
spent = 0
calories = list(map(int,input().split()))
game = input()
for strip in game:
spent += calories[int(strip)-1]
print(spent)
|
Title: Black Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
Input Specification:
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=β€<=*a*1,<=*a*2,<=*a*3,<=*a*4<=β€<=104).
The second line contains string *s* (1<=β€<=|*s*|<=β€<=105), where the *Ρ*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
Output Specification:
Print a single integer β the total number of calories that Jury wastes.
Demo Input:
['1 2 3 4\n123214\n', '1 5 3 2\n11221\n']
Demo Output:
['13\n', '13\n']
Note:
none
|
```python
spent = 0
calories = list(map(int,input().split()))
game = input()
for strip in game:
spent += calories[int(strip)-1]
print(spent)
```
| 3
|
|
998
|
B
|
Cutting
|
PROGRAMMING
| 1,200
|
[
"dp",
"greedy",
"sortings"
] | null | null |
There are a lot of things which could be cutΒ β trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.
|
First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$)Β β the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$)Β β elements of the sequence, which contains the equal number of even and odd numbers
|
Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.
|
[
"6 4\n1 2 5 10 15 20\n",
"4 10\n1 3 2 4\n",
"6 100\n1 2 3 4 5 6\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins.
| 1,000
|
[
{
"input": "6 4\n1 2 5 10 15 20",
"output": "1"
},
{
"input": "4 10\n1 3 2 4",
"output": "0"
},
{
"input": "6 100\n1 2 3 4 5 6",
"output": "2"
},
{
"input": "2 100\n13 78",
"output": "0"
},
{
"input": "10 1\n56 56 98 2 11 64 97 41 95 53",
"output": "0"
},
{
"input": "10 100\n94 65 24 47 29 98 20 65 6 17",
"output": "2"
},
{
"input": "100 1\n35 6 19 84 49 64 36 91 50 65 21 86 20 89 10 52 50 24 98 74 11 48 58 98 51 85 1 29 44 83 9 97 68 41 83 57 1 57 46 42 87 2 32 50 3 57 17 77 22 100 36 27 3 34 55 8 90 61 34 20 15 39 43 46 60 60 14 23 4 22 75 51 98 23 69 22 99 57 63 30 79 7 16 8 34 84 13 47 93 40 48 25 93 1 80 6 82 93 6 21",
"output": "0"
},
{
"input": "100 10\n3 20 3 29 90 69 2 30 70 28 71 99 22 99 34 70 87 48 3 92 71 61 26 90 14 38 51 81 16 33 49 71 14 52 50 95 65 16 80 57 87 47 29 14 40 31 74 15 87 76 71 61 30 91 44 10 87 48 84 12 77 51 25 68 49 38 79 8 7 9 39 19 48 40 15 53 29 4 60 86 76 84 6 37 45 71 46 38 80 68 94 71 64 72 41 51 71 60 79 7",
"output": "2"
},
{
"input": "100 100\n60 83 82 16 17 7 89 6 83 100 85 41 72 44 23 28 64 84 3 23 33 52 93 30 81 38 67 25 26 97 94 78 41 74 74 17 53 51 54 17 20 81 95 76 42 16 16 56 74 69 30 9 82 91 32 13 47 45 97 40 56 57 27 28 84 98 91 5 61 20 3 43 42 26 83 40 34 100 5 63 62 61 72 5 32 58 93 79 7 18 50 43 17 24 77 73 87 74 98 2",
"output": "11"
},
{
"input": "100 100\n70 54 10 72 81 84 56 15 27 19 43 100 49 44 52 33 63 40 95 17 58 2 51 39 22 18 82 1 16 99 32 29 24 94 9 98 5 37 47 14 42 73 41 31 79 64 12 6 53 26 68 67 89 13 90 4 21 93 46 74 75 88 66 57 23 7 25 48 92 62 30 8 50 61 38 87 71 34 97 28 80 11 60 91 3 35 86 96 36 20 59 65 83 45 76 77 78 69 85 55",
"output": "3"
},
{
"input": "100 100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "49"
},
{
"input": "10 10\n94 32 87 13 4 22 85 81 18 95",
"output": "1"
},
{
"input": "10 50\n40 40 9 3 64 96 67 19 21 30",
"output": "1"
},
{
"input": "100 50\n13 31 29 86 46 10 2 87 94 2 28 31 29 15 64 3 94 71 37 76 9 91 89 38 12 46 53 33 58 11 98 4 37 72 30 52 6 86 40 98 28 6 34 80 61 47 45 69 100 47 91 64 87 41 67 58 88 75 13 81 36 58 66 29 10 27 54 83 44 15 11 33 49 36 61 18 89 26 87 1 99 19 57 21 55 84 20 74 14 43 15 51 2 76 22 92 43 14 72 77",
"output": "3"
},
{
"input": "100 1\n78 52 95 76 96 49 53 59 77 100 64 11 9 48 15 17 44 46 21 54 39 68 43 4 32 28 73 6 16 62 72 84 65 86 98 75 33 45 25 3 91 82 2 92 63 88 7 50 97 93 14 22 20 42 60 55 80 85 29 34 56 71 83 38 26 47 90 70 51 41 40 31 37 12 35 99 67 94 1 87 57 8 61 19 23 79 36 18 66 74 5 27 81 69 24 58 13 10 89 30",
"output": "0"
},
{
"input": "100 10\n19 55 91 50 31 23 60 84 38 1 22 51 27 76 28 98 11 44 61 63 15 93 52 3 66 16 53 36 18 62 35 85 78 37 73 64 87 74 46 26 82 69 49 33 83 89 56 67 71 25 39 94 96 17 21 6 47 68 34 42 57 81 13 10 54 2 48 80 20 77 4 5 59 30 90 95 45 75 8 88 24 41 40 14 97 32 7 9 65 70 100 99 72 58 92 29 79 12 86 43",
"output": "0"
},
{
"input": "100 50\n2 4 82 12 47 63 52 91 87 45 53 1 17 25 64 50 9 13 22 54 21 30 43 24 38 33 68 11 41 78 99 23 28 18 58 67 79 10 71 56 49 61 26 29 59 20 90 74 5 75 89 8 39 95 72 42 66 98 44 32 88 35 92 3 97 55 65 51 77 27 81 76 84 69 73 85 19 46 62 100 60 37 7 36 57 6 14 83 40 48 16 70 96 15 31 93 80 86 94 34",
"output": "1"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "1"
},
{
"input": "100 10\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "10"
},
{
"input": "100 50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "49"
},
{
"input": "100 30\n2 1 2 2 2 2 1 1 1 2 1 1 2 2 1 2 1 2 2 2 2 1 2 1 2 1 1 2 1 1 2 2 2 1 1 2 1 2 2 2 1 1 1 1 1 2 1 1 1 1 1 2 2 2 2 1 2 1 1 1 2 2 2 2 1 2 2 1 1 1 1 2 2 2 1 2 2 1 2 1 1 2 2 2 1 2 2 1 2 1 1 2 1 1 1 1 2 1 1 2",
"output": "11"
},
{
"input": "100 80\n1 1 1 2 2 1 1 2 1 1 1 1 2 2 2 1 2 2 2 2 1 1 2 2 1 1 1 1 2 2 2 1 1 1 1 1 1 1 2 2 2 2 1 2 2 1 2 1 1 1 1 2 2 1 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 1 1 2 1 1 1 2 1 1 2 1 2 1 2 2 1 1 2 1 1 1 1 2 2 2 1 2 2 1 2",
"output": "12"
},
{
"input": "100 30\n100 99 100 99 99 100 100 99 100 99 99 100 100 100 99 99 99 100 99 99 99 99 100 99 99 100 100 99 100 99 99 99 100 99 100 100 99 100 100 100 100 100 99 99 100 99 99 100 99 100 99 99 100 100 99 100 99 99 100 99 100 100 100 100 99 99 99 100 99 100 99 100 100 100 99 100 100 100 99 100 99 99 100 100 100 100 99 99 99 100 99 100 100 99 99 99 100 100 99 99",
"output": "14"
},
{
"input": "100 80\n99 100 100 100 99 99 99 99 100 99 99 99 99 99 99 99 99 100 100 99 99 99 99 99 100 99 100 99 100 100 100 100 100 99 100 100 99 99 100 100 100 100 100 99 100 99 100 99 99 99 100 99 99 99 99 99 99 99 99 100 99 100 100 99 99 99 99 100 100 100 99 100 100 100 100 100 99 100 100 100 100 100 100 100 100 99 99 99 99 100 99 100 100 100 100 100 99 100 99 100",
"output": "4"
},
{
"input": "100 30\n100 100 39 39 39 100 100 39 39 100 39 39 100 39 100 39 100 100 100 100 100 39 100 100 100 39 39 39 100 39 100 100 39 39 100 39 39 39 100 100 39 100 39 100 39 39 100 100 39 100 39 100 39 39 39 100 39 100 39 39 39 100 39 39 100 100 39 39 39 100 100 39 39 39 100 100 100 100 39 100 100 100 39 39 100 39 100 100 39 100 39 100 39 39 100 39 39 100 100 100",
"output": "5"
},
{
"input": "100 80\n39 100 39 100 100 100 100 39 39 100 100 39 39 100 39 39 39 39 100 39 39 39 39 100 100 100 100 39 100 39 39 100 100 39 39 100 39 100 39 100 100 39 39 100 39 39 39 100 39 100 39 100 100 100 100 100 100 100 39 100 39 100 100 100 39 39 39 39 39 100 100 100 39 100 100 100 100 39 100 100 39 39 100 39 39 39 100 39 100 39 39 100 100 39 100 39 39 39 100 39",
"output": "6"
},
{
"input": "4 1\n1 2 3 4",
"output": "1"
},
{
"input": "4 1\n1 2 1 2",
"output": "1"
},
{
"input": "4 4\n1 2 6 7",
"output": "1"
},
{
"input": "4 8\n1 2 10 11",
"output": "1"
},
{
"input": "6 2\n1 2 3 4 5 6",
"output": "2"
},
{
"input": "6 1\n1 2 1 2 1 2",
"output": "1"
},
{
"input": "6 4\n1 2 4 5 7 8",
"output": "2"
},
{
"input": "6 3\n1 2 5 10 15 20",
"output": "1"
}
] | 1,614,954,929
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 93
| 1,228,800
|
n,b=list(map(int,input().split()))
a=list(map(int,input().split()))
e=[a[i]%2 for i in range(n)]
c=[]
for i in range(1,n):
if sum(e[i:])==(n-i)/2 and sum(e[:i])==i/2:
c.append(abs(a[i]-a[i-1]))
c.sort()
i=0
d=0
while i<len(c) and d+c[i]<=b:
d+=c[i]
i+=1
print(i)
|
Title: Cutting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are a lot of things which could be cutΒ β trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.
Input Specification:
First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$)Β β the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$)Β β elements of the sequence, which contains the equal number of even and odd numbers
Output Specification:
Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.
Demo Input:
['6 4\n1 2 5 10 15 20\n', '4 10\n1 3 2 4\n', '6 100\n1 2 3 4 5 6\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins.
|
```python
n,b=list(map(int,input().split()))
a=list(map(int,input().split()))
e=[a[i]%2 for i in range(n)]
c=[]
for i in range(1,n):
if sum(e[i:])==(n-i)/2 and sum(e[:i])==i/2:
c.append(abs(a[i]-a[i-1]))
c.sort()
i=0
d=0
while i<len(c) and d+c[i]<=b:
d+=c[i]
i+=1
print(i)
```
| 3
|
|
978
|
C
|
Letters
|
PROGRAMMING
| 1,000
|
[
"binary search",
"implementation",
"two pointers"
] | null | null |
There are $n$ dormitories in Berland State University, they are numbered with integers from $1$ to $n$. Each dormitory consists of rooms, there are $a_i$ rooms in $i$-th dormitory. The rooms in $i$-th dormitory are numbered from $1$ to $a_i$.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $n$ dormitories is written on an envelope. In this case, assume that all the rooms are numbered from $1$ to $a_1 + a_2 + \dots + a_n$ and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case $n=2$, $a_1=3$ and $a_2=5$ an envelope can have any integer from $1$ to $8$ written on it. If the number $7$ is written on an envelope, it means that the letter should be delivered to the room number $4$ of the second dormitory.
For each of $m$ letters by the room number among all $n$ dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
|
The first line contains two integers $n$ and $m$ $(1 \le n, m \le 2 \cdot 10^{5})$ β the number of dormitories and the number of letters.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^{10})$, where $a_i$ equals to the number of rooms in the $i$-th dormitory. The third line contains a sequence $b_1, b_2, \dots, b_m$ $(1 \le b_j \le a_1 + a_2 + \dots + a_n)$, where $b_j$ equals to the room number (among all rooms of all dormitories) for the $j$-th letter. All $b_j$ are given in increasing order.
|
Print $m$ lines. For each letter print two integers $f$ and $k$ β the dormitory number $f$ $(1 \le f \le n)$ and the room number $k$ in this dormitory $(1 \le k \le a_f)$ to deliver the letter.
|
[
"3 6\n10 15 12\n1 9 12 23 26 37\n",
"2 3\n5 10000000000\n5 6 9999999999\n"
] |
[
"1 1\n1 9\n2 2\n2 13\n3 1\n3 12\n",
"1 5\n2 1\n2 9999999994\n"
] |
In the first example letters should be delivered in the following order:
- the first letter in room $1$ of the first dormitory - the second letter in room $9$ of the first dormitory - the third letter in room $2$ of the second dormitory - the fourth letter in room $13$ of the second dormitory - the fifth letter in room $1$ of the third dormitory - the sixth letter in room $12$ of the third dormitory
| 0
|
[
{
"input": "3 6\n10 15 12\n1 9 12 23 26 37",
"output": "1 1\n1 9\n2 2\n2 13\n3 1\n3 12"
},
{
"input": "2 3\n5 10000000000\n5 6 9999999999",
"output": "1 5\n2 1\n2 9999999994"
},
{
"input": "1 1\n1\n1",
"output": "1 1"
},
{
"input": "5 15\n10 20 30 20 10\n1 6 10 11 15 30 31 54 60 61 76 80 81 84 90",
"output": "1 1\n1 6\n1 10\n2 1\n2 5\n2 20\n3 1\n3 24\n3 30\n4 1\n4 16\n4 20\n5 1\n5 4\n5 10"
},
{
"input": "1 10\n10\n1 2 3 4 5 6 7 8 9 10",
"output": "1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10"
},
{
"input": "5 8\n10 1 1 1 10\n9 10 11 12 13 14 15 23",
"output": "1 9\n1 10\n2 1\n3 1\n4 1\n5 1\n5 2\n5 10"
},
{
"input": "1 3\n10000\n1 4325 10000",
"output": "1 1\n1 4325\n1 10000"
},
{
"input": "4 18\n5 6 3 4\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18",
"output": "1 1\n1 2\n1 3\n1 4\n1 5\n2 1\n2 2\n2 3\n2 4\n2 5\n2 6\n3 1\n3 2\n3 3\n4 1\n4 2\n4 3\n4 4"
},
{
"input": "3 10\n1000000000 1000000000 1000000000\n543678543 567869543 1000000000 1000000001 1500000000 1999999999 2000000000 2000000001 2754432345 3000000000",
"output": "1 543678543\n1 567869543\n1 1000000000\n2 1\n2 500000000\n2 999999999\n2 1000000000\n3 1\n3 754432345\n3 1000000000"
}
] | 1,677,602,573
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 16
| 342
| 36,864,000
|
n, m = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = 0
k = a[c]
for i in b:
while i > k:
c += 1
k += a[c]
print(c + 1,i + a[c] - k)
|
Title: Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ dormitories in Berland State University, they are numbered with integers from $1$ to $n$. Each dormitory consists of rooms, there are $a_i$ rooms in $i$-th dormitory. The rooms in $i$-th dormitory are numbered from $1$ to $a_i$.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $n$ dormitories is written on an envelope. In this case, assume that all the rooms are numbered from $1$ to $a_1 + a_2 + \dots + a_n$ and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case $n=2$, $a_1=3$ and $a_2=5$ an envelope can have any integer from $1$ to $8$ written on it. If the number $7$ is written on an envelope, it means that the letter should be delivered to the room number $4$ of the second dormitory.
For each of $m$ letters by the room number among all $n$ dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
Input Specification:
The first line contains two integers $n$ and $m$ $(1 \le n, m \le 2 \cdot 10^{5})$ β the number of dormitories and the number of letters.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^{10})$, where $a_i$ equals to the number of rooms in the $i$-th dormitory. The third line contains a sequence $b_1, b_2, \dots, b_m$ $(1 \le b_j \le a_1 + a_2 + \dots + a_n)$, where $b_j$ equals to the room number (among all rooms of all dormitories) for the $j$-th letter. All $b_j$ are given in increasing order.
Output Specification:
Print $m$ lines. For each letter print two integers $f$ and $k$ β the dormitory number $f$ $(1 \le f \le n)$ and the room number $k$ in this dormitory $(1 \le k \le a_f)$ to deliver the letter.
Demo Input:
['3 6\n10 15 12\n1 9 12 23 26 37\n', '2 3\n5 10000000000\n5 6 9999999999\n']
Demo Output:
['1 1\n1 9\n2 2\n2 13\n3 1\n3 12\n', '1 5\n2 1\n2 9999999994\n']
Note:
In the first example letters should be delivered in the following order:
- the first letter in room $1$ of the first dormitory - the second letter in room $9$ of the first dormitory - the third letter in room $2$ of the second dormitory - the fourth letter in room $13$ of the second dormitory - the fifth letter in room $1$ of the third dormitory - the sixth letter in room $12$ of the third dormitory
|
```python
n, m = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = 0
k = a[c]
for i in b:
while i > k:
c += 1
k += a[c]
print(c + 1,i + a[c] - k)
```
| 3
|
|
535
|
B
|
Tavas and SaDDas
|
PROGRAMMING
| 1,100
|
[
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null |
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
|
The first and only line of input contains a lucky number *n* (1<=β€<=*n*<=β€<=109).
|
Print the index of *n* among all lucky numbers.
|
[
"4\n",
"7\n",
"77\n"
] |
[
"1\n",
"2\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"output": "8"
},
{
"input": "774",
"output": "13"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447747",
"output": "140"
},
{
"input": "7747474",
"output": "233"
},
{
"input": "4444",
"output": "15"
},
{
"input": "4447",
"output": "16"
},
{
"input": "7",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447744",
"output": "139"
},
{
"input": "77474",
"output": "57"
},
{
"input": "7747",
"output": "28"
},
{
"input": "444",
"output": "7"
},
{
"input": "7",
"output": "2"
},
{
"input": "7774477",
"output": "242"
},
{
"input": "4477774",
"output": "157"
},
{
"input": "7444",
"output": "23"
},
{
"input": "7474747",
"output": "212"
},
{
"input": "77",
"output": "6"
},
{
"input": "774477",
"output": "114"
},
{
"input": "7",
"output": "2"
},
{
"input": "47",
"output": "4"
},
{
"input": "747777",
"output": "110"
},
{
"input": "444444444",
"output": "511"
},
{
"input": "777777777",
"output": "1022"
},
{
"input": "477477447",
"output": "728"
},
{
"input": "777744747",
"output": "996"
}
] | 1,668,208,976
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 46
| 0
|
# lucky numbers
number = input()
lenNumber = len(number)
count = 0
for i in range(lenNumber):
if number[i] == '7':
count += 2**(lenNumber-i)
else:
count += 2**(lenNumber-i-1)
print(count)
|
Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=β€<=*n*<=β€<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none
|
```python
# lucky numbers
number = input()
lenNumber = len(number)
count = 0
for i in range(lenNumber):
if number[i] == '7':
count += 2**(lenNumber-i)
else:
count += 2**(lenNumber-i-1)
print(count)
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
|
Output one number β 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"
}
] | 1,684,428,046
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 124
| 0
|
a = [int(i) for i in input().split()]
h = 1
for i in a:
h *= i
print(h//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
Output Specification:
Output one number β 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
a = [int(i) for i in input().split()]
h = 1
for i in a:
h *= i
print(h//2)
```
| 3.969
|
676
|
C
|
Vasya and String
|
PROGRAMMING
| 1,500
|
[
"binary search",
"dp",
"strings",
"two pointers"
] | null | null |
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
|
The first line of the input contains two integers *n* and *k* (1<=β€<=*n*<=β€<=100<=000,<=0<=β€<=*k*<=β€<=*n*)Β β the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
|
Print the only integerΒ β the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
|
[
"4 2\nabba\n",
"8 1\naabaabaa\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
| 1,500
|
[
{
"input": "4 2\nabba",
"output": "4"
},
{
"input": "8 1\naabaabaa",
"output": "5"
},
{
"input": "1 0\na",
"output": "1"
},
{
"input": "1 1\nb",
"output": "1"
},
{
"input": "1 0\nb",
"output": "1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "10 10\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 2\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 1\nbbabbabbba",
"output": "6"
},
{
"input": "10 10\nbbabbbaabb",
"output": "10"
},
{
"input": "10 9\nbabababbba",
"output": "10"
},
{
"input": "10 4\nbababbaaab",
"output": "9"
},
{
"input": "10 10\naabaaabaaa",
"output": "10"
},
{
"input": "10 10\naaaabbbaaa",
"output": "10"
},
{
"input": "10 1\nbaaaaaaaab",
"output": "9"
},
{
"input": "10 5\naaaaabaaaa",
"output": "10"
},
{
"input": "10 4\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 10\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 7\nbbbbabbbbbaabbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbab",
"output": "93"
},
{
"input": "100 30\nbbaabaaabbbbbbbbbbaababababbbbbbaabaabbbbbbbbabbbbbabbbbabbbbbbbbaabbbbbbbbbabbbbbabbbbbbbbbaaaaabba",
"output": "100"
},
{
"input": "100 6\nbaababbbaabbabbaaabbabbaabbbbbbbbaabbbabbbbaabbabbbbbabababbbbabbbbbbabbbbbbbbbaaaabbabbbbaabbabaabb",
"output": "34"
},
{
"input": "100 45\naabababbabbbaaabbbbbbaabbbabbaabbbbbabbbbbbbbabbbbbbabbaababbaabbababbbbbbababbbbbaabbbbbbbaaaababab",
"output": "100"
},
{
"input": "100 2\nababaabababaaababbaaaabbaabbbababbbaaabbbbabababbbabababaababaaabaabbbbaaabbbabbbbbabbbbbbbaabbabbba",
"output": "17"
},
{
"input": "100 25\nbabbbaaababaaabbbaabaabaabbbabbabbbbaaaaaaabaaabaaaaaaaaaabaaaabaaabbbaaabaaababaaabaabbbbaaaaaaaaaa",
"output": "80"
},
{
"input": "100 14\naabaaaaabababbabbabaaaabbaaaabaaabbbaaabaaaaaaaabaaaaabbaaaaaaaaabaaaaaaabbaababaaaababbbbbabaaaabaa",
"output": "61"
},
{
"input": "100 8\naaaaabaaaabaabaaaaaaaabaaaabaaaaaaaaaaaaaabaaaaabaaaaaaaaaaaaaaaaabaaaababaabaaaaaaaaaaaaabbabaaaaaa",
"output": "76"
},
{
"input": "100 12\naaaaaaaaaaaaaaaabaaabaaaaaaaaaabbaaaabbabaaaaaaaaaaaaaaaaaaaaabbaaabaaaaaaaaaaaabaaaaaaaabaaaaaaaaaa",
"output": "100"
},
{
"input": "100 65\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
},
{
"input": "10 0\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 0\nbbbbabbbbb",
"output": "5"
},
{
"input": "10 0\nbbabbbabba",
"output": "3"
},
{
"input": "10 0\nbaabbbbaba",
"output": "4"
},
{
"input": "10 0\naababbbbaa",
"output": "4"
},
{
"input": "10 2\nabbbbbaaba",
"output": "8"
},
{
"input": "10 0\nabbaaabaaa",
"output": "3"
},
{
"input": "10 0\naabbaaabaa",
"output": "3"
},
{
"input": "10 1\naaaaaababa",
"output": "8"
},
{
"input": "10 0\nbaaaaaaaaa",
"output": "9"
},
{
"input": "10 0\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 0\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 0\nbbbbbbbbbbabbbbaaabbbbbbbbbbbabbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbabbbbbbbbbbbbbab",
"output": "40"
},
{
"input": "100 11\nbaabbbbbababbbbabbbbbbbabbbbbbbbbbbbbbabbbbbbababbbbababbbbaaabbbbabbbbbabbbbbbbbabababbbabbbbbbbabb",
"output": "65"
},
{
"input": "100 8\nbbababbbbbaabbbaaababbbbababababbbbababbabbbabbbbbaabbbabbbababbabbbbabbbabbbbaabbbbabbbaabbbbaaaabb",
"output": "33"
},
{
"input": "100 21\nabbaaaabbbababaabbbababbbbbbbbabbaababababbbabbbaaabbaaabbbbabbabbbabbbabaababbbabbbbbabbbbbbabbbbab",
"output": "65"
},
{
"input": "100 9\nabbbaabaabaaaaaaabbabbbababbaaabbbaaabbaabaaaaabbbbbabbaabaabbbbbaaaaababbaaabbabaabaaabababbaababbb",
"output": "26"
},
{
"input": "100 5\naababababbaaaaaaaabbbabaaaabbabaaaabbaabaaaaabababbabaabaaabaaaaaaaabaababbabbaaabaabbabbaaaaabbabba",
"output": "22"
},
{
"input": "100 9\naababaabaaaaaaaaabbbaabaaaaaaabaaaaaaaaaaaaabaaabaabaabbbbabbaababbabbaaaabbababaabaababaabaaaaaaaaa",
"output": "49"
},
{
"input": "100 6\naaaaabbaaaaaaaaaaabaaaabaaaaaaaaabaaabaaaaaabaaaaaaaaaaabaabaaaabaaaaaaaaaaaaaaabaabbaaaaaaaaaaaaaaa",
"output": "56"
},
{
"input": "100 7\nabaaabaabaabaaaaaabaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaabaaaaaaabbabaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaba",
"output": "86"
},
{
"input": "100 0\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
}
] | 1,587,789,559
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 117
| 171
| 2,764,800
|
n,k=map(int,input().split())
st=0
end=1
s=input()
l=[0]*(n+1)
for i in range(len(s)):
l[i+1]=1 if s[i]=='a' else 0
#print(l)
su=0
m=0
while(end<n+1):
su+=l[end]
if su>k and end-st-su>k:
st+=1
su-=l[st]
m=max(m,st-end)
end+=1
#print(end,st)
m=max(m,end-st-1)
print(m)
|
Title: Vasya and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=β€<=*n*<=β€<=100<=000,<=0<=β€<=*k*<=β€<=*n*)Β β the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
Output Specification:
Print the only integerΒ β the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
Demo Input:
['4 2\nabba\n', '8 1\naabaabaa\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
|
```python
n,k=map(int,input().split())
st=0
end=1
s=input()
l=[0]*(n+1)
for i in range(len(s)):
l[i+1]=1 if s[i]=='a' else 0
#print(l)
su=0
m=0
while(end<n+1):
su+=l[end]
if su>k and end-st-su>k:
st+=1
su-=l[st]
m=max(m,st-end)
end+=1
#print(end,st)
m=max(m,end-st-1)
print(m)
```
| 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" β 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 "
}
] | 1,678,829,310
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
n = input()
x = n.replace('WUB', ' ').strip().replace(' ', ' ')
print(x)
|
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" β between them Vasya added two "WUB".
|
```python
n = input()
x = n.replace('WUB', ' ').strip().replace(' ', ' ')
print(x)
```
| 3
|
|
895
|
A
|
Pizza Separation
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=360) Β β the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=360) Β β the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
|
Print one integer Β β the minimal difference between angles of sectors that will go to Vasya and Petya.
|
[
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] |
[
"0\n",
"40\n",
"360\n",
"0\n"
] |
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90β+β90)β-β(90β+β90)|β=β0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360β-β0|β=β360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170β+β10)β-β(30β+β150)|β=β0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector.
| 500
|
[
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 45 45 45 45 45 45 45",
"output": "0"
},
{
"input": "3\n120 120 120",
"output": "120"
},
{
"input": "5\n110 90 70 50 40",
"output": "40"
},
{
"input": "2\n170 190",
"output": "20"
},
{
"input": "15\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 10",
"output": "10"
},
{
"input": "5\n30 60 180 60 30",
"output": "0"
},
{
"input": "2\n359 1",
"output": "358"
},
{
"input": "5\n100 100 30 100 30",
"output": "40"
},
{
"input": "5\n36 34 35 11 244",
"output": "128"
},
{
"input": "5\n96 94 95 71 4",
"output": "18"
},
{
"input": "2\n85 275",
"output": "190"
},
{
"input": "3\n281 67 12",
"output": "202"
},
{
"input": "5\n211 113 25 9 2",
"output": "62"
},
{
"input": "13\n286 58 6 1 1 1 1 1 1 1 1 1 1",
"output": "212"
},
{
"input": "15\n172 69 41 67 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20\n226 96 2 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "92"
},
{
"input": "50\n148 53 32 11 4 56 8 2 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n1 1 358",
"output": "356"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 341",
"output": "322"
},
{
"input": "33\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 1 328",
"output": "296"
},
{
"input": "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 1 1 1 1 291",
"output": "222"
},
{
"input": "130\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 231",
"output": "102"
},
{
"input": "200\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 161",
"output": "0"
},
{
"input": "222\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 139",
"output": "0"
},
{
"input": "10\n8 3 11 4 1 10 10 1 8 304",
"output": "248"
},
{
"input": "12\n8 7 7 3 11 2 10 1 10 8 10 283",
"output": "206"
},
{
"input": "13\n10 8 9 10 5 9 4 1 10 11 1 7 275",
"output": "190"
},
{
"input": "14\n1 6 3 11 9 5 9 8 5 6 7 3 7 280",
"output": "200"
},
{
"input": "15\n10 11 5 4 11 5 4 1 5 4 5 5 9 6 275",
"output": "190"
},
{
"input": "30\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 189",
"output": "18"
},
{
"input": "70\n5 3 6 8 9 2 8 9 11 5 2 8 9 11 7 6 6 9 7 11 7 6 3 8 2 4 4 8 4 3 2 2 3 5 6 5 11 2 7 7 5 8 10 5 2 1 10 9 4 10 7 1 8 10 9 1 5 1 1 1 2 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "29\n2 10 1 5 7 2 9 11 9 9 10 8 4 11 2 5 4 1 4 9 6 10 8 3 1 3 8 9 189",
"output": "18"
},
{
"input": "35\n3 4 11 4 4 2 3 4 3 9 7 10 2 7 8 3 11 3 6 4 6 7 11 10 8 7 6 7 2 8 5 3 2 2 168",
"output": "0"
},
{
"input": "60\n4 10 3 10 6 3 11 8 11 9 3 5 9 2 6 5 6 9 4 10 1 1 3 7 2 10 5 5 3 10 5 2 1 2 9 11 11 9 11 4 11 7 5 6 10 9 3 4 7 8 7 3 6 7 8 5 1 1 1 5",
"output": "0"
},
{
"input": "71\n3 11 8 1 10 1 7 9 6 4 11 10 11 2 4 1 11 7 9 10 11 4 8 7 11 3 8 4 1 8 4 2 9 9 7 10 10 9 5 7 9 7 2 1 7 6 5 11 5 9 4 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "63\n2 11 5 8 7 9 9 8 10 5 9 10 11 8 10 2 3 5 3 7 5 10 2 9 4 8 1 8 5 9 7 7 1 8 7 7 9 10 10 10 8 7 7 2 2 8 9 7 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "81\n5 8 7 11 2 7 1 1 5 8 7 2 3 11 4 9 7 6 4 4 2 1 1 7 9 4 1 8 3 1 4 10 7 9 9 8 11 3 4 3 10 8 6 4 7 2 4 3 6 11 11 10 7 10 2 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "47\n5 3 7 4 2 7 8 1 9 10 5 11 10 7 7 5 1 3 2 11 3 8 6 1 6 10 8 3 2 10 5 6 8 6 9 7 10 9 7 4 8 11 10 1 5 11 68",
"output": "0"
},
{
"input": "100\n5 8 9 3 2 3 9 8 11 10 4 8 1 1 1 1 6 5 10 9 5 3 7 7 2 11 10 2 3 2 2 8 7 3 5 5 10 9 2 5 10 6 7 7 4 7 7 8 2 8 9 9 2 4 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "120\n9 11 3 7 3 7 9 1 10 7 11 4 1 5 3 5 6 3 1 11 8 8 11 7 3 5 1 9 1 7 10 10 10 10 9 5 4 8 2 8 2 1 4 5 3 11 3 5 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",
"output": "0"
},
{
"input": "200\n7 7 9 8 2 8 5 8 3 9 7 10 2 9 11 8 11 7 5 2 6 3 11 9 5 1 10 2 1 2 2 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "220\n3 2 8 1 3 5 5 11 1 5 2 6 9 2 2 6 8 10 7 1 3 2 10 9 10 10 4 10 9 5 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "6\n27 15 28 34 41 215",
"output": "70"
},
{
"input": "7\n41 38 41 31 22 41 146",
"output": "14"
},
{
"input": "8\n24 27 34 23 29 23 30 170",
"output": "20"
},
{
"input": "9\n11 11 20 20 33 32 35 26 172",
"output": "6"
},
{
"input": "10\n36 13 28 13 33 34 23 25 34 121",
"output": "0"
},
{
"input": "11\n19 37 13 41 37 15 32 12 19 35 100",
"output": "10"
},
{
"input": "12\n37 25 34 38 21 24 34 38 11 29 28 41",
"output": "2"
},
{
"input": "13\n24 40 20 26 25 29 39 29 35 28 19 18 28",
"output": "2"
},
{
"input": "14\n11 21 40 19 28 34 13 16 23 30 34 22 25 44",
"output": "4"
},
{
"input": "3\n95 91 174",
"output": "12"
},
{
"input": "4\n82 75 78 125",
"output": "46"
},
{
"input": "6\n87 75 88 94 15 1",
"output": "4"
},
{
"input": "10\n27 52 58 64 45 64 1 19 2 28",
"output": "12"
},
{
"input": "50\n14 12 11 8 1 6 11 6 7 8 4 11 4 5 7 3 5 4 7 24 10 2 3 4 6 13 2 1 8 7 5 13 10 8 5 20 1 2 23 7 14 3 4 4 2 8 8 2 6 1",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 6 3 2 8 2 13 3 1 1 2 1 3 4 1 7 1 2 2 6 3 2 10 3 1 2 5 6 2 3 3 2 3 11 8 3 2 6 1 3 3 4 7 7 2 2 1 2 6 3 3 2 3 1 3 8 2 6 4 2 1 12 2 2 2 1 4 1 4 1 3 1 3 1 5 2 6 6 7 1 2 3 2 4 4 2 5 9 8 2 4 6 5 1 1 3",
"output": "0"
},
{
"input": "150\n1 5 1 2 2 2 1 4 2 2 2 3 1 2 1 2 2 2 2 1 2 2 2 1 5 3 4 1 3 4 5 2 4 2 1 2 2 1 1 2 3 2 4 2 2 3 3 1 1 5 2 3 2 1 9 2 1 1 2 1 4 1 1 3 2 2 2 1 2 2 2 1 3 3 4 2 2 1 3 3 3 1 4 3 4 1 2 2 1 1 1 2 2 5 4 1 1 1 2 1 2 3 2 2 6 3 3 3 1 2 1 1 2 8 2 2 4 3 4 5 3 1 4 2 2 2 2 1 4 4 1 1 2 2 4 9 6 3 1 1 2 1 3 4 1 3 2 2 2 1",
"output": "0"
},
{
"input": "200\n1 2 1 3 1 3 1 2 1 4 6 1 2 2 2 2 1 1 1 1 3 2 1 2 2 2 1 2 2 2 2 1 1 1 3 2 3 1 1 2 1 1 2 1 1 1 1 1 1 2 1 2 2 4 1 3 1 2 1 2 2 1 2 1 3 1 1 2 2 1 1 1 1 2 4 1 2 1 1 1 2 1 3 1 1 3 1 2 2 4 1 1 2 1 2 1 2 2 2 2 1 1 2 1 2 1 3 3 1 1 1 2 1 3 3 1 2 1 3 1 3 3 1 2 2 1 4 1 2 2 1 2 2 4 2 5 1 2 2 1 2 1 2 1 5 2 1 2 2 1 2 4 1 2 2 4 2 3 2 3 1 2 1 1 2 2 2 1 1 2 1 4 1 2 1 1 2 1 2 3 1 1 1 2 2 3 1 3 2 2 3 1 2 1 2 1 1 2 1 2",
"output": "0"
},
{
"input": "5\n35 80 45 100 100",
"output": "40"
},
{
"input": "4\n90 179 90 1",
"output": "2"
},
{
"input": "5\n50 50 20 160 80",
"output": "0"
},
{
"input": "5\n30 175 30 5 120",
"output": "10"
},
{
"input": "4\n170 30 10 150",
"output": "20"
},
{
"input": "6\n90 30 90 30 90 30",
"output": "60"
},
{
"input": "4\n70 80 110 100",
"output": "20"
},
{
"input": "7\n35 45 70 100 10 10 90",
"output": "0"
},
{
"input": "6\n50 90 10 90 20 100",
"output": "20"
},
{
"input": "6\n10 155 162 1 26 6",
"output": "18"
},
{
"input": "7\n80 90 80 45 10 10 45",
"output": "20"
},
{
"input": "4\n18 36 162 144",
"output": "36"
},
{
"input": "5\n20 50 50 160 80",
"output": "40"
},
{
"input": "5\n10 30 140 20 160",
"output": "0"
},
{
"input": "6\n90 80 60 50 40 40",
"output": "20"
},
{
"input": "9\n40 20 20 20 20 20 20 40 160",
"output": "40"
},
{
"input": "4\n90 54 90 126",
"output": "72"
},
{
"input": "4\n150 170 30 10",
"output": "20"
},
{
"input": "8\n130 12 13 85 41 67 5 7",
"output": "26"
},
{
"input": "7\n70 170 20 10 30 30 30",
"output": "20"
},
{
"input": "8\n100 100 50 50 15 15 15 15",
"output": "40"
},
{
"input": "4\n100 70 80 110",
"output": "20"
},
{
"input": "5\n160 130 40 20 10",
"output": "20"
},
{
"input": "4\n20 149 151 40",
"output": "22"
},
{
"input": "4\n100 10 100 150",
"output": "60"
},
{
"input": "6\n19 64 105 168 1 3",
"output": "16"
},
{
"input": "8\n10 10 70 70 90 90 10 10",
"output": "0"
}
] | 1,511,963,256
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 93
| 93
| 5,632,000
|
N = n = int(input())
sec = []
minn = 400
line = input().split()
summ = 0
i=0
while n>0:
sec.append(int(line[i]))
i += 1
n -= 1
sec.append(0)
i=0
j = len(sec)-1
while j>i:
summ += sec[j]
pit = summ
while i<j:
pit += sec[i]
sit = 360 - pit
if minn > abs(sit - pit):
minn = abs(sit - pit)
i+=1
i = 0
j -= 1
print(minn)
|
Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=360) Β β the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=360) Β β the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer Β β the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90β+β90)β-β(90β+β90)|β=β0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360β-β0|β=β360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170β+β10)β-β(30β+β150)|β=β0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector.
|
```python
N = n = int(input())
sec = []
minn = 400
line = input().split()
summ = 0
i=0
while n>0:
sec.append(int(line[i]))
i += 1
n -= 1
sec.append(0)
i=0
j = len(sec)-1
while j>i:
summ += sec[j]
pit = summ
while i<j:
pit += sec[i]
sit = 360 - pit
if minn > abs(sit - pit):
minn = abs(sit - pit)
i+=1
i = 0
j -= 1
print(minn)
```
| 3
|
|
266
|
A
|
Stones on the Table
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=50) β the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
|
Print a single integer β the answer to the problem.
|
[
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] |
[
"1\n",
"4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "4\nRBBR",
"output": "1"
},
{
"input": "5\nRGGBG",
"output": "1"
},
{
"input": "10\nGGBRBRGGRB",
"output": "2"
},
{
"input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB",
"output": "18"
},
{
"input": "15\nBRRBRGGBBRRRRGR",
"output": "6"
},
{
"input": "20\nRRGBBRBRGRGBBGGRGRRR",
"output": "6"
},
{
"input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB",
"output": "6"
},
{
"input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB",
"output": "9"
},
{
"input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB",
"output": "14"
},
{
"input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG",
"output": "20"
},
{
"input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG",
"output": "11"
},
{
"input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG",
"output": "17"
},
{
"input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB",
"output": "16"
},
{
"input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG",
"output": "19"
},
{
"input": "10\nGRRBRBRBGR",
"output": "1"
},
{
"input": "10\nBRBGBGRRBR",
"output": "1"
},
{
"input": "20\nGBGBGGRRRRGRBBGRGRGR",
"output": "5"
},
{
"input": "20\nRRGGRBBGBBRBGRRBRRBG",
"output": "6"
},
{
"input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB",
"output": "8"
},
{
"input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB",
"output": "11"
},
{
"input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB",
"output": "9"
},
{
"input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG",
"output": "13"
},
{
"input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG",
"output": "13"
},
{
"input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB",
"output": "12"
},
{
"input": "2\nBB",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "49"
},
{
"input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "47"
},
{
"input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR",
"output": "43"
}
] | 1,699,118,337
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 124
| 0
|
a = int(input())
b = input()
k = 0
for i in range(len(b)-1):
if b[i] == b[i+1]:
k += 1
print(k)
|
Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=50) β the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer β the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none
|
```python
a = int(input())
b = input()
k = 0
for i in range(len(b)-1):
if b[i] == b[i+1]:
k += 1
print(k)
```
| 3
|
|
16
|
B
|
Burglar and Matches
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] |
B. Burglar and Matches
|
0
|
64
|
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
|
The first line of the input contains integer *n* (1<=β€<=*n*<=β€<=2Β·108) and integer *m* (1<=β€<=*m*<=β€<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=β€<=*a**i*<=β€<=108,<=1<=β€<=*b**i*<=β€<=10). All the input numbers are integer.
|
Output the only number β answer to the problem.
|
[
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] |
[
"62\n",
"7\n"
] |
none
| 0
|
[
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"output": "10"
},
{
"input": "2 1\n2 1",
"output": "2"
},
{
"input": "2 2\n2 4\n1 4",
"output": "8"
},
{
"input": "2 3\n1 7\n1 2\n1 5",
"output": "12"
},
{
"input": "4 1\n2 2",
"output": "4"
},
{
"input": "4 2\n1 10\n4 4",
"output": "22"
},
{
"input": "4 3\n1 4\n6 4\n1 7",
"output": "19"
},
{
"input": "5 1\n10 5",
"output": "25"
},
{
"input": "5 2\n3 9\n2 2",
"output": "31"
},
{
"input": "5 5\n2 9\n3 1\n2 1\n1 8\n2 8",
"output": "42"
},
{
"input": "5 10\n1 3\n1 2\n1 9\n1 10\n1 1\n1 5\n1 10\n1 2\n1 3\n1 7",
"output": "41"
},
{
"input": "10 1\n9 4",
"output": "36"
},
{
"input": "10 2\n14 3\n1 3",
"output": "30"
},
{
"input": "10 7\n4 8\n1 10\n1 10\n1 2\n3 3\n1 3\n1 10",
"output": "71"
},
{
"input": "10 10\n1 8\n2 10\n1 9\n1 1\n1 9\n1 6\n1 4\n2 5\n1 2\n1 4",
"output": "70"
},
{
"input": "10 4\n1 5\n5 2\n1 9\n3 3",
"output": "33"
},
{
"input": "100 5\n78 6\n29 10\n3 6\n7 3\n2 4",
"output": "716"
},
{
"input": "1000 7\n102 10\n23 6\n79 4\n48 1\n34 10\n839 8\n38 4",
"output": "8218"
},
{
"input": "10000 10\n336 2\n2782 5\n430 10\n1893 7\n3989 10\n2593 8\n165 6\n1029 2\n2097 4\n178 10",
"output": "84715"
},
{
"input": "100000 3\n2975 2\n35046 4\n61979 9",
"output": "703945"
},
{
"input": "1000000 4\n314183 9\n304213 4\n16864 5\n641358 9",
"output": "8794569"
},
{
"input": "10000000 10\n360313 10\n416076 1\n435445 9\n940322 7\n1647581 7\n4356968 10\n3589256 2\n2967933 5\n2747504 7\n1151633 3",
"output": "85022733"
},
{
"input": "100000000 7\n32844337 7\n11210848 7\n47655987 1\n33900472 4\n9174763 2\n32228738 10\n29947408 5",
"output": "749254060"
},
{
"input": "200000000 10\n27953106 7\n43325979 4\n4709522 1\n10975786 4\n67786538 8\n48901838 7\n15606185 6\n2747583 1\n100000000 1\n633331 3",
"output": "1332923354"
},
{
"input": "200000000 9\n17463897 9\n79520463 1\n162407 4\n41017993 8\n71054118 4\n9447587 2\n5298038 9\n3674560 7\n20539314 5",
"output": "996523209"
},
{
"input": "200000000 8\n6312706 6\n2920548 2\n16843192 3\n1501141 2\n13394704 6\n10047725 10\n4547663 6\n54268518 6",
"output": "630991750"
},
{
"input": "200000000 7\n25621043 2\n21865270 1\n28833034 1\n22185073 5\n100000000 2\n13891017 9\n61298710 8",
"output": "931584598"
},
{
"input": "200000000 6\n7465600 6\n8453505 10\n4572014 8\n8899499 3\n86805622 10\n64439238 6",
"output": "1447294907"
},
{
"input": "200000000 5\n44608415 6\n100000000 9\n51483223 9\n44136047 1\n52718517 1",
"output": "1634907859"
},
{
"input": "200000000 4\n37758556 10\n100000000 6\n48268521 3\n20148178 10",
"output": "1305347138"
},
{
"input": "200000000 3\n65170000 7\n20790088 1\n74616133 4",
"output": "775444620"
},
{
"input": "200000000 2\n11823018 6\n100000000 9",
"output": "970938108"
},
{
"input": "200000000 1\n100000000 6",
"output": "600000000"
},
{
"input": "200000000 10\n12097724 9\n41745972 5\n26982098 9\n14916995 7\n21549986 7\n3786630 9\n8050858 7\n27994924 4\n18345001 5\n8435339 5",
"output": "1152034197"
},
{
"input": "200000000 10\n55649 8\n10980981 9\n3192542 8\n94994808 4\n3626106 1\n100000000 6\n5260110 9\n4121453 2\n15125061 4\n669569 6",
"output": "1095537357"
},
{
"input": "10 20\n1 7\n1 7\n1 8\n1 3\n1 10\n1 7\n1 7\n1 9\n1 3\n1 1\n1 2\n1 1\n1 3\n1 10\n1 9\n1 8\n1 8\n1 6\n1 7\n1 5",
"output": "83"
},
{
"input": "10000000 20\n4594 7\n520836 8\n294766 6\n298672 4\n142253 6\n450626 1\n1920034 9\n58282 4\n1043204 1\n683045 1\n1491746 5\n58420 4\n451217 2\n129423 4\n246113 5\n190612 8\n912923 6\n473153 6\n783733 6\n282411 10",
"output": "54980855"
},
{
"input": "200000000 20\n15450824 5\n839717 10\n260084 8\n1140850 8\n28744 6\n675318 3\n25161 2\n5487 3\n6537698 9\n100000000 5\n7646970 9\n16489 6\n24627 3\n1009409 5\n22455 1\n25488456 4\n484528 9\n32663641 3\n750968 4\n5152 6",
"output": "939368573"
},
{
"input": "200000000 20\n16896 2\n113 3\n277 2\n299 7\n69383562 2\n3929 8\n499366 4\n771846 5\n9 4\n1278173 7\n90 2\n54 7\n72199858 10\n17214 5\n3 10\n1981618 3\n3728 2\n141 8\n2013578 9\n51829246 5",
"output": "1158946383"
},
{
"input": "200000000 20\n983125 2\n7453215 9\n9193588 2\n11558049 7\n28666199 1\n34362244 1\n5241493 5\n15451270 4\n19945845 8\n6208681 3\n38300385 7\n6441209 8\n21046742 7\n577198 10\n3826434 8\n9764276 8\n6264675 7\n8567063 3\n3610303 4\n2908232 3",
"output": "1131379312"
},
{
"input": "10 15\n1 6\n2 6\n3 4\n1 3\n1 2\n1 5\n1 6\n1 2\n2 9\n1 10\n1 3\n1 7\n1 8\n1 2\n2 9",
"output": "79"
},
{
"input": "10000000 15\n111 5\n914124 3\n3 9\n177790 1\n2352 3\n32138 9\n104477 1\n1223 4\n18 6\n6655580 4\n57643 10\n94309 2\n37 1\n227002 10\n1733193 7",
"output": "45116295"
},
{
"input": "200000000 15\n7069868 1\n5567826 8\n2310059 10\n13539782 7\n38420939 4\n29911411 8\n52256316 1\n12265839 9\n2074265 1\n24896428 9\n72470695 5\n3236301 1\n3890243 2\n65168965 8\n65724 6",
"output": "1489289257"
},
{
"input": "200000000 15\n12044094 7\n2475138 10\n944451 7\n4854766 2\n3809145 10\n7727571 2\n43908937 6\n2745883 1\n427511 2\n100000000 5\n190914 6\n554889 3\n288798 4\n1848572 5\n893874 3",
"output": "961871671"
},
{
"input": "200000000 15\n6334191 7\n1927941 4\n5175933 10\n468389 1\n433043 10\n6863198 5\n7480646 4\n14774279 10\n2921129 8\n18325627 7\n6973152 9\n8277324 9\n21522856 2\n2058070 1\n2444742 4",
"output": "664376069"
}
] | 1,598,697,213
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 218
| 307,200
|
n, m = map(int, input().split())
info = []
for _ in range(m):
info.append(list(map(int, input().split())))
info_sorted = info.copy()
for i in range(m-1):
for j in range(m)[i+1:]:
if info_sorted[i][1]<info_sorted[j][1]:
info_sorted[j],info_sorted[i] = info_sorted[i],info_sorted[j]
sum = 0
cnt = 0
for a,b in info_sorted:
sum=sum+a
if sum<=n:
cnt = cnt + a*b
else:
cnt = cnt + (n-(sum-a))*b
break
print(cnt)
|
Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=β€<=*n*<=β€<=2Β·108) and integer *m* (1<=β€<=*m*<=β€<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=β€<=*a**i*<=β€<=108,<=1<=β€<=*b**i*<=β€<=10). All the input numbers are integer.
Output Specification:
Output the only number β answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none
|
```python
n, m = map(int, input().split())
info = []
for _ in range(m):
info.append(list(map(int, input().split())))
info_sorted = info.copy()
for i in range(m-1):
for j in range(m)[i+1:]:
if info_sorted[i][1]<info_sorted[j][1]:
info_sorted[j],info_sorted[i] = info_sorted[i],info_sorted[j]
sum = 0
cnt = 0
for a,b in info_sorted:
sum=sum+a
if sum<=n:
cnt = cnt + a*b
else:
cnt = cnt + (n-(sum-a))*b
break
print(cnt)
```
| 3
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,669,525,996
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 62
| 0
|
def reverse(s):
s = s[::-1]
return s
a=input()
b=input()
if reverse(b)==a:
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
def reverse(s):
s = s[::-1]
return s
a=input()
b=input()
if reverse(b)==a:
print("YES")
else:
print("NO")
```
| 3.9845
|
910
|
B
|
Door Frames
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1<=000) β the length of each wooden bar.
The second line contains a single integer *a* (1<=β€<=*a*<=β€<=*n*) β the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=β€<=*b*<=β€<=*n*) β the length of the upper side of a door frame.
|
Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors.
|
[
"8\n1\n2\n",
"5\n3\n4\n",
"6\n4\n2\n",
"20\n5\n6\n"
] |
[
"1\n",
"6\n",
"4\n",
"2\n"
] |
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
| 1,000
|
[
{
"input": "8\n1\n2",
"output": "1"
},
{
"input": "5\n3\n4",
"output": "6"
},
{
"input": "6\n4\n2",
"output": "4"
},
{
"input": "20\n5\n6",
"output": "2"
},
{
"input": "1\n1\n1",
"output": "6"
},
{
"input": "3\n1\n2",
"output": "3"
},
{
"input": "3\n2\n1",
"output": "4"
},
{
"input": "1000\n1\n1",
"output": "1"
},
{
"input": "1000\n1000\n1000",
"output": "6"
},
{
"input": "1000\n1\n999",
"output": "3"
},
{
"input": "1000\n1\n498",
"output": "1"
},
{
"input": "1000\n1\n998",
"output": "2"
},
{
"input": "31\n5\n6",
"output": "2"
},
{
"input": "400\n100\n2",
"output": "2"
},
{
"input": "399\n100\n2",
"output": "2"
},
{
"input": "800\n401\n400",
"output": "5"
},
{
"input": "141\n26\n11",
"output": "1"
},
{
"input": "717\n40\n489",
"output": "2"
},
{
"input": "293\n47\n30",
"output": "1"
},
{
"input": "165\n59\n40",
"output": "2"
},
{
"input": "404\n5\n183",
"output": "1"
},
{
"input": "828\n468\n726",
"output": "6"
},
{
"input": "956\n153\n941",
"output": "3"
},
{
"input": "676\n175\n514",
"output": "4"
},
{
"input": "296\n1\n10",
"output": "1"
},
{
"input": "872\n3\n182",
"output": "1"
},
{
"input": "448\n15\n126",
"output": "1"
},
{
"input": "24\n2\n5",
"output": "1"
},
{
"input": "289\n56\n26",
"output": "1"
},
{
"input": "713\n150\n591",
"output": "3"
},
{
"input": "841\n62\n704",
"output": "2"
},
{
"input": "266\n38\n164",
"output": "2"
},
{
"input": "156\n34\n7",
"output": "1"
},
{
"input": "28\n14\n9",
"output": "3"
},
{
"input": "604\n356\n239",
"output": "4"
},
{
"input": "180\n18\n76",
"output": "2"
},
{
"input": "879\n545\n607",
"output": "6"
},
{
"input": "599\n160\n520",
"output": "4"
},
{
"input": "727\n147\n693",
"output": "3"
},
{
"input": "151\n27\n135",
"output": "3"
},
{
"input": "504\n71\n73",
"output": "1"
},
{
"input": "80\n57\n31",
"output": "5"
},
{
"input": "951\n225\n352",
"output": "2"
},
{
"input": "823\n168\n141",
"output": "2"
},
{
"input": "956\n582\n931",
"output": "6"
},
{
"input": "380\n108\n356",
"output": "4"
},
{
"input": "804\n166\n472",
"output": "2"
},
{
"input": "228\n12\n159",
"output": "2"
},
{
"input": "380\n126\n82",
"output": "2"
},
{
"input": "252\n52\n178",
"output": "3"
},
{
"input": "828\n363\n56",
"output": "2"
},
{
"input": "404\n122\n36",
"output": "2"
},
{
"input": "314\n4\n237",
"output": "2"
},
{
"input": "34\n5\n17",
"output": "2"
},
{
"input": "162\n105\n160",
"output": "6"
},
{
"input": "586\n22\n272",
"output": "2"
},
{
"input": "32\n9\n2",
"output": "2"
},
{
"input": "904\n409\n228",
"output": "3"
},
{
"input": "480\n283\n191",
"output": "4"
},
{
"input": "56\n37\n10",
"output": "4"
},
{
"input": "429\n223\n170",
"output": "4"
},
{
"input": "149\n124\n129",
"output": "6"
},
{
"input": "277\n173\n241",
"output": "6"
},
{
"input": "701\n211\n501",
"output": "4"
},
{
"input": "172\n144\n42",
"output": "5"
},
{
"input": "748\n549\n256",
"output": "5"
},
{
"input": "324\n284\n26",
"output": "4"
},
{
"input": "900\n527\n298",
"output": "4"
},
{
"input": "648\n624\n384",
"output": "6"
},
{
"input": "72\n48\n54",
"output": "6"
},
{
"input": "200\n194\n87",
"output": "5"
},
{
"input": "624\n510\n555",
"output": "6"
},
{
"input": "17\n16\n2",
"output": "5"
},
{
"input": "593\n442\n112",
"output": "4"
},
{
"input": "169\n158\n11",
"output": "4"
},
{
"input": "41\n38\n17",
"output": "5"
},
{
"input": "762\n609\n442",
"output": "6"
},
{
"input": "186\n98\n104",
"output": "6"
},
{
"input": "314\n304\n294",
"output": "6"
},
{
"input": "35\n35\n33",
"output": "6"
},
{
"input": "8\n3\n5",
"output": "3"
},
{
"input": "11\n3\n5",
"output": "2"
},
{
"input": "5\n4\n2",
"output": "5"
},
{
"input": "41\n5\n36",
"output": "3"
},
{
"input": "7\n4\n1",
"output": "4"
},
{
"input": "6\n1\n4",
"output": "2"
},
{
"input": "597\n142\n484",
"output": "3"
},
{
"input": "6\n6\n1",
"output": "5"
},
{
"input": "8\n4\n2",
"output": "3"
},
{
"input": "4\n1\n4",
"output": "3"
},
{
"input": "7\n2\n3",
"output": "2"
},
{
"input": "100\n100\n50",
"output": "5"
},
{
"input": "5\n1\n3",
"output": "2"
},
{
"input": "10\n4\n6",
"output": "3"
},
{
"input": "8\n8\n2",
"output": "5"
},
{
"input": "5\n2\n4",
"output": "4"
},
{
"input": "11\n5\n3",
"output": "3"
},
{
"input": "668\n248\n336",
"output": "3"
},
{
"input": "2\n2\n1",
"output": "5"
},
{
"input": "465\n126\n246",
"output": "3"
},
{
"input": "5\n1\n5",
"output": "3"
},
{
"input": "132\n34\n64",
"output": "2"
},
{
"input": "11\n1\n6",
"output": "2"
},
{
"input": "8\n4\n5",
"output": "4"
},
{
"input": "4\n2\n4",
"output": "4"
},
{
"input": "576\n238\n350",
"output": "4"
},
{
"input": "6\n1\n5",
"output": "3"
},
{
"input": "5\n1\n4",
"output": "3"
},
{
"input": "9\n2\n8",
"output": "3"
},
{
"input": "7\n3\n4",
"output": "3"
},
{
"input": "9\n4\n5",
"output": "3"
},
{
"input": "10\n3\n4",
"output": "2"
},
{
"input": "18\n5\n8",
"output": "2"
},
{
"input": "2\n1\n1",
"output": "3"
},
{
"input": "100\n40\n60",
"output": "3"
},
{
"input": "6\n4\n4",
"output": "6"
},
{
"input": "3\n1\n1",
"output": "2"
},
{
"input": "10\n3\n7",
"output": "3"
},
{
"input": "9\n2\n5",
"output": "2"
},
{
"input": "6\n2\n3",
"output": "3"
}
] | 1,645,691,803
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 120
| 77
| 0
|
n = int(input())
a = int(input())
b = int(input())
temp = n
res1, res2 = 1, 1
for i in range(2):
if temp < a:
res1 += 1
temp = n
temp -= a
if temp < b:
res1 += 1
temp = n
temp -= b
for i in range(2):
if temp < a:
res1 += 1
temp = n
temp -= a
temp = n
for i in range(4):
if temp < a:
res2 += 1
temp = n
temp -= a
for i in range(2):
if temp < b:
res2 += 1
temp = n
temp -= b
ans = min(res1, res2)
print(ans)
|
Title: Door Frames
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1<=000) β the length of each wooden bar.
The second line contains a single integer *a* (1<=β€<=*a*<=β€<=*n*) β the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=β€<=*b*<=β€<=*n*) β the length of the upper side of a door frame.
Output Specification:
Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors.
Demo Input:
['8\n1\n2\n', '5\n3\n4\n', '6\n4\n2\n', '20\n5\n6\n']
Demo Output:
['1\n', '6\n', '4\n', '2\n']
Note:
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
|
```python
n = int(input())
a = int(input())
b = int(input())
temp = n
res1, res2 = 1, 1
for i in range(2):
if temp < a:
res1 += 1
temp = n
temp -= a
if temp < b:
res1 += 1
temp = n
temp -= b
for i in range(2):
if temp < a:
res1 += 1
temp = n
temp -= a
temp = n
for i in range(4):
if temp < a:
res2 += 1
temp = n
temp -= a
for i in range(2):
if temp < b:
res2 += 1
temp = n
temp -= b
ans = min(res1, res2)
print(ans)
```
| 3
|
|
119
|
A
|
Epic Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
|
The only string contains space-separated integers *a*, *b* and *n* (1<=β€<=*a*,<=*b*,<=*n*<=β€<=100) β the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
|
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
|
[
"3 5 9\n",
"1 1 100\n"
] |
[
"0",
"1"
] |
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*,β*b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*,β0)β=β*gcd*(0,β*x*)β=β*x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3,β9)β=β3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5,β6)β=β1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3,β5)β=β1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5,β4)β=β1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3,β3)β=β3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5,β0)β=β5 stones from the heap. As 0β<β5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
| 500
|
[
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,542,296,900
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 248
| 0
|
[a,b,n] = [int(x) for x in input().split()]
def gcd(x, y):
"""θ―₯ε½ζ°θΏεδΈ€δΈͺζ°ηζ倧ε
¬ηΊ¦ζ°"""
# θ·εζε°εΌ
a = min(x,y)
b = max(x,y)
for i in range(1,a+1):
if((a % i == 0) and (b % i == 0)):
gcd = i
return gcd
step = 0
while n > 0:
step += 1
n -= gcd(a,n)
if n == 0:
break
else:
step += 1
n -= gcd(b,n)
if step % 2==0:
print('1')
else:
print('0')
|
Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=β€<=*a*,<=*b*,<=*n*<=β€<=100) β the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*,β*b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*,β0)β=β*gcd*(0,β*x*)β=β*x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3,β9)β=β3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5,β6)β=β1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3,β5)β=β1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5,β4)β=β1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3,β3)β=β3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5,β0)β=β5 stones from the heap. As 0β<β5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
|
```python
[a,b,n] = [int(x) for x in input().split()]
def gcd(x, y):
"""θ―₯ε½ζ°θΏεδΈ€δΈͺζ°ηζ倧ε
¬ηΊ¦ζ°"""
# θ·εζε°εΌ
a = min(x,y)
b = max(x,y)
for i in range(1,a+1):
if((a % i == 0) and (b % i == 0)):
gcd = i
return gcd
step = 0
while n > 0:
step += 1
n -= gcd(a,n)
if n == 0:
break
else:
step += 1
n -= gcd(b,n)
if step % 2==0:
print('1')
else:
print('0')
```
| 3
|
|
272
|
A
|
Dima and Friends
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
|
In a single line print the answer to the problem.
|
[
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] |
[
"3\n",
"2\n",
"3\n"
] |
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
| 500
|
[
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 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",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,643,481,836
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 186
| 0
|
n = int(input())
list1 = [int(num) for num in input().split()]
sumi = sum(list1)
ways=0
for i in range(1,6):
if((sumi+i)%(n+1)!=1):
ways+=1
print(ways)
|
Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
|
```python
n = int(input())
list1 = [int(num) for num in input().split()]
sumi = sum(list1)
ways=0
for i in range(1,6):
if((sumi+i)%(n+1)!=1):
ways+=1
print(ways)
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length β (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line β the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,611,213,129
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 102
| 109
| 0
|
a=input()
b=input()
s=[]
k=0
for i in range(len(a)):
s.append(int(a[i])^int(b[i]))
for i in range(len(s)):
print(s[i],end='')
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length β (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line β the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
a=input()
b=input()
s=[]
k=0
for i in range(len(a)):
s.append(int(a[i])^int(b[i]))
for i in range(len(s)):
print(s[i],end='')
```
| 3.97275
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=β€<=*n*<=β€<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line β the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,656,390,546
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 77
| 0
|
a=int(input())
d=["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
s=5
while a>s:
a-=s
s*=2
s//=5
if a%s==0:
a//=s
else:
a//=s
a+=1
print(d[a-1])
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=β€<=*n*<=β€<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line β the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
a=int(input())
d=["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
s=5
while a>s:
a-=s
s*=2
s//=5
if a%s==0:
a//=s
else:
a//=s
a+=1
print(d[a-1])
```
| 3.9615
|
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"
}
] | 1,699,840,444
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
n=input()
n = list(n)
n[0] = n[0].upper()
for i in n:
print(i,end="")
|
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
n=input()
n = list(n)
n[0] = n[0].upper()
for i in n:
print(i,end="")
```
| 3
|
|
463
|
B
|
Caisa and Pylons
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math"
] | null | null |
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=β€<=<=*h**i*<=<=β€<=<=105) representing the heights of the pylons.
|
Print a single number representing the minimum number of dollars paid by Caisa.
|
[
"5\n3 4 3 2 4\n",
"3\n4 4 4\n"
] |
[
"4\n",
"4\n"
] |
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
| 1,000
|
[
{
"input": "5\n3 4 3 2 4",
"output": "4"
},
{
"input": "3\n4 4 4",
"output": "4"
},
{
"input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730",
"output": "4000"
},
{
"input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621",
"output": "3993"
},
{
"input": "30\n30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "30"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n69",
"output": "69"
}
] | 1,645,361,051
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 49
| 140
| 11,264,000
|
n = int(input())
h = list(map(int, input().split()))
a = [0]
a.extend(h)
min_cost = 0
for i in range(1, n + 1):
d = a[0] - a[i]
if d < 0:
a[0] += abs(d)
min_cost += abs(d)
print(min_cost)
|
Title: Caisa and Pylons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=β€<=<=*h**i*<=<=β€<=<=105) representing the heights of the pylons.
Output Specification:
Print a single number representing the minimum number of dollars paid by Caisa.
Demo Input:
['5\n3 4 3 2 4\n', '3\n4 4 4\n']
Demo Output:
['4\n', '4\n']
Note:
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
|
```python
n = int(input())
h = list(map(int, input().split()))
a = [0]
a.extend(h)
min_cost = 0
for i in range(1, n + 1):
d = a[0] - a[i]
if d < 0:
a[0] += abs(d)
min_cost += abs(d)
print(min_cost)
```
| 3
|
|
53
|
A
|
Autocomplete
|
PROGRAMMING
| 1,100
|
[
"implementation"
] |
A. Autocomplete
|
2
|
256
|
Autocomplete is a program function that enables inputting the text (in editors, command line shells, browsers etc.) completing the text by its inputted part. Vasya is busy working on a new browser called 'BERowser'. He happens to be working on the autocomplete function in the address line at this very moment. A list consisting of *n* last visited by the user pages and the inputted part *s* are known. Your task is to complete *s* to make it an address of one of the pages from the list. You have to find the lexicographically smallest address having a prefix *s*.
|
The first line contains the *s* line which is the inputted part. The second line contains an integer *n* (1<=β€<=*n*<=β€<=100) which is the number of visited pages. Then follow *n* lines which are the visited pages, one on each line. All the lines have lengths of from 1 to 100 symbols inclusively and consist of lowercase Latin letters only.
|
If *s* is not the beginning of any of *n* addresses of the visited pages, print *s*. Otherwise, print the lexicographically minimal address of one of the visited pages starting from *s*.
The lexicographical order is the order of words in a dictionary. The lexicographical comparison of lines is realized by the '<' operator in the modern programming languages.
|
[
"next\n2\nnextpermutation\nnextelement\n",
"find\n4\nfind\nfindfirstof\nfindit\nfand\n",
"find\n4\nfondfind\nfondfirstof\nfondit\nfand\n"
] |
[
"nextelement\n",
"find\n",
"find\n"
] |
none
| 500
|
[
{
"input": "next\n2\nnextpermutation\nnextelement",
"output": "nextelement"
},
{
"input": "find\n4\nfind\nfindfirstof\nfindit\nfand",
"output": "find"
},
{
"input": "find\n4\nfondfind\nfondfirstof\nfondit\nfand",
"output": "find"
},
{
"input": "kudljmxcse\n4\nkudljmxcse\nszjebdoad\nchz\na",
"output": "kudljmxcse"
},
{
"input": "ntqwpa\n5\nvvepyowvn\nntqwpakay\nhh\nygiafasda\nntqwpadm",
"output": "ntqwpadm"
},
{
"input": "aflb\n6\nsaej\nujxsiijg\npp\nhgoprw\ncp\nnt",
"output": "aflb"
},
{
"input": "dzwzyj\n7\nwvixktp\ndzwzyjuhn\ndzwzyjqrbd\ndzwzyji\ndzwzyjyfys\ndzwzyjrcb\nxptb",
"output": "dzwzyji"
},
{
"input": "wmblbphwdjjskzmlsyiznluiudelhlvcpyrooajvbwudnnstdhesauyxjugdwhrrwg\n1\nwjhsbxrrhadgtnybsugdtprncwerwezxuaxnqfpnosbispmnymnaqssdkjeynrnn",
"output": "wmblbphwdjjskzmlsyiznluiudelhlvcpyrooajvbwudnnstdhesauyxjugdwhrrwg"
},
{
"input": "hzkqvwliymwjbejfpnydrbwskhyrtrlsdinfrgwmnbdpwytcnjeoowxrfgfuaffzayjylvzu\n1\nhzkqvwliymwjbejfpnydrbwskhyrtrlsdinfrgwmnbdpwytcnjeoowxrfgfuaffzayjylvzubwjlvhhsfurqb",
"output": "hzkqvwliymwjbejfpnydrbwskhyrtrlsdinfrgwmnbdpwytcnjeoowxrfgfuaffzayjylvzubwjlvhhsfurqb"
},
{
"input": "msjnqudojxtzvpc\n2\nvlxclsvqbucmbrkwwtoxek\nmsjnqudojxtzvpcldwjyystsxrtexfhllzhnkidmhmyxpld",
"output": "msjnqudojxtzvpcldwjyystsxrtexfhllzhnkidmhmyxpld"
}
] | 1,571,541,510
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 280
| 0
|
s = input()
n = int(input())
t = "z" * 101
for i in range(n):
s1 = input()
if s1[:len(s)] == s:
if s1[len(s):] < t: t = s1[len(s):]
if t == "z" * 101: print(s)
else: print(s + t)
|
Title: Autocomplete
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Autocomplete is a program function that enables inputting the text (in editors, command line shells, browsers etc.) completing the text by its inputted part. Vasya is busy working on a new browser called 'BERowser'. He happens to be working on the autocomplete function in the address line at this very moment. A list consisting of *n* last visited by the user pages and the inputted part *s* are known. Your task is to complete *s* to make it an address of one of the pages from the list. You have to find the lexicographically smallest address having a prefix *s*.
Input Specification:
The first line contains the *s* line which is the inputted part. The second line contains an integer *n* (1<=β€<=*n*<=β€<=100) which is the number of visited pages. Then follow *n* lines which are the visited pages, one on each line. All the lines have lengths of from 1 to 100 symbols inclusively and consist of lowercase Latin letters only.
Output Specification:
If *s* is not the beginning of any of *n* addresses of the visited pages, print *s*. Otherwise, print the lexicographically minimal address of one of the visited pages starting from *s*.
The lexicographical order is the order of words in a dictionary. The lexicographical comparison of lines is realized by the '<' operator in the modern programming languages.
Demo Input:
['next\n2\nnextpermutation\nnextelement\n', 'find\n4\nfind\nfindfirstof\nfindit\nfand\n', 'find\n4\nfondfind\nfondfirstof\nfondit\nfand\n']
Demo Output:
['nextelement\n', 'find\n', 'find\n']
Note:
none
|
```python
s = input()
n = int(input())
t = "z" * 101
for i in range(n):
s1 = input()
if s1[:len(s)] == s:
if s1[len(s):] < t: t = s1[len(s):]
if t == "z" * 101: print(s)
else: print(s + t)
```
| 3.93
|
577
|
A
|
Multiplication Table
|
PROGRAMMING
| 1,000
|
[
"implementation",
"number theory"
] | null | null |
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=Γ<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
|
The single line contains numbers *n* and *x* (1<=β€<=*n*<=β€<=105, 1<=β€<=*x*<=β€<=109) β the size of the table and the number that we are looking for in the table.
|
Print a single number: the number of times *x* occurs in the table.
|
[
"10 5\n",
"6 12\n",
"5 13\n"
] |
[
"2\n",
"4\n",
"0\n"
] |
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
| 500
|
[
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 1000000000",
"output": "0"
},
{
"input": "100000 1000000000",
"output": "16"
},
{
"input": "100000 362880",
"output": "154"
},
{
"input": "1 4",
"output": "0"
},
{
"input": "9 12",
"output": "4"
},
{
"input": "10 123",
"output": "0"
},
{
"input": "9551 975275379",
"output": "0"
},
{
"input": "17286 948615687",
"output": "0"
},
{
"input": "58942 936593001",
"output": "0"
},
{
"input": "50000 989460910",
"output": "4"
},
{
"input": "22741 989460910",
"output": "0"
},
{
"input": "22740 989460910",
"output": "0"
},
{
"input": "100000 989460910",
"output": "4"
},
{
"input": "100000 98280",
"output": "128"
},
{
"input": "100000 997920",
"output": "222"
},
{
"input": "100000 720720",
"output": "226"
},
{
"input": "100000 2162160",
"output": "282"
},
{
"input": "100000 4324320",
"output": "320"
},
{
"input": "100000 8648640",
"output": "348"
},
{
"input": "100000 183783600",
"output": "438"
},
{
"input": "100000 551350800",
"output": "392"
},
{
"input": "40000 551350800",
"output": "150"
},
{
"input": "20000 400000000",
"output": "1"
},
{
"input": "19999 400000000",
"output": "0"
},
{
"input": "19999 399960001",
"output": "1"
},
{
"input": "31621 999887641",
"output": "1"
},
{
"input": "31622 999887641",
"output": "1"
},
{
"input": "31620 999887641",
"output": "0"
},
{
"input": "100000 999887641",
"output": "3"
},
{
"input": "100000 25",
"output": "3"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "100000 3628800",
"output": "220"
},
{
"input": "100000 39916800",
"output": "328"
},
{
"input": "100000 479001600",
"output": "254"
},
{
"input": "4 9",
"output": "1"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "20 100",
"output": "3"
},
{
"input": "10 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "89874 1",
"output": "1"
}
] | 1,698,098,410
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 47
| 62
| 1,638,400
|
def solve():
n, x = map(int, input().split())
cnt = 0
for i in range(n+1):
cnt += x <= i*n and x % i == 0
print(cnt)
# t = int(input())
t = 1
while t:
solve()
t -= 1
|
Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=Γ<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=β€<=*n*<=β€<=105, 1<=β€<=*x*<=β€<=109) β the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
|
```python
def solve():
n, x = map(int, input().split())
cnt = 0
for i in range(n+1):
cnt += x <= i*n and x % i == 0
print(cnt)
# t = int(input())
t = 1
while t:
solve()
t -= 1
```
| 3
|
|
946
|
A
|
Partition
|
PROGRAMMING
| 800
|
[
"greedy"
] | null | null |
You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100) β the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=β€<=*a**i*<=β€<=100) β the elements of sequence *a*.
|
Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.
|
[
"3\n1 -2 0\n",
"6\n16 23 16 15 42 8\n"
] |
[
"3\n",
"120\n"
] |
In the first example we may choose *b*β=β{1,β0}, *c*β=β{β-β2}. Then *B*β=β1, *C*β=ββ-β2, *B*β-β*C*β=β3.
In the second example we choose *b*β=β{16,β23,β16,β15,β42,β8}, *c*β=β{} (an empty sequence). Then *B*β=β120, *C*β=β0, *B*β-β*C*β=β120.
| 0
|
[
{
"input": "3\n1 -2 0",
"output": "3"
},
{
"input": "6\n16 23 16 15 42 8",
"output": "120"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "100\n-100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -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": "2\n-1 5",
"output": "6"
},
{
"input": "3\n-2 0 1",
"output": "3"
},
{
"input": "12\n-1 -2 -3 4 4 -6 -6 56 3 3 -3 3",
"output": "94"
},
{
"input": "4\n1 -1 1 -1",
"output": "4"
},
{
"input": "4\n100 -100 100 -100",
"output": "400"
},
{
"input": "3\n-2 -5 10",
"output": "17"
},
{
"input": "5\n1 -2 3 -4 5",
"output": "15"
},
{
"input": "3\n-100 100 -100",
"output": "300"
},
{
"input": "6\n1 -1 1 -1 1 -1",
"output": "6"
},
{
"input": "6\n2 -2 2 -2 2 -2",
"output": "12"
},
{
"input": "9\n12 93 -2 0 0 0 3 -3 -9",
"output": "122"
},
{
"input": "6\n-1 2 4 -5 -3 55",
"output": "70"
},
{
"input": "6\n-12 8 68 -53 1 -15",
"output": "157"
},
{
"input": "2\n-2 1",
"output": "3"
},
{
"input": "3\n100 -100 100",
"output": "300"
},
{
"input": "5\n100 100 -1 -100 2",
"output": "303"
},
{
"input": "6\n-5 -4 -3 -2 -1 0",
"output": "15"
},
{
"input": "6\n4 4 4 -3 -3 2",
"output": "20"
},
{
"input": "2\n-1 2",
"output": "3"
},
{
"input": "1\n100",
"output": "100"
},
{
"input": "5\n-1 -2 3 1 2",
"output": "9"
},
{
"input": "5\n100 -100 100 -100 100",
"output": "500"
},
{
"input": "5\n1 -1 1 -1 1",
"output": "5"
},
{
"input": "4\n0 0 0 -1",
"output": "1"
},
{
"input": "5\n100 -100 -1 2 100",
"output": "303"
},
{
"input": "2\n75 0",
"output": "75"
},
{
"input": "4\n55 56 -59 -58",
"output": "228"
},
{
"input": "2\n9 71",
"output": "80"
},
{
"input": "2\n9 70",
"output": "79"
},
{
"input": "2\n9 69",
"output": "78"
},
{
"input": "2\n100 -100",
"output": "200"
},
{
"input": "4\n-9 4 -9 5",
"output": "27"
},
{
"input": "42\n91 -27 -79 -56 80 -93 -23 10 80 94 61 -89 -64 81 34 99 31 -32 -69 92 79 -9 73 66 -8 64 99 99 58 -19 -40 21 1 -33 93 -23 -62 27 55 41 57 36",
"output": "2348"
},
{
"input": "7\n-1 2 2 2 -1 2 -1",
"output": "11"
},
{
"input": "6\n-12 8 17 -69 7 -88",
"output": "201"
},
{
"input": "3\n1 -2 5",
"output": "8"
},
{
"input": "6\n-2 3 -4 5 6 -1",
"output": "21"
},
{
"input": "2\n-5 1",
"output": "6"
},
{
"input": "4\n2 2 -2 4",
"output": "10"
},
{
"input": "68\n21 47 -75 -25 64 83 83 -21 89 24 43 44 -35 34 -42 92 -96 -52 -66 64 14 -87 25 -61 -78 83 -96 -18 95 83 -93 -28 75 49 87 65 -93 -69 -2 95 -24 -36 -61 -71 88 -53 -93 -51 -81 -65 -53 -46 -56 6 65 58 19 100 57 61 -53 44 -58 48 -8 80 -88 72",
"output": "3991"
},
{
"input": "5\n5 5 -10 -1 1",
"output": "22"
},
{
"input": "3\n-1 2 3",
"output": "6"
},
{
"input": "76\n57 -38 -48 -81 93 -32 96 55 -44 2 38 -46 42 64 71 -73 95 31 -39 -62 -1 75 -17 57 28 52 12 -11 82 -84 59 -86 73 -97 34 97 -57 -85 -6 39 -5 -54 95 24 -44 35 -18 9 91 7 -22 -61 -80 54 -40 74 -90 15 -97 66 -52 -49 -24 65 21 -93 -29 -24 -4 -1 76 -93 7 -55 -53 1",
"output": "3787"
},
{
"input": "5\n-1 -2 1 2 3",
"output": "9"
},
{
"input": "4\n2 2 -2 -2",
"output": "8"
},
{
"input": "6\n100 -100 100 -100 100 -100",
"output": "600"
},
{
"input": "100\n-59 -33 34 0 69 24 -22 58 62 -36 5 45 -19 -73 61 -9 95 42 -73 -64 91 -96 2 53 -8 82 -79 16 18 -5 -53 26 71 38 -31 12 -33 -1 -65 -6 3 -89 22 33 -27 -36 41 11 -47 -32 47 -56 -38 57 -63 -41 23 41 29 78 16 -65 90 -58 -12 6 -60 42 -36 -52 -54 -95 -10 29 70 50 -94 1 93 48 -71 -77 -16 54 56 -60 66 76 31 8 44 -61 -74 23 37 38 18 -18 29 41",
"output": "4362"
},
{
"input": "2\n-1 1",
"output": "2"
},
{
"input": "3\n1 -2 100",
"output": "103"
},
{
"input": "5\n1 -2 3 1 2",
"output": "9"
},
{
"input": "10\n100 -10 -100 10 10 10 10 10 10 10",
"output": "280"
},
{
"input": "4\n2 0 -2 4",
"output": "8"
},
{
"input": "4\n3 -3 1 -1",
"output": "8"
},
{
"input": "3\n1 -1 1",
"output": "3"
},
{
"input": "4\n2 5 -2 4",
"output": "13"
},
{
"input": "2\n-2 2",
"output": "4"
},
{
"input": "3\n1 -2 1",
"output": "4"
},
{
"input": "5\n-1 -2 1 1 -1",
"output": "6"
},
{
"input": "4\n-2 0 2 4",
"output": "8"
},
{
"input": "8\n-42 7 87 -16 -5 65 -88 1",
"output": "311"
},
{
"input": "3\n1 -3 4",
"output": "8"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n-1 2 -1",
"output": "4"
},
{
"input": "18\n-21 12 65 66 -24 62 82 35 -45 -47 28 37 5 -32 22 -14 -69 -95",
"output": "761"
},
{
"input": "4\n-1 1 -1 1",
"output": "4"
},
{
"input": "5\n-1 2 1 1 1",
"output": "6"
},
{
"input": "3\n1 1 1",
"output": "3"
}
] | 1,637,589,656
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 72
| 77
| 0
|
n = int(input())
a = list(map(int, input().split()))
sum1 = 0
for i in range(n):
a[i] = abs(a[i])
sum1 = sum1 + a[i]
print(sum1)
|
Title: Partition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100) β the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=β€<=*a**i*<=β€<=100) β the elements of sequence *a*.
Output Specification:
Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.
Demo Input:
['3\n1 -2 0\n', '6\n16 23 16 15 42 8\n']
Demo Output:
['3\n', '120\n']
Note:
In the first example we may choose *b*β=β{1,β0}, *c*β=β{β-β2}. Then *B*β=β1, *C*β=ββ-β2, *B*β-β*C*β=β3.
In the second example we choose *b*β=β{16,β23,β16,β15,β42,β8}, *c*β=β{} (an empty sequence). Then *B*β=β120, *C*β=β0, *B*β-β*C*β=β120.
|
```python
n = int(input())
a = list(map(int, input().split()))
sum1 = 0
for i in range(n):
a[i] = abs(a[i])
sum1 = sum1 + a[i]
print(sum1)
```
| 3
|
|
630
|
L
|
Cracking the Code
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
The protection of a popular program developed by one of IT City companies is organized the following way. After installation it outputs a random five digit number which should be sent in SMS to a particular phone number. In response an SMS activation code arrives.
A young hacker Vasya disassembled the program and found the algorithm that transforms the shown number into the activation code. Note: it is clear that Vasya is a law-abiding hacker, and made it for a noble purpose β to show the developer the imperfection of their protection.
The found algorithm looks the following way. At first the digits of the number are shuffled in the following order <first digit><third digit><fifth digit><fourth digit><second digit>. For example the shuffle of 12345 should lead to 13542. On the second stage the number is raised to the fifth power. The result of the shuffle and exponentiation of the number 12345 is 455Β 422Β 043Β 125Β 550Β 171Β 232. The answer is the 5 last digits of this result. For the number 12345 the answer should be 71232.
Vasya is going to write a keygen program implementing this algorithm. Can you do the same?
|
The only line of the input contains a positive integer five digit number for which the activation code should be found.
|
Output exactly 5 digits without spaces between them β the found activation code of the program.
|
[
"12345\n"
] |
[
"71232"
] |
none
| 0
|
[
{
"input": "12345",
"output": "71232"
},
{
"input": "13542",
"output": "84443"
},
{
"input": "71232",
"output": "10151"
},
{
"input": "11111",
"output": "36551"
},
{
"input": "10000",
"output": "00000"
},
{
"input": "99999",
"output": "99999"
},
{
"input": "91537",
"output": "27651"
},
{
"input": "70809",
"output": "00000"
},
{
"input": "41675",
"output": "61851"
},
{
"input": "32036",
"output": "82432"
}
] | 1,691,310,587
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 10
| 31
| 0
|
n = input()
coded = str(int(n[0] + n[2] + n[4] + n[3] + n[1]) ** 5)
print(coded[len(coded) - 5:])
|
Title: Cracking the Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The protection of a popular program developed by one of IT City companies is organized the following way. After installation it outputs a random five digit number which should be sent in SMS to a particular phone number. In response an SMS activation code arrives.
A young hacker Vasya disassembled the program and found the algorithm that transforms the shown number into the activation code. Note: it is clear that Vasya is a law-abiding hacker, and made it for a noble purpose β to show the developer the imperfection of their protection.
The found algorithm looks the following way. At first the digits of the number are shuffled in the following order <first digit><third digit><fifth digit><fourth digit><second digit>. For example the shuffle of 12345 should lead to 13542. On the second stage the number is raised to the fifth power. The result of the shuffle and exponentiation of the number 12345 is 455Β 422Β 043Β 125Β 550Β 171Β 232. The answer is the 5 last digits of this result. For the number 12345 the answer should be 71232.
Vasya is going to write a keygen program implementing this algorithm. Can you do the same?
Input Specification:
The only line of the input contains a positive integer five digit number for which the activation code should be found.
Output Specification:
Output exactly 5 digits without spaces between them β the found activation code of the program.
Demo Input:
['12345\n']
Demo Output:
['71232']
Note:
none
|
```python
n = input()
coded = str(int(n[0] + n[2] + n[4] + n[3] + n[1]) ** 5)
print(coded[len(coded) - 5:])
```
| 3
|
|
165
|
A
|
Supercentral Point
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
|
The first input line contains the only integer *n* (1<=β€<=*n*<=β€<=200) β 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*|<=β€<=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 β 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,β1) and (1,β2).
In the second sample there is one supercental point β point (0,β0).
| 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"
}
] | 1,693,425,189
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 26
| 124
| 2,150,400
|
n=int(input())
x=[]
y=[]
c=0
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
for i in range(n):
r=l=lo=u=0
for j in range(n):
if(i==j):
continue
else:
if(x[j]>x[i] and y[j]==y[i]):
r+=1
elif(x[j]<x[i] and y[j]==y[i]):
l+=1
elif(x[j]==x[i] and y[j]<y[i]):
lo+=1
elif(x[j]==x[i] and y[j]>y[i]):
u+=1
if(r>=1 and l>=1 and lo>=1 and u>=1):
c+=1
print(c)
|
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<=β€<=*n*<=β€<=200) β 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*|<=β€<=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 β 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,β1) and (1,β2).
In the second sample there is one supercental point β point (0,β0).
|
```python
n=int(input())
x=[]
y=[]
c=0
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
for i in range(n):
r=l=lo=u=0
for j in range(n):
if(i==j):
continue
else:
if(x[j]>x[i] and y[j]==y[i]):
r+=1
elif(x[j]<x[i] and y[j]==y[i]):
l+=1
elif(x[j]==x[i] and y[j]<y[i]):
lo+=1
elif(x[j]==x[i] and y[j]>y[i]):
u+=1
if(r>=1 and l>=1 and lo>=1 and u>=1):
c+=1
print(c)
```
| 3
|
|
221
|
A
|
Little Elephant and Function
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
The Little Elephant enjoys recursive functions.
This time he enjoys the sorting function. Let *a* is a permutation of an integers from 1 to *n*, inclusive, and *a**i* denotes the *i*-th element of the permutation. The Little Elephant's recursive function *f*(*x*), that sorts the first *x* permutation's elements, works as follows:
- If *x*<==<=1, exit the function. - Otherwise, call *f*(*x*<=-<=1), and then make *swap*(*a**x*<=-<=1,<=*a**x*) (swap the *x*-th and (*x*<=-<=1)-th elements of *a*).
The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to *n*, such that after performing the Little Elephant's function (that is call *f*(*n*)), the permutation will be sorted in ascending order.
|
A single line contains integer *n* (1<=β€<=*n*<=β€<=1000) β the size of permutation.
|
In a single line print *n* distinct integers from 1 to *n* β the required permutation. Numbers in a line should be separated by spaces.
It is guaranteed that the answer exists.
|
[
"1\n",
"2\n"
] |
[
"1 ",
"2 1 "
] |
none
| 500
|
[
{
"input": "1",
"output": "1 "
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "3",
"output": "3 1 2 "
},
{
"input": "4",
"output": "4 1 2 3 "
},
{
"input": "5",
"output": "5 1 2 3 4 "
},
{
"input": "6",
"output": "6 1 2 3 4 5 "
},
{
"input": "7",
"output": "7 1 2 3 4 5 6 "
},
{
"input": "1000",
"output": "1000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "999",
"output": "999 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "998",
"output": "998 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "997",
"output": "997 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "900",
"output": "900 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "101",
"output": "101 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "3",
"output": "3 1 2 "
},
{
"input": "67",
"output": "67 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 "
},
{
"input": "779",
"output": "779 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "195",
"output": "195 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "907",
"output": "907 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "971",
"output": "971 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "683",
"output": "683 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "99",
"output": "99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 "
},
{
"input": "811",
"output": "811 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "875",
"output": "875 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "291",
"output": "291 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "3",
"output": "3 1 2 "
},
{
"input": "419",
"output": "419 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "779",
"output": "779 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "195",
"output": "195 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "907",
"output": "907 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "21",
"output": "21 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 "
},
{
"input": "22",
"output": "22 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 "
},
{
"input": "23",
"output": "23 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 "
},
{
"input": "24",
"output": "24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 "
},
{
"input": "25",
"output": "25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 "
}
] | 1,570,826,395
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 248
| 307,200
|
n = int(input())
ans = [n]+list(range(1, n))
print(*ans)
|
Title: Little Elephant and Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant enjoys recursive functions.
This time he enjoys the sorting function. Let *a* is a permutation of an integers from 1 to *n*, inclusive, and *a**i* denotes the *i*-th element of the permutation. The Little Elephant's recursive function *f*(*x*), that sorts the first *x* permutation's elements, works as follows:
- If *x*<==<=1, exit the function. - Otherwise, call *f*(*x*<=-<=1), and then make *swap*(*a**x*<=-<=1,<=*a**x*) (swap the *x*-th and (*x*<=-<=1)-th elements of *a*).
The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to *n*, such that after performing the Little Elephant's function (that is call *f*(*n*)), the permutation will be sorted in ascending order.
Input Specification:
A single line contains integer *n* (1<=β€<=*n*<=β€<=1000) β the size of permutation.
Output Specification:
In a single line print *n* distinct integers from 1 to *n* β the required permutation. Numbers in a line should be separated by spaces.
It is guaranteed that the answer exists.
Demo Input:
['1\n', '2\n']
Demo Output:
['1 ', '2 1 ']
Note:
none
|
```python
n = int(input())
ans = [n]+list(range(1, n))
print(*ans)
```
| 3
|
|
222
|
A
|
Shooshuns and Sequence
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
|
The first line contains two space-separated integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=105) β the sequence that the shooshuns found.
|
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
|
[
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] |
[
"1\n",
"-1\n"
] |
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
| 500
|
[
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "4 3\n1 2 3 4",
"output": "-1"
},
{
"input": "5 3\n2 1 1 1 1",
"output": "1"
},
{
"input": "5 3\n1 1 1 1 1",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 1\n2 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 2\n2 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 2\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 3\n3 2 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "-1"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "20 10\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 11\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "20 9\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20 20\n20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10",
"output": "19"
}
] | 1,556,120,192
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 47
| 404
| 10,547,200
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
k = k-1
if len(set(a)) == 1:
print(0)
else:
if len(set(a[k:])) > 1:
print(-1)
else:
pos = k
while pos > 0 and a[pos-1] == a[k]:
pos -= 1
print(pos)
|
Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=105) β the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
|
```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
k = k-1
if len(set(a)) == 1:
print(0)
else:
if len(set(a[k:])) > 1:
print(-1)
else:
pos = k
while pos > 0 and a[pos-1] == a[k]:
pos -= 1
print(pos)
```
| 3
|
|
270
|
A
|
Fancy Fence
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation",
"math"
] | null | null |
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
|
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) β the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) β the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
|
[
"3\n30\n60\n90\n"
] |
[
"NO\nYES\nYES\n"
] |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case β a square.
| 500
|
[
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,638,466,147
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 3
| 62
| 0
|
for _ in range(int(input())):
a = int(input())
ans = 360 / (180 - a)
if ans == int(ans):
print("YES")
else:
print("NO")
|
Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) β the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) β the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case β a square.
|
```python
for _ in range(int(input())):
a = int(input())
ans = 360 / (180 - a)
if ans == int(ans):
print("YES")
else:
print("NO")
```
| 3
|
|
873
|
B
|
Balanced Substring
|
PROGRAMMING
| 1,500
|
[
"dp",
"implementation"
] | null | null |
You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*.
|
The first line contains *n* (1<=β€<=*n*<=β€<=100000) β the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.
|
If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.
|
[
"8\n11010111\n",
"3\n111\n"
] |
[
"4\n",
"0\n"
] |
In the first example you can choose the substring [3,β6]. It is balanced, and its length is 4. Choosing the substring [2,β5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
| 0
|
[
{
"input": "8\n11010111",
"output": "4"
},
{
"input": "3\n111",
"output": "0"
},
{
"input": "11\n00001000100",
"output": "2"
},
{
"input": "10\n0100000000",
"output": "2"
},
{
"input": "13\n0001000011010",
"output": "6"
},
{
"input": "14\n00000100101011",
"output": "10"
},
{
"input": "14\n01111101111111",
"output": "2"
},
{
"input": "18\n110010101101111111",
"output": "10"
},
{
"input": "11\n00010000011",
"output": "4"
},
{
"input": "10\n1000010110",
"output": "6"
},
{
"input": "15\n100000100000011",
"output": "4"
},
{
"input": "18\n011010101110111101",
"output": "8"
},
{
"input": "10\n0011011111",
"output": "6"
},
{
"input": "3\n011",
"output": "2"
},
{
"input": "14\n11111111111111",
"output": "0"
},
{
"input": "65\n11001000110001001011011110111100000000101001001010101111000100010",
"output": "48"
},
{
"input": "10\n0011111000",
"output": "10"
},
{
"input": "13\n1110000001110",
"output": "12"
},
{
"input": "3\n110",
"output": "2"
},
{
"input": "4\n1110",
"output": "2"
},
{
"input": "9\n001011001",
"output": "8"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "12\n110000110000",
"output": "8"
},
{
"input": "3\n001",
"output": "2"
},
{
"input": "14\n11000011000000",
"output": "8"
},
{
"input": "19\n0000011111111100000",
"output": "18"
},
{
"input": "45\n011010001100001111110001011100000001101100111",
"output": "44"
},
{
"input": "18\n000001111111100000",
"output": "16"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "12\n000101011001",
"output": "10"
}
] | 1,679,054,635
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 53
| 108
| 9,216,000
|
input()
d={0:-1}
a=b=0
for i,c in enumerate(input()):
b+=2*int(c)-1
d[b]=x=d.get(b,i)
a=max(a,i-x)
print(a)
|
Title: Balanced Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*.
Input Specification:
The first line contains *n* (1<=β€<=*n*<=β€<=100000) β the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.
Output Specification:
If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.
Demo Input:
['8\n11010111\n', '3\n111\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example you can choose the substring [3,β6]. It is balanced, and its length is 4. Choosing the substring [2,β5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
|
```python
input()
d={0:-1}
a=b=0
for i,c in enumerate(input()):
b+=2*int(c)-1
d[b]=x=d.get(b,i)
a=max(a,i-x)
print(a)
```
| 3
|
|
549
|
A
|
Face Detection
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them.
In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2<=Γ<=2 square, such that from the four letters of this square you can make word "face".
You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap.
|
The first line contains two space-separated integers, *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=50) β the height and the width of the image, respectively.
Next *n* lines define the image. Each line contains *m* lowercase Latin letters.
|
In the single line print the number of faces on the image.
|
[
"4 4\nxxxx\nxfax\nxcex\nxxxx\n",
"4 2\nxx\ncf\nae\nxx\n",
"2 3\nfac\ncef\n",
"1 4\nface\n"
] |
[
"1\n",
"1\n",
"2\n",
"0\n"
] |
In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column:
In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column.
In the third sample two faces are shown:
In the fourth sample the image has no faces on it.
| 250
|
[
{
"input": "4 4\nxxxx\nxfax\nxcex\nxxxx",
"output": "1"
},
{
"input": "4 2\nxx\ncf\nae\nxx",
"output": "1"
},
{
"input": "2 3\nfac\ncef",
"output": "2"
},
{
"input": "1 4\nface",
"output": "0"
},
{
"input": "5 5\nwmmwn\nlurcm\nkeetd\nfokon\ncxxgx",
"output": "0"
},
{
"input": "5 5\nkjxbw\neacra\nxefhx\nucmcz\npgtjk",
"output": "1"
},
{
"input": "1 1\np",
"output": "0"
},
{
"input": "2 5\nacdmw\nefazb",
"output": "1"
},
{
"input": "5 2\ndz\nda\nsx\nyu\nzz",
"output": "0"
},
{
"input": "5 5\nxeljd\nwriac\nveief\nlcacf\nbqefn",
"output": "2"
},
{
"input": "5 5\nacnbx\nefacp\nlrefa\norqce\nzvbay",
"output": "3"
},
{
"input": "5 5\nbyjvu\nkmaca\nalefe\nwcacg\nrefez",
"output": "5"
},
{
"input": "5 5\npuxac\nbbaef\naccfa\nefaec\nligsr",
"output": "5"
},
{
"input": "37 4\nacjo\nefac\nacef\nefac\nwpef\nicac\naefe\ncfac\naece\ncfaf\nyqce\nmiaf\nirce\nycaf\naefc\ncfae\nrsnc\nbacz\nqefb\npdhs\nffac\nfaef\nacfd\nacmi\nefvm\nacaz\nefpn\nacao\nefer\nacap\nefec\nacaf\nefef\nacbj\nefac\nacef\nefoz",
"output": "49"
},
{
"input": "7 3\njac\naef\ncfa\naec\ncfq\ndig\nxyq",
"output": "5"
},
{
"input": "35 1\ny\na\nk\ng\ni\nd\nv\nn\nl\nx\nu\nx\nu\no\nd\nf\nk\nj\nr\nm\nq\ns\nc\nd\nc\nm\nv\nh\nn\ne\nl\nt\nz\ny\no",
"output": "0"
},
{
"input": "9 46\nuuexbaacesjclggslacermcbkxlcxhdgqtacdwfryxzuxc\naclrsaefakndbnzlkefenuphgcgoedhkaxefjtnkgfeaca\nefuqunpmfxdyyffyhvracozzrxlpekhtsrfhlilfmyhefg\numyacfzffvicqtdpiulefnwcojuwtfbvlxkfsiapdnzpqo\nactefvuxqptremlqjhdbdwacjxdxitxjktecvefacamjcz\neflarseklqrkayhosverpfefzirqigzlxezabhzeferkwm\nztpypwxyohtacsnnvoigqwcachtaceyxlkhfhgwbnfmklb\nfzpqeymhljiefctvqtbgcdjefkxefarbciyosmktmoziac\nbssxizyrzyhacszttwhrzspbfkgfcejybdjcpuczhrkqef",
"output": "20"
},
{
"input": "7 30\nmjfracgaacacctacrreyrlkacuacay\nrlacefacefeftaeftkacacaefcefev\nacefacefacraccfaeaefefecaeacaf\nefacefacefaefaecfcfacacaecfefa\nncefacefacecacfaeaecefefcaeace\nfafaceacuafaefadcfcafacaefcfea\nzsvefafukcecfarkaeaecefecailgu",
"output": "95"
},
{
"input": "2 2\nff\nee",
"output": "0"
},
{
"input": "2 2\nfa\ncc",
"output": "0"
},
{
"input": "2 2\nfa\naa",
"output": "0"
},
{
"input": "2 2\nfa\nea",
"output": "0"
},
{
"input": "2 2\nfa\ndd",
"output": "0"
},
{
"input": "2 2\nee\nff",
"output": "0"
},
{
"input": "2 2\ncc\ncf",
"output": "0"
},
{
"input": "2 2\naa\nae",
"output": "0"
},
{
"input": "2 2\nef\nac",
"output": "1"
},
{
"input": "2 2\nfe\nca",
"output": "1"
}
] | 1,434,126,828
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 62
| 0
|
m, n = map(int, input().split(' '))
pic = []
count = 0
if m < 2 or n < 2:
print(0)
exit()
for i in range(m):
pic.append(input())
for i in range(m-1):
for j in range(n-1):
#print(pic[i+1][j:j+2])
s = pic[i][j:j+2] + pic[i+1][j:j+2]
if 'f' in s and 'a' in s and 'c' in s and 'e' in s:
count += 1
print(count)
|
Title: Face Detection
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them.
In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2<=Γ<=2 square, such that from the four letters of this square you can make word "face".
You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap.
Input Specification:
The first line contains two space-separated integers, *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=50) β the height and the width of the image, respectively.
Next *n* lines define the image. Each line contains *m* lowercase Latin letters.
Output Specification:
In the single line print the number of faces on the image.
Demo Input:
['4 4\nxxxx\nxfax\nxcex\nxxxx\n', '4 2\nxx\ncf\nae\nxx\n', '2 3\nfac\ncef\n', '1 4\nface\n']
Demo Output:
['1\n', '1\n', '2\n', '0\n']
Note:
In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column:
In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column.
In the third sample two faces are shown:
In the fourth sample the image has no faces on it.
|
```python
m, n = map(int, input().split(' '))
pic = []
count = 0
if m < 2 or n < 2:
print(0)
exit()
for i in range(m):
pic.append(input())
for i in range(m-1):
for j in range(n-1):
#print(pic[i+1][j:j+2])
s = pic[i][j:j+2] + pic[i+1][j:j+2]
if 'f' in s and 'a' in s and 'c' in s and 'e' in s:
count += 1
print(count)
```
| 3
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=100) β the number of red and blue socks that Vasya's got.
|
Print two space-separated integers β the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,679,898,507
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 46
| 0
|
a, b = map(int, input().split(' '))
min1 = int(min(a, b))
a -= min1
b -= min1
max1 = int(max(a, b))
print(min1, max1 // 2)
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=100) β the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers β the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
a, b = map(int, input().split(' '))
min1 = int(min(a, b))
a -= min1
b -= min1
max1 = int(max(a, b))
print(min1, max1 // 2)
```
| 3
|
|
545
|
D
|
Queue
|
PROGRAMMING
| 1,300
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are *n* people in the queue. For each person we know time *t**i* needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=105).
The next line contains *n* integers *t**i* (1<=β€<=*t**i*<=β€<=109), separated by spaces.
|
Print a single number β the maximum number of not disappointed people in the queue.
|
[
"5\n15 2 1 5 3\n"
] |
[
"4\n"
] |
Value 4 is achieved at such an arrangement, for example: 1,β2,β3,β5,β15. Thus, you can make everything feel not disappointed except for the person with time 5.
| 1,750
|
[
{
"input": "5\n15 2 1 5 3",
"output": "4"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10\n13 2 5 55 21 34 1 8 1 3",
"output": "6"
},
{
"input": "10\n8 256 16 1 2 1 64 4 128 32",
"output": "10"
},
{
"input": "10\n10000 40000 10000 50000 20000 100000 10000 100 30000 500",
"output": "6"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "15\n9 11 45 86 52 65 35 3 93 7 21 45 15 11 39",
"output": "6"
},
{
"input": "20\n16839799 17525904 91276752 42650694 60106463 12243176 54892123 25142243 16015971 41250998 11150057 6994983 67700784 16562412 82163675 46178521 33914268 91966607 93976858 84100064",
"output": "5"
},
{
"input": "26\n1000 4110030 64221 131521030 942374833 1003 2055015 32110 513757 16440130 263042057 32880256 128439 557559573 16051 8220066 469240078 65760513 256878 790176315 4012 2005 1027508 928528684 8030 805074697",
"output": "21"
}
] | 1,698,134,290
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 108
| 13,107,200
|
num=int(input())
listtemp=[int(x) for x in input().split()]
listtemp.sort()
oklist=listtemp[0]
ans=1
for i in range(1,num):
if listtemp[i]>=oklist:
oklist+=listtemp[i]
ans+=1
print(ans)
|
Title: Queue
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are *n* people in the queue. For each person we know time *t**i* needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=105).
The next line contains *n* integers *t**i* (1<=β€<=*t**i*<=β€<=109), separated by spaces.
Output Specification:
Print a single number β the maximum number of not disappointed people in the queue.
Demo Input:
['5\n15 2 1 5 3\n']
Demo Output:
['4\n']
Note:
Value 4 is achieved at such an arrangement, for example: 1,β2,β3,β5,β15. Thus, you can make everything feel not disappointed except for the person with time 5.
|
```python
num=int(input())
listtemp=[int(x) for x in input().split()]
listtemp.sort()
oklist=listtemp[0]
ans=1
for i in range(1,num):
if listtemp[i]>=oklist:
oklist+=listtemp[i]
ans+=1
print(ans)
```
| 3
|
|
318
|
A
|
Even Odds
|
PROGRAMMING
| 900
|
[
"math"
] | null | null |
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
|
The only line of input contains integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
Print the number that will stand at the position number *k* after Volodya's manipulations.
|
[
"10 3\n",
"7 7\n"
] |
[
"5",
"6"
] |
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
| 500
|
[
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output": "7"
},
{
"input": "1000000000000 500000000001",
"output": "2"
},
{
"input": "999999999997 499999999999",
"output": "999999999997"
},
{
"input": "999999999999 999999999999",
"output": "999999999998"
},
{
"input": "1000000000000 1",
"output": "1"
},
{
"input": "999999999999 1",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000 1000000000000",
"output": "1000000000000"
},
{
"input": "1000000000000 500000000000",
"output": "999999999999"
},
{
"input": "1000000000000 499999999999",
"output": "999999999997"
},
{
"input": "999999999997 499999999998",
"output": "999999999995"
},
{
"input": "619234238 556154835",
"output": "493075432"
},
{
"input": "38151981 36650624",
"output": "35149266"
},
{
"input": "680402465 442571217",
"output": "204739968"
},
{
"input": "109135284 9408714",
"output": "18817427"
},
{
"input": "603701841 56038951",
"output": "112077901"
},
{
"input": "356764822 321510177",
"output": "286255532"
},
{
"input": "284911189 142190783",
"output": "284381565"
},
{
"input": "91028405 61435545",
"output": "31842684"
}
] | 1,694,933,552
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 122
| 0
|
n, k = map(int, input().split())
if n%2 == 0:
if k <= n//2:
g = k+k-1
else:
g = k-(n//2-(k-n//2))
else:
if k <= (n+1)//2:
g = k+k-1
else:
g = k-((n+1)//2-(k-(n+1)//2))
print(g)
|
Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
|
```python
n, k = map(int, input().split())
if n%2 == 0:
if k <= n//2:
g = k+k-1
else:
g = k-(n//2-(k-n//2))
else:
if k <= (n+1)//2:
g = k+k-1
else:
g = k-((n+1)//2-(k-(n+1)//2))
print(g)
```
| 3
|
|
538
|
B
|
Quasi Binary
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null |
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011Β β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=106).
|
In the first line print a single integer *k*Β β the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
|
[
"9\n",
"32\n"
] |
[
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] |
none
| 1,000
|
[
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201",
"output": "2\n100 10101 "
},
{
"input": "314159",
"output": "9\n1 1 1 1 11 1011 101011 101011 111111 "
},
{
"input": "999999",
"output": "9\n111111 111111 111111 111111 111111 111111 111111 111111 111111 "
},
{
"input": "2",
"output": "2\n1 1 "
},
{
"input": "10",
"output": "1\n10 "
},
{
"input": "21",
"output": "2\n10 11 "
},
{
"input": "98",
"output": "9\n10 11 11 11 11 11 11 11 11 "
},
{
"input": "102030",
"output": "3\n10 1010 101010 "
},
{
"input": "909090",
"output": "9\n101010 101010 101010 101010 101010 101010 101010 101010 101010 "
},
{
"input": "909823",
"output": "9\n101000 101100 101100 101100 101100 101100 101101 101111 101111 "
},
{
"input": "1000000",
"output": "1\n1000000 "
},
{
"input": "111111",
"output": "1\n111111 "
},
{
"input": "123456",
"output": "6\n1 11 111 1111 11111 111111 "
},
{
"input": "987654",
"output": "9\n100000 110000 111000 111100 111110 111111 111111 111111 111111 "
},
{
"input": "908172",
"output": "9\n100000 101000 101010 101010 101010 101010 101010 101011 101111 "
},
{
"input": "8",
"output": "8\n1 1 1 1 1 1 1 1 "
},
{
"input": "100009",
"output": "9\n1 1 1 1 1 1 1 1 100001 "
},
{
"input": "900000",
"output": "9\n100000 100000 100000 100000 100000 100000 100000 100000 100000 "
},
{
"input": "1435",
"output": "5\n1 101 111 111 1111 "
},
{
"input": "1453",
"output": "5\n10 110 111 111 1111 "
}
] | 1,611,989,389
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 62
| 307,200
|
def QuasiBinary(n):
if n==415:
return ([1, 101, 101, 101, 111])
l = len(str(n))
a = int("1"*l)
ans = []
while n>0:
if a<=n:
# print("Appending a:",a," n:",n)
ans.append(a)
n = n - a
else:
# print("13 a:",a," n:",n)
if a%2==0:
a = a - 9
else:
a = a - 1
return ans
s=input()
#ans = QuasiBinary(s)
m=int(max(s))
print(m)
for i in range(m):
ans = []
for c in s:
if int(c)>i:
ans.append(1)
else:
ans.append(0)
print("".join(str(i) for i in ans[ans.index(1):]), end = " ")
|
Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011Β β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=106).
Output Specification:
In the first line print a single integer *k*Β β the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none
|
```python
def QuasiBinary(n):
if n==415:
return ([1, 101, 101, 101, 111])
l = len(str(n))
a = int("1"*l)
ans = []
while n>0:
if a<=n:
# print("Appending a:",a," n:",n)
ans.append(a)
n = n - a
else:
# print("13 a:",a," n:",n)
if a%2==0:
a = a - 9
else:
a = a - 1
return ans
s=input()
#ans = QuasiBinary(s)
m=int(max(s))
print(m)
for i in range(m):
ans = []
for c in s:
if int(c)>i:
ans.append(1)
else:
ans.append(0)
print("".join(str(i) for i in ans[ans.index(1):]), end = " ")
```
| 3
|
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