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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,678,290,714
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
a,b,c = map(int,input().split())
print(min((2*a+2*b),(a+b+c)))
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
a,b,c = map(int,input().split())
print(min((2*a+2*b),(a+b+c)))
```
| 0
|
|
433
|
A
|
Kitahara Haruki's Gift
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
|
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
|
[
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] |
[
"YES\n",
"NO\n"
] |
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.
| 500
|
[
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "52\n200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 100 200 100 200 200 200 100 200 200",
"output": "YES"
},
{
"input": "2\n100 200",
"output": "NO"
},
{
"input": "2\n200 100",
"output": "NO"
},
{
"input": "3\n100 100 100",
"output": "NO"
},
{
"input": "3\n200 200 200",
"output": "NO"
},
{
"input": "3\n200 100 200",
"output": "NO"
},
{
"input": "4\n100 100 100 100",
"output": "YES"
},
{
"input": "4\n200 200 200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 100 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "100\n100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "100\n100 100 100 100 100 100 100 100 200 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "100\n100 100 200 100 100 200 200 200 200 100 200 100 100 100 200 100 100 100 100 200 100 100 100 100 100 100 200 100 100 200 200 100 100 100 200 200 200 100 200 200 100 200 100 100 200 100 200 200 100 200 200 100 100 200 200 100 200 200 100 100 200 100 200 100 200 200 200 200 200 100 200 200 200 200 200 200 100 100 200 200 200 100 100 100 200 100 100 200 100 100 100 200 200 100 100 200 200 200 200 100",
"output": "YES"
},
{
"input": "100\n100 100 200 200 100 200 100 100 100 100 100 100 200 100 200 200 200 100 100 200 200 200 200 200 100 200 100 200 100 100 100 200 100 100 200 100 200 100 100 100 200 200 100 100 100 200 200 200 200 200 100 200 200 100 100 100 100 200 100 100 200 100 100 100 100 200 200 200 100 200 100 200 200 200 100 100 200 200 200 200 100 200 100 200 200 100 200 100 200 200 200 200 200 200 100 100 100 200 200 100",
"output": "NO"
},
{
"input": "100\n100 200 100 100 200 200 200 200 100 200 200 200 200 200 200 200 200 200 100 100 100 200 200 200 200 200 100 200 200 200 200 100 200 200 100 100 200 100 100 100 200 100 100 100 200 100 200 100 200 200 200 100 100 200 100 200 100 200 100 100 100 200 100 200 100 100 100 100 200 200 200 200 100 200 200 100 200 100 100 100 200 100 100 100 100 100 200 100 100 100 200 200 200 100 200 100 100 100 200 200",
"output": "YES"
},
{
"input": "99\n100 200 200 200 100 200 100 200 200 100 100 100 100 200 100 100 200 100 200 100 100 200 100 100 200 200 100 100 100 100 200 200 200 200 200 100 100 200 200 100 100 100 100 200 200 100 100 100 100 100 200 200 200 100 100 100 200 200 200 100 200 100 100 100 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 100 200 100 200 200 200 200 100 200 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "99\n100 200 100 100 100 100 200 200 100 200 100 100 200 100 100 100 100 100 100 200 100 100 100 100 100 100 100 200 100 200 100 100 100 100 100 100 100 200 200 200 200 200 200 200 100 200 100 200 100 200 100 200 100 100 200 200 200 100 200 200 200 200 100 200 100 200 200 200 200 100 200 100 200 200 100 200 200 200 200 200 100 100 200 100 100 100 100 200 200 200 100 100 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n200 100 100 100 200 200 200 100 100 100 100 100 100 100 100 100 200 200 100 200 200 100 200 100 100 200 200 200 100 200 100 200 200 100 200 100 200 200 200 100 100 200 200 200 200 100 100 100 100 200 200 200 200 100 200 200 200 100 100 100 200 200 200 100 200 100 200 100 100 100 200 100 200 200 100 200 200 200 100 100 100 200 200 200 100 200 200 200 100 100 100 200 100 200 100 100 100 200 200",
"output": "YES"
},
{
"input": "56\n100 200 200 200 200 200 100 200 100 100 200 100 100 100 100 100 200 200 200 100 200 100 100 200 200 200 100 200 100 200 200 100 100 100 100 100 200 100 200 100 200 200 200 100 100 200 200 200 200 200 200 200 200 200 200 100",
"output": "YES"
},
{
"input": "72\n200 100 200 200 200 100 100 200 200 100 100 100 100 200 100 200 100 100 100 100 200 100 200 100 100 200 100 100 200 100 200 100 100 200 100 200 100 100 200 200 200 200 200 100 100 200 200 200 200 100 100 100 200 200 100 100 100 100 100 200 100 100 200 100 100 200 200 100 100 200 100 200",
"output": "YES"
},
{
"input": "32\n200 200 200 100 100 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "48\n200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 100 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "60\n100 100 200 200 100 200 100 200 100 100 100 100 100 100 200 100 100 100 200 100 200 100 100 100 100 100 200 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100",
"output": "YES"
},
{
"input": "24\n200 200 100 100 200 100 200 200 100 200 200 200 200 200 100 200 200 200 200 200 200 200 200 100",
"output": "YES"
},
{
"input": "40\n100 100 200 200 200 200 100 100 100 200 100 100 200 200 100 100 100 100 100 200 100 200 200 100 200 200 200 100 100 100 100 100 200 200 100 200 100 100 200 100",
"output": "NO"
},
{
"input": "5\n200 200 200 200 200",
"output": "NO"
},
{
"input": "9\n100 100 100 200 100 100 200 100 200",
"output": "YES"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "7\n200 200 200 100 200 200 200",
"output": "NO"
},
{
"input": "4\n100 100 200 200",
"output": "YES"
},
{
"input": "6\n100 100 100 200 200 200",
"output": "NO"
},
{
"input": "4\n200 100 100 200",
"output": "YES"
},
{
"input": "5\n100 100 100 100 200",
"output": "YES"
}
] | 1,627,566,774
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 77
| 6,758,400
|
input()
s = list(map(int,input().split()))
n = sum(s)/ 2
if n % 100 or n % 200 or not s.count(100):
print('NO')
else:
print('YES')
|
Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.
|
```python
input()
s = list(map(int,input().split()))
n = sum(s)/ 2
if n % 100 or n % 200 or not s.count(100):
print('NO')
else:
print('YES')
```
| 0
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,686,517,931
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 50
| 92
| 0
|
s=input()
v="WUB"
if v in s:
s=s.replace("WUB"," ")
print(s)
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
s=input()
v="WUB"
if v in s:
s=s.replace("WUB"," ")
print(s)
```
| 0
|
|
219
|
B
|
Special Offer! Super Price 999 Bourles!
|
PROGRAMMING
| 1,400
|
[
"implementation"
] | null | null |
Polycarpus is an amateur businessman. Recently he was surprised to find out that the market for paper scissors is completely free! Without further ado, Polycarpus decided to start producing and selling such scissors.
Polycaprus calculated that the optimal celling price for such scissors would be *p* bourles. However, he read somewhere that customers are attracted by prices that say something like "Special Offer! Super price 999 bourles!". So Polycarpus decided to lower the price a little if it leads to the desired effect.
Polycarpus agrees to lower the price by no more than *d* bourles so that the number of nines at the end of the resulting price is maximum. If there are several ways to do it, he chooses the maximum possible price.
Note, Polycarpus counts only the trailing nines in a price.
|
The first line contains two integers *p* and *d* (1<=≤<=*p*<=≤<=1018; 0<=≤<=*d*<=<<=*p*) — the initial price of scissors and the maximum possible price reduction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
Print the required price — the maximum price that ends with the largest number of nines and that is less than *p* by no more than *d*.
The required number shouldn't have leading zeroes.
|
[
"1029 102\n",
"27191 17\n"
] |
[
"999\n",
"27189\n"
] |
none
| 1,000
|
[
{
"input": "1029 102",
"output": "999"
},
{
"input": "27191 17",
"output": "27189"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "9 0",
"output": "9"
},
{
"input": "20 1",
"output": "19"
},
{
"input": "100 23",
"output": "99"
},
{
"input": "10281 1",
"output": "10281"
},
{
"input": "2111 21",
"output": "2099"
},
{
"input": "3021 112",
"output": "2999"
},
{
"input": "1000000000000000000 999999999999999999",
"output": "999999999999999999"
},
{
"input": "29287101 301",
"output": "29286999"
},
{
"input": "302918113 8113",
"output": "302917999"
},
{
"input": "23483247283432 47283432",
"output": "23483239999999"
},
{
"input": "47283432 7283432",
"output": "46999999"
},
{
"input": "7283432 7283431",
"output": "6999999"
},
{
"input": "2304324853947 5853947",
"output": "2304319999999"
},
{
"input": "2485348653485 123483",
"output": "2485348599999"
},
{
"input": "29845435345 34543",
"output": "29845429999"
},
{
"input": "2348723847234234 234829384234",
"output": "2348699999999999"
},
{
"input": "2348723847234234 234829384234",
"output": "2348699999999999"
},
{
"input": "596383801524465437 13997918422040",
"output": "596379999999999999"
},
{
"input": "621306590487786841 47851896849379",
"output": "621299999999999999"
},
{
"input": "990575220328844835 100861359807341",
"output": "990499999999999999"
},
{
"input": "403277728241895842 15097810739041",
"output": "403269999999999999"
},
{
"input": "287854791214303304 98046359947548",
"output": "287799999999999999"
},
{
"input": "847222126505823289 115713658562976",
"output": "847199999999999999"
},
{
"input": "991096248227872657 181679439312637",
"output": "990999999999999999"
},
{
"input": "954402996235787062 354162450334047",
"output": "954399999999999999"
},
{
"input": "220466716596033408 44952575147901",
"output": "220459999999999999"
},
{
"input": "559198116944738707 844709119308273",
"output": "558999999999999999"
},
{
"input": "363980380443991024 4242310030748",
"output": "363979999999999999"
},
{
"input": "733498827000355608 13253459808159",
"output": "733489999999999999"
},
{
"input": "757663489894439901 139905688448459",
"output": "757599999999999999"
},
{
"input": "30528581170507487 1199546082507",
"output": "30527999999999999"
},
{
"input": "534463403123444176 67776394133861",
"output": "534399999999999999"
},
{
"input": "399943891120381720 89545256475298",
"output": "399899999999999999"
},
{
"input": "697076786191991245 95935185412097",
"output": "696999999999999999"
},
{
"input": "495773842562930245 17116719198640",
"output": "495769999999999999"
},
{
"input": "343540186435799067 48368225269792",
"output": "343499999999999999"
},
{
"input": "393776794010351632 4138311260892",
"output": "393775999999999999"
},
{
"input": "830005749156754342 157633405415940",
"output": "829999999999999999"
},
{
"input": "735716632509713228 109839072010906",
"output": "735699999999999999"
},
{
"input": "925835698451819219 232827103605000",
"output": "925799999999999999"
},
{
"input": "362064657893189225 54298707317247",
"output": "362059999999999999"
},
{
"input": "286739242579659245 61808986676984",
"output": "286699999999999999"
},
{
"input": "234522568185994645 14536016333590",
"output": "234519999999999999"
},
{
"input": "989980699593228598 382407804389880",
"output": "989899999999999999"
},
{
"input": "953287447601143003 367647762226264",
"output": "952999999999999999"
},
{
"input": "369834331957505226 421031521866991",
"output": "369799999999999999"
},
{
"input": "433225528653135646 16671330805568",
"output": "433219999999999999"
},
{
"input": "664584428369850915 516656201621892",
"output": "664499999999999999"
},
{
"input": "100813383516253625 468493737928751",
"output": "100799999999999999"
},
{
"input": "63600749936231318 12287109070881",
"output": "63599999999999999"
},
{
"input": "196643334958802150 3659421793154",
"output": "196639999999999999"
},
{
"input": "803015192835672406 14043666502157",
"output": "803009999999999999"
},
{
"input": "43201857567928862 5891486380570",
"output": "43199999999999999"
},
{
"input": "142195487377202511 32209508975060",
"output": "142189999999999999"
},
{
"input": "159171676706847083 28512592184962",
"output": "159169999999999999"
},
{
"input": "377788117133266645 12127036235155",
"output": "377779999999999999"
},
{
"input": "949501478909148807 31763408418934",
"output": "949499999999999999"
},
{
"input": "955412075341421601 220849506773896",
"output": "955399999999999999"
},
{
"input": "652742935922718161 11045914932687",
"output": "652739999999999999"
},
{
"input": "371621017875752909 511452352707014",
"output": "371599999999999999"
},
{
"input": "979748686171802330 281906901894586",
"output": "979699999999999999"
},
{
"input": "987860891213585005 85386263418762",
"output": "987799999999999999"
},
{
"input": "59225847802373220 8605552735740",
"output": "59219999999999999"
},
{
"input": "22532595810287625 1459945485391",
"output": "22531999999999999"
},
{
"input": "191654878233371957 258451919478343",
"output": "191599999999999999"
},
{
"input": "796937674525939896 892734175683845",
"output": "796899999999999999"
},
{
"input": "166564871934000326 22888347028438",
"output": "166559999999999999"
},
{
"input": "559198116944738707 84470911930827",
"output": "559189999999999999"
},
{
"input": "559198116944738707 8447091193082",
"output": "559189999999999999"
},
{
"input": "559198116944738707 844709119308",
"output": "559197999999999999"
},
{
"input": "559198116944738707 84470911930",
"output": "559198099999999999"
},
{
"input": "559198116944738707 8447091193",
"output": "559198109999999999"
},
{
"input": "559198116944738707 844709119",
"output": "559198116899999999"
},
{
"input": "559198116944738707 84470911",
"output": "559198116899999999"
},
{
"input": "559198116944738707 8447091",
"output": "559198116939999999"
},
{
"input": "559198116944738707 844709",
"output": "559198116943999999"
},
{
"input": "559198116944738707 84470",
"output": "559198116944699999"
},
{
"input": "559198116944738707 8447",
"output": "559198116944737999"
},
{
"input": "559198116944738707 844",
"output": "559198116944737999"
},
{
"input": "559198116944738707 84",
"output": "559198116944738699"
},
{
"input": "559198116944738707 8",
"output": "559198116944738699"
},
{
"input": "559198116944738707 7",
"output": "559198116944738707"
},
{
"input": "559198116944738707 6",
"output": "559198116944738707"
},
{
"input": "559198116944738707 1",
"output": "559198116944738707"
},
{
"input": "559198116944738707 0",
"output": "559198116944738707"
},
{
"input": "559198116944738700 1",
"output": "559198116944738699"
},
{
"input": "559198116944738700 0",
"output": "559198116944738700"
},
{
"input": "559198116944738999 0",
"output": "559198116944738999"
},
{
"input": "559198116944738999 1",
"output": "559198116944738999"
},
{
"input": "199 100",
"output": "199"
},
{
"input": "99 10",
"output": "99"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "18 17",
"output": "9"
},
{
"input": "199 198",
"output": "199"
},
{
"input": "1000000000000000000 0",
"output": "1000000000000000000"
},
{
"input": "59 3",
"output": "59"
},
{
"input": "9999 10",
"output": "9999"
},
{
"input": "999999999999999998 999999999999999997",
"output": "899999999999999999"
},
{
"input": "8 7",
"output": "8"
}
] | 1,666,446,829
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 154
| 135,270,400
|
from collections import defaultdict, deque
from functools import lru_cache
from heapq import heappush, heappop
from bisect import bisect_right, bisect_left
from fractions import Fraction as frac
import math
hpop = heappop
hpush = heappush
MOD = 10**9 + 7
def calc(x):
d_sum = sum(map(int,str(x)))
return x*x + d_sum *x
def solution():
n,d = map(int, input().split())
str_n = list(str(n))
ans = n
end = len(str_n) - 1
while end >= 0 and str_n[end] == "9":
end -= 1
for i in range(end)[::-1]:
cur_val = int(str_n[i])
if cur_val != 0:
length = len(str_n) - (i+1)
nines = "9"*length
new_str_n = "".join(str_n[:i]) + str(cur_val - 1) + nines
new_n = int(new_str_n)
if n - int(new_n) > d:
break
ans = new_n
return print(ans)
def main():
t = 1
#t = int(input())
for _ in range(t):
solution()
import sys
import threading
sys.setrecursionlimit(1 << 30)
threading.stack_size(1 << 27)
thread = threading.Thread(target=main)
thread.start(); thread.join()
#main()
|
Title: Special Offer! Super Price 999 Bourles!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is an amateur businessman. Recently he was surprised to find out that the market for paper scissors is completely free! Without further ado, Polycarpus decided to start producing and selling such scissors.
Polycaprus calculated that the optimal celling price for such scissors would be *p* bourles. However, he read somewhere that customers are attracted by prices that say something like "Special Offer! Super price 999 bourles!". So Polycarpus decided to lower the price a little if it leads to the desired effect.
Polycarpus agrees to lower the price by no more than *d* bourles so that the number of nines at the end of the resulting price is maximum. If there are several ways to do it, he chooses the maximum possible price.
Note, Polycarpus counts only the trailing nines in a price.
Input Specification:
The first line contains two integers *p* and *d* (1<=≤<=*p*<=≤<=1018; 0<=≤<=*d*<=<<=*p*) — the initial price of scissors and the maximum possible price reduction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Output Specification:
Print the required price — the maximum price that ends with the largest number of nines and that is less than *p* by no more than *d*.
The required number shouldn't have leading zeroes.
Demo Input:
['1029 102\n', '27191 17\n']
Demo Output:
['999\n', '27189\n']
Note:
none
|
```python
from collections import defaultdict, deque
from functools import lru_cache
from heapq import heappush, heappop
from bisect import bisect_right, bisect_left
from fractions import Fraction as frac
import math
hpop = heappop
hpush = heappush
MOD = 10**9 + 7
def calc(x):
d_sum = sum(map(int,str(x)))
return x*x + d_sum *x
def solution():
n,d = map(int, input().split())
str_n = list(str(n))
ans = n
end = len(str_n) - 1
while end >= 0 and str_n[end] == "9":
end -= 1
for i in range(end)[::-1]:
cur_val = int(str_n[i])
if cur_val != 0:
length = len(str_n) - (i+1)
nines = "9"*length
new_str_n = "".join(str_n[:i]) + str(cur_val - 1) + nines
new_n = int(new_str_n)
if n - int(new_n) > d:
break
ans = new_n
return print(ans)
def main():
t = 1
#t = int(input())
for _ in range(t):
solution()
import sys
import threading
sys.setrecursionlimit(1 << 30)
threading.stack_size(1 << 27)
thread = threading.Thread(target=main)
thread.start(); thread.join()
#main()
```
| 3
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,560,342,053
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 218
| 0
|
n=int(input())
inp = list(map(int,input().split()))
even=0
odd=0
for i in range(len(inp)):
if(inp[i]%2==0):
even+=1
else:
odd+=1
if((odd>1 and even==1) or (even>1 and odd==1)):
print(i+1)
break
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n=int(input())
inp = list(map(int,input().split()))
even=0
odd=0
for i in range(len(inp)):
if(inp[i]%2==0):
even+=1
else:
odd+=1
if((odd>1 and even==1) or (even>1 and odd==1)):
print(i+1)
break
```
| 0
|
115
|
A
|
Party
|
PROGRAMMING
| 900
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
|
Print a single integer denoting the minimum number of groups that will be formed in the party.
|
[
"5\n-1\n1\n2\n1\n-1\n"
] |
[
"3\n"
] |
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
| 500
|
[
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": "2"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n2\n-1",
"output": "2"
},
{
"input": "2\n-1\n-1",
"output": "1"
},
{
"input": "3\n2\n-1\n1",
"output": "3"
},
{
"input": "3\n-1\n-1\n-1",
"output": "1"
},
{
"input": "5\n4\n5\n1\n-1\n4",
"output": "3"
},
{
"input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7",
"output": "4"
},
{
"input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4",
"output": "5"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1",
"output": "2"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1",
"output": "1"
},
{
"input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11",
"output": "12"
},
{
"input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2",
"output": "11"
},
{
"input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6",
"output": "6"
},
{
"input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1",
"output": "3"
},
{
"input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1",
"output": "2"
},
{
"input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1",
"output": "3"
},
{
"input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8",
"output": "5"
},
{
"input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10",
"output": "6"
},
{
"input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1",
"output": "4"
},
{
"input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9",
"output": "4"
},
{
"input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8",
"output": "3"
},
{
"input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11",
"output": "7"
},
{
"input": "7\n7\n6\n-1\n-1\n4\n4\n-1",
"output": "3"
},
{
"input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34",
"output": "6"
},
{
"input": "5\n2\n3\n4\n5\n-1",
"output": "5"
}
] | 1,645,008,943
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 106
| 904
| 0
|
n = int(input())
Ps = []
for i in range(n):
Ps.append(int(input()))
maxLength = 1
for i in range(n):
if Ps[i] != -1:
j = i
length = 0
while j >= 0:
j = Ps[j] - 1
length += 1
maxLength = max(length, maxLength)
print(maxLength)
|
Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
|
```python
n = int(input())
Ps = []
for i in range(n):
Ps.append(int(input()))
maxLength = 1
for i in range(n):
if Ps[i] != -1:
j = i
length = 0
while j >= 0:
j = Ps[j] - 1
length += 1
maxLength = max(length, maxLength)
print(maxLength)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 0
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,573,987,755
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 104
| 468
| 1,843,200
|
def lhalf(txt):
return txt[0:len(txt)//2]
def rhalf(txt):
return txt[len(txt)//2:]
def rsort(txt):
if len(txt)%2 == 1:
return txt
else:
subtxt1 = rsort(lhalf(txt))
subtxt2 = rsort(rhalf(txt))
return ''.join(sorted([subtxt1,subtxt2]))
txt1 = input()
txt2 = input()
if rsort(txt1) == rsort(txt2):
print('YES')
else:
print('NO')
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
def lhalf(txt):
return txt[0:len(txt)//2]
def rhalf(txt):
return txt[len(txt)//2:]
def rsort(txt):
if len(txt)%2 == 1:
return txt
else:
subtxt1 = rsort(lhalf(txt))
subtxt2 = rsort(rhalf(txt))
return ''.join(sorted([subtxt1,subtxt2]))
txt1 = input()
txt2 = input()
if rsort(txt1) == rsort(txt2):
print('YES')
else:
print('NO')
```
| 3
|
|
772
|
A
|
Voltage Keepsake
|
PROGRAMMING
| 1,800
|
[
"binary search",
"math"
] | null | null |
You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
|
The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning.
|
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
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\n2 2\n2 1000\n",
"1 100\n1 1\n",
"3 5\n4 3\n5 2\n6 1\n"
] |
[
"2.0000000000",
"-1\n",
"0.5000000000"
] |
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.
| 500
|
[
{
"input": "2 1\n2 2\n2 1000",
"output": "2.0000000000"
},
{
"input": "1 100\n1 1",
"output": "-1"
},
{
"input": "3 5\n4 3\n5 2\n6 1",
"output": "0.5000000000"
},
{
"input": "1 1\n1 87",
"output": "-1"
},
{
"input": "1 1\n100 77",
"output": "0.7777777778"
},
{
"input": "5 10\n3 81\n3 49\n1 20\n1 12\n1 30",
"output": "-1"
},
{
"input": "5 10\n4 3\n1 54\n2 57\n2 31\n1 99",
"output": "-1"
},
{
"input": "5 10\n2 81\n3 31\n4 49\n1 35\n1 67",
"output": "263.0000000000"
},
{
"input": "10 1\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92",
"output": "102.2222222222"
},
{
"input": "10 1\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16",
"output": "17.7777777778"
},
{
"input": "10 1\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40",
"output": "44.4444444444"
},
{
"input": "2 1\n1 10\n1 10",
"output": "20.0000000000"
},
{
"input": "20 16807\n75250 50074\n43659 8931\n11273 27545\n50879 77924\n37710 64441\n38166 84493\n43043 7988\n22504 82328\n31730 78841\n42613 44304\n33170 17710\n97158 29561\n70934 93100\n80279 51817\n95336 99098\n7827 13513\n29268 23811\n77634 80980\n79150 36580\n58822 11968",
"output": "0.2244225704"
}
] | 1,591,459,828
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 2,000
| 6,348,800
|
line = input().split()
n = int(line[0])
recharge_power = int(line[1])
chargers = []
total_discharge = -recharge_power
total_power = 0
for i in range(n):
line = input().split()
draining_power = int(line[0])
curr_power = int(line[1])
chargers.append([draining_power, curr_power])
total_discharge += draining_power
total_power += curr_power
def solve(time):
available_charging = recharge_power * time
charging_required = 0
for charger in chargers:
needed_power = charger[0] * time - charger[1]
charging_required += max(needed_power, 0)
delta = charging_required - available_charging
delta /= recharge_power
return delta
def binary_search(left, right):
mid = (left + right) / 2
if solve(mid) <= 0.0001 and solve(mid) >= -0.0001:
return mid
elif solve(mid) > 0:
return binary_search(left, mid)
else:
return binary_search(mid, right)
if total_discharge <= 0:
print(-1)
else:
maximum_time = total_power / total_discharge
print(binary_search(0, maximum_time + 7))
|
Title: Voltage Keepsake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Input Specification:
The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning.
Output Specification:
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
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\n2 2\n2 1000\n', '1 100\n1 1\n', '3 5\n4 3\n5 2\n6 1\n']
Demo Output:
['2.0000000000', '-1\n', '0.5000000000']
Note:
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.
|
```python
line = input().split()
n = int(line[0])
recharge_power = int(line[1])
chargers = []
total_discharge = -recharge_power
total_power = 0
for i in range(n):
line = input().split()
draining_power = int(line[0])
curr_power = int(line[1])
chargers.append([draining_power, curr_power])
total_discharge += draining_power
total_power += curr_power
def solve(time):
available_charging = recharge_power * time
charging_required = 0
for charger in chargers:
needed_power = charger[0] * time - charger[1]
charging_required += max(needed_power, 0)
delta = charging_required - available_charging
delta /= recharge_power
return delta
def binary_search(left, right):
mid = (left + right) / 2
if solve(mid) <= 0.0001 and solve(mid) >= -0.0001:
return mid
elif solve(mid) > 0:
return binary_search(left, mid)
else:
return binary_search(mid, right)
if total_discharge <= 0:
print(-1)
else:
maximum_time = total_power / total_discharge
print(binary_search(0, maximum_time + 7))
```
| 0
|
|
174
|
A
|
Problem About Equation
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
A group of *n* merry programmers celebrate Robert Floyd's birthday. Polucarpus has got an honourable task of pouring Ber-Cola to everybody. Pouring the same amount of Ber-Cola to everybody is really important. In other words, the drink's volume in each of the *n* mugs must be the same.
Polycarpus has already began the process and he partially emptied the Ber-Cola bottle. Now the first mug has *a*1 milliliters of the drink, the second one has *a*2 milliliters and so on. The bottle has *b* milliliters left and Polycarpus plans to pour them into the mugs so that the main equation was fulfilled.
Write a program that would determine what volume of the drink Polycarpus needs to add into each mug to ensure that the following two conditions were fulfilled simultaneously:
- there were *b* milliliters poured in total. That is, the bottle need to be emptied; - after the process is over, the volumes of the drink in the mugs should be equal.
|
The first line contains a pair of integers *n*, *b* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*b*<=≤<=100), where *n* is the total number of friends in the group and *b* is the current volume of drink in the bottle. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the current volume of drink in the *i*-th mug.
|
Print a single number "-1" (without the quotes), if there is no solution. Otherwise, print *n* float numbers *c*1,<=*c*2,<=...,<=*c**n*, where *c**i* is the volume of the drink to add in the *i*-th mug. Print the numbers with no less than 6 digits after the decimal point, print each *c**i* on a single line. Polycarpus proved that if a solution exists then it is unique.
Russian locale is installed by default on the testing computer. Make sure that your solution use the point to separate the integer part of a real number from the decimal, not a comma.
|
[
"5 50\n1 2 3 4 5\n",
"2 2\n1 100\n"
] |
[
"12.000000\n11.000000\n10.000000\n9.000000\n8.000000\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5 50\n1 2 3 4 5",
"output": "12.000000\n11.000000\n10.000000\n9.000000\n8.000000"
},
{
"input": "2 2\n1 100",
"output": "-1"
},
{
"input": "2 2\n1 1",
"output": "1.000000\n1.000000"
},
{
"input": "3 2\n1 2 1",
"output": "1.000000\n0.000000\n1.000000"
},
{
"input": "3 5\n1 2 1",
"output": "2.000000\n1.000000\n2.000000"
},
{
"input": "10 95\n0 0 0 0 0 1 1 1 1 1",
"output": "10.000000\n10.000000\n10.000000\n10.000000\n10.000000\n9.000000\n9.000000\n9.000000\n9.000000\n9.000000"
},
{
"input": "3 5\n1 2 3",
"output": "2.666667\n1.666667\n0.666667"
},
{
"input": "3 5\n1 3 2",
"output": "2.666667\n0.666667\n1.666667"
},
{
"input": "3 5\n2 1 3",
"output": "1.666667\n2.666667\n0.666667"
},
{
"input": "3 5\n2 3 1",
"output": "1.666667\n0.666667\n2.666667"
},
{
"input": "3 5\n3 1 2",
"output": "0.666667\n2.666667\n1.666667"
},
{
"input": "3 5\n3 2 1",
"output": "0.666667\n1.666667\n2.666667"
},
{
"input": "2 1\n1 1",
"output": "0.500000\n0.500000"
},
{
"input": "2 1\n2 2",
"output": "0.500000\n0.500000"
},
{
"input": "3 2\n2 1 2",
"output": "0.333333\n1.333333\n0.333333"
},
{
"input": "3 3\n2 2 1",
"output": "0.666667\n0.666667\n1.666667"
},
{
"input": "3 3\n3 1 2",
"output": "0.000000\n2.000000\n1.000000"
},
{
"input": "100 100\n37 97 75 52 33 29 51 22 33 37 45 96 96 60 82 58 86 71 28 73 38 50 6 6 90 17 26 76 13 41 100 47 17 93 4 1 56 16 41 74 25 17 69 61 39 37 96 73 49 93 52 14 62 24 91 30 9 97 52 100 6 16 85 8 12 26 10 3 94 63 80 27 29 78 9 48 79 64 60 18 98 75 81 35 24 81 2 100 23 70 21 60 98 38 29 29 58 37 49 72",
"output": "-1"
},
{
"input": "100 100\n1 3 7 7 9 5 9 3 7 8 10 1 3 10 10 6 1 3 10 4 3 9 4 9 5 4 9 2 8 7 4 3 3 3 5 10 8 9 10 1 9 2 4 8 3 10 9 2 3 9 8 2 4 4 4 7 1 1 7 3 7 8 9 5 1 2 6 7 1 10 9 10 5 10 1 10 5 2 4 3 10 1 6 5 6 7 8 9 3 8 6 10 8 7 2 3 8 6 3 6",
"output": "-1"
},
{
"input": "100 61\n81 80 83 72 87 76 91 92 77 93 77 94 76 73 71 88 88 76 87 73 89 73 85 81 79 90 76 73 82 93 79 93 71 75 72 71 78 85 92 89 88 93 74 87 71 94 74 87 85 89 90 93 86 94 92 87 90 91 75 73 90 84 92 94 92 79 74 85 74 74 89 76 84 84 84 83 86 84 82 71 76 74 83 81 89 73 73 74 71 77 90 94 73 94 73 75 93 89 84 92",
"output": "-1"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1..."
},
{
"input": "100 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1.000000\n1..."
},
{
"input": "100 100\n99 100 99 100 100 100 99 99 99 100 100 100 99 100 99 100 100 100 100 100 99 99 99 99 100 99 100 99 100 99 99 100 100 100 100 100 99 99 99 100 99 99 100 99 100 99 100 99 99 99 99 100 100 99 99 99 100 100 99 100 100 100 99 99 100 100 100 100 100 100 99 99 99 99 99 100 99 99 100 99 100 100 100 99 100 99 99 100 99 100 100 100 99 100 99 100 100 100 100 99",
"output": "1.530000\n0.530000\n1.530000\n0.530000\n0.530000\n0.530000\n1.530000\n1.530000\n1.530000\n0.530000\n0.530000\n0.530000\n1.530000\n0.530000\n1.530000\n0.530000\n0.530000\n0.530000\n0.530000\n0.530000\n1.530000\n1.530000\n1.530000\n1.530000\n0.530000\n1.530000\n0.530000\n1.530000\n0.530000\n1.530000\n1.530000\n0.530000\n0.530000\n0.530000\n0.530000\n0.530000\n1.530000\n1.530000\n1.530000\n0.530000\n1.530000\n1.530000\n0.530000\n1.530000\n0.530000\n1.530000\n0.530000\n1.530000\n1.530000\n1.530000\n1.530000\n0..."
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100",
"output": "0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n1.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n1.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0.940000\n0..."
},
{
"input": "100 100\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99",
"output": "1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n1.020000\n0.020000\n1.020000\n1..."
},
{
"input": "10 100\n52 52 51 52 52 52 51 51 52 52",
"output": "9.700000\n9.700000\n10.700000\n9.700000\n9.700000\n9.700000\n10.700000\n10.700000\n9.700000\n9.700000"
},
{
"input": "10 100\n13 13 13 13 12 13 12 13 12 12",
"output": "9.600000\n9.600000\n9.600000\n9.600000\n10.600000\n9.600000\n10.600000\n9.600000\n10.600000\n10.600000"
},
{
"input": "10 100\n50 51 47 51 48 46 49 51 46 51",
"output": "9.000000\n8.000000\n12.000000\n8.000000\n11.000000\n13.000000\n10.000000\n8.000000\n13.000000\n8.000000"
},
{
"input": "10 100\n13 13 9 12 12 11 13 8 10 13",
"output": "8.400000\n8.400000\n12.400000\n9.400000\n9.400000\n10.400000\n8.400000\n13.400000\n11.400000\n8.400000"
},
{
"input": "93 91\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0.978495\n0..."
},
{
"input": "93 97\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1.043011\n1..."
},
{
"input": "91 99\n99 100 100 100 99 100 100 100 99 100 99 99 100 99 100 100 100 99 99 100 99 100 100 100 100 100 99 99 100 99 100 99 99 100 100 100 100 99 99 100 100 100 99 100 100 99 100 100 99 100 99 99 99 100 99 99 99 100 99 100 99 100 99 100 99 99 100 100 100 100 99 100 99 100 99 99 100 100 99 100 100 100 100 99 99 100 100 99 99 100 99",
"output": "1.648352\n0.648352\n0.648352\n0.648352\n1.648352\n0.648352\n0.648352\n0.648352\n1.648352\n0.648352\n1.648352\n1.648352\n0.648352\n1.648352\n0.648352\n0.648352\n0.648352\n1.648352\n1.648352\n0.648352\n1.648352\n0.648352\n0.648352\n0.648352\n0.648352\n0.648352\n1.648352\n1.648352\n0.648352\n1.648352\n0.648352\n1.648352\n1.648352\n0.648352\n0.648352\n0.648352\n0.648352\n1.648352\n1.648352\n0.648352\n0.648352\n0.648352\n1.648352\n0.648352\n0.648352\n1.648352\n0.648352\n0.648352\n1.648352\n0.648352\n1.648352\n1..."
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{
"input": "99 98\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n1.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0.979798\n0..."
},
{
"input": "98 99\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99 100 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 100 99 99 99 99 99",
"output": "1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n0.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n1.051020\n0.051020\n1.051020\n1..."
},
{
"input": "13 97\n52 52 51 51 52 52 51 52 51 51 52 52 52",
"output": "7.076923\n7.076923\n8.076923\n8.076923\n7.076923\n7.076923\n8.076923\n7.076923\n8.076923\n8.076923\n7.076923\n7.076923\n7.076923"
},
{
"input": "17 99\n13 13 12 13 11 12 12 12 13 13 11 13 13 13 13 12 13",
"output": "5.294118\n5.294118\n6.294118\n5.294118\n7.294118\n6.294118\n6.294118\n6.294118\n5.294118\n5.294118\n7.294118\n5.294118\n5.294118\n5.294118\n5.294118\n6.294118\n5.294118"
},
{
"input": "9 91\n52 51 50 52 52 51 50 48 51",
"output": "8.888889\n9.888889\n10.888889\n8.888889\n8.888889\n9.888889\n10.888889\n12.888889\n9.888889"
},
{
"input": "17 91\n13 13 13 13 12 12 13 13 12 13 12 13 10 12 13 13 12",
"output": "4.823529\n4.823529\n4.823529\n4.823529\n5.823529\n5.823529\n4.823529\n4.823529\n5.823529\n4.823529\n5.823529\n4.823529\n7.823529\n5.823529\n4.823529\n4.823529\n5.823529"
},
{
"input": "2 3\n1 1",
"output": "1.500000\n1.500000"
},
{
"input": "2 90\n0 89",
"output": "89.500000\n0.500000"
},
{
"input": "4 17\n3 4 8 1",
"output": "5.250000\n4.250000\n0.250000\n7.250000"
},
{
"input": "2 9\n5 5",
"output": "4.500000\n4.500000"
},
{
"input": "7 28\n1 3 9 10 9 6 10",
"output": "9.857143\n7.857143\n1.857143\n0.857143\n1.857143\n4.857143\n0.857143"
},
{
"input": "5 11\n1 2 3 4 5",
"output": "4.200000\n3.200000\n2.200000\n1.200000\n0.200000"
},
{
"input": "2 1\n1 1",
"output": "0.500000\n0.500000"
},
{
"input": "5 3\n1 1 1 1 1",
"output": "0.600000\n0.600000\n0.600000\n0.600000\n0.600000"
},
{
"input": "3 1\n100 100 100",
"output": "0.333333\n0.333333\n0.333333"
},
{
"input": "5 50\n2 2 3 2 2",
"output": "10.200000\n10.200000\n9.200000\n10.200000\n10.200000"
},
{
"input": "3 3\n2 2 3",
"output": "1.333333\n1.333333\n0.333333"
},
{
"input": "2 52\n2 100",
"output": "-1"
},
{
"input": "3 2\n2 2 3",
"output": "1.000000\n1.000000\n0.000000"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "0.200000\n0.200000\n0.200000\n0.200000\n0.200000"
},
{
"input": "2 4\n1 2",
"output": "2.500000\n1.500000"
},
{
"input": "5 49\n1 2 3 4 5",
"output": "11.800000\n10.800000\n9.800000\n8.800000\n7.800000"
}
] | 1,556,307,840
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 14
| 218
| 0
|
r = lambda: int(input())
ra = lambda: [*map(int, input().split())]
p = lambda a: print('{:.6f}'.format(a))
n, m = ra()
a = ra()
b = []
e = (m+sum(a))/n
for i in range(n):
b.append(e-a[i])
m-=abs(b[len(b)-1])
a[i]+=b[len(b)-1]
if m<0:
print(-1)
else:
for i in b:
p(i)
|
Title: Problem About Equation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A group of *n* merry programmers celebrate Robert Floyd's birthday. Polucarpus has got an honourable task of pouring Ber-Cola to everybody. Pouring the same amount of Ber-Cola to everybody is really important. In other words, the drink's volume in each of the *n* mugs must be the same.
Polycarpus has already began the process and he partially emptied the Ber-Cola bottle. Now the first mug has *a*1 milliliters of the drink, the second one has *a*2 milliliters and so on. The bottle has *b* milliliters left and Polycarpus plans to pour them into the mugs so that the main equation was fulfilled.
Write a program that would determine what volume of the drink Polycarpus needs to add into each mug to ensure that the following two conditions were fulfilled simultaneously:
- there were *b* milliliters poured in total. That is, the bottle need to be emptied; - after the process is over, the volumes of the drink in the mugs should be equal.
Input Specification:
The first line contains a pair of integers *n*, *b* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*b*<=≤<=100), where *n* is the total number of friends in the group and *b* is the current volume of drink in the bottle. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the current volume of drink in the *i*-th mug.
Output Specification:
Print a single number "-1" (without the quotes), if there is no solution. Otherwise, print *n* float numbers *c*1,<=*c*2,<=...,<=*c**n*, where *c**i* is the volume of the drink to add in the *i*-th mug. Print the numbers with no less than 6 digits after the decimal point, print each *c**i* on a single line. Polycarpus proved that if a solution exists then it is unique.
Russian locale is installed by default on the testing computer. Make sure that your solution use the point to separate the integer part of a real number from the decimal, not a comma.
Demo Input:
['5 50\n1 2 3 4 5\n', '2 2\n1 100\n']
Demo Output:
['12.000000\n11.000000\n10.000000\n9.000000\n8.000000\n', '-1\n']
Note:
none
|
```python
r = lambda: int(input())
ra = lambda: [*map(int, input().split())]
p = lambda a: print('{:.6f}'.format(a))
n, m = ra()
a = ra()
b = []
e = (m+sum(a))/n
for i in range(n):
b.append(e-a[i])
m-=abs(b[len(b)-1])
a[i]+=b[len(b)-1]
if m<0:
print(-1)
else:
for i in b:
p(i)
```
| 0
|
|
609
|
B
|
The Best Gift
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"implementation"
] | null | null |
Emily's birthday is next week and Jack has decided to buy a present for her. He knows she loves books so he goes to the local bookshop, where there are *n* books on sale from one of *m* genres.
In the bookshop, Jack decides to buy two books of different genres.
Based on the genre of books on sale in the shop, find the number of options available to Jack for choosing two books of different genres for Emily. Options are considered different if they differ in at least one book.
The books are given by indices of their genres. The genres are numbered from 1 to *m*.
|
The first line contains two positive integers *n* and *m* (2<=≤<=*n*<=≤<=2·105,<=2<=≤<=*m*<=≤<=10) — the number of books in the bookstore and the number of genres.
The second line contains a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (1<=≤<=*a**i*<=≤<=*m*) equals the genre of the *i*-th book.
It is guaranteed that for each genre there is at least one book of that genre.
|
Print the only integer — the number of ways in which Jack can choose books.
It is guaranteed that the answer doesn't exceed the value 2·109.
|
[
"4 3\n2 1 3 1\n",
"7 4\n4 2 3 1 2 4 3\n"
] |
[
"5\n",
"18\n"
] |
The answer to the first test sample equals 5 as Sasha can choose:
1. the first and second books, 1. the first and third books, 1. the first and fourth books, 1. the second and third books, 1. the third and fourth books.
| 0
|
[
{
"input": "4 3\n2 1 3 1",
"output": "5"
},
{
"input": "7 4\n4 2 3 1 2 4 3",
"output": "18"
},
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "3 2\n1 2 2",
"output": "2"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "45"
},
{
"input": "9 2\n1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "12 3\n1 2 3 1 2 3 1 2 3 1 2 3",
"output": "48"
},
{
"input": "100 3\n2 1 1 1 3 2 3 3 2 3 3 1 3 3 1 3 3 1 1 1 2 3 1 2 3 1 2 3 3 1 3 1 1 2 3 2 3 3 2 3 3 1 2 2 1 2 3 2 3 2 2 1 1 3 1 3 2 1 3 1 3 1 3 1 1 3 3 3 2 3 2 2 2 2 1 3 3 3 1 2 1 2 3 2 1 3 1 3 2 1 3 1 2 1 2 3 1 3 2 3",
"output": "3296"
},
{
"input": "100 5\n5 5 2 4 5 4 4 4 4 2 5 3 4 2 4 4 1 1 5 3 2 2 1 3 3 2 5 3 4 5 1 3 5 4 4 4 3 1 4 4 3 4 5 2 5 4 2 1 2 2 3 5 5 5 1 4 5 3 1 4 2 2 5 1 5 3 4 1 5 1 2 2 3 5 1 3 2 4 2 4 2 2 4 1 3 5 2 2 2 3 3 4 3 2 2 5 5 4 2 5",
"output": "3953"
},
{
"input": "100 10\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "4428"
}
] | 1,535,509,737
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 202
| 3,993,600
|
n, m = map(int, input().split())
books_per_genre = [0]*m
books = list(map(int, input().split()))
for book in books:
books_per_genre[book-1] += 1
result = 0
used = 0
for genre in books_per_genre:
result += (genre*(n-genre-used))
used += genre
print(result)
|
Title: The Best Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emily's birthday is next week and Jack has decided to buy a present for her. He knows she loves books so he goes to the local bookshop, where there are *n* books on sale from one of *m* genres.
In the bookshop, Jack decides to buy two books of different genres.
Based on the genre of books on sale in the shop, find the number of options available to Jack for choosing two books of different genres for Emily. Options are considered different if they differ in at least one book.
The books are given by indices of their genres. The genres are numbered from 1 to *m*.
Input Specification:
The first line contains two positive integers *n* and *m* (2<=≤<=*n*<=≤<=2·105,<=2<=≤<=*m*<=≤<=10) — the number of books in the bookstore and the number of genres.
The second line contains a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (1<=≤<=*a**i*<=≤<=*m*) equals the genre of the *i*-th book.
It is guaranteed that for each genre there is at least one book of that genre.
Output Specification:
Print the only integer — the number of ways in which Jack can choose books.
It is guaranteed that the answer doesn't exceed the value 2·109.
Demo Input:
['4 3\n2 1 3 1\n', '7 4\n4 2 3 1 2 4 3\n']
Demo Output:
['5\n', '18\n']
Note:
The answer to the first test sample equals 5 as Sasha can choose:
1. the first and second books, 1. the first and third books, 1. the first and fourth books, 1. the second and third books, 1. the third and fourth books.
|
```python
n, m = map(int, input().split())
books_per_genre = [0]*m
books = list(map(int, input().split()))
for book in books:
books_per_genre[book-1] += 1
result = 0
used = 0
for genre in books_per_genre:
result += (genre*(n-genre-used))
used += genre
print(result)
```
| 3
|
|
120
|
C
|
Winnie-the-Pooh and honey
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] | null | null |
As we all know, Winnie-the-Pooh just adores honey. Ones he and the Piglet found out that the Rabbit has recently gotten hold of an impressive amount of this sweet and healthy snack. As you may guess, Winnie and the Piglet asked to come at the Rabbit's place. Thus, there are *n* jars of honey lined up in front of Winnie-the-Pooh, jar number *i* contains *a**i* kilos of honey. Winnie-the-Pooh eats the honey like that: each time he chooses a jar containing most honey. If the jar has less that *k* kilos of honey or if Winnie-the-Pooh has already eaten from it three times, he gives the jar to Piglet. Otherwise he eats exactly *k* kilos of honey from the jar and puts it back. Winnie does so until he gives all jars to the Piglet. Count how much honey Piglet will overall get after Winnie satisfies his hunger.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, separated by spaces (1<=≤<=*a**i*<=≤<=100).
|
Print a single number — how many kilos of honey gets Piglet.
|
[
"3 3\n15 8 10\n"
] |
[
"9\n"
] |
none
| 0
|
[
{
"input": "3 3\n15 8 10",
"output": "9"
},
{
"input": "1 3\n3",
"output": "0"
},
{
"input": "3 4\n3 8 2",
"output": "5"
},
{
"input": "3 2\n95 25 49",
"output": "151"
},
{
"input": "3 1\n8 3 2",
"output": "5"
},
{
"input": "5 1\n4 7 9 5 7",
"output": "17"
},
{
"input": "8 6\n19 15 1 14 7 2 10 14",
"output": "16"
},
{
"input": "8 5\n5 2 17 12 16 12 17 3",
"output": "14"
},
{
"input": "10 7\n26 11 10 8 5 20 9 27 30 9",
"output": "43"
},
{
"input": "10 10\n20 82 19 82 18 96 40 99 87 2",
"output": "325"
},
{
"input": "10 10\n75 52 78 83 60 31 46 28 33 17",
"output": "233"
},
{
"input": "20 5\n33 45 36 13 46 40 15 11 29 44 43 50 14 19 46 46 46 26 42 6",
"output": "375"
},
{
"input": "20 2\n4 2 6 9 8 4 4 7 2 3 7 7 10 6 3 5 2 9 8 5",
"output": "21"
},
{
"input": "30 3\n20 37 89 77 74 6 52 87 19 58 3 38 40 38 42 12 1 23 29 38 12 65 15 1 92 45 23 94 61 73",
"output": "1021"
},
{
"input": "30 2\n10 5 46 30 28 18 24 35 73 2 10 24 72 86 97 95 71 12 14 57 27 94 81 59 43 77 22 58 16 96",
"output": "1208"
},
{
"input": "50 13\n53 55 51 81 59 22 11 20 30 80 38 17 8 38 69 52 11 74 16 38 80 97 39 74 78 56 75 28 4 58 80 88 78 89 95 8 13 70 36 29 49 15 74 44 19 52 42 59 92 37",
"output": "1012"
},
{
"input": "100 33\n84 70 12 53 10 38 4 66 42 1 100 98 42 10 31 26 22 94 19 43 86 5 37 64 77 98 81 40 17 66 52 43 5 7 79 92 44 78 9 95 10 86 42 56 34 91 12 17 26 16 24 99 11 37 89 100 60 74 32 66 13 29 3 24 41 99 93 87 85 74 5 3 70 46 23 12 43 10 24 32 95 2 57 86 29 100 29 62 17 24 4 40 40 73 29 11 69 89 10 31",
"output": "1467"
},
{
"input": "100 12\n90 59 100 12 82 31 66 28 7 13 43 42 48 94 60 32 20 92 37 39 22 55 14 23 77 56 21 55 10 89 93 79 5 80 40 80 6 15 56 82 68 61 32 100 23 7 13 92 32 82 17 85 49 85 13 75 4 7 42 14 84 22 50 12 11 75 4 85 32 96 56 13 34 100 66 37 58 58 24 58 81 63 59 55 89 97 90 69 29 11 71 58 58 43 72 96 81 72 14 64",
"output": "2377"
},
{
"input": "100 3\n92 51 91 67 32 97 40 24 78 72 93 90 57 77 66 13 55 24 17 90 65 35 69 97 79 31 85 2 90 80 49 44 73 77 34 45 74 75 35 14 34 56 60 54 36 55 16 49 12 54 92 68 70 71 5 79 67 89 64 80 20 84 4 39 55 84 53 59 68 11 89 89 15 73 67 86 91 78 49 24 58 83 55 34 82 97 27 29 78 15 20 48 9 87 41 34 91 76 82 44",
"output": "4839"
},
{
"input": "100 1\n85 75 55 65 39 26 47 16 9 11 3 4 70 23 56 64 36 34 16 13 18 28 32 80 8 79 76 4 21 75 93 51 85 86 100 88 91 71 97 28 66 22 47 87 91 95 3 56 81 53 88 90 21 30 74 45 58 73 36 40 70 51 28 75 20 54 69 52 14 48 52 53 91 81 32 68 16 48 60 66 91 32 55 42 93 51 3 46 23 64 39 1 30 42 20 83 99 29 35 59",
"output": "4801"
}
] | 1,589,602,793
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 6,656,000
|
n,k=map(int,input().split(' '))
ans=0
arr=list(map(int,input().split(' ')))
for t in arr:
# t=int(input())
val=max(t-3*k,t%k)
ans+=val
print(ans)
#k
|
Title: Winnie-the-Pooh and honey
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As we all know, Winnie-the-Pooh just adores honey. Ones he and the Piglet found out that the Rabbit has recently gotten hold of an impressive amount of this sweet and healthy snack. As you may guess, Winnie and the Piglet asked to come at the Rabbit's place. Thus, there are *n* jars of honey lined up in front of Winnie-the-Pooh, jar number *i* contains *a**i* kilos of honey. Winnie-the-Pooh eats the honey like that: each time he chooses a jar containing most honey. If the jar has less that *k* kilos of honey or if Winnie-the-Pooh has already eaten from it three times, he gives the jar to Piglet. Otherwise he eats exactly *k* kilos of honey from the jar and puts it back. Winnie does so until he gives all jars to the Piglet. Count how much honey Piglet will overall get after Winnie satisfies his hunger.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, separated by spaces (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Print a single number — how many kilos of honey gets Piglet.
Demo Input:
['3 3\n15 8 10\n']
Demo Output:
['9\n']
Note:
none
|
```python
n,k=map(int,input().split(' '))
ans=0
arr=list(map(int,input().split(' ')))
for t in arr:
# t=int(input())
val=max(t-3*k,t%k)
ans+=val
print(ans)
#k
```
| -1
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,662,469,851
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 512,000
|
from fractions import Fraction
a, b = map(int, input().split())
x, y = 0, []
if a >= b:
x = a
else:
x = a
for i in range(x):
if x <= 6:
y.append(x)
x += 1
print(Fraction(len(y), 6))
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
from fractions import Fraction
a, b = map(int, input().split())
x, y = 0, []
if a >= b:
x = a
else:
x = a
for i in range(x):
if x <= 6:
y.append(x)
x += 1
print(Fraction(len(y), 6))
```
| 0
|
545
|
C
|
Woodcutters
|
PROGRAMMING
| 1,500
|
[
"dp",
"greedy"
] | null | null |
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
|
Print a single number — the maximum number of trees that you can cut down by the given rules.
|
[
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] |
[
"3\n",
"4\n"
] |
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
| 1,750
|
[
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10\n37 3\n39 4\n40 5\n41 1\n44 1\n47 7\n48 11\n50 6\n52 5\n57 2\n58 7\n60 4\n62 1\n67 3\n68 12\n69 8\n70 1\n71 5\n72 5\n73 6\n74 4",
"output": "10"
},
{
"input": "40\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1",
"output": "2"
},
{
"input": "67\n1 1\n3 8\n4 10\n7 8\n9 2\n10 1\n11 5\n12 8\n13 4\n16 6\n18 3\n19 3\n22 5\n24 6\n27 5\n28 3\n29 3\n30 5\n32 5\n33 10\n34 7\n35 8\n36 5\n41 3\n42 2\n43 5\n46 4\n48 4\n49 9\n52 4\n53 9\n55 1\n56 4\n59 7\n68 7\n69 4\n71 9\n72 10\n74 5\n76 4\n77 9\n80 7\n81 9\n82 5\n83 5\n84 9\n85 7\n86 9\n87 4\n88 7\n89 10\n90 3\n91 5\n92 10\n93 5\n94 8\n95 4\n96 2\n97 10\n98 1\n99 3\n100 1\n101 5\n102 4\n103 8\n104 8\n105 8",
"output": "5"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "10\n7 12\n10 2\n12 2\n15 1\n19 2\n20 1\n53 25\n63 10\n75 12\n87 1",
"output": "9"
},
{
"input": "3\n1 1\n1000 1000\n1000000000 1000000000",
"output": "3"
},
{
"input": "2\n1 999999999\n1000000000 1000000000",
"output": "2"
},
{
"input": "10\n999999900 1000000000\n999999901 1000000000\n999999902 1000000000\n999999903 1000000000\n999999904 1000000000\n999999905 1000000000\n999999906 1000000000\n999999907 1000000000\n999999908 1000000000\n999999909 1000000000",
"output": "2"
},
{
"input": "2\n100000000 1000000000\n1000000000 1000000000",
"output": "2"
}
] | 1,684,945,390
| 1,390
|
PyPy 3
|
OK
|
TESTS
| 67
| 171
| 9,728,000
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
x, h = [], []
for _ in range(n):
x0, h0 = map(int, input().split())
x.append(x0)
h.append(h0)
dp = [1, 0, 1]
for i in range(1, n):
u = [x[i - 1], x[i - 1], x[i - 1] + h[i - 1]]
v = [x[i] - h[i], x[i], x[i]]
dp0 = [0] * 3
for j in range(3):
for k in range(3):
if u[j] < v[k]:
dp0[k] = max(dp0[k], dp[j] + min(k ^ 1, 1))
dp = dp0
ans = max(dp)
print(ans)
|
Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
|
```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
x, h = [], []
for _ in range(n):
x0, h0 = map(int, input().split())
x.append(x0)
h.append(h0)
dp = [1, 0, 1]
for i in range(1, n):
u = [x[i - 1], x[i - 1], x[i - 1] + h[i - 1]]
v = [x[i] - h[i], x[i], x[i]]
dp0 = [0] * 3
for j in range(3):
for k in range(3):
if u[j] < v[k]:
dp0[k] = max(dp0[k], dp[j] + min(k ^ 1, 1))
dp = dp0
ans = max(dp)
print(ans)
```
| 3
|
|
386
|
A
|
Second-Price Auction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
|
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
|
[
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] |
[
"2 5\n",
"1 8\n",
"6 9\n"
] |
none
| 500
|
[
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,590,826,330
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 93
| 0
|
n=int(input())
c=input()
ca=c.split()
a1=max(ca)
i=ca.index(a1)+1
ca.remove(max(ca))
a2=max(ca)
print(i,a2)
|
Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none
|
```python
n=int(input())
c=input()
ca=c.split()
a1=max(ca)
i=ca.index(a1)+1
ca.remove(max(ca))
a2=max(ca)
print(i,a2)
```
| 0
|
|
501
|
A
|
Contest
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
|
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
|
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
|
[
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] |
[
"Vasya\n",
"Tie\n",
"Misha\n"
] |
none
| 500
|
[
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
"input": "750 1000 54 103",
"output": "Tie"
},
{
"input": "2000 1250 176 130",
"output": "Tie"
},
{
"input": "1250 1750 145 179",
"output": "Tie"
},
{
"input": "2000 2000 176 179",
"output": "Tie"
},
{
"input": "1500 1500 148 148",
"output": "Tie"
},
{
"input": "2750 1750 134 147",
"output": "Misha"
},
{
"input": "3250 250 175 173",
"output": "Misha"
},
{
"input": "500 500 170 176",
"output": "Misha"
},
{
"input": "250 1000 179 178",
"output": "Vasya"
},
{
"input": "3250 1000 160 138",
"output": "Misha"
},
{
"input": "3000 2000 162 118",
"output": "Tie"
},
{
"input": "1500 1250 180 160",
"output": "Tie"
},
{
"input": "1250 2500 100 176",
"output": "Tie"
},
{
"input": "3500 3500 177 178",
"output": "Tie"
},
{
"input": "3000 3250 16 34",
"output": "Tie"
},
{
"input": "1750 3000 137 49",
"output": "Vasya"
},
{
"input": "500 1500 179 71",
"output": "Vasya"
},
{
"input": "1250 2000 101 180",
"output": "Misha"
},
{
"input": "250 750 180 176",
"output": "Vasya"
},
{
"input": "2250 2250 163 145",
"output": "Vasya"
},
{
"input": "3000 3000 176 78",
"output": "Vasya"
},
{
"input": "250 3500 8 178",
"output": "Vasya"
},
{
"input": "1750 1250 179 180",
"output": "Misha"
},
{
"input": "2750 1750 13 164",
"output": "Misha"
},
{
"input": "1750 2250 178 53",
"output": "Vasya"
},
{
"input": "2500 2750 73 179",
"output": "Misha"
},
{
"input": "1000 3500 178 175",
"output": "Vasya"
},
{
"input": "1000 500 7 162",
"output": "Misha"
},
{
"input": "1000 250 175 48",
"output": "Misha"
},
{
"input": "1750 500 166 177",
"output": "Misha"
},
{
"input": "250 250 0 0",
"output": "Tie"
},
{
"input": "250 3500 0 0",
"output": "Vasya"
},
{
"input": "250 3500 0 180",
"output": "Vasya"
},
{
"input": "3500 3500 180 180",
"output": "Tie"
},
{
"input": "3500 250 0 180",
"output": "Misha"
}
] | 1,623,246,715
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 109
| 0
|
misha_pt, vasya_pt, misha_time, vasya_time = map(int, input().split())
misha_pt = max((3 * misha_pt) // 10, misha_pt - ((misha_pt * misha_time) // 250))
vasya_pt = max((3 * vasya_pt) // 10, vasya_pt - ((vasya_pt * vasya_time) // 250))
if misha_pt > vasya_pt:
print("Misha")
elif vasya_pt > misha_pt:
print("Vasya")
else:
print("Tie")
|
Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none
|
```python
misha_pt, vasya_pt, misha_time, vasya_time = map(int, input().split())
misha_pt = max((3 * misha_pt) // 10, misha_pt - ((misha_pt * misha_time) // 250))
vasya_pt = max((3 * vasya_pt) // 10, vasya_pt - ((vasya_pt * vasya_time) // 250))
if misha_pt > vasya_pt:
print("Misha")
elif vasya_pt > misha_pt:
print("Vasya")
else:
print("Tie")
```
| 3
|
|
371
|
C
|
Hamburgers
|
PROGRAMMING
| 1,600
|
[
"binary search",
"brute force"
] | null | null |
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
|
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
|
[
"BBBSSC\n6 4 1\n1 2 3\n4\n",
"BBC\n1 10 1\n1 10 1\n21\n",
"BSC\n1 1 1\n1 1 3\n1000000000000\n"
] |
[
"2\n",
"7\n",
"200000000001\n"
] |
none
| 1,500
|
[
{
"input": "BBBSSC\n6 4 1\n1 2 3\n4",
"output": "2"
},
{
"input": "BBC\n1 10 1\n1 10 1\n21",
"output": "7"
},
{
"input": "BSC\n1 1 1\n1 1 3\n1000000000000",
"output": "200000000001"
},
{
"input": "B\n1 1 1\n1 1 1\n381",
"output": "382"
},
{
"input": "BSC\n3 5 6\n7 3 9\n100",
"output": "10"
},
{
"input": "BSC\n100 1 1\n100 1 1\n100",
"output": "51"
},
{
"input": "SBBCCSBB\n1 50 100\n31 59 21\n100000",
"output": "370"
},
{
"input": "BBBBCCCCCCCCCCCCCCCCCCCCSSSSBBBBBBBBSS\n100 100 100\n1 1 1\n3628800",
"output": "95502"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n200",
"output": "0"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n2000",
"output": "1"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n300",
"output": "0"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n300000000",
"output": "42858"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n914159265358",
"output": "130594181"
},
{
"input": "SSSSSSSSSSBBBBBBBBBCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSBB\n31 53 97\n13 17 31\n914159265358",
"output": "647421579"
},
{
"input": "BBBCSBSBBSSSSCCCCBBCSBBBBSSBBBCBSCCSSCSSCSBSSSCCCCBSCSSBSSSCCCBBCCCSCBCBBCCSCCCCSBBCCBBBBCCCCCCBSSCB\n91 87 17\n64 44 43\n958532915587",
"output": "191668251"
},
{
"input": "CSSCBBCCCSBSCBBBCSBBBCBSBCSCBCSCBCBSBCBCSSBBSBBCBBBBSCSBBCCBCCBCBBSBSBCSCSBBSSBBCSSBCSCSCCSSBCBBCBSB\n56 34 48\n78 6 96\n904174875419",
"output": "140968956"
},
{
"input": "CCSCCCSBBBSCBSCSCCSSBBBSSBBBSBBBCBCSSBCSCBBCCCBCBCBCCCSSBSBBCCCCCBBSCBSCBCBBCBBCSSBCSBSSCCSCCSCCBBBS\n33 73 67\n4 56 42\n886653164314",
"output": "277425898"
},
{
"input": "SBCSSCBBSSBCSSBBBSSBSCBSSSCBBSBBBBCSBCSBSCBSCBSCBSBSSCCCCBSBCCBCBSCCCBSCCBSBBCBSSCCCCSBSBBBSSSBCSCBC\n94 16 85\n14 18 91\n836590091442",
"output": "217522127"
},
{
"input": "BSCSBSCCSCSSCCCSBCSSBCBBSCCBSCCSSSSSSSSSCCSBSCCBBCBBSBSCCCCBCSBSBSSBBBBBSSBSSCBCCSSBSSSCBBCSBBSBCCCB\n67 54 8\n36 73 37\n782232051273",
"output": "154164772"
},
{
"input": "CBBCBSBCCSCBSSCCBCSBCSBBSCBBCSCCBSCCSCSBBSSBSBSCBBSBBCSSSSBBBBSBBCBCSBBCBCSSBBCSBSCCSCSBCSCBSCCBBCSC\n71 71 52\n52 88 3\n654400055575",
"output": "137826467"
},
{
"input": "CBBCBSBCCSCBSSCCBCSBCSBBSCBBCSCCBSCCSCSBBSBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBCBBCSC\n100 1 1\n1 17 23\n954400055575",
"output": "1355681897"
},
{
"input": "C\n100 100 100\n1 1 1\n1000000000000",
"output": "1000000000100"
},
{
"input": "SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n100 100 100\n100 100 100\n1000000000000",
"output": "100000001"
},
{
"input": "B\n100 100 100\n1 1 1\n1",
"output": "101"
},
{
"input": "SC\n2 1 1\n1 1 1\n100000000000",
"output": "50000000001"
},
{
"input": "B\n100 1 1\n1 1 1\n1000000000000",
"output": "1000000000100"
},
{
"input": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\n1 1 1\n100 100 100\n1000000000000",
"output": "100000000"
},
{
"input": "CC\n1 1 1\n100 100 100\n1",
"output": "0"
},
{
"input": "B\n100 100 100\n1 1 1\n1000000000000",
"output": "1000000000100"
},
{
"input": "BSC\n100 100 100\n1 1 1\n1000000000000",
"output": "333333333433"
},
{
"input": "BSC\n100 100 100\n1 1 1\n1",
"output": "100"
}
] | 1,627,986,217
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 124
| 20,172,800
|
recipe=input()
nb,ns,nc=map(int,input().split())
pb,ps,pc=map(int,input().split())
money=int(input())
bread=0
sausages=0
cheese=0
for item in recipe:
if item=="B":
bread+=1
elif item=="S":
sausages+=1
else:
cheese+=1
i=0
j=10000000000000
while i<j-1:
mid=(i+j)//2
price=pb*max(0,mid*bread-nb)+ps*max(0,mid*sausages-ns)+pc*max(0,mid*cheese-nc)
if price<=money:
i=mid
else:
j=mid
else:
print(i)
|
Title: Hamburgers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
Input Specification:
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
Demo Input:
['BBBSSC\n6 4 1\n1 2 3\n4\n', 'BBC\n1 10 1\n1 10 1\n21\n', 'BSC\n1 1 1\n1 1 3\n1000000000000\n']
Demo Output:
['2\n', '7\n', '200000000001\n']
Note:
none
|
```python
recipe=input()
nb,ns,nc=map(int,input().split())
pb,ps,pc=map(int,input().split())
money=int(input())
bread=0
sausages=0
cheese=0
for item in recipe:
if item=="B":
bread+=1
elif item=="S":
sausages+=1
else:
cheese+=1
i=0
j=10000000000000
while i<j-1:
mid=(i+j)//2
price=pb*max(0,mid*bread-nb)+ps*max(0,mid*sausages-ns)+pc*max(0,mid*cheese-nc)
if price<=money:
i=mid
else:
j=mid
else:
print(i)
```
| 3
|
|
411
|
A
|
Password Check
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
|
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
|
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
|
[
"abacaba\n",
"X12345\n",
"CONTEST_is_STARTED!!11\n"
] |
[
"Too weak\n",
"Too weak\n",
"Correct\n"
] |
none
| 0
|
[
{
"input": "abacaba",
"output": "Too weak"
},
{
"input": "X12345",
"output": "Too weak"
},
{
"input": "CONTEST_is_STARTED!!11",
"output": "Correct"
},
{
"input": "1zA__",
"output": "Correct"
},
{
"input": "1zA_",
"output": "Too weak"
},
{
"input": "zA___",
"output": "Too weak"
},
{
"input": "1A___",
"output": "Too weak"
},
{
"input": "z1___",
"output": "Too weak"
},
{
"input": "0",
"output": "Too weak"
},
{
"input": "_",
"output": "Too weak"
},
{
"input": "a",
"output": "Too weak"
},
{
"input": "D",
"output": "Too weak"
},
{
"input": "_",
"output": "Too weak"
},
{
"input": "?",
"output": "Too weak"
},
{
"input": "?",
"output": "Too weak"
},
{
"input": "._,.!.,...?_,!.",
"output": "Too weak"
},
{
"input": "!_?_,?,?.,.,_!!!.!,.__,?!!,_!,?_,!??,?!..._!?_,?_!,?_.,._,,_.,.",
"output": "Too weak"
},
{
"input": "?..!.,,?,__.,...????_???__!,?...?.,,,,___!,.!,_,,_,??!_?_,!!?_!_??.?,.!!?_?_.,!",
"output": "Too weak"
},
{
"input": "XZX",
"output": "Too weak"
},
{
"input": "R",
"output": "Too weak"
},
{
"input": "H.FZ",
"output": "Too weak"
},
{
"input": "KSHMICWPK,LSBM_JVZ!IPDYDG_GOPCHXFJTKJBIFY,FPHMY,CB?PZEAG..,X,.GFHPIDBB,IQ?MZ",
"output": "Too weak"
},
{
"input": "EFHI,,Y?HMMUI,,FJGAY?FYPBJQMYM!DZHLFCTFWT?JOPDW,S_!OR?ATT?RWFBMAAKUHIDMHSD?LCZQY!UD_CGYGBAIRDPICYS",
"output": "Too weak"
},
{
"input": "T,NDMUYCCXH_L_FJHMCCAGX_XSCPGOUZSY?D?CNDSYRITYS,VAT!PJVKNTBMXGGRYKACLYU.RJQ_?UWKXYIDE_AE",
"output": "Too weak"
},
{
"input": "y",
"output": "Too weak"
},
{
"input": "qgw",
"output": "Too weak"
},
{
"input": "g",
"output": "Too weak"
},
{
"input": "loaray",
"output": "Too weak"
},
{
"input": "d_iymyvxolmjayhwpedocopqwmy.oalrdg!_n?.lrxpamhygps?kkzxydsbcaihfs.j?eu!oszjsy.vzu?!vs.bprz_j",
"output": "Too weak"
},
{
"input": "txguglvclyillwnono",
"output": "Too weak"
},
{
"input": "FwX",
"output": "Too weak"
},
{
"input": "Zi",
"output": "Too weak"
},
{
"input": "PodE",
"output": "Too weak"
},
{
"input": "SdoOuJ?nj_wJyf",
"output": "Too weak"
},
{
"input": "MhnfZjsUyXYw?f?ubKA",
"output": "Too weak"
},
{
"input": "CpWxDVzwHfYFfoXNtXMFuAZr",
"output": "Too weak"
},
{
"input": "9.,0",
"output": "Too weak"
},
{
"input": "5,8",
"output": "Too weak"
},
{
"input": "7",
"output": "Too weak"
},
{
"input": "34__39_02!,!,82!129!2!566",
"output": "Too weak"
},
{
"input": "96156027.65935663!_87!,44,..7914_!0_1,.4!!62!.8350!17_282!!9.2584,!!7__51.526.7",
"output": "Too weak"
},
{
"input": "90328_",
"output": "Too weak"
},
{
"input": "B9",
"output": "Too weak"
},
{
"input": "P1H",
"output": "Too weak"
},
{
"input": "J2",
"output": "Too weak"
},
{
"input": "M6BCAKW!85OSYX1D?.53KDXP42F",
"output": "Too weak"
},
{
"input": "C672F429Y8X6XU7S,.K9111UD3232YXT81S4!729ER7DZ.J7U1R_7VG6.FQO,LDH",
"output": "Too weak"
},
{
"input": "W2PI__!.O91H8OFY6AB__R30L9XOU8800?ZUD84L5KT99818NFNE35V.8LJJ5P2MM.B6B",
"output": "Too weak"
},
{
"input": "z1",
"output": "Too weak"
},
{
"input": "p1j",
"output": "Too weak"
},
{
"input": "j9",
"output": "Too weak"
},
{
"input": "v8eycoylzv0qkix5mfs_nhkn6k!?ovrk9!b69zy!4frc?k",
"output": "Too weak"
},
{
"input": "l4!m_44kpw8.jg!?oh,?y5oraw1tg7_x1.osl0!ny?_aihzhtt0e2!mr92tnk0es!1f,9he40_usa6c50l",
"output": "Too weak"
},
{
"input": "d4r!ak.igzhnu!boghwd6jl",
"output": "Too weak"
},
{
"input": "It0",
"output": "Too weak"
},
{
"input": "Yb1x",
"output": "Too weak"
},
{
"input": "Qf7",
"output": "Too weak"
},
{
"input": "Vu7jQU8.!FvHBYTsDp6AphaGfnEmySP9te",
"output": "Correct"
},
{
"input": "Ka4hGE,vkvNQbNolnfwp",
"output": "Correct"
},
{
"input": "Ee9oluD?amNItsjeQVtOjwj4w_ALCRh7F3eaZah",
"output": "Correct"
},
{
"input": "Um3Fj?QLhNuRE_Gx0cjMLOkGCm",
"output": "Correct"
},
{
"input": "Oq2LYmV9HmlaW",
"output": "Correct"
},
{
"input": "Cq7r3Wrb.lDb_0wsf7!ruUUGSf08RkxD?VsBEDdyE?SHK73TFFy0f8gmcATqGafgTv8OOg8or2HyMPIPiQ2Hsx8q5rn3_WZe",
"output": "Correct"
},
{
"input": "Wx4p1fOrEMDlQpTlIx0p.1cnFD7BnX2K8?_dNLh4cQBx_Zqsv83BnL5hGKNcBE9g3QB,!fmSvgBeQ_qiH7",
"output": "Correct"
},
{
"input": "k673,",
"output": "Too weak"
},
{
"input": "LzuYQ",
"output": "Too weak"
},
{
"input": "Pasq!",
"output": "Too weak"
},
{
"input": "x5hve",
"output": "Too weak"
},
{
"input": "b27fk",
"output": "Too weak"
},
{
"input": "h6y1l",
"output": "Too weak"
},
{
"input": "i9nij",
"output": "Too weak"
},
{
"input": "Gf5Q6",
"output": "Correct"
},
{
"input": "Uf24o",
"output": "Correct"
},
{
"input": "Oj9vu",
"output": "Correct"
},
{
"input": "c7jqaudcqmv8o7zvb5x_gp6zcgl6nwr7tz5or!28.tj8s1m2.wxz5a4id03!rq07?662vy.7.p5?vk2f2mc7ag8q3861rgd0rmbr",
"output": "Too weak"
},
{
"input": "i6a.,8jb,n0kv4.1!7h?p.96pnhhgy6cl7dg7e4o6o384ys3z.t71kkq,,w,oqi4?u,,m5!rzu6wym_4hm,ohjy!.vvksl?pt,,1",
"output": "Too weak"
},
{
"input": "M10V_MN_1K8YX2LA!89EYV7!5V9?,.IDHDP6JEC.OGLY.180LMZ6KW3Z5E17IT94ZNHS!79GN09Q6LH0,F3AYNKP?KM,QP_?XRD6",
"output": "Too weak"
},
{
"input": "Hi7zYuVXCPhaho68YgCMzzgLILM6toQTJq8akMqqrnUn6ZCD36iA1yVVpvlsIiMpCu!1QZd4ycIrQ5Kcrhk5k0jTrwdAAEEP_T2f",
"output": "Correct"
},
{
"input": "Bk2Q38vDSW5JqYu.077iYC.9YoiPc!Dh6FJWOVze6?YXiFjPNa4F1RG?154m9mY2jQobBnbxM,cDV8l1UX1?v?p.tTYIyJO!NYmE",
"output": "Correct"
},
{
"input": "Ro1HcZ.piN,JRR88DLh,WtW!pbFM076?wCSbqfK7N2s5zUySFBtzk7HV,BxHXR0zALAr016z5jvvB.WUdEcKgYFav5TygwHQC..C",
"output": "Correct"
},
{
"input": "!?.,_",
"output": "Too weak"
}
] | 1,640,456,093
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 46
| 0
|
sifre=input()
ozel=["!", "?", ".", ",", "_"]
bOzel=False
bDigit=False
for n in sifre:
if n in ozel:
bOzel=True
elif n.isnumeric():
bDigit=True
if bOzel and bDigit:
break
if len(sifre)>4 and sifre.upper()!=sifre and bOzel and bDigit:
print("Correct")
else:
print("Too weak")
|
Title: Password Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input Specification:
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output Specification:
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Demo Input:
['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n']
Demo Output:
['Too weak\n', 'Too weak\n', 'Correct\n']
Note:
none
|
```python
sifre=input()
ozel=["!", "?", ".", ",", "_"]
bOzel=False
bDigit=False
for n in sifre:
if n in ozel:
bOzel=True
elif n.isnumeric():
bDigit=True
if bOzel and bDigit:
break
if len(sifre)>4 and sifre.upper()!=sifre and bOzel and bDigit:
print("Correct")
else:
print("Too weak")
```
| 0
|
|
785
|
A
|
Anton and Polyhedrons
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
|
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
|
[
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] |
[
"42\n",
"28\n"
] |
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
| 500
|
[
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,694,288,954
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 249
| 16,076,800
|
n = int(input())
t = 1
my_list = []
while t <= n:
i = input()
my_list.append(i)
t += 1
polyhedra_faces = {
"Tetrahedron": 4,
"Cube": 6,
"Octahedron": 8,
"Dodecahedron": 12,
"Icosahedron": 20
}
total_faces = sum(polyhedra_faces.get(polyhedron, 0) for polyhedron in my_list)
print(total_faces)
|
Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
|
```python
n = int(input())
t = 1
my_list = []
while t <= n:
i = input()
my_list.append(i)
t += 1
polyhedra_faces = {
"Tetrahedron": 4,
"Cube": 6,
"Octahedron": 8,
"Dodecahedron": 12,
"Icosahedron": 20
}
total_faces = sum(polyhedra_faces.get(polyhedron, 0) for polyhedron in my_list)
print(total_faces)
```
| 3
|
|
940
|
B
|
Our Tanya is Crying Out Loud
|
PROGRAMMING
| 1,400
|
[
"dp",
"greedy"
] | null | null |
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
|
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
|
[
"9\n2\n3\n1\n",
"5\n5\n2\n20\n",
"19\n3\n4\n2\n"
] |
[
"6\n",
"8\n",
"12\n"
] |
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total.
| 1,250
|
[
{
"input": "9\n2\n3\n1",
"output": "6"
},
{
"input": "5\n5\n2\n20",
"output": "8"
},
{
"input": "19\n3\n4\n2",
"output": "12"
},
{
"input": "1845999546\n999435865\n1234234\n2323423",
"output": "1044857680578777"
},
{
"input": "1604353664\n1604353665\n9993432\n1",
"output": "16032999235141416"
},
{
"input": "777888456\n1\n98\n43",
"output": "76233068590"
},
{
"input": "1162261467\n3\n1\n2000000000",
"output": "1162261466"
},
{
"input": "1000000000\n1999999999\n789987\n184569875",
"output": "789986999210013"
},
{
"input": "2000000000\n2\n1\n2000000000",
"output": "1999999999"
},
{
"input": "1999888325\n3\n2\n2000000000",
"output": "3333258884"
},
{
"input": "1897546487\n687\n89798979\n879876541",
"output": "110398404423"
},
{
"input": "20\n1\n20\n1",
"output": "380"
},
{
"input": "16\n5\n17\n3",
"output": "54"
},
{
"input": "19\n19\n19\n1",
"output": "1"
},
{
"input": "18\n2\n3\n16",
"output": "40"
},
{
"input": "1\n11\n8\n9",
"output": "0"
},
{
"input": "9\n10\n1\n20",
"output": "8"
},
{
"input": "19\n10\n19\n2",
"output": "173"
},
{
"input": "16\n9\n14\n2",
"output": "100"
},
{
"input": "15\n2\n5\n2",
"output": "21"
},
{
"input": "14\n7\n13\n1",
"output": "14"
},
{
"input": "43\n3\n45\n3",
"output": "189"
},
{
"input": "99\n1\n98\n1",
"output": "9604"
},
{
"input": "77\n93\n100\n77",
"output": "7600"
},
{
"input": "81\n3\n91\n95",
"output": "380"
},
{
"input": "78\n53\n87\n34",
"output": "2209"
},
{
"input": "80\n3\n15\n1",
"output": "108"
},
{
"input": "97\n24\n4\n24",
"output": "40"
},
{
"input": "100\n100\n1\n100",
"output": "99"
},
{
"input": "87\n4\n17\n7",
"output": "106"
},
{
"input": "65\n2\n3\n6",
"output": "36"
},
{
"input": "1000000\n1435\n3\n999999",
"output": "1005804"
},
{
"input": "783464\n483464\n2\n966928",
"output": "1566926"
},
{
"input": "248035\n11\n3\n20",
"output": "202"
},
{
"input": "524287\n2\n945658\n999756",
"output": "34963354"
},
{
"input": "947352\n78946\n85\n789654",
"output": "790589"
},
{
"input": "1000000\n1\n999899\n60",
"output": "999898000101"
},
{
"input": "753687\n977456\n6547\n456",
"output": "4934382242"
},
{
"input": "1000000\n500000\n1\n999997",
"output": "999998"
},
{
"input": "997458\n843596\n1\n843596",
"output": "997457"
},
{
"input": "821109\n92\n6547\n98787",
"output": "394566"
},
{
"input": "1073741823\n2\n9543\n8923453",
"output": "188412866"
},
{
"input": "1000999777\n1934999345\n2356346\n34534565",
"output": "2358701818178496"
},
{
"input": "2000000000\n1\n2000000000\n98",
"output": "3999999998000000000"
},
{
"input": "1999324353\n978435356\n1\n978435356",
"output": "1020888998"
},
{
"input": "2000000000\n2\n2000000000\n2000000000",
"output": "84000000000"
},
{
"input": "241375690\n17\n2\n1998789654",
"output": "482751378"
},
{
"input": "171507000\n350\n789\n6548687",
"output": "14216965"
},
{
"input": "1100220011\n10001\n2\n1999778654",
"output": "1999998674"
},
{
"input": "1867622656\n43216\n789644\n12315468",
"output": "24630936"
},
{
"input": "1867622656\n43216\n1\n1879865413",
"output": "1867622655"
},
{
"input": "1999999999\n1000000000\n789987\n184569875",
"output": "789987183779888"
},
{
"input": "1987987897\n103546\n7\n98754563",
"output": "98946650"
},
{
"input": "10\n2\n2\n5",
"output": "13"
},
{
"input": "7\n2\n1\n100000",
"output": "6"
},
{
"input": "7\n2\n3\n1",
"output": "8"
},
{
"input": "2000000000\n666666667\n1\n1",
"output": "666666668"
},
{
"input": "1999999997\n666666666\n2\n2",
"output": "1333333334"
}
] | 1,699,599,822
| 2,147,483,647
|
Python 3
|
MEMORY_LIMIT_EXCEEDED
|
TESTS
| 9
| 249
| 268,390,400
|
def solve(n, k, A, B):
x = n
if n < k or k == 1:
return A * (n - 1)
allsub = A * (n - 1)
divonce = (x % k) * A + B + A * (x // k - 1)
if allsub <= divonce:
return allsub
ret = (x % k) * A + B
x = x // k
costs = [None] * (x + 1)
costs[1] = 0
for i in range(2, len(costs)):
costs[i] = costs[i - 1] + A
if i % k == 0:
costs[i] = min(costs[i], costs[i // k] + B)
ret += costs[-1]
return ret
n, k, A, B = [int(input()) for _ in range(4)]
print(solve(n, k, A, B))
|
Title: Our Tanya is Crying Out Loud
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
Output Specification:
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
Demo Input:
['9\n2\n3\n1\n', '5\n5\n2\n20\n', '19\n3\n4\n2\n']
Demo Output:
['6\n', '8\n', '12\n']
Note:
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total.
|
```python
def solve(n, k, A, B):
x = n
if n < k or k == 1:
return A * (n - 1)
allsub = A * (n - 1)
divonce = (x % k) * A + B + A * (x // k - 1)
if allsub <= divonce:
return allsub
ret = (x % k) * A + B
x = x // k
costs = [None] * (x + 1)
costs[1] = 0
for i in range(2, len(costs)):
costs[i] = costs[i - 1] + A
if i % k == 0:
costs[i] = min(costs[i], costs[i // k] + B)
ret += costs[-1]
return ret
n, k, A, B = [int(input()) for _ in range(4)]
print(solve(n, k, A, B))
```
| 0
|
|
1
|
B
|
Spreadsheet
|
PROGRAMMING
| 1,600
|
[
"implementation",
"math"
] |
B. Spreadsheets
|
10
|
64
|
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
|
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=105), the number of coordinates in the test. Then there follow *n* lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .
|
Write *n* lines, each line should contain a cell coordinates in the other numeration system.
|
[
"2\nR23C55\nBC23\n"
] |
[
"BC23\nR23C55\n"
] |
none
| 0
|
[
{
"input": "2\nR23C55\nBC23",
"output": "BC23\nR23C55"
},
{
"input": "1\nA1",
"output": "R1C1"
},
{
"input": "5\nR8C3\nD1\nR7C2\nR8C9\nR8C9",
"output": "C8\nR1C4\nB7\nI8\nI8"
},
{
"input": "4\nR4C25\nR90C35\nAP55\nX83",
"output": "Y4\nAI90\nR55C42\nR83C24"
},
{
"input": "10\nR50C12\nR23C47\nY96\nR44C13\nR19C21\nR95C73\nBK12\nR51C74\nAY34\nR63C25",
"output": "L50\nAU23\nR96C25\nM44\nU19\nBU95\nR12C63\nBV51\nR34C51\nY63"
}
] | 1,691,404,699
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print("_RANDOM_GUESS_1691404699.2787287")# 1691404699.278747
|
Title: Spreadsheet
Time Limit: 10 seconds
Memory Limit: 64 megabytes
Problem Description:
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
Input Specification:
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=105), the number of coordinates in the test. Then there follow *n* lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .
Output Specification:
Write *n* lines, each line should contain a cell coordinates in the other numeration system.
Demo Input:
['2\nR23C55\nBC23\n']
Demo Output:
['BC23\nR23C55\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1691404699.2787287")# 1691404699.278747
```
| 0
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,512,984,512
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 124
| 5,529,600
|
n=int(input())
list1=[int(i) for i in input().split()]
for i in range(0,n,1):
if i==0:
if (list1[i+1]+list1[i+2])%2==0:
print(1)
elif i==n-1:
if (list1[i-1]+list1[i-2])%2==0:
print(n)
else:
if (list1[i-1]+list1[i+1])%2==0:
print(i+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n=int(input())
list1=[int(i) for i in input().split()]
for i in range(0,n,1):
if i==0:
if (list1[i+1]+list1[i+2])%2==0:
print(1)
elif i==n-1:
if (list1[i-1]+list1[i-2])%2==0:
print(n)
else:
if (list1[i-1]+list1[i+1])%2==0:
print(i+1)
```
| 0
|
618
|
D
|
Hamiltonian Spanning Tree
|
PROGRAMMING
| 2,200
|
[
"dfs and similar",
"dp",
"graph matchings",
"greedy",
"trees"
] | null | null |
A group of *n* cities is connected by a network of roads. There is an undirected road between every pair of cities, so there are roads in total. It takes exactly *y* seconds to traverse any single road.
A spanning tree is a set of roads containing exactly *n*<=-<=1 roads such that it's possible to travel between any two cities using only these roads.
Some spanning tree of the initial network was chosen. For every road in this tree the time one needs to traverse this road was changed from *y* to *x* seconds. Note that it's not guaranteed that *x* is smaller than *y*.
You would like to travel through all the cities using the shortest path possible. Given *n*, *x*, *y* and a description of the spanning tree that was chosen, find the cost of the shortest path that starts in any city, ends in any city and visits all cities exactly once.
|
The first line of the input contains three integers *n*, *x* and *y* (2<=≤<=*n*<=≤<=200<=000,<=1<=≤<=*x*,<=*y*<=≤<=109).
Each of the next *n*<=-<=1 lines contains a description of a road in the spanning tree. The *i*-th of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*) — indices of the cities connected by the *i*-th road. It is guaranteed that these roads form a spanning tree.
|
Print a single integer — the minimum number of seconds one needs to spend in order to visit all the cities exactly once.
|
[
"5 2 3\n1 2\n1 3\n3 4\n5 3\n",
"5 3 2\n1 2\n1 3\n3 4\n5 3\n"
] |
[
"9\n",
"8\n"
] |
In the first sample, roads of the spanning tree have cost 2, while other roads have cost 3. One example of an optimal path is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3a11f64ac0349d4ecd3a2b4c3443aeb7ac3b28b9.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample, we have the same spanning tree, but roads in the spanning tree cost 3, while other roads cost 2. One example of an optimal path is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3fdb844c44665567f5addf82820eb6f96a060920.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 1,750
|
[
{
"input": "5 2 3\n1 2\n1 3\n3 4\n5 3",
"output": "9"
},
{
"input": "5 3 2\n1 2\n1 3\n3 4\n5 3",
"output": "8"
},
{
"input": "50 23129 410924\n18 28\n17 23\n21 15\n18 50\n50 11\n32 3\n44 41\n50 31\n50 34\n5 14\n36 13\n22 40\n20 9\n9 43\n19 47\n48 40\n20 22\n33 45\n35 22\n33 24\n9 6\n13 1\n13 24\n49 20\n1 20\n29 38\n10 35\n25 23\n49 30\n42 8\n20 18\n32 15\n32 1\n27 10\n20 47\n41 7\n20 14\n18 26\n4 20\n20 2\n46 37\n41 16\n46 41\n12 20\n8 40\n18 37\n29 3\n32 39\n23 37",
"output": "8113631"
},
{
"input": "2 3 4\n1 2",
"output": "3"
},
{
"input": "50 491238 12059\n42 3\n5 9\n11 9\n41 15\n42 34\n11 6\n40 16\n23 8\n41 7\n22 6\n24 29\n7 17\n31 2\n17 33\n39 42\n42 6\n41 50\n21 45\n19 41\n1 21\n42 1\n2 25\n17 28\n49 42\n30 13\n4 12\n10 32\n48 35\n21 2\n14 6\n49 29\n18 20\n38 22\n19 37\n20 47\n3 36\n1 44\n20 7\n4 11\n39 26\n30 40\n6 7\n25 46\n2 27\n30 42\n10 11\n8 21\n42 43\n35 8",
"output": "590891"
},
{
"input": "2 4 1\n1 2",
"output": "4"
},
{
"input": "5 2 2\n1 2\n1 3\n1 4\n1 5",
"output": "8"
},
{
"input": "4 100 1\n1 2\n1 3\n1 4",
"output": "102"
},
{
"input": "3 2 1\n1 2\n1 3",
"output": "3"
},
{
"input": "5 6 1\n1 2\n1 3\n1 4\n1 5",
"output": "9"
},
{
"input": "3 100 1\n1 2\n2 3",
"output": "101"
},
{
"input": "2 2 1\n1 2",
"output": "2"
},
{
"input": "5 3 2\n1 2\n1 3\n1 4\n1 5",
"output": "9"
},
{
"input": "4 1000 1\n1 2\n1 3\n1 4",
"output": "1002"
},
{
"input": "4 100 1\n1 2\n2 3\n3 4",
"output": "3"
},
{
"input": "2 3 1\n1 2",
"output": "3"
},
{
"input": "5 4 3\n1 2\n1 3\n1 4\n1 5",
"output": "13"
}
] | 1,454,093,631
| 6,231
|
Python 3
|
RUNTIME_ERROR
|
PRETESTS
| 2
| 1,060
| 43,929,600
|
from collections import defaultdict
from functools import reduce
n, x, y = [int(x) for x in input().split()]
E = defaultdict(set)
for i in range(n-1):
u, v = [int(x) for x in input().split()]
E[u].add(v)
E[v].add(u)
def poplarge(v):
mindeg = n
for u in E[v]:
if len(E[u]) < mindeg:
mindeg = len(E[u])
argmin = u
E[v].remove(argmin)
mindeg = n
stat2 = None
for u in E[v]:
if len(E[u]) < mindeg:
mindeg = len(E[u])
if stat2 is not None:
E[stat2].remove(v)
stat2 = u
else:
E[u].remove(v)
E[v].clear()
E[v].add(argmin)
E[v].add(stat2)
if x < y:
ans = 0
for v in E:
ans += max(0, len(E[v])-2)
poplarge(v)
print(ans*y + (n-1-ans)*x)
else:
for v in E:
if len(E[v]) == n-1:
print((n-2)*y + x)
break
elif len(E[v]) > 1:
print((n-1)*y)
break
|
Title: Hamiltonian Spanning Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A group of *n* cities is connected by a network of roads. There is an undirected road between every pair of cities, so there are roads in total. It takes exactly *y* seconds to traverse any single road.
A spanning tree is a set of roads containing exactly *n*<=-<=1 roads such that it's possible to travel between any two cities using only these roads.
Some spanning tree of the initial network was chosen. For every road in this tree the time one needs to traverse this road was changed from *y* to *x* seconds. Note that it's not guaranteed that *x* is smaller than *y*.
You would like to travel through all the cities using the shortest path possible. Given *n*, *x*, *y* and a description of the spanning tree that was chosen, find the cost of the shortest path that starts in any city, ends in any city and visits all cities exactly once.
Input Specification:
The first line of the input contains three integers *n*, *x* and *y* (2<=≤<=*n*<=≤<=200<=000,<=1<=≤<=*x*,<=*y*<=≤<=109).
Each of the next *n*<=-<=1 lines contains a description of a road in the spanning tree. The *i*-th of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*) — indices of the cities connected by the *i*-th road. It is guaranteed that these roads form a spanning tree.
Output Specification:
Print a single integer — the minimum number of seconds one needs to spend in order to visit all the cities exactly once.
Demo Input:
['5 2 3\n1 2\n1 3\n3 4\n5 3\n', '5 3 2\n1 2\n1 3\n3 4\n5 3\n']
Demo Output:
['9\n', '8\n']
Note:
In the first sample, roads of the spanning tree have cost 2, while other roads have cost 3. One example of an optimal path is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3a11f64ac0349d4ecd3a2b4c3443aeb7ac3b28b9.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample, we have the same spanning tree, but roads in the spanning tree cost 3, while other roads cost 2. One example of an optimal path is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3fdb844c44665567f5addf82820eb6f96a060920.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
from collections import defaultdict
from functools import reduce
n, x, y = [int(x) for x in input().split()]
E = defaultdict(set)
for i in range(n-1):
u, v = [int(x) for x in input().split()]
E[u].add(v)
E[v].add(u)
def poplarge(v):
mindeg = n
for u in E[v]:
if len(E[u]) < mindeg:
mindeg = len(E[u])
argmin = u
E[v].remove(argmin)
mindeg = n
stat2 = None
for u in E[v]:
if len(E[u]) < mindeg:
mindeg = len(E[u])
if stat2 is not None:
E[stat2].remove(v)
stat2 = u
else:
E[u].remove(v)
E[v].clear()
E[v].add(argmin)
E[v].add(stat2)
if x < y:
ans = 0
for v in E:
ans += max(0, len(E[v])-2)
poplarge(v)
print(ans*y + (n-1-ans)*x)
else:
for v in E:
if len(E[v]) == n-1:
print((n-2)*y + x)
break
elif len(E[v]) > 1:
print((n-1)*y)
break
```
| -1
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,696,964,096
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
def main():
n = int(input())
sum = 0
for i in range(n):
row_sum = 0
for j in range(3):
a = int(input())
if a == 1:
row_sum += 1
if row_sum >= 2:
sum += 1
print(sum)
if __name__ == "__main__":
main()
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
def main():
n = int(input())
sum = 0
for i in range(n):
row_sum = 0
for j in range(3):
a = int(input())
if a == 1:
row_sum += 1
if row_sum >= 2:
sum += 1
print(sum)
if __name__ == "__main__":
main()
```
| -1
|
|
807
|
A
|
Is it rated?
|
PROGRAMMING
| 900
|
[
"implementation",
"sortings"
] | null | null |
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
|
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
|
[
"6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n",
"4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n",
"5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n"
] |
[
"rated\n",
"unrated\n",
"maybe\n"
] |
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
| 500
|
[
{
"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884",
"output": "rated"
},
{
"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699",
"output": "maybe"
},
{
"input": "2\n1 1\n1 1",
"output": "maybe"
},
{
"input": "2\n4126 4126\n4126 4126",
"output": "maybe"
},
{
"input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423",
"output": "rated"
},
{
"input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110",
"output": "unrated"
},
{
"input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143",
"output": "maybe"
},
{
"input": "2\n3936 3936\n2967 2967",
"output": "maybe"
},
{
"input": "2\n1 1\n2 2",
"output": "unrated"
},
{
"input": "2\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n2 1\n1 2",
"output": "rated"
},
{
"input": "2\n2967 2967\n3936 3936",
"output": "unrated"
},
{
"input": "3\n1200 1200\n1200 1200\n1300 1300",
"output": "unrated"
},
{
"input": "3\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "3\n1 1\n1 1\n2 2",
"output": "unrated"
},
{
"input": "2\n3 2\n3 2",
"output": "rated"
},
{
"input": "3\n5 5\n4 4\n3 4",
"output": "rated"
},
{
"input": "3\n200 200\n200 200\n300 300",
"output": "unrated"
},
{
"input": "3\n1 1\n2 2\n3 3",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699",
"output": "maybe"
},
{
"input": "2\n10 10\n8 8",
"output": "maybe"
},
{
"input": "3\n1500 1500\n1500 1500\n1600 1600",
"output": "unrated"
},
{
"input": "3\n1500 1500\n1500 1500\n1700 1700",
"output": "unrated"
},
{
"input": "4\n100 100\n100 100\n70 70\n80 80",
"output": "unrated"
},
{
"input": "2\n1 2\n2 1",
"output": "rated"
},
{
"input": "3\n5 5\n4 3\n3 3",
"output": "rated"
},
{
"input": "3\n1600 1650\n1500 1550\n1400 1450",
"output": "rated"
},
{
"input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700",
"output": "unrated"
},
{
"input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700",
"output": "unrated"
},
{
"input": "2\n1600 1600\n1400 1400",
"output": "maybe"
},
{
"input": "2\n3 1\n9 8",
"output": "rated"
},
{
"input": "2\n2 1\n1 1",
"output": "rated"
},
{
"input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670",
"output": "unrated"
},
{
"input": "2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "2\n10 11\n5 4",
"output": "rated"
},
{
"input": "2\n15 14\n13 12",
"output": "rated"
},
{
"input": "2\n2 1\n2 2",
"output": "rated"
},
{
"input": "3\n2670 2670\n3670 3670\n4106 4106",
"output": "unrated"
},
{
"input": "3\n4 5\n3 3\n2 2",
"output": "rated"
},
{
"input": "2\n10 9\n10 10",
"output": "rated"
},
{
"input": "3\n1011 1011\n1011 999\n2200 2100",
"output": "rated"
},
{
"input": "2\n3 3\n5 5",
"output": "unrated"
},
{
"input": "2\n1500 1500\n3000 2000",
"output": "rated"
},
{
"input": "2\n5 6\n5 5",
"output": "rated"
},
{
"input": "3\n2000 2000\n1500 1501\n500 500",
"output": "rated"
},
{
"input": "2\n2 3\n2 2",
"output": "rated"
},
{
"input": "2\n3 3\n2 2",
"output": "maybe"
},
{
"input": "2\n1 2\n1 1",
"output": "rated"
},
{
"input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699",
"output": "maybe"
},
{
"input": "2\n15 14\n14 13",
"output": "rated"
},
{
"input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900",
"output": "unrated"
},
{
"input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884",
"output": "rated"
},
{
"input": "2\n100 99\n100 100",
"output": "rated"
},
{
"input": "4\n2 2\n1 1\n1 1\n2 2",
"output": "unrated"
},
{
"input": "3\n100 101\n100 100\n100 100",
"output": "rated"
},
{
"input": "4\n1000 1001\n900 900\n950 950\n890 890",
"output": "rated"
},
{
"input": "2\n2 3\n1 1",
"output": "rated"
},
{
"input": "2\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n3 2\n2 2",
"output": "rated"
},
{
"input": "2\n3 2\n3 3",
"output": "rated"
},
{
"input": "2\n1 1\n2 2",
"output": "unrated"
},
{
"input": "3\n3 2\n3 3\n3 3",
"output": "rated"
},
{
"input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400",
"output": "rated"
},
{
"input": "3\n1000 1000\n500 500\n400 300",
"output": "rated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000",
"output": "unrated"
},
{
"input": "2\n1 1\n2 3",
"output": "rated"
},
{
"input": "2\n6 2\n6 2",
"output": "rated"
},
{
"input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246",
"output": "unrated"
},
{
"input": "2\n1500 1500\n1600 1600",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699",
"output": "maybe"
},
{
"input": "2\n20 30\n10 5",
"output": "rated"
},
{
"input": "3\n1 1\n2 2\n1 1",
"output": "unrated"
},
{
"input": "2\n1 2\n3 3",
"output": "rated"
},
{
"input": "5\n5 5\n4 4\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n2 2\n2 1",
"output": "rated"
},
{
"input": "2\n100 100\n90 89",
"output": "rated"
},
{
"input": "2\n1000 900\n2000 2000",
"output": "rated"
},
{
"input": "2\n50 10\n10 50",
"output": "rated"
},
{
"input": "2\n200 200\n100 100",
"output": "maybe"
},
{
"input": "3\n2 2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "3\n1000 1000\n300 300\n100 100",
"output": "maybe"
},
{
"input": "4\n2 2\n2 2\n3 3\n4 4",
"output": "unrated"
},
{
"input": "2\n5 3\n6 3",
"output": "rated"
},
{
"input": "2\n1200 1100\n1200 1000",
"output": "rated"
},
{
"input": "2\n5 5\n4 4",
"output": "maybe"
},
{
"input": "2\n5 5\n3 3",
"output": "maybe"
},
{
"input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100",
"output": "unrated"
},
{
"input": "5\n10 10\n9 9\n8 8\n7 7\n6 6",
"output": "maybe"
},
{
"input": "3\n1000 1000\n300 300\n10 10",
"output": "maybe"
},
{
"input": "5\n6 6\n5 5\n4 4\n3 3\n2 2",
"output": "maybe"
},
{
"input": "2\n3 3\n1 1",
"output": "maybe"
},
{
"input": "4\n2 2\n2 2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "2\n1000 1000\n700 700",
"output": "maybe"
},
{
"input": "2\n4 3\n5 3",
"output": "rated"
},
{
"input": "2\n1000 1000\n1100 1100",
"output": "unrated"
},
{
"input": "4\n5 5\n4 4\n3 3\n2 2",
"output": "maybe"
},
{
"input": "3\n1 1\n2 3\n2 2",
"output": "rated"
},
{
"input": "2\n1 2\n1 3",
"output": "rated"
},
{
"input": "2\n3 3\n1 2",
"output": "rated"
},
{
"input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "rated"
},
{
"input": "5\n1 1\n2 2\n3 3\n4 4\n5 5",
"output": "unrated"
},
{
"input": "2\n10 10\n1 2",
"output": "rated"
},
{
"input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900",
"output": "unrated"
},
{
"input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699",
"output": "unrated"
},
{
"input": "2\n100 100\n110 110",
"output": "unrated"
},
{
"input": "3\n3 3\n3 3\n4 4",
"output": "unrated"
},
{
"input": "3\n3 3\n3 2\n4 4",
"output": "rated"
},
{
"input": "3\n5 2\n4 4\n3 3",
"output": "rated"
},
{
"input": "4\n4 4\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n1 1\n3 2",
"output": "rated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699",
"output": "unrated"
},
{
"input": "3\n3 3\n3 3\n3 4",
"output": "rated"
},
{
"input": "3\n1 2\n2 2\n3 3",
"output": "rated"
},
{
"input": "3\n1 2\n1 2\n1 2",
"output": "rated"
},
{
"input": "2\n2 1\n2 1",
"output": "rated"
},
{
"input": "2\n1 2\n3 4",
"output": "rated"
},
{
"input": "2\n3 2\n2 3",
"output": "rated"
},
{
"input": "3\n1500 1500\n1600 1600\n1600 1600",
"output": "unrated"
},
{
"input": "3\n1 1\n3 3\n4 4",
"output": "unrated"
},
{
"input": "3\n1 1\n2 2\n2 2",
"output": "unrated"
},
{
"input": "2\n10 12\n8 8",
"output": "rated"
},
{
"input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500",
"output": "unrated"
},
{
"input": "2\n1 2\n2 2",
"output": "rated"
},
{
"input": "3\n1500 1400\n1200 1200\n1100 1100",
"output": "rated"
},
{
"input": "2\n10 12\n10 10",
"output": "rated"
},
{
"input": "3\n1500 1500\n1400 1400\n1300 1300",
"output": "maybe"
},
{
"input": "3\n3 3\n4 4\n5 5",
"output": "unrated"
},
{
"input": "3\n2 6\n3 5\n4 4",
"output": "rated"
},
{
"input": "2\n5 6\n4 6",
"output": "rated"
},
{
"input": "4\n10 10\n10 10\n7 7\n8 8",
"output": "unrated"
},
{
"input": "2\n4 4\n3 3",
"output": "maybe"
}
] | 1,612,794,242
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 150
| 124
| 1,433,600
|
n = int(input())
a = []
for i in range(n):
b, c = map(int, input().split())
a.append(b)
if b != c:
print("rated")
break
else:
for i in range(1, n):
if a[i] > a[i - 1]:
print("unrated")
break
else:
print("maybe")
|
Title: Is it rated?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output Specification:
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Demo Input:
['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n']
Demo Output:
['rated\n', 'unrated\n', 'maybe\n']
Note:
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
```python
n = int(input())
a = []
for i in range(n):
b, c = map(int, input().split())
a.append(b)
if b != c:
print("rated")
break
else:
for i in range(1, n):
if a[i] > a[i - 1]:
print("unrated")
break
else:
print("maybe")
```
| 3
|
|
733
|
D
|
Kostya the Sculptor
|
PROGRAMMING
| 1,600
|
[
"data structures",
"hashing"
] | null | null |
Kostya is a genial sculptor, he has an idea: to carve a marble sculpture in the shape of a sphere. Kostya has a friend Zahar who works at a career. Zahar knows about Kostya's idea and wants to present him a rectangular parallelepiped of marble from which he can carve the sphere.
Zahar has *n* stones which are rectangular parallelepipeds. The edges sizes of the *i*-th of them are *a**i*, *b**i* and *c**i*. He can take no more than two stones and present them to Kostya.
If Zahar takes two stones, he should glue them together on one of the faces in order to get a new piece of rectangular parallelepiped of marble. Thus, it is possible to glue a pair of stones together if and only if two faces on which they are glued together match as rectangles. In such gluing it is allowed to rotate and flip the stones in any way.
Help Zahar choose such a present so that Kostya can carve a sphere of the maximum possible volume and present it to Zahar.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=105).
*n* lines follow, in the *i*-th of which there are three integers *a**i*,<=*b**i* and *c**i* (1<=≤<=*a**i*,<=*b**i*,<=*c**i*<=≤<=109) — the lengths of edges of the *i*-th stone. Note, that two stones may have exactly the same sizes, but they still will be considered two different stones.
|
In the first line print *k* (1<=≤<=*k*<=≤<=2) the number of stones which Zahar has chosen. In the second line print *k* distinct integers from 1 to *n* — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from 1 to *n* in the order as they are given in the input data.
You can print the stones in arbitrary order. If there are several answers print any of them.
|
[
"6\n5 5 5\n3 2 4\n1 4 1\n2 1 3\n3 2 4\n3 3 4\n",
"7\n10 7 8\n5 10 3\n4 2 6\n5 5 5\n10 2 8\n4 2 1\n7 7 7\n"
] |
[
"1\n1\n",
"2\n1 5\n"
] |
In the first example we can connect the pairs of stones:
- 2 and 4, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1 - 2 and 5, the size of the parallelepiped: 3 × 2 × 8 or 6 × 2 × 4 or 3 × 4 × 4, the radius of the inscribed sphere 1, or 1, or 1.5 respectively. - 2 and 6, the size of the parallelepiped: 3 × 5 × 4, the radius of the inscribed sphere 1.5 - 4 and 5, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1 - 5 and 6, the size of the parallelepiped: 3 × 4 × 5, the radius of the inscribed sphere 1.5
Or take only one stone:
- 1 the size of the parallelepiped: 5 × 5 × 5, the radius of the inscribed sphere 2.5 - 2 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1 - 3 the size of the parallelepiped: 1 × 4 × 1, the radius of the inscribed sphere 0.5 - 4 the size of the parallelepiped: 2 × 1 × 3, the radius of the inscribed sphere 0.5 - 5 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1 - 6 the size of the parallelepiped: 3 × 3 × 4, the radius of the inscribed sphere 1.5
It is most profitable to take only the first stone.
| 2,000
|
[
{
"input": "6\n5 5 5\n3 2 4\n1 4 1\n2 1 3\n3 2 4\n3 3 4",
"output": "1\n1"
},
{
"input": "7\n10 7 8\n5 10 3\n4 2 6\n5 5 5\n10 2 8\n4 2 1\n7 7 7",
"output": "2\n1 5"
},
{
"input": "1\n1 1 1",
"output": "1\n1"
},
{
"input": "2\n2 3 1\n2 2 3",
"output": "2\n2 1"
},
{
"input": "1\n1000000000 1000000000 1000000000",
"output": "1\n1"
},
{
"input": "3\n100 100 100\n25 63 11\n63 15 11",
"output": "1\n1"
},
{
"input": "2\n999999999 1000000000 1000000000\n1000000000 1000000000 1000000000",
"output": "2\n2 1"
},
{
"input": "3\n1 1 2\n1 2 2\n1 1 1",
"output": "1\n1"
},
{
"input": "3\n500 1000 1000\n1000 499 1000\n999 999 999",
"output": "2\n1 2"
},
{
"input": "3\n500 1000 1000\n1000 499 1000\n1000 1001 1001",
"output": "1\n3"
},
{
"input": "9\n1 3 2\n3 3 1\n3 1 2\n3 3 2\n2 2 2\n3 2 1\n3 3 1\n3 3 1\n2 1 2",
"output": "2\n4 8"
},
{
"input": "3\n20 30 5\n20 30 6\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n5 20 30\n6 20 30\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 5 30\n20 6 30\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 30 5\n30 20 6\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 30 5\n6 20 30\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 30 5\n6 30 20\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 30 5\n20 6 30\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 30 5\n30 6 20\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 5 30\n20 30 6\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 5 30\n30 20 6\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 5 30\n6 20 30\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 5 30\n6 30 20\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n20 5 30\n30 6 20\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n5 20 30\n20 30 6\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n5 20 30\n30 20 6\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n5 20 30\n6 30 20\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n5 20 30\n20 6 30\n10 10 10",
"output": "2\n2 1"
},
{
"input": "3\n5 20 30\n30 6 20\n10 10 10",
"output": "2\n2 1"
}
] | 1,541,622,950
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 5
| 139
| 409,600
|
lines = []
#for line in f.readlines():
n = int(input())
for i in range(n):
lines.append([int(x) for x in input().split() if x.isdigit()])
def del_dublicates(m):
dublicates = []
result = []
for i in range(len(m)):
if m[i][0] not in dublicates:
result.append(m[i])
dublicates.append(m[i][0])
return result
def new_box_analysis(mylist):
hash_box = []
hash_box.append([int((mylist[0]+mylist[1])*(mylist[0]+mylist[1]+1)/2+max(mylist[0],mylist[1])),2])
hash_box.append([int((mylist[2]+mylist[1])*(mylist[2]+mylist[1]+1)/2+max(mylist[2],mylist[1])),0])
hash_box.append([int((mylist[2]+mylist[0])*(mylist[2]+mylist[0]+1)/2+max(mylist[2],mylist[0])),1])
hash_box = del_dublicates(hash_box)
m = min(mylist[0],mylist[1],mylist[2])
#print(new)
return hash_box,m
def generate_new_box(i,num_vertix_for_i, list_j):
new_verix = lines[i][num_vertix_for_i] + lines[list_j[0]][list_j[1]]
vertices = []
vertices.append(new_verix)
for k in range(3):
if k != num_vertix_for_i:
vertices.append(lines[i][k])
return min(vertices)
T = []
list_box = []
diameter = 0
N = 10000
T = [[] for g in range(N+1)]
for i in range(n):
hash_for_box = [0 for g in range(3)]
hash_for_box, d_k = new_box_analysis(lines[i])
if d_k>diameter:
diameter = d_k
for ll in range(len(list_box)):
_ = list_box.pop()
list_box.append(i+1)
for k in range(len(hash_for_box)):
if len(T[hash_for_box[k][0]]) == 0:
T[hash_for_box[k][0]].append(i)
T[hash_for_box[k][0]].append(hash_for_box[k][1]) #### добавляем номер box и третью сторону
else:
d_m = generate_new_box(i,hash_for_box[k][1],T[hash_for_box[k][0]])
if d_m > diameter:
diameter = d_m
for ll in range(len(list_box)):
_ = list_box.pop()
list_box.append(i+1)
list_box.append(T[hash_for_box[k][0]][0]+1)
if lines[i][hash_for_box[k][1]] > T[hash_for_box[k][0]][1]:
T[hash_for_box[k][0]].pop()
T[hash_for_box[k][0]].pop()
T[hash_for_box[k][0]].append(i)
T[hash_for_box[k][0]].append(hash_for_box[k][1])
print(len(list_box))
print(str(min(list_box[0],list_box[1])) + " " + str(max(list_box[0],list_box[1]))) if len(list_box) > 1 else print(str(list_box[0]))
|
Title: Kostya the Sculptor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kostya is a genial sculptor, he has an idea: to carve a marble sculpture in the shape of a sphere. Kostya has a friend Zahar who works at a career. Zahar knows about Kostya's idea and wants to present him a rectangular parallelepiped of marble from which he can carve the sphere.
Zahar has *n* stones which are rectangular parallelepipeds. The edges sizes of the *i*-th of them are *a**i*, *b**i* and *c**i*. He can take no more than two stones and present them to Kostya.
If Zahar takes two stones, he should glue them together on one of the faces in order to get a new piece of rectangular parallelepiped of marble. Thus, it is possible to glue a pair of stones together if and only if two faces on which they are glued together match as rectangles. In such gluing it is allowed to rotate and flip the stones in any way.
Help Zahar choose such a present so that Kostya can carve a sphere of the maximum possible volume and present it to Zahar.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=105).
*n* lines follow, in the *i*-th of which there are three integers *a**i*,<=*b**i* and *c**i* (1<=≤<=*a**i*,<=*b**i*,<=*c**i*<=≤<=109) — the lengths of edges of the *i*-th stone. Note, that two stones may have exactly the same sizes, but they still will be considered two different stones.
Output Specification:
In the first line print *k* (1<=≤<=*k*<=≤<=2) the number of stones which Zahar has chosen. In the second line print *k* distinct integers from 1 to *n* — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from 1 to *n* in the order as they are given in the input data.
You can print the stones in arbitrary order. If there are several answers print any of them.
Demo Input:
['6\n5 5 5\n3 2 4\n1 4 1\n2 1 3\n3 2 4\n3 3 4\n', '7\n10 7 8\n5 10 3\n4 2 6\n5 5 5\n10 2 8\n4 2 1\n7 7 7\n']
Demo Output:
['1\n1\n', '2\n1 5\n']
Note:
In the first example we can connect the pairs of stones:
- 2 and 4, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1 - 2 and 5, the size of the parallelepiped: 3 × 2 × 8 or 6 × 2 × 4 or 3 × 4 × 4, the radius of the inscribed sphere 1, or 1, or 1.5 respectively. - 2 and 6, the size of the parallelepiped: 3 × 5 × 4, the radius of the inscribed sphere 1.5 - 4 and 5, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1 - 5 and 6, the size of the parallelepiped: 3 × 4 × 5, the radius of the inscribed sphere 1.5
Or take only one stone:
- 1 the size of the parallelepiped: 5 × 5 × 5, the radius of the inscribed sphere 2.5 - 2 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1 - 3 the size of the parallelepiped: 1 × 4 × 1, the radius of the inscribed sphere 0.5 - 4 the size of the parallelepiped: 2 × 1 × 3, the radius of the inscribed sphere 0.5 - 5 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1 - 6 the size of the parallelepiped: 3 × 3 × 4, the radius of the inscribed sphere 1.5
It is most profitable to take only the first stone.
|
```python
lines = []
#for line in f.readlines():
n = int(input())
for i in range(n):
lines.append([int(x) for x in input().split() if x.isdigit()])
def del_dublicates(m):
dublicates = []
result = []
for i in range(len(m)):
if m[i][0] not in dublicates:
result.append(m[i])
dublicates.append(m[i][0])
return result
def new_box_analysis(mylist):
hash_box = []
hash_box.append([int((mylist[0]+mylist[1])*(mylist[0]+mylist[1]+1)/2+max(mylist[0],mylist[1])),2])
hash_box.append([int((mylist[2]+mylist[1])*(mylist[2]+mylist[1]+1)/2+max(mylist[2],mylist[1])),0])
hash_box.append([int((mylist[2]+mylist[0])*(mylist[2]+mylist[0]+1)/2+max(mylist[2],mylist[0])),1])
hash_box = del_dublicates(hash_box)
m = min(mylist[0],mylist[1],mylist[2])
#print(new)
return hash_box,m
def generate_new_box(i,num_vertix_for_i, list_j):
new_verix = lines[i][num_vertix_for_i] + lines[list_j[0]][list_j[1]]
vertices = []
vertices.append(new_verix)
for k in range(3):
if k != num_vertix_for_i:
vertices.append(lines[i][k])
return min(vertices)
T = []
list_box = []
diameter = 0
N = 10000
T = [[] for g in range(N+1)]
for i in range(n):
hash_for_box = [0 for g in range(3)]
hash_for_box, d_k = new_box_analysis(lines[i])
if d_k>diameter:
diameter = d_k
for ll in range(len(list_box)):
_ = list_box.pop()
list_box.append(i+1)
for k in range(len(hash_for_box)):
if len(T[hash_for_box[k][0]]) == 0:
T[hash_for_box[k][0]].append(i)
T[hash_for_box[k][0]].append(hash_for_box[k][1]) #### добавляем номер box и третью сторону
else:
d_m = generate_new_box(i,hash_for_box[k][1],T[hash_for_box[k][0]])
if d_m > diameter:
diameter = d_m
for ll in range(len(list_box)):
_ = list_box.pop()
list_box.append(i+1)
list_box.append(T[hash_for_box[k][0]][0]+1)
if lines[i][hash_for_box[k][1]] > T[hash_for_box[k][0]][1]:
T[hash_for_box[k][0]].pop()
T[hash_for_box[k][0]].pop()
T[hash_for_box[k][0]].append(i)
T[hash_for_box[k][0]].append(hash_for_box[k][1])
print(len(list_box))
print(str(min(list_box[0],list_box[1])) + " " + str(max(list_box[0],list_box[1]))) if len(list_box) > 1 else print(str(list_box[0]))
```
| -1
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,640,886,875
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 46
| 0
|
st = list(input())
k = "hello"
j = 0
for i in st:
if k[j] == i:
j = j+1
if j==5:
break;
if j==5:
print("YES")
else:
print("NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
st = list(input())
k = "hello"
j = 0
for i in st:
if k[j] == i:
j = j+1
if j==5:
break;
if j==5:
print("YES")
else:
print("NO")
```
| 3.977
|
785
|
A
|
Anton and Polyhedrons
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
|
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
|
[
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] |
[
"42\n",
"28\n"
] |
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
| 500
|
[
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,695,332,135
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 249
| 0
|
n = int(input())
total_faces = 0
for _ in range(n):
shape = input().capitalize()
if shape == "Tetrahedron" :
total_faces +=4
if shape == "Cube" :
total_faces +=6
if shape == "Octahedron" :
total_faces +=8
if shape == "Dodecahedron" :
total_faces +=12
if shape == "Icosahedron" :
total_faces +=20
print(total_faces)
|
Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
|
```python
n = int(input())
total_faces = 0
for _ in range(n):
shape = input().capitalize()
if shape == "Tetrahedron" :
total_faces +=4
if shape == "Cube" :
total_faces +=6
if shape == "Octahedron" :
total_faces +=8
if shape == "Dodecahedron" :
total_faces +=12
if shape == "Icosahedron" :
total_faces +=20
print(total_faces)
```
| 3
|
|
350
|
A
|
TL
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
|
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
|
If there is a valid TL value, print it. Otherwise, print -1.
|
[
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] |
[
"5",
"-1\n"
] |
none
| 500
|
[
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43 42 38 40 22 41 1 49 47 9 8 19 15 29 30 16 18\n91 58 86 51 94 94 73 84 98 69 74 56 52 80 88 61 53 99 88 50 55 95 65 84 87 79 51 52 69 60 74 73 93 61 73 59 64 56 95 78 86 72 79 70 93 78 54 61 71 50",
"output": "49"
},
{
"input": "55 44\n93 17 74 15 34 16 41 80 26 54 94 94 86 93 20 44 63 72 39 43 67 4 37 49 76 94 5 51 64 74 11 47 77 97 57 30 42 72 71 26 8 14 67 64 49 57 30 23 40 4 76 78 87 78 79\n38 55 17 65 26 7 36 65 48 28 49 93 18 98 31 90 26 57 1 26 88 56 48 56 23 13 8 67 80 2 51 3 21 33 20 54 2 45 21 36 3 98 62 2",
"output": "-1"
},
{
"input": "32 100\n30 8 4 35 18 41 18 12 33 39 39 18 39 19 33 46 45 33 34 27 14 39 40 21 38 9 42 35 27 10 14 14\n65 49 89 64 47 78 59 52 73 51 84 82 88 63 91 99 67 87 53 99 75 47 85 82 58 47 80 50 65 91 83 90 77 52 100 88 97 74 98 99 50 93 65 61 65 65 65 96 61 51 84 67 79 90 92 83 100 100 100 95 80 54 77 51 98 64 74 62 60 96 73 74 94 55 89 60 92 65 74 79 66 81 53 47 71 51 54 85 74 97 68 72 88 94 100 85 65 63 65 90",
"output": "46"
},
{
"input": "1 50\n7\n65 52 99 78 71 19 96 72 80 15 50 94 20 35 79 95 44 41 45 53 77 50 74 66 59 96 26 84 27 48 56 84 36 78 89 81 67 34 79 74 99 47 93 92 90 96 72 28 78 66",
"output": "14"
},
{
"input": "1 1\n4\n9",
"output": "8"
},
{
"input": "1 1\n2\n4",
"output": "-1"
},
{
"input": "22 56\n49 20 42 68 15 46 98 78 82 8 7 33 50 30 75 96 36 88 35 99 19 87\n15 18 81 24 35 89 25 32 23 3 48 24 52 69 18 32 23 61 48 98 50 38 5 17 70 20 38 32 49 54 68 11 51 81 46 22 19 59 29 38 45 83 18 13 91 17 84 62 25 60 97 32 23 13 83 58",
"output": "-1"
},
{
"input": "1 1\n50\n100",
"output": "-1"
},
{
"input": "1 1\n49\n100",
"output": "98"
},
{
"input": "1 1\n100\n100",
"output": "-1"
},
{
"input": "1 1\n99\n100",
"output": "-1"
},
{
"input": "8 4\n1 2 49 99 99 95 78 98\n100 100 100 100",
"output": "99"
},
{
"input": "68 85\n43 55 2 4 72 45 19 56 53 81 18 90 11 87 47 8 94 88 24 4 67 9 21 70 25 66 65 27 46 13 8 51 65 99 37 43 71 59 71 79 32 56 49 43 57 85 95 81 40 28 60 36 72 81 60 40 16 78 61 37 29 26 15 95 70 27 50 97\n6 6 48 72 54 31 1 50 29 64 93 9 29 93 66 63 25 90 52 1 66 13 70 30 24 87 32 90 84 72 44 13 25 45 31 16 92 60 87 40 62 7 20 63 86 78 73 88 5 36 74 100 64 34 9 5 62 29 58 48 81 46 84 56 27 1 60 14 54 88 31 93 62 7 9 69 27 48 10 5 33 10 53 66 2",
"output": "-1"
},
{
"input": "5 100\n1 1 1 1 1\n77 53 38 29 97 33 64 17 78 100 27 12 42 44 20 24 44 68 58 57 65 90 8 24 4 6 74 68 61 43 25 69 8 62 36 85 67 48 69 30 35 41 42 12 87 66 50 92 53 76 38 67 85 7 80 78 53 76 94 8 37 50 4 100 4 71 10 48 34 47 83 42 25 81 64 72 25 51 53 75 43 98 53 77 94 38 81 15 89 91 72 76 7 36 27 41 88 18 19 75",
"output": "2"
},
{
"input": "3 3\n2 3 4\n8 9 10",
"output": "4"
},
{
"input": "2 1\n2 3\n15",
"output": "4"
},
{
"input": "2 1\n2 4\n4",
"output": "-1"
},
{
"input": "2 3\n4 5\n10 11 12",
"output": "8"
},
{
"input": "3 1\n2 3 3\n5",
"output": "4"
},
{
"input": "2 1\n9 10\n100",
"output": "18"
},
{
"input": "3 3\n3 12 15\n7 8 9",
"output": "-1"
},
{
"input": "2 2\n3 5\n7 8",
"output": "6"
},
{
"input": "3 3\n4 5 6\n10 11 12",
"output": "8"
},
{
"input": "3 5\n2 3 3\n6 6 6 6 2",
"output": "-1"
},
{
"input": "3 6\n4 5 3\n8 9 7 10 7 11",
"output": "6"
},
{
"input": "3 6\n4 5 2\n8 9 6 10 7 4",
"output": "-1"
},
{
"input": "2 1\n4 6\n10",
"output": "8"
},
{
"input": "1 2\n1\n3 1",
"output": "-1"
},
{
"input": "2 1\n5 6\n20",
"output": "10"
},
{
"input": "2 1\n1 5\n5",
"output": "-1"
},
{
"input": "3 2\n10 20 30\n30 40",
"output": "-1"
},
{
"input": "2 2\n5 6\n7 100",
"output": "-1"
},
{
"input": "2 1\n2 5\n7",
"output": "5"
},
{
"input": "1 1\n5\n20",
"output": "10"
},
{
"input": "2 1\n10 11\n100",
"output": "20"
},
{
"input": "1 1\n1\n10",
"output": "2"
},
{
"input": "1 1\n10\n100",
"output": "20"
}
] | 1,574,697,328
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 7
| 248
| 0
|
# Lang: pypy3.6-v7.1.0-win32\pypy3.exe
# Problem Name: TL
# Problem Serial No: 350
# Problem Type: A
# Problem Url: https://codeforces.com/problemset/problem/350/A
# Solution Generated at: 2019-11-25 21:43:16.146243 UTC
i = int
cin = input
le = len
s = str
_= input()
correct = list(map(int, cin().split()))
min_tl = max(correct)
incorrect = list(map(lambda x: i(x) <= min_tl, cin().split()))
if any(incorrect):
print(-1)
exit()
exist_extra = 0
for k in correct:
if k * 2 <= min_tl:
exist_extra += 1
break
if exist_extra == 0:
print(-1)
exit()
print(min_tl)
# Accepted
|
Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none
|
```python
# Lang: pypy3.6-v7.1.0-win32\pypy3.exe
# Problem Name: TL
# Problem Serial No: 350
# Problem Type: A
# Problem Url: https://codeforces.com/problemset/problem/350/A
# Solution Generated at: 2019-11-25 21:43:16.146243 UTC
i = int
cin = input
le = len
s = str
_= input()
correct = list(map(int, cin().split()))
min_tl = max(correct)
incorrect = list(map(lambda x: i(x) <= min_tl, cin().split()))
if any(incorrect):
print(-1)
exit()
exist_extra = 0
for k in correct:
if k * 2 <= min_tl:
exist_extra += 1
break
if exist_extra == 0:
print(-1)
exit()
print(min_tl)
# Accepted
```
| 0
|
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\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..\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..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,615,364,408
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 154
| 307,200
|
n, m = map(int, input().split())
mat = []
cor = []
for _ in range(n):
mat.append(input())
for i in range(n):
for j in range(m):
if mat[i][j] == '*':
cor.append((i, j))
if cor[0][1] == cor[1][1]: # Right Top Corner Missing
print(cor[0][0] + 1, cor[2][1] + 1)
elif cor[0][1] == cor[2][1]: # Left Top / Right Bottom Corner Missing
if cor[0][0] == cor[1][0]: # Right Bottom Corner Missing
print(cor[2][0] + 1, cor[1][1] + 1)
else: # Left Top Corner Missing
print(cor[0][0] + 1, cor[1][1] + 1)
else: # Left Bottom Corner Missing
print(cor[2][0]+1, cor[0][1]+1)
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
n, m = map(int, input().split())
mat = []
cor = []
for _ in range(n):
mat.append(input())
for i in range(n):
for j in range(m):
if mat[i][j] == '*':
cor.append((i, j))
if cor[0][1] == cor[1][1]: # Right Top Corner Missing
print(cor[0][0] + 1, cor[2][1] + 1)
elif cor[0][1] == cor[2][1]: # Left Top / Right Bottom Corner Missing
if cor[0][0] == cor[1][0]: # Right Bottom Corner Missing
print(cor[2][0] + 1, cor[1][1] + 1)
else: # Left Top Corner Missing
print(cor[0][0] + 1, cor[1][1] + 1)
else: # Left Bottom Corner Missing
print(cor[2][0]+1, cor[0][1]+1)
```
| 3
|
|
799
|
C
|
Fountains
|
PROGRAMMING
| 1,800
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. There are *n* available fountains, for each fountain its beauty and cost are known. There are two types of money in the game: coins and diamonds, so each fountain cost can be either in coins or diamonds. No money changes between the types are allowed.
Help Arkady to find two fountains with maximum total beauty so that he can buy both at the same time.
|
The first line contains three integers *n*, *c* and *d* (2<=≤<=*n*<=≤<=100<=000, 0<=≤<=*c*,<=*d*<=≤<=100<=000) — the number of fountains, the number of coins and diamonds Arkady has.
The next *n* lines describe fountains. Each of these lines contain two integers *b**i* and *p**i* (1<=≤<=*b**i*,<=*p**i*<=≤<=100<=000) — the beauty and the cost of the *i*-th fountain, and then a letter "C" or "D", describing in which type of money is the cost of fountain *i*: in coins or in diamonds, respectively.
|
Print the maximum total beauty of exactly two fountains Arkady can build. If he can't build two fountains, print 0.
|
[
"3 7 6\n10 8 C\n4 3 C\n5 6 D\n",
"2 4 5\n2 5 C\n2 1 D\n",
"3 10 10\n5 5 C\n5 5 C\n10 11 D\n"
] |
[
"9\n",
"0\n",
"10\n"
] |
In the first example Arkady should build the second fountain with beauty 4, which costs 3 coins. The first fountain he can't build because he don't have enough coins. Also Arkady should build the third fountain with beauty 5 which costs 6 diamonds. Thus the total beauty of built fountains is 9.
In the second example there are two fountains, but Arkady can't build both of them, because he needs 5 coins for the first fountain, and Arkady has only 4 coins.
| 1,500
|
[
{
"input": "3 7 6\n10 8 C\n4 3 C\n5 6 D",
"output": "9"
},
{
"input": "2 4 5\n2 5 C\n2 1 D",
"output": "0"
},
{
"input": "3 10 10\n5 5 C\n5 5 C\n10 11 D",
"output": "10"
},
{
"input": "6 68 40\n1 18 D\n6 16 D\n11 16 D\n7 23 D\n16 30 D\n2 20 D",
"output": "18"
},
{
"input": "6 4 9\n6 6 D\n1 4 D\n6 7 C\n7 6 D\n5 7 D\n2 5 D",
"output": "3"
},
{
"input": "52 38 22\n9 25 D\n28 29 C\n29 25 D\n4 28 D\n23 29 D\n24 25 D\n17 12 C\n11 19 C\n13 14 C\n12 15 D\n7 25 C\n2 25 C\n6 17 C\n2 20 C\n15 23 D\n8 21 C\n13 15 D\n29 15 C\n25 20 D\n22 20 C\n2 13 D\n13 22 D\n27 20 C\n1 21 D\n22 17 C\n14 21 D\n4 25 D\n5 23 C\n9 21 C\n2 20 C\n14 18 C\n29 24 C\n14 29 D\n9 27 C\n23 21 D\n18 26 D\n7 23 C\n13 25 C\n21 26 C\n30 24 C\n21 24 C\n28 22 C\n8 29 C\n3 12 C\n21 22 D\n22 26 C\n13 17 D\n12 12 D\n11 11 C\n18 24 D\n7 13 D\n3 11 C",
"output": "57"
},
{
"input": "6 68 40\n6 16 D\n11 16 D\n1 18 D\n2 20 D\n7 23 D\n16 30 D",
"output": "18"
},
{
"input": "2 1 1\n1 1 C\n1 1 D",
"output": "2"
},
{
"input": "2 100000 100000\n100000 100000 C\n100000 100000 D",
"output": "200000"
},
{
"input": "4 15 9\n5 10 C\n5 10 D\n6 10 D\n7 5 C",
"output": "12"
}
] | 1,666,802,527
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
def maxa(A,c):
b = -1
p = 0
for i in range(len(A)):
if A[i][1]<=c:
if(A[i][0]>b):
b = A[i][0]
return b
def maxad(A,c):
f = [-1,0]
s = [-1,0]
for i in range(len(A)):
if(A[i][0]>f[0]):
if(A[i][1]+f[1]<=c):
t = f
f = A[i]
s = t
elif(A[i][1]+s[1]<=c):
f = A[i]
elif(A[i][0]>s[0]):
if(A[i][1]+f[1]<=c):
s = A[i]
if(f[0]==-1 or s[0]==-1):
return 0
return f[0]+s[0]
def f(C,D,c,d):
if(len(C)==0):
return maxad(D,d)
elif(len(D)==0):
return maxad(C,c)
else:
a = maxa(C,c)
b = maxa(D,d)
if(a==-1 and b==-1):
return 0
elif(a==-1):
return maxad(D,d)
elif(b==-1):
return maxad(C,c)
else:
return a+b
n,c,d = list(map(int,input().split()))
C = []
D = []
for i in range(n):
b,p,t = input().split()
if(t=='C'):
C.append([int(b),int(p)])
else:
D.append([int(b),int(p)])
ans = f(C,D,c,d)
print(ans)
|
Title: Fountains
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. There are *n* available fountains, for each fountain its beauty and cost are known. There are two types of money in the game: coins and diamonds, so each fountain cost can be either in coins or diamonds. No money changes between the types are allowed.
Help Arkady to find two fountains with maximum total beauty so that he can buy both at the same time.
Input Specification:
The first line contains three integers *n*, *c* and *d* (2<=≤<=*n*<=≤<=100<=000, 0<=≤<=*c*,<=*d*<=≤<=100<=000) — the number of fountains, the number of coins and diamonds Arkady has.
The next *n* lines describe fountains. Each of these lines contain two integers *b**i* and *p**i* (1<=≤<=*b**i*,<=*p**i*<=≤<=100<=000) — the beauty and the cost of the *i*-th fountain, and then a letter "C" or "D", describing in which type of money is the cost of fountain *i*: in coins or in diamonds, respectively.
Output Specification:
Print the maximum total beauty of exactly two fountains Arkady can build. If he can't build two fountains, print 0.
Demo Input:
['3 7 6\n10 8 C\n4 3 C\n5 6 D\n', '2 4 5\n2 5 C\n2 1 D\n', '3 10 10\n5 5 C\n5 5 C\n10 11 D\n']
Demo Output:
['9\n', '0\n', '10\n']
Note:
In the first example Arkady should build the second fountain with beauty 4, which costs 3 coins. The first fountain he can't build because he don't have enough coins. Also Arkady should build the third fountain with beauty 5 which costs 6 diamonds. Thus the total beauty of built fountains is 9.
In the second example there are two fountains, but Arkady can't build both of them, because he needs 5 coins for the first fountain, and Arkady has only 4 coins.
|
```python
def maxa(A,c):
b = -1
p = 0
for i in range(len(A)):
if A[i][1]<=c:
if(A[i][0]>b):
b = A[i][0]
return b
def maxad(A,c):
f = [-1,0]
s = [-1,0]
for i in range(len(A)):
if(A[i][0]>f[0]):
if(A[i][1]+f[1]<=c):
t = f
f = A[i]
s = t
elif(A[i][1]+s[1]<=c):
f = A[i]
elif(A[i][0]>s[0]):
if(A[i][1]+f[1]<=c):
s = A[i]
if(f[0]==-1 or s[0]==-1):
return 0
return f[0]+s[0]
def f(C,D,c,d):
if(len(C)==0):
return maxad(D,d)
elif(len(D)==0):
return maxad(C,c)
else:
a = maxa(C,c)
b = maxa(D,d)
if(a==-1 and b==-1):
return 0
elif(a==-1):
return maxad(D,d)
elif(b==-1):
return maxad(C,c)
else:
return a+b
n,c,d = list(map(int,input().split()))
C = []
D = []
for i in range(n):
b,p,t = input().split()
if(t=='C'):
C.append([int(b),int(p)])
else:
D.append([int(b),int(p)])
ans = f(C,D,c,d)
print(ans)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $n - 2$ intermediate planets. Formally: we number all the planets from $1$ to $n$. $1$ is Earth, $n$ is Mars. Natasha will make exactly $n$ flights: $1 \to 2 \to \ldots n \to 1$.
Flight from $x$ to $y$ consists of two phases: take-off from planet $x$ and landing to planet $y$. This way, the overall itinerary of the trip will be: the $1$-st planet $\to$ take-off from the $1$-st planet $\to$ landing to the $2$-nd planet $\to$ $2$-nd planet $\to$ take-off from the $2$-nd planet $\to$ $\ldots$ $\to$ landing to the $n$-th planet $\to$ the $n$-th planet $\to$ take-off from the $n$-th planet $\to$ landing to the $1$-st planet $\to$ the $1$-st planet.
The mass of the rocket together with all the useful cargo (but without fuel) is $m$ tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that $1$ ton of fuel can lift off $a_i$ tons of rocket from the $i$-th planet or to land $b_i$ tons of rocket onto the $i$-th planet.
For example, if the weight of rocket is $9$ tons, weight of fuel is $3$ tons and take-off coefficient is $8$ ($a_i = 8$), then $1.5$ tons of fuel will be burnt (since $1.5 \cdot 8 = 9 + 3$). The new weight of fuel after take-off will be $1.5$ tons.
Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.
Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.
|
The first line contains a single integer $n$ ($2 \le n \le 1000$) — number of planets.
The second line contains the only integer $m$ ($1 \le m \le 1000$) — weight of the payload.
The third line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 1000$), where $a_i$ is the number of tons, which can be lifted off by one ton of fuel.
The fourth line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 1000$), where $b_i$ is the number of tons, which can be landed by one ton of fuel.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
|
If Natasha can fly to Mars through $(n - 2)$ planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number $-1$.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
The answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Formally, let your answer be $p$, and the jury's answer be $q$. Your answer is considered correct if $\frac{|p - q|}{\max{(1, |q|)}} \le 10^{-6}$.
|
[
"2\n12\n11 8\n7 5\n",
"3\n1\n1 4 1\n2 5 3\n",
"6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3\n"
] |
[
"10.0000000000\n",
"-1\n",
"85.4800000000\n"
] |
Let's consider the first example.
Initially, the mass of a rocket with fuel is $22$ tons.
- At take-off from Earth one ton of fuel can lift off $11$ tons of cargo, so to lift off $22$ tons you need to burn $2$ tons of fuel. Remaining weight of the rocket with fuel is $20$ tons.- During landing on Mars, one ton of fuel can land $5$ tons of cargo, so for landing $20$ tons you will need to burn $4$ tons of fuel. There will be $16$ tons of the rocket with fuel remaining.- While taking off from Mars, one ton of fuel can raise $8$ tons of cargo, so to lift off $16$ tons you will need to burn $2$ tons of fuel. There will be $14$ tons of rocket with fuel after that.- During landing on Earth, one ton of fuel can land $7$ tons of cargo, so for landing $14$ tons you will need to burn $2$ tons of fuel. Remaining weight is $12$ tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth.
| 0
|
[
{
"input": "2\n12\n11 8\n7 5",
"output": "10.0000000000"
},
{
"input": "3\n1\n1 4 1\n2 5 3",
"output": "-1"
},
{
"input": "6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3",
"output": "85.4800000000"
},
{
"input": "3\n3\n1 2 1\n2 2 2",
"output": "-1"
},
{
"input": "4\n4\n2 3 2 2\n2 3 4 3",
"output": "284.0000000000"
},
{
"input": "5\n2\n1 2 2 1 2\n4 5 1 4 1",
"output": "-1"
},
{
"input": "7\n7\n3 2 6 2 2 2 5\n4 7 5 6 2 2 2",
"output": "4697.0000000000"
},
{
"input": "2\n1000\n12 34\n56 78",
"output": "159.2650775220"
},
{
"input": "8\n4\n1 1 4 1 3 1 8 1\n1 1 1 1 1 3 1 2",
"output": "-1"
},
{
"input": "9\n2\n8 7 1 1 3 7 1 2 4\n4 1 1 8 7 7 1 1 5",
"output": "-1"
},
{
"input": "10\n10\n9 8 8 7 2 10 2 9 2 4\n3 10 6 2 6 6 5 9 4 5",
"output": "3075.7142857143"
},
{
"input": "20\n12\n3 9 12 13 16 18 9 9 19 7 2 5 17 14 7 7 15 16 5 7\n16 9 13 5 14 10 4 3 16 16 12 20 17 11 4 5 5 14 6 15",
"output": "4670.8944493007"
},
{
"input": "30\n5\n25 1 28 1 27 25 24 1 28 1 12 1 29 16 1 1 1 1 27 1 24 1 1 1 1 1 1 1 30 3\n1 22 1 1 24 2 13 1 16 21 1 27 14 16 1 1 7 1 1 18 1 23 10 1 15 16 16 15 10 1",
"output": "-1"
},
{
"input": "40\n13\n1 1 1 23 21 1 1 1 1 1 40 32 1 21 1 8 1 1 36 15 33 1 30 1 1 37 22 1 4 39 7 1 9 37 1 1 1 28 1 1\n1 34 17 1 38 20 8 14 1 18 29 3 21 21 18 14 1 11 1 1 23 1 25 1 14 1 7 31 9 20 25 1 1 1 1 8 26 12 1 1",
"output": "-1"
},
{
"input": "50\n19\n17 7 13 42 19 25 10 25 2 36 17 40 30 48 34 43 34 20 5 15 8 7 43 35 21 40 40 19 30 11 49 7 24 23 43 30 38 49 10 8 30 11 28 50 48 25 25 20 48 24\n49 35 10 22 24 50 50 7 6 13 16 35 12 43 50 44 35 33 38 49 26 18 23 37 7 38 23 20 28 48 41 16 6 32 32 34 11 39 38 9 38 23 16 31 37 47 33 20 46 30",
"output": "7832.1821424977"
},
{
"input": "60\n21\n11 35 1 28 39 13 19 56 13 13 21 25 1 1 23 1 52 26 53 1 1 1 30 39 1 7 1 1 3 1 1 10 1 1 37 1 1 25 1 1 1 53 1 3 48 1 6 5 4 15 1 14 25 53 25 38 27 1 1 1\n1 1 1 35 40 58 10 22 1 56 1 59 1 6 33 1 1 1 1 18 14 1 1 40 25 47 1 34 1 1 53 1 1 25 1 45 1 1 25 34 3 1 1 1 53 27 11 58 1 1 1 10 12 1 1 1 31 52 1 1",
"output": "-1"
},
{
"input": "70\n69\n70 66 57 58 24 60 39 2 48 61 65 22 10 26 68 62 48 25 12 14 45 57 6 30 48 15 46 33 42 28 69 42 64 25 24 8 62 12 68 53 55 20 32 70 3 5 41 49 16 26 2 34 34 20 39 65 18 47 62 31 39 28 61 67 7 14 31 31 53 54\n40 33 24 20 68 20 22 39 53 56 48 38 59 45 47 46 7 69 11 58 61 40 35 38 62 66 18 36 44 48 67 24 14 27 67 63 68 30 50 6 58 7 6 35 20 58 6 12 12 23 14 2 63 27 29 22 49 16 55 40 70 27 27 70 42 38 66 55 69 47",
"output": "217989.4794743629"
},
{
"input": "80\n21\n65 4 26 25 1 1 1 1 1 1 60 1 29 43 48 6 48 13 29 1 1 62 1 1 1 1 1 1 1 26 9 1 22 1 35 13 66 36 1 1 1 38 55 21 70 1 58 70 1 1 38 1 1 20 1 1 51 1 1 28 1 23 11 1 39 47 1 52 41 1 63 1 1 52 1 45 11 10 80 1\n1 1 25 30 1 1 55 54 1 48 10 37 22 1 74 1 78 13 1 65 32 1 1 1 1 69 5 59 1 1 65 1 40 1 31 1 1 75 54 1 60 1 1 1 1 1 1 1 11 29 36 1 72 71 52 1 1 1 37 1 1 75 43 9 53 1 62 1 29 1 40 27 59 74 41 53 19 30 1 73",
"output": "-1"
},
{
"input": "90\n35\n1 68 16 30 24 1 1 1 35 1 1 67 1 1 1 1 33 16 37 77 83 1 77 26 1 1 68 67 70 62 1 47 1 1 1 84 1 65 1 32 83 1 1 1 28 1 71 76 84 1 1 5 1 74 10 1 1 1 38 87 13 1 7 66 81 49 1 9 1 11 1 25 1 1 1 1 7 1 1 36 61 47 51 1 1 69 40 1 37 1\n40 1 21 1 19 51 37 52 64 1 86 1 5 24 1 1 1 19 36 1 1 77 24 4 1 18 89 1 1 1 1 1 29 22 1 80 32 36 6 1 63 1 30 1 1 1 86 79 73 52 9 1 1 11 7 1 25 20 1 20 1 49 1 37 1 41 1 1 1 1 54 55 1 10 1 1 1 1 1 1 66 1 68 1 1 1 1 53 1 1",
"output": "-1"
},
{
"input": "2\n1\n1 1\n1 1",
"output": "-1"
},
{
"input": "2\n1\n1 1\n2 2",
"output": "-1"
},
{
"input": "2\n1\n2 2\n1 1",
"output": "-1"
},
{
"input": "2\n1\n2 2\n2 2",
"output": "15.0000000000"
},
{
"input": "2\n2\n1 1\n1 1",
"output": "-1"
},
{
"input": "2\n2\n1 1\n2 2",
"output": "-1"
},
{
"input": "2\n2\n2 2\n1 1",
"output": "-1"
},
{
"input": "2\n2\n2 2\n2 2",
"output": "30.0000000000"
},
{
"input": "40\n55\n1 382 1 1 1 629 111 689 396 614 1 1 995 148 7 820 913 1 1 169 157 1 702 1 159 1 1 226 1 253 1 319 1 130 1 1 1 466 1 756\n1 23 555 1 412 1 1 373 316 234 888 1 112 818 33 443 313 1 235 1 1 610 110 535 1 445 1 386 1 1 758 1 292 1 862 1 244 428 530 1",
"output": "-1"
},
{
"input": "49\n1\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100\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",
"output": "695580114.6380882263"
},
{
"input": "2\n12\n11 8\n1 1",
"output": "-1"
},
{
"input": "3\n3\n7 11 17\n19 31 33",
"output": "1.6012429470"
}
] | 1,614,016,764
| 1,164
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
n=int(input())
m=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if((1 in a) or (1 in b)):
print(-1)
else:
#print(m)
temp = m*(b[0]/(b[0]-1))
#print(temp)
for i in range(n-1,0,-1):
temp = temp*(a[i]/(a[i]-1))
#print(temp)
temp = temp*(b[i]/(b[i]-1))
#print(temp)
temp = temp*(a[0]/(a[0]-1))
print(temp)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $n - 2$ intermediate planets. Formally: we number all the planets from $1$ to $n$. $1$ is Earth, $n$ is Mars. Natasha will make exactly $n$ flights: $1 \to 2 \to \ldots n \to 1$.
Flight from $x$ to $y$ consists of two phases: take-off from planet $x$ and landing to planet $y$. This way, the overall itinerary of the trip will be: the $1$-st planet $\to$ take-off from the $1$-st planet $\to$ landing to the $2$-nd planet $\to$ $2$-nd planet $\to$ take-off from the $2$-nd planet $\to$ $\ldots$ $\to$ landing to the $n$-th planet $\to$ the $n$-th planet $\to$ take-off from the $n$-th planet $\to$ landing to the $1$-st planet $\to$ the $1$-st planet.
The mass of the rocket together with all the useful cargo (but without fuel) is $m$ tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that $1$ ton of fuel can lift off $a_i$ tons of rocket from the $i$-th planet or to land $b_i$ tons of rocket onto the $i$-th planet.
For example, if the weight of rocket is $9$ tons, weight of fuel is $3$ tons and take-off coefficient is $8$ ($a_i = 8$), then $1.5$ tons of fuel will be burnt (since $1.5 \cdot 8 = 9 + 3$). The new weight of fuel after take-off will be $1.5$ tons.
Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.
Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.
Input Specification:
The first line contains a single integer $n$ ($2 \le n \le 1000$) — number of planets.
The second line contains the only integer $m$ ($1 \le m \le 1000$) — weight of the payload.
The third line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 1000$), where $a_i$ is the number of tons, which can be lifted off by one ton of fuel.
The fourth line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 1000$), where $b_i$ is the number of tons, which can be landed by one ton of fuel.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
Output Specification:
If Natasha can fly to Mars through $(n - 2)$ planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number $-1$.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
The answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Formally, let your answer be $p$, and the jury's answer be $q$. Your answer is considered correct if $\frac{|p - q|}{\max{(1, |q|)}} \le 10^{-6}$.
Demo Input:
['2\n12\n11 8\n7 5\n', '3\n1\n1 4 1\n2 5 3\n', '6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3\n']
Demo Output:
['10.0000000000\n', '-1\n', '85.4800000000\n']
Note:
Let's consider the first example.
Initially, the mass of a rocket with fuel is $22$ tons.
- At take-off from Earth one ton of fuel can lift off $11$ tons of cargo, so to lift off $22$ tons you need to burn $2$ tons of fuel. Remaining weight of the rocket with fuel is $20$ tons.- During landing on Mars, one ton of fuel can land $5$ tons of cargo, so for landing $20$ tons you will need to burn $4$ tons of fuel. There will be $16$ tons of the rocket with fuel remaining.- While taking off from Mars, one ton of fuel can raise $8$ tons of cargo, so to lift off $16$ tons you will need to burn $2$ tons of fuel. There will be $14$ tons of rocket with fuel after that.- During landing on Earth, one ton of fuel can land $7$ tons of cargo, so for landing $14$ tons you will need to burn $2$ tons of fuel. Remaining weight is $12$ tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth.
|
```python
n=int(input())
m=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if((1 in a) or (1 in b)):
print(-1)
else:
#print(m)
temp = m*(b[0]/(b[0]-1))
#print(temp)
for i in range(n-1,0,-1):
temp = temp*(a[i]/(a[i]-1))
#print(temp)
temp = temp*(b[i]/(b[i]-1))
#print(temp)
temp = temp*(a[0]/(a[0]-1))
print(temp)
```
| 0
|
|
447
|
B
|
DZY Loves Strings
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
|
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
|
Print a single integer — the largest possible value of the resulting string DZY could get.
|
[
"abc\n3\n1 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\n"
] |
[
"41\n"
] |
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.
| 1,000
|
[
{
"input": "abc\n3\n1 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",
"output": "41"
},
{
"input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453",
"output": "29978"
},
{
"input": "ajeeseerqnpaujubmajpibxrccazaawetywxmifzehojf\n23\n359 813 772 413 733 654 33 87 890 433 395 311 801 852 376 148 914 420 636 695 583 733 664 394 407 314",
"output": "1762894"
},
{
"input": "uahngxejpomhbsebcxvelfsojbaouynnlsogjyvktpwwtcyddkcdqcqs\n34\n530 709 150 660 947 830 487 142 208 276 885 542 138 214 76 184 273 753 30 195 722 236 82 691 572 585",
"output": "2960349"
},
{
"input": "xnzeqmouqyzvblcidmhbkqmtusszuczadpooslqxegldanwopilmdwzbczvrwgnwaireykwpugvpnpafbxlyggkgawghysufuegvmzvpgcqyjkoadcreaguzepbendwnowsuekxxivkziibxvxfoilofxcgnxvfefyezfhevfvtetsuhwtyxdlkccdkvqjl\n282\n170 117 627 886 751 147 414 187 150 960 410 70 576 681 641 729 798 877 611 108 772 643 683 166 305 933",
"output": "99140444"
},
{
"input": "pplkqmluhfympkjfjnfdkwrkpumgdmbkfbbldpepicbbmdgafttpopzdxsevlqbtywzkoxyviglbbxsohycbdqksrhlumsldiwzjmednbkcjishkiekfrchzuztkcxnvuykhuenqojrmzaxlaoxnljnvqgnabtmcftisaazzgbmubmpsorygyusmeonrhrgphnfhlaxrvyhuxsnnezjxmdoklpquzpvjbxgbywppmegzxknhfzyygrmejleesoqfwheulmqhonqaukyuejtwxskjldplripyihbfpookxkuehiwqthbfafyrgmykuxglpplozycgydyecqkgfjljfqvigqhuxssqqtfanwszduwbsoytnrtgc\n464\n838 95 473 955 690 84 436 19 179 437 674 626 377 365 781 4 733 776 462 203 119 256 381 668 855 686",
"output": "301124161"
},
{
"input": "qkautnuilwlhjsldfcuwhiqtgtoihifszlyvfaygrnivzgvwthkrzzdtfjcirrjjlrmjtbjlzmjeqmuffsjorjyggzefwgvmblvotvzffnwjhqxorpowzdcnfksdibezdtfjjxfozaghieksbmowrbeehuxlesmvqjsphlvauxiijm\n98\n121 622 0 691 616 959 838 161 581 862 876 830 267 812 598 106 337 73 588 323 999 17 522 399 657 495",
"output": "30125295"
},
{
"input": "tghyxqfmhz\n8\n191 893 426 203 780 326 148 259 182 140 847 636 778 97 167 773 219 891 758 993 695 603 223 779 368 165",
"output": "136422"
},
{
"input": "nyawbfjxnxjiyhwkydaruozobpphgjqdpfdqzezcsoyvurnapu\n30\n65 682 543 533 990 148 815 821 315 916 632 771 332 513 472 864 12 73 548 687 660 572 507 192 226 348",
"output": "2578628"
},
{
"input": "pylrnkrbcjgoytvdnhmlvnkknijkdgdhworlvtwuonrkhrilkewcnofodaumgvnsisxooswgrgtvdeauyxhkipfoxrrtysuepjcf\n60\n894 206 704 179 272 337 413 828 119 182 330 46 440 102 250 191 242 539 678 783 843 431 612 567 33 338",
"output": "9168707"
},
{
"input": "vhjnkrxbyhjhnjrxvwxmhxwoxttbtqosfxtcuvhfjlkyfspeypthsdkkwnqdpxdlnxsgtzvkrgqosgfjrwetqbxgoarkjhrjbspzgblsapifltkfxbfdbxqwoohlgyzijmiwnpmveybyzvasoctxsmgjehpyysmqblwnmkappbecklqjfmxhlyceordroflnposohfplrvijxbwvqdtvzhobtrumiujnyrfbwthvciinuveoizkccelxtaveiiagryqnyvsgfnipnavrtmdqlcnldepocbpzmqnarkdvykds\n276\n364 244 798 82 582 9 309 950 286 547 892 371 569 159 705 975 740 845 655 179 130 993 255 552 882 657",
"output": "144901921"
},
{
"input": "gsaddmezrnttfalbwlqbnedumvikplfosw\n12\n290 850 872 361 483 895 152 118 974 619 701 154 899 285 328 712 669 984 407 340 851 775 324 892 554 860",
"output": "809931"
},
{
"input": "a\n0\n5 1 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": "5"
},
{
"input": "lol\n3\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",
"output": "21"
}
] | 1,550,424,255
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
s = input()
k = int(input())
wi = list(map(int, input().split()))
res = 0
for i in range(len(s)):
res += wi[ord(s[i])-97]*(i+1)
print(res+max(wi)*(k*(k+1)//2 + len(s)*k))
|
Title: DZY Loves Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
Input Specification:
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
Output Specification:
Print a single integer — the largest possible value of the resulting string DZY could get.
Demo Input:
['abc\n3\n1 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\n']
Demo Output:
['41\n']
Note:
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.
|
```python
s = input()
k = int(input())
wi = list(map(int, input().split()))
res = 0
for i in range(len(s)):
res += wi[ord(s[i])-97]*(i+1)
print(res+max(wi)*(k*(k+1)//2 + len(s)*k))
```
| -1
|
|
573
|
A
|
Bear and Poker
|
PROGRAMMING
| 1,300
|
[
"implementation",
"math",
"number theory"
] | null | null |
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
|
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
|
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
|
[
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] |
[
"Yes\n",
"No\n"
] |
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal.
| 500
|
[
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724664764 101294996 614007760 202904092 555293973 707655552 108023967 73123445 612562357 552908390 914853758 915004122 466129205 122853497 814592742 373389439 818473058",
"output": "No"
},
{
"input": "2\n1 1",
"output": "Yes"
},
{
"input": "2\n72 72",
"output": "Yes"
},
{
"input": "2\n49 42",
"output": "No"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "Yes"
},
{
"input": "6\n162000 96000 648000 1000 864000 432000",
"output": "Yes"
},
{
"input": "8\n600000 100000 100000 100000 900000 600000 900000 600000",
"output": "Yes"
},
{
"input": "12\n2048 1024 6144 1024 3072 3072 6144 1024 4096 2048 6144 3072",
"output": "Yes"
},
{
"input": "20\n246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246",
"output": "Yes"
},
{
"input": "50\n840868705 387420489 387420489 795385082 634350497 206851546 536870912 536870912 414927754 387420489 387420489 536870912 387420489 149011306 373106005 536870912 700746206 387420489 777952883 847215247 176645254 576664386 387420489 230876513 536870912 536870912 536870912 387420489 387420489 536870912 460495524 528643722 387420489 536870912 470369206 899619085 387420489 631148352 387420489 387420489 536870912 414666674 521349938 776784669 387420489 102428009 536870912 387420489 536870912 718311009",
"output": "No"
},
{
"input": "2\n5 6",
"output": "No"
},
{
"input": "3\n536870912 387420489 257407169",
"output": "No"
},
{
"input": "4\n2 2 5 2",
"output": "No"
},
{
"input": "2\n33554432 59049",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 387420489",
"output": "Yes"
},
{
"input": "2\n1 5",
"output": "No"
},
{
"input": "18\n2 3 5 7 11 13 17 19 23 29 31 37 43 47 53 59 67 71",
"output": "No"
},
{
"input": "2\n1 30",
"output": "No"
},
{
"input": "3\n335544320 71744535 71744535",
"output": "Yes"
},
{
"input": "5\n1000000000 999999999 999999998 999999997 999999996",
"output": "No"
},
{
"input": "2\n25 5",
"output": "No"
},
{
"input": "4\n75 150 75 5",
"output": "No"
},
{
"input": "3\n536870912 387420489 362797056",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 89",
"output": "No"
},
{
"input": "4\n547 2606459 222763549 143466789",
"output": "No"
},
{
"input": "3\n129140163 33554432 1",
"output": "Yes"
},
{
"input": "10\n244140625 244140625 244140625 244140625 244140625 244140625 244140625 244140625 536870912 387420489",
"output": "No"
},
{
"input": "3\n5 5 1",
"output": "No"
},
{
"input": "5\n3 7 29 36760123 823996703",
"output": "No"
}
] | 1,442,638,243
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 280
| 3,788,800
|
n = int(input())
def inp(s):
j = 0
a = []
for i in range(len(s)):
if s[i] == " ":
a.append(int(s[j:i]))
j = i+1
if i == len(s)-1:
a.append(int(s[j:]))
return a
s = input()
a = inp(s)
m = max(a)
f = 0
for i in a:
if i == m:
continue
else:
if m%i != 0:
f = 1
break
else:
if (m/i)%2 == 0 or (m/i)%3 == 0:
continue
else:
f = 1
break
if f == 0:
print ("Yes")
else:
print ("No")
|
Title: Bear and Poker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal.
|
```python
n = int(input())
def inp(s):
j = 0
a = []
for i in range(len(s)):
if s[i] == " ":
a.append(int(s[j:i]))
j = i+1
if i == len(s)-1:
a.append(int(s[j:]))
return a
s = input()
a = inp(s)
m = max(a)
f = 0
for i in a:
if i == m:
continue
else:
if m%i != 0:
f = 1
break
else:
if (m/i)%2 == 0 or (m/i)%3 == 0:
continue
else:
f = 1
break
if f == 0:
print ("Yes")
else:
print ("No")
```
| 0
|
|
161
|
D
|
Distance in Tree
|
PROGRAMMING
| 1,800
|
[
"dfs and similar",
"dp",
"trees"
] | null | null |
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with *n* vertices and a positive number *k*. Find the number of distinct pairs of the vertices which have a distance of exactly *k* between them. Note that pairs (*v*, *u*) and (*u*, *v*) are considered to be the same pair.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 1<=≤<=*k*<=≤<=500) — the number of vertices and the required distance between the vertices.
Next *n*<=-<=1 lines describe the edges as "*a**i* *b**i*" (without the quotes) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), where *a**i* and *b**i* are the vertices connected by the *i*-th edge. All given edges are different.
|
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly *k* between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"5 2\n1 2\n2 3\n3 4\n2 5\n",
"5 3\n1 2\n2 3\n3 4\n4 5\n"
] |
[
"4\n",
"2\n"
] |
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
| 2,000
|
[
{
"input": "5 2\n1 2\n2 3\n3 4\n2 5",
"output": "4"
},
{
"input": "5 3\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10 1\n2 1\n3 1\n4 3\n5 4\n6 5\n7 1\n8 6\n9 2\n10 6",
"output": "9"
},
{
"input": "10 2\n2 1\n3 1\n4 3\n5 4\n6 5\n7 1\n8 6\n9 2\n10 6",
"output": "10"
},
{
"input": "10 3\n2 1\n3 1\n4 3\n5 4\n6 5\n7 1\n8 6\n9 2\n10 6",
"output": "8"
},
{
"input": "50 3\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13\n15 14\n16 15\n17 16\n18 17\n19 18\n20 19\n21 20\n22 21\n23 22\n24 23\n25 24\n26 25\n27 26\n28 27\n29 28\n30 29\n31 30\n32 31\n33 32\n34 33\n35 34\n36 35\n37 36\n38 37\n39 38\n40 39\n41 40\n42 41\n43 42\n44 43\n45 44\n46 45\n47 46\n48 47\n49 48\n50 49",
"output": "47"
},
{
"input": "50 4\n2 1\n3 1\n4 2\n5 2\n6 3\n7 3\n8 4\n9 4\n10 5\n11 5\n12 6\n13 6\n14 7\n15 7\n16 8\n17 8\n18 9\n19 9\n20 10\n21 10\n22 11\n23 11\n24 12\n25 12\n26 13\n27 13\n28 14\n29 14\n30 15\n31 15\n32 16\n33 16\n34 17\n35 17\n36 18\n37 18\n38 19\n39 19\n40 20\n41 20\n42 21\n43 21\n44 22\n45 22\n46 23\n47 23\n48 24\n49 24\n50 25",
"output": "124"
},
{
"input": "50 5\n2 1\n3 1\n4 2\n5 2\n6 4\n7 4\n8 6\n9 6\n10 8\n11 8\n12 10\n13 10\n14 12\n15 12\n16 14\n17 14\n18 16\n19 16\n20 18\n21 18\n22 20\n23 20\n24 22\n25 22\n26 24\n27 24\n28 26\n29 26\n30 28\n31 28\n32 30\n33 30\n34 32\n35 32\n36 34\n37 34\n38 36\n39 36\n40 38\n41 38\n42 40\n43 40\n44 42\n45 42\n46 44\n47 44\n48 46\n49 46\n50 48",
"output": "84"
},
{
"input": "10 20\n2 1\n3 2\n4 1\n5 3\n6 4\n7 1\n8 2\n9 7\n10 2",
"output": "0"
}
] | 1,699,419,698
| 2,147,483,647
|
PyPy 3-64
|
MEMORY_LIMIT_EXCEEDED
|
TESTS
| 11
| 1,964
| 536,883,200
|
import sys
sys.setrecursionlimit(100000)
n, k = map(int, input().split())
v = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
v[a] += [b]
v[b] += [a]
dp = [[0 for i in range(k + 1)] for i in range(n + 1)]
dp1 = [[0 for i in range(k + 1)] for i in range(n + 1)]
ans = 0
def dfs(x, last):
global ans
dp[x][0] = 1
for i in v[x]:
if i == last:continue
dfs(i, x)
for j in range(k):
dp[x][j + 1] += dp[i][j]
ans += dp[x][k]
dfs(1, 0)
ff = 0
def pre_calc(x, last):
global ff
for j in range(1, k):
dp1[x][j + 1] += dp1[last][j]
for i in v[x]:
if i == last:continue
for j in range(k):
dp[x][j + 1] -= dp[i][j]
for j in range(1, k):
dp1[i][j + 1] += dp[x][j]
for j in range(k):
dp[x][j + 1] += dp[i][j]
pre_calc(i, x)
# print('!',x, dp1[x][k])
ff += dp1[x][k]
pre_calc(1, 0)
print(ans + ff // 2)
|
Title: Distance in Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with *n* vertices and a positive number *k*. Find the number of distinct pairs of the vertices which have a distance of exactly *k* between them. Note that pairs (*v*, *u*) and (*u*, *v*) are considered to be the same pair.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 1<=≤<=*k*<=≤<=500) — the number of vertices and the required distance between the vertices.
Next *n*<=-<=1 lines describe the edges as "*a**i* *b**i*" (without the quotes) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), where *a**i* and *b**i* are the vertices connected by the *i*-th edge. All given edges are different.
Output Specification:
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly *k* between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['5 2\n1 2\n2 3\n3 4\n2 5\n', '5 3\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['4\n', '2\n']
Note:
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
|
```python
import sys
sys.setrecursionlimit(100000)
n, k = map(int, input().split())
v = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
v[a] += [b]
v[b] += [a]
dp = [[0 for i in range(k + 1)] for i in range(n + 1)]
dp1 = [[0 for i in range(k + 1)] for i in range(n + 1)]
ans = 0
def dfs(x, last):
global ans
dp[x][0] = 1
for i in v[x]:
if i == last:continue
dfs(i, x)
for j in range(k):
dp[x][j + 1] += dp[i][j]
ans += dp[x][k]
dfs(1, 0)
ff = 0
def pre_calc(x, last):
global ff
for j in range(1, k):
dp1[x][j + 1] += dp1[last][j]
for i in v[x]:
if i == last:continue
for j in range(k):
dp[x][j + 1] -= dp[i][j]
for j in range(1, k):
dp1[i][j + 1] += dp[x][j]
for j in range(k):
dp[x][j + 1] += dp[i][j]
pre_calc(i, x)
# print('!',x, dp1[x][k])
ff += dp1[x][k]
pre_calc(1, 0)
print(ans + ff // 2)
```
| 0
|
|
56
|
A
|
Bar
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Bar
|
2
|
256
|
According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol.
|
Print a single number which is the number of people Vasya should check to guarantee the law enforcement.
|
[
"5\n18\nVODKA\nCOKE\n19\n17\n"
] |
[
"2\n"
] |
In the sample test the second and fifth clients should be checked.
| 500
|
[
{
"input": "5\n18\nVODKA\nCOKE\n19\n17",
"output": "2"
},
{
"input": "2\n2\nGIN",
"output": "2"
},
{
"input": "3\nWHISKEY\n3\nGIN",
"output": "3"
},
{
"input": "4\n813\nIORBQITQXMPTFAEMEQDQIKFGKGOTNKTOSZCBRPXJLUKVLVHJYNRUJXK\nRUM\nRHVRWGODYWWTYZFLFYKCVUFFRTQDINKNWPKFHZBFWBHWINWJW",
"output": "1"
},
{
"input": "4\nSAKE\nSAKE\n13\n2",
"output": "4"
},
{
"input": "2\n0\n17",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
}
] | 1,607,203,339
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 28
| 310
| 0
|
ALCOHOL = { "ABSINTH", "BEER", "BRANDY", "CHAMPAGNE", "GIN", "RUM",
"SAKE", "TEQUILA", "VODKA", "WHISKEY", "WINE" }
n = int(input())
ans = 0
for _ in range(n):
s = input()
if (s[0].isdigit() and int(s) < 18) or s in ALCOHOL:
ans += 1
print(ans)
|
Title: Bar
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol.
Output Specification:
Print a single number which is the number of people Vasya should check to guarantee the law enforcement.
Demo Input:
['5\n18\nVODKA\nCOKE\n19\n17\n']
Demo Output:
['2\n']
Note:
In the sample test the second and fifth clients should be checked.
|
```python
ALCOHOL = { "ABSINTH", "BEER", "BRANDY", "CHAMPAGNE", "GIN", "RUM",
"SAKE", "TEQUILA", "VODKA", "WHISKEY", "WINE" }
n = int(input())
ans = 0
for _ in range(n):
s = input()
if (s[0].isdigit() and int(s) < 18) or s in ALCOHOL:
ans += 1
print(ans)
```
| 3.9225
|
785
|
A
|
Anton and Polyhedrons
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
|
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
|
[
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] |
[
"42\n",
"28\n"
] |
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
| 500
|
[
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,689,770,041
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n=int(input())
l={'Tetrahedron':4,'Cube':6, 'Octahedron':8,'Dodecahedron':12,'Icosahedron':20}
for i in range(n):
m=0
s = input()
for key, value in l.items():
if s.lower() == value.lower():
m += key
print(m)
|
Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
|
```python
n=int(input())
l={'Tetrahedron':4,'Cube':6, 'Octahedron':8,'Dodecahedron':12,'Icosahedron':20}
for i in range(n):
m=0
s = input()
for key, value in l.items():
if s.lower() == value.lower():
m += key
print(m)
```
| -1
|
|
312
|
B
|
Archer
|
PROGRAMMING
| 1,300
|
[
"math",
"probabilities"
] | null | null |
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
|
A single line contains four integers .
|
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"1 2 1 2\n"
] |
[
"0.666666666667"
] |
none
| 1,000
|
[
{
"input": "1 2 1 2",
"output": "0.666666666667"
},
{
"input": "1 3 1 3",
"output": "0.600000000000"
},
{
"input": "1 3 2 3",
"output": "0.428571428571"
},
{
"input": "3 4 3 4",
"output": "0.800000000000"
},
{
"input": "1 2 10 11",
"output": "0.523809523810"
},
{
"input": "4 5 4 5",
"output": "0.833333333333"
},
{
"input": "466 701 95 721",
"output": "0.937693791148"
},
{
"input": "268 470 444 885",
"output": "0.725614009325"
},
{
"input": "632 916 713 821",
"output": "0.719292895126"
},
{
"input": "269 656 918 992",
"output": "0.428937461623"
},
{
"input": "71 657 187 695",
"output": "0.310488463257"
},
{
"input": "435 852 973 978",
"output": "0.511844133157"
},
{
"input": "518 816 243 359",
"output": "0.719734031025"
},
{
"input": "882 962 311 811",
"output": "0.966386645447"
},
{
"input": "684 774 580 736",
"output": "0.906051574446"
},
{
"input": "486 868 929 999",
"output": "0.577723252958"
},
{
"input": "132 359 996 998",
"output": "0.368154532345"
},
{
"input": "933 977 266 450",
"output": "0.972879407907"
},
{
"input": "298 833 615 872",
"output": "0.441270817024"
},
{
"input": "34 554 14 958",
"output": "0.817324099167"
},
{
"input": "836 934 800 905",
"output": "0.906105535462"
},
{
"input": "482 815 69 509",
"output": "0.914365577772"
},
{
"input": "284 423 137 521",
"output": "0.885974839378"
},
{
"input": "648 881 486 703",
"output": "0.800911421248"
},
{
"input": "450 885 755 836",
"output": "0.533901011176"
},
{
"input": "533 773 823 998",
"output": "0.729222130525"
},
{
"input": "897 957 92 898",
"output": "0.993193806364"
},
{
"input": "699 925 441 928",
"output": "0.866816866175"
},
{
"input": "64 704 148 603",
"output": "0.289486317811"
},
{
"input": "719 735 626 990",
"output": "0.986124079764"
},
{
"input": "1 1000 1 1000",
"output": "0.500250125063"
}
] | 1,633,945,411
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 92
| 20,172,800
|
a,b,c,d=map(int, input().split())
smallR,Zanoes=a/b,c/d
a1,b1=1-smallR,1-Zanoes
ans = smallR / (1-(a1 * b1))
print (ans)
|
Title: Archer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input Specification:
A single line contains four integers .
Output Specification:
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['1 2 1 2\n']
Demo Output:
['0.666666666667']
Note:
none
|
```python
a,b,c,d=map(int, input().split())
smallR,Zanoes=a/b,c/d
a1,b1=1-smallR,1-Zanoes
ans = smallR / (1-(a1 * b1))
print (ans)
```
| 3
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,696,075,857
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 62
| 0
|
x =input()
sol_num=0
for i in range(int(x)):
y =input().split(" ")
count=y.count("1")
if count>=2:
sol_num+=1
print(sol_num)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
x =input()
sol_num=0
for i in range(int(x)):
y =input().split(" ")
count=y.count("1")
if count>=2:
sol_num+=1
print(sol_num)
```
| 3
|
|
888
|
A
|
Local Extrema
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000) — the elements of array *a*.
|
Print the number of local extrema in the given array.
|
[
"3\n1 2 3\n",
"4\n1 5 2 5\n"
] |
[
"0\n",
"2\n"
] |
none
| 0
|
[
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "4\n1 5 2 5",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n548",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n3 2 3",
"output": "1"
},
{
"input": "2\n1 1000",
"output": "0"
},
{
"input": "3\n1 2 1",
"output": "1"
}
] | 1,690,337,021
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
a = int(input())
list = map(int, input().split())
lmx = 0
lmn = 0
for i in range(a):
if list[i] == 0:
pass
elif list[i] > list[i-1] and list[i] > list[i+1]:
lmx += 1
elif list[i] < list[i-1] and list[i] < list[i+1]:
lmn += 1
else:
pass
print(lmx+lmn)
|
Title: Local Extrema
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000) — the elements of array *a*.
Output Specification:
Print the number of local extrema in the given array.
Demo Input:
['3\n1 2 3\n', '4\n1 5 2 5\n']
Demo Output:
['0\n', '2\n']
Note:
none
|
```python
a = int(input())
list = map(int, input().split())
lmx = 0
lmn = 0
for i in range(a):
if list[i] == 0:
pass
elif list[i] > list[i-1] and list[i] > list[i+1]:
lmx += 1
elif list[i] < list[i-1] and list[i] < list[i+1]:
lmn += 1
else:
pass
print(lmx+lmn)
```
| -1
|
|
572
|
A
|
Arrays
|
PROGRAMMING
| 900
|
[
"sortings"
] | null | null |
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
|
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
|
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
|
[
"3 3\n2 1\n1 2 3\n3 4 5\n",
"3 3\n3 3\n1 2 3\n3 4 5\n",
"5 2\n3 1\n1 1 1 1 1\n2 2\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 500
|
[
{
"input": "3 3\n2 1\n1 2 3\n3 4 5",
"output": "YES"
},
{
"input": "3 3\n3 3\n1 2 3\n3 4 5",
"output": "NO"
},
{
"input": "5 2\n3 1\n1 1 1 1 1\n2 2",
"output": "YES"
},
{
"input": "3 5\n1 1\n5 5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n1 1\n1\n1",
"output": "NO"
},
{
"input": "3 3\n1 1\n1 2 3\n1 2 3",
"output": "YES"
},
{
"input": "3 3\n1 2\n1 2 3\n1 2 3",
"output": "YES"
},
{
"input": "3 3\n2 2\n1 2 3\n1 2 3",
"output": "NO"
},
{
"input": "10 15\n10 1\n1 1 5 17 22 29 32 36 39 48\n9 10 20 23 26 26 32 32 33 39 43 45 47 49 49",
"output": "YES"
},
{
"input": "10 15\n1 15\n91 91 91 92 92 94 94 95 98 100\n92 92 93 93 93 94 95 96 97 98 98 99 99 100 100",
"output": "YES"
},
{
"input": "15 10\n12 5\n9 25 25 32 32 38 40 41 46 46 48 51 64 64 73\n5 14 30 35 50 52 67 79 89 99",
"output": "YES"
},
{
"input": "15 10\n4 10\n22 32 35 45 45 50 51 55 79 80 83 88 90 92 93\n46 48 52 55 60 60 68 75 80 81",
"output": "YES"
},
{
"input": "20 30\n2 8\n6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 10\n1 1 2 2 2 2 2 2 2 3 3 4 5 5 5 5 6 6 6 6 6 6 7 7 7 8 8 9 10 10",
"output": "NO"
},
{
"input": "20 30\n19 29\n1 1 2 2 2 3 4 4 7 7 7 8 8 8 8 8 9 9 9 9\n6 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "1 1\n1 1\n-1000000000\n30",
"output": "YES"
},
{
"input": "3 3\n1 3\n1 3 3\n3 3 3",
"output": "YES"
},
{
"input": "3 3\n1 1\n3 3 3\n2 2 2",
"output": "NO"
},
{
"input": "5 5\n3 3\n1 5 6 7 8\n1 2 5 6 7",
"output": "NO"
},
{
"input": "3 4\n2 2\n5 6 7\n1 2 3 4",
"output": "NO"
},
{
"input": "3 3\n3 3\n1 2 3\n4 5 6",
"output": "YES"
},
{
"input": "5 5\n4 5\n2 2 3 4 5\n5 6 7 8 9",
"output": "YES"
}
] | 1,612,459,724
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 109
| 2,048,000
|
import sys,math
try:
sys.stdin=open("input.txt","r")
finally:
#t=int(input())
t=1
for _ in range(t):
a,b=map(int,input().split())
k,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if m==1:
if max(b)!=max(a):
print("YES")
else:
print("NO")
else:
print("NO")
|
Title: Arrays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
Input Specification:
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
Output Specification:
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
Demo Input:
['3 3\n2 1\n1 2 3\n3 4 5\n', '3 3\n3 3\n1 2 3\n3 4 5\n', '5 2\n3 1\n1 1 1 1 1\n2 2\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
import sys,math
try:
sys.stdin=open("input.txt","r")
finally:
#t=int(input())
t=1
for _ in range(t):
a,b=map(int,input().split())
k,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if m==1:
if max(b)!=max(a):
print("YES")
else:
print("NO")
else:
print("NO")
```
| -1
|
|
268
|
A
|
Games
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
|
In a single line print the number of games where the host team is going to play in the guest uniform.
|
[
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] |
[
"1\n",
"5\n",
"0\n"
] |
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
| 500
|
[
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28",
"output": "1"
},
{
"input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77",
"output": "6"
},
{
"input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35",
"output": "4"
},
{
"input": "2\n46 6\n6 46",
"output": "2"
},
{
"input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11",
"output": "10"
},
{
"input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33",
"output": "6"
},
{
"input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73",
"output": "0"
},
{
"input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2",
"output": "1"
},
{
"input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100",
"output": "8"
},
{
"input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "450"
},
{
"input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3",
"output": "0"
},
{
"input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5",
"output": "20"
},
{
"input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2",
"output": "108"
},
{
"input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2",
"output": "312"
},
{
"input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6",
"output": "96"
},
{
"input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97",
"output": "210"
},
{
"input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78",
"output": "73"
},
{
"input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80",
"output": "277"
},
{
"input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21",
"output": "154"
},
{
"input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17",
"output": "418"
},
{
"input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80",
"output": "74"
},
{
"input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9",
"output": "59"
},
{
"input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67",
"output": "100"
},
{
"input": "4\n8 7\n8 7\n7 8\n7 8",
"output": "8"
},
{
"input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1",
"output": "10"
},
{
"input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "72"
},
{
"input": "4\n1 2\n1 2\n2 1\n2 1",
"output": "8"
}
] | 1,689,375,847
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
x=int(input())
a=[]
b=[]
v=0
for i in range(x):
y=list(map(int,input().split()))
a.append(y[0])
b.append(y[1])
for i in a:
for c in b:
if i==c:
v+=1
print(v)
|
Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
|
```python
x=int(input())
a=[]
b=[]
v=0
for i in range(x):
y=list(map(int,input().split()))
a.append(y[0])
b.append(y[1])
for i in a:
for c in b:
if i==c:
v+=1
print(v)
```
| 3
|
|
233
|
A
|
Perfect Permutation
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
|
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
|
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
|
[
"1\n",
"2\n",
"4\n"
] |
[
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] |
none
| 500
|
[
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
"output": "-1"
},
{
"input": "20",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 "
},
{
"input": "8",
"output": "2 1 4 3 6 5 8 7 "
},
{
"input": "9",
"output": "-1"
},
{
"input": "10",
"output": "2 1 4 3 6 5 8 7 10 9 "
},
{
"input": "11",
"output": "-1"
},
{
"input": "21",
"output": "-1"
},
{
"input": "50",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 "
},
{
"input": "51",
"output": "-1"
},
{
"input": "52",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 "
},
{
"input": "84",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 "
},
{
"input": "86",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 "
},
{
"input": "100",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 "
},
{
"input": "98",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 "
},
{
"input": "96",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 "
},
{
"input": "33",
"output": "-1"
},
{
"input": "34",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 "
},
{
"input": "36",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 "
},
{
"input": "38",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 "
},
{
"input": "40",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 "
},
{
"input": "42",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 "
},
{
"input": "44",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 "
},
{
"input": "46",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 "
},
{
"input": "48",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 "
}
] | 1,682,955,138
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n=int(input())
if (n%2==0):
for i in range (n):
|
Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none
|
```python
n=int(input())
if (n%2==0):
for i in range (n):
```
| -1
|
|
591
|
A
|
Wizards' Duel
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
|
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
|
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
|
[
"100\n50\n50\n",
"199\n60\n40\n"
] |
[
"50\n",
"119.4\n"
] |
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.
| 500
|
[
{
"input": "100\n50\n50",
"output": "50"
},
{
"input": "199\n60\n40",
"output": "119.4"
},
{
"input": "1\n1\n1",
"output": "0.5"
},
{
"input": "1\n1\n500",
"output": "0.001996007984"
},
{
"input": "1\n500\n1",
"output": "0.998003992"
},
{
"input": "1\n500\n500",
"output": "0.5"
},
{
"input": "1000\n1\n1",
"output": "500"
},
{
"input": "1000\n1\n500",
"output": "1.996007984"
},
{
"input": "1000\n500\n1",
"output": "998.003992"
},
{
"input": "1000\n500\n500",
"output": "500"
},
{
"input": "101\n11\n22",
"output": "33.66666667"
},
{
"input": "987\n1\n3",
"output": "246.75"
},
{
"input": "258\n25\n431",
"output": "14.14473684"
},
{
"input": "979\n39\n60",
"output": "385.6666667"
},
{
"input": "538\n479\n416",
"output": "287.9351955"
},
{
"input": "583\n112\n248",
"output": "181.3777778"
},
{
"input": "978\n467\n371",
"output": "545.0190931"
},
{
"input": "980\n322\n193",
"output": "612.7378641"
},
{
"input": "871\n401\n17",
"output": "835.576555"
},
{
"input": "349\n478\n378",
"output": "194.885514"
},
{
"input": "425\n458\n118",
"output": "337.9340278"
},
{
"input": "919\n323\n458",
"output": "380.0729834"
},
{
"input": "188\n59\n126",
"output": "59.95675676"
},
{
"input": "644\n428\n484",
"output": "302.2280702"
},
{
"input": "253\n80\n276",
"output": "56.85393258"
},
{
"input": "745\n152\n417",
"output": "199.0158172"
},
{
"input": "600\n221\n279",
"output": "265.2"
},
{
"input": "690\n499\n430",
"output": "370.6243272"
},
{
"input": "105\n68\n403",
"output": "15.15923567"
},
{
"input": "762\n462\n371",
"output": "422.6218487"
},
{
"input": "903\n460\n362",
"output": "505.3284672"
},
{
"input": "886\n235\n95",
"output": "630.9393939"
},
{
"input": "655\n203\n18",
"output": "601.6515837"
},
{
"input": "718\n29\n375",
"output": "51.53960396"
},
{
"input": "296\n467\n377",
"output": "163.7819905"
},
{
"input": "539\n61\n56",
"output": "281.017094"
},
{
"input": "133\n53\n124",
"output": "39.82485876"
},
{
"input": "998\n224\n65",
"output": "773.5363322"
},
{
"input": "961\n173\n47",
"output": "755.6954545"
},
{
"input": "285\n468\n62",
"output": "251.6603774"
},
{
"input": "496\n326\n429",
"output": "214.1668874"
},
{
"input": "627\n150\n285",
"output": "216.2068966"
},
{
"input": "961\n443\n50",
"output": "863.535497"
},
{
"input": "623\n422\n217",
"output": "411.4334898"
},
{
"input": "678\n295\n29",
"output": "617.3148148"
}
] | 1,445,765,038
| 1,438
|
Python 3
|
OK
|
TESTS
| 45
| 46
| 0
|
l = int(input())
p = int(input())
q = int(input())
ans = (l/(p+q))*p
print(ans)
|
Title: Wizards' Duel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
Input Specification:
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
Output Specification:
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['100\n50\n50\n', '199\n60\n40\n']
Demo Output:
['50\n', '119.4\n']
Note:
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.
|
```python
l = int(input())
p = int(input())
q = int(input())
ans = (l/(p+q))*p
print(ans)
```
| 3
|
|
805
|
A
|
Fake NP
|
PROGRAMMING
| 1,000
|
[
"greedy",
"math"
] | null | null |
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem.
|
The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109).
|
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
|
[
"19 29\n",
"3 6\n"
] |
[
"2\n",
"3\n"
] |
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html)
The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
| 500
|
[
{
"input": "19 29",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "39 91",
"output": "2"
},
{
"input": "76 134",
"output": "2"
},
{
"input": "93 95",
"output": "2"
},
{
"input": "17 35",
"output": "2"
},
{
"input": "94 95",
"output": "2"
},
{
"input": "51 52",
"output": "2"
},
{
"input": "47 52",
"output": "2"
},
{
"input": "38 98",
"output": "2"
},
{
"input": "30 37",
"output": "2"
},
{
"input": "56 92",
"output": "2"
},
{
"input": "900000000 1000000000",
"output": "2"
},
{
"input": "37622224 162971117",
"output": "2"
},
{
"input": "760632746 850720703",
"output": "2"
},
{
"input": "908580370 968054552",
"output": "2"
},
{
"input": "951594860 953554446",
"output": "2"
},
{
"input": "347877978 913527175",
"output": "2"
},
{
"input": "620769961 988145114",
"output": "2"
},
{
"input": "820844234 892579936",
"output": "2"
},
{
"input": "741254764 741254768",
"output": "2"
},
{
"input": "80270976 80270977",
"output": "2"
},
{
"input": "392602363 392602367",
"output": "2"
},
{
"input": "519002744 519002744",
"output": "519002744"
},
{
"input": "331900277 331900277",
"output": "331900277"
},
{
"input": "419873015 419873018",
"output": "2"
},
{
"input": "349533413 349533413",
"output": "349533413"
},
{
"input": "28829775 28829776",
"output": "2"
},
{
"input": "568814539 568814539",
"output": "568814539"
},
{
"input": "720270740 720270743",
"output": "2"
},
{
"input": "871232720 871232722",
"output": "2"
},
{
"input": "305693653 305693653",
"output": "305693653"
},
{
"input": "634097178 634097179",
"output": "2"
},
{
"input": "450868287 450868290",
"output": "2"
},
{
"input": "252662256 252662260",
"output": "2"
},
{
"input": "575062045 575062049",
"output": "2"
},
{
"input": "273072892 273072894",
"output": "2"
},
{
"input": "770439256 770439256",
"output": "770439256"
},
{
"input": "2 1000000000",
"output": "2"
},
{
"input": "6 8",
"output": "2"
},
{
"input": "2 879190747",
"output": "2"
},
{
"input": "5 5",
"output": "5"
},
{
"input": "999999937 999999937",
"output": "999999937"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "5 100",
"output": "2"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 18",
"output": "2"
},
{
"input": "7 7",
"output": "7"
},
{
"input": "39916801 39916801",
"output": "39916801"
},
{
"input": "3 8",
"output": "2"
},
{
"input": "13 13",
"output": "13"
},
{
"input": "4 8",
"output": "2"
},
{
"input": "3 12",
"output": "2"
},
{
"input": "6 12",
"output": "2"
},
{
"input": "999999103 999999103",
"output": "999999103"
},
{
"input": "100000007 100000007",
"output": "100000007"
},
{
"input": "3 99",
"output": "2"
},
{
"input": "999999733 999999733",
"output": "999999733"
},
{
"input": "5 10",
"output": "2"
},
{
"input": "982451653 982451653",
"output": "982451653"
},
{
"input": "999900001 1000000000",
"output": "2"
},
{
"input": "999727999 999727999",
"output": "999727999"
},
{
"input": "2 999999999",
"output": "2"
},
{
"input": "242 244",
"output": "2"
},
{
"input": "3 10",
"output": "2"
},
{
"input": "15 27",
"output": "2"
},
{
"input": "998244353 998244353",
"output": "998244353"
},
{
"input": "5 15",
"output": "2"
},
{
"input": "999999797 999999797",
"output": "999999797"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "999999929 999999929",
"output": "999999929"
},
{
"input": "3 111111",
"output": "2"
},
{
"input": "12 18",
"output": "2"
},
{
"input": "479001599 479001599",
"output": "479001599"
},
{
"input": "10000019 10000019",
"output": "10000019"
},
{
"input": "715827883 715827883",
"output": "715827883"
},
{
"input": "999992977 999992977",
"output": "999992977"
},
{
"input": "11 11",
"output": "11"
},
{
"input": "29 29",
"output": "29"
},
{
"input": "1000003 1000003",
"output": "1000003"
},
{
"input": "6 15",
"output": "2"
},
{
"input": "1200007 1200007",
"output": "1200007"
},
{
"input": "3 1000000000",
"output": "2"
},
{
"input": "990000023 990000023",
"output": "990000023"
},
{
"input": "1717 1717",
"output": "1717"
},
{
"input": "141650963 141650963",
"output": "141650963"
},
{
"input": "1002523 1002523",
"output": "1002523"
},
{
"input": "900000011 900000011",
"output": "900000011"
},
{
"input": "104729 104729",
"output": "104729"
},
{
"input": "4 12",
"output": "2"
},
{
"input": "100003 100003",
"output": "100003"
},
{
"input": "17 17",
"output": "17"
},
{
"input": "10 100",
"output": "2"
}
] | 1,632,157,325
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 93
| 109
| 20,172,800
|
l, r = map(int,input().split())
if(l == r):
print(l)
else:
print('2')
|
Title: Fake NP
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem.
Input Specification:
The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109).
Output Specification:
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
Demo Input:
['19 29\n', '3 6\n']
Demo Output:
['2\n', '3\n']
Note:
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html)
The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
|
```python
l, r = map(int,input().split())
if(l == r):
print(l)
else:
print('2')
```
| 3
|
|
172
|
A
|
Phone Code
|
PROGRAMMING
| 800
|
[
"*special",
"brute force",
"implementation"
] | null | null |
Polycarpus has *n* friends in Tarasov city. Polycarpus knows phone numbers of all his friends: they are strings *s*1,<=*s*2,<=...,<=*s**n*. All these strings consist only of digits and have the same length.
Once Polycarpus needed to figure out Tarasov city phone code. He assumed that the phone code of the city is the longest common prefix of all phone numbers of his friends. In other words, it is the longest string *c* which is a prefix (the beginning) of each *s**i* for all *i* (1<=≤<=*i*<=≤<=*n*). Help Polycarpus determine the length of the city phone code.
|
The first line of the input contains an integer *n* (2<=≤<=*n*<=≤<=3·104) — the number of Polycarpus's friends. The following *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* — the phone numbers of Polycarpus's friends. It is guaranteed that all strings consist only of digits and have the same length from 1 to 20, inclusive. It is also guaranteed that all strings are different.
|
Print the number of digits in the city phone code.
|
[
"4\n00209\n00219\n00999\n00909\n",
"2\n1\n2\n",
"3\n77012345678999999999\n77012345678901234567\n77012345678998765432\n"
] |
[
"2\n",
"0\n",
"12\n"
] |
A prefix of string *t* is a string that is obtained by deleting zero or more digits from the end of string *t*. For example, string "00209" has 6 prefixes: "" (an empty prefix), "0", "00", "002", "0020", "00209".
In the first sample the city phone code is string "00".
In the second sample the city phone code is an empty string.
In the third sample the city phone code is string "770123456789".
| 1,000
|
[
{
"input": "4\n00209\n00219\n00999\n00909",
"output": "2"
},
{
"input": "2\n1\n2",
"output": "0"
},
{
"input": "3\n77012345678999999999\n77012345678901234567\n77012345678998765432",
"output": "12"
},
{
"input": "5\n4491183345\n4491184811\n4491162340\n4491233399\n4491449214",
"output": "4"
},
{
"input": "10\n15424\n10953\n19176\n15514\n16284\n18680\n19305\n13816\n16168\n15924",
"output": "1"
},
{
"input": "10\n4906361343\n8985777485\n1204265609\n7088384855\n4127287014\n7904807820\n3032139021\n5999959109\n6477458281\n3244359368",
"output": "0"
},
{
"input": "10\n3717208309\n3717208306\n3717208302\n3717208301\n3717208303\n3717208308\n3717208304\n3717208307\n3717208300\n3717208305",
"output": "9"
},
{
"input": "9\n2881\n2808\n2868\n2874\n2894\n2870\n2818\n2896\n2890",
"output": "2"
},
{
"input": "2\n4\n9",
"output": "0"
},
{
"input": "2\n29867863763143509570\n59261213969200291523",
"output": "0"
},
{
"input": "2\n84\n62",
"output": "0"
},
{
"input": "2\n75970434466248453472\n75970434466248453476",
"output": "19"
},
{
"input": "10\n17254072509168593435\n17254072509168593433\n17254072509168593430\n17254072509168593432\n17254072509168593439\n17254072509168593436\n17254072509168593438\n17254072509168593437\n17254072509168593431\n17254072509168593434",
"output": "19"
}
] | 1,699,469,804
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 124
| 2,867,200
|
x = [input() for i in range(int(input()))]
a, b = min(x), max(x)
while a != b:
a, b = a[:-1], b[:-1]
print(len(a))
|
Title: Phone Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has *n* friends in Tarasov city. Polycarpus knows phone numbers of all his friends: they are strings *s*1,<=*s*2,<=...,<=*s**n*. All these strings consist only of digits and have the same length.
Once Polycarpus needed to figure out Tarasov city phone code. He assumed that the phone code of the city is the longest common prefix of all phone numbers of his friends. In other words, it is the longest string *c* which is a prefix (the beginning) of each *s**i* for all *i* (1<=≤<=*i*<=≤<=*n*). Help Polycarpus determine the length of the city phone code.
Input Specification:
The first line of the input contains an integer *n* (2<=≤<=*n*<=≤<=3·104) — the number of Polycarpus's friends. The following *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* — the phone numbers of Polycarpus's friends. It is guaranteed that all strings consist only of digits and have the same length from 1 to 20, inclusive. It is also guaranteed that all strings are different.
Output Specification:
Print the number of digits in the city phone code.
Demo Input:
['4\n00209\n00219\n00999\n00909\n', '2\n1\n2\n', '3\n77012345678999999999\n77012345678901234567\n77012345678998765432\n']
Demo Output:
['2\n', '0\n', '12\n']
Note:
A prefix of string *t* is a string that is obtained by deleting zero or more digits from the end of string *t*. For example, string "00209" has 6 prefixes: "" (an empty prefix), "0", "00", "002", "0020", "00209".
In the first sample the city phone code is string "00".
In the second sample the city phone code is an empty string.
In the third sample the city phone code is string "770123456789".
|
```python
x = [input() for i in range(int(input()))]
a, b = min(x), max(x)
while a != b:
a, b = a[:-1], b[:-1]
print(len(a))
```
| 3
|
|
342
|
E
|
Xenia and Tree
|
PROGRAMMING
| 2,400
|
[
"data structures",
"divide and conquer",
"trees"
] | null | null |
Xenia the programmer has a tree consisting of *n* nodes. We will consider the tree nodes indexed from 1 to *n*. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.
The distance between two tree nodes *v* and *u* is the number of edges in the shortest path between *v* and *u*.
Xenia needs to learn how to quickly execute queries of two types:
1. paint a specified blue node in red; 1. calculate which red node is the closest to the given one and print the shortest distance to the closest red node.
Your task is to write a program which will execute the described queries.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105) — the number of nodes in the tree and the number of queries. Next *n*<=-<=1 lines contain the tree edges, the *i*-th line contains a pair of integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*) — an edge of the tree.
Next *m* lines contain queries. Each query is specified as a pair of integers *t**i*,<=*v**i* (1<=≤<=*t**i*<=≤<=2,<=1<=≤<=*v**i*<=≤<=*n*). If *t**i*<==<=1, then as a reply to the query we need to paint a blue node *v**i* in red. If *t**i*<==<=2, then we should reply to the query by printing the shortest distance from some red node to node *v**i*.
It is guaranteed that the given graph is a tree and that all queries are correct.
|
For each second type query print the reply in a single line.
|
[
"5 4\n1 2\n2 3\n2 4\n4 5\n2 1\n2 5\n1 2\n2 5\n"
] |
[
"0\n3\n2\n"
] |
none
| 2,500
|
[
{
"input": "5 4\n1 2\n2 3\n2 4\n4 5\n2 1\n2 5\n1 2\n2 5",
"output": "0\n3\n2"
}
] | 1,693,245,766
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1693245766.1947157")# 1693245766.194732
|
Title: Xenia and Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the programmer has a tree consisting of *n* nodes. We will consider the tree nodes indexed from 1 to *n*. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.
The distance between two tree nodes *v* and *u* is the number of edges in the shortest path between *v* and *u*.
Xenia needs to learn how to quickly execute queries of two types:
1. paint a specified blue node in red; 1. calculate which red node is the closest to the given one and print the shortest distance to the closest red node.
Your task is to write a program which will execute the described queries.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105) — the number of nodes in the tree and the number of queries. Next *n*<=-<=1 lines contain the tree edges, the *i*-th line contains a pair of integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*) — an edge of the tree.
Next *m* lines contain queries. Each query is specified as a pair of integers *t**i*,<=*v**i* (1<=≤<=*t**i*<=≤<=2,<=1<=≤<=*v**i*<=≤<=*n*). If *t**i*<==<=1, then as a reply to the query we need to paint a blue node *v**i* in red. If *t**i*<==<=2, then we should reply to the query by printing the shortest distance from some red node to node *v**i*.
It is guaranteed that the given graph is a tree and that all queries are correct.
Output Specification:
For each second type query print the reply in a single line.
Demo Input:
['5 4\n1 2\n2 3\n2 4\n4 5\n2 1\n2 5\n1 2\n2 5\n']
Demo Output:
['0\n3\n2\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1693245766.1947157")# 1693245766.194732
```
| 0
|
|
26
|
A
|
Almost Prime
|
PROGRAMMING
| 900
|
[
"number theory"
] |
A. Almost Prime
|
2
|
256
|
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
|
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
|
Output the amount of almost prime numbers between 1 and *n*, inclusive.
|
[
"10\n",
"21\n"
] |
[
"2\n",
"8\n"
] |
none
| 500
|
[
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,637,837,919
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 1,652
| 0
|
n = int(input())
s = []
sum1 = 0
# for i in range(2, n + 1):
# for j in range(2, i // 2 + 1):
# if i % j == 0:
# sum1 = sum1 + 1
# if sum1 == 0:
# s.append(i)
# sum1 = 0
# print(s)
sum0 = 0
sum2 = 0
for i in range(2, n + 1):
for j in range(2, i // 2 + 1):
if i % j == 0:
sum1 = sum1 + 1
if sum1 == 0:
s.append(i)
sum1 = 0
for j in range(2, i + 1):
if i % j == 0 and j in s:
sum0 = sum0 + 1
if sum0 == 2:
sum2 = sum2 + 1
sum0 = 0
print(sum2)
|
Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none
|
```python
n = int(input())
s = []
sum1 = 0
# for i in range(2, n + 1):
# for j in range(2, i // 2 + 1):
# if i % j == 0:
# sum1 = sum1 + 1
# if sum1 == 0:
# s.append(i)
# sum1 = 0
# print(s)
sum0 = 0
sum2 = 0
for i in range(2, n + 1):
for j in range(2, i // 2 + 1):
if i % j == 0:
sum1 = sum1 + 1
if sum1 == 0:
s.append(i)
sum1 = 0
for j in range(2, i + 1):
if i % j == 0 and j in s:
sum0 = sum0 + 1
if sum0 == 2:
sum2 = sum2 + 1
sum0 = 0
print(sum2)
```
| 3.587
|
811
|
B
|
Vladik and Complicated Book
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] | null | null |
Vladik had started reading a complicated book about algorithms containing *n* pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation *P*<==<=[*p*1,<=*p*2,<=...,<=*p**n*], where *p**i* denotes the number of page that should be read *i*-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation *P* from position *l* to position *r* inclusive, because she loves the order. For every of such sorting Vladik knows number *x* — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has *p**x* changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.
|
First line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — permutation *P*. Note that elements in permutation are distinct.
Each of the next *m* lines contains three space-separated integers *l**i*, *r**i*, *x**i* (1<=≤<=*l**i*<=≤<=*x**i*<=≤<=*r**i*<=≤<=*n*) — left and right borders of sorted subsegment in *i*-th sorting and position that is interesting to Vladik.
|
For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.
|
[
"5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n",
"6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3\n"
] |
[
"Yes\nNo\nYes\nYes\nNo\n",
"Yes\nNo\nYes\nNo\nYes\n"
] |
Explanation of first test case:
1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 1. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 1. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No".
| 1,000
|
[
{
"input": "5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3",
"output": "Yes\nNo\nYes\nYes\nNo"
},
{
"input": "6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3",
"output": "Yes\nNo\nYes\nNo\nYes"
},
{
"input": "10 10\n10 1 6 7 9 8 4 3 5 2\n1 1 1\n4 4 4\n7 7 7\n3 3 3\n1 6 5\n2 6 2\n6 8 7\n1 1 1\n7 9 9\n2 9 4",
"output": "Yes\nYes\nYes\nYes\nYes\nYes\nYes\nYes\nYes\nNo"
},
{
"input": "20 20\n18 17 2 3 16 15 1 9 12 8 20 11 13 14 4 5 19 7 10 6\n13 15 15\n1 1 1\n2 2 2\n11 14 13\n10 11 10\n2 8 6\n12 18 16\n4 8 8\n2 2 2\n5 11 11\n4 9 9\n5 6 6\n3 20 12\n8 8 8\n6 16 11\n9 18 18\n8 18 17\n1 1 1\n2 6 5\n1 4 3",
"output": "No\nYes\nYes\nNo\nYes\nYes\nNo\nNo\nYes\nYes\nNo\nNo\nNo\nYes\nNo\nNo\nYes\nYes\nYes\nNo"
},
{
"input": "5 10\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3",
"output": "Yes\nNo\nYes\nYes\nNo\nYes\nNo\nYes\nYes\nNo"
}
] | 1,614,828,644
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 61
| 204,800
|
[permutaionLenght, numberOfTestCases] = list(map(int, input().rstrip().split()))
basePermutation = list(map(int, input().rstrip().split()))
basePermutation.insert(0, 0)
basePermutationCopy = basePermutation;
permutation = [0]
def removeEquals(number):
if (number in permutation):
return False
else:
return True
def generateSequence(left, right, permutation):
rightValue = basePermutation[right]
leftValue = basePermutation[left]
if (left != 1 and right != 1 and rightValue - leftValue != (permutaionLenght - 1) and leftValue - rightValue != (permutaionLenght - 1)):
permutation.insert(1, 5)
if (rightValue < leftValue and leftValue):
tmp = list(reversed(list(range(leftValue, rightValue + 1))))
permutation.extend(tmp)
else:
tmp = list(range(leftValue, rightValue + 1))
permutation.extend(tmp)
return permutation
for i in range(numberOfTestCases+1):
[right, left, x] = list(map(int, input().rstrip().split()))
if(basePermutation[left] == basePermutation[right]):
permutation = basePermutation
else:
permutation = generateSequence(left, right, permutation)
basePermutationCopy = list(filter(removeEquals, basePermutationCopy))
permutation.extend(basePermutationCopy)
if(basePermutation[x] == permutation[x] ):
print("Yes")
else:
print("No")
permutation = [0]
basePermutationCopy = basePermutation
""""
5 1
5 4 3 2 1
1 3 1
5 1
5 4 3 2 1
4 4 4
5 5
5 4 3 2 1
1 5 3
1 3 1
2 4 3
4 4 4
2 5 3
"""
"""
should be
"""
|
Title: Vladik and Complicated Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vladik had started reading a complicated book about algorithms containing *n* pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation *P*<==<=[*p*1,<=*p*2,<=...,<=*p**n*], where *p**i* denotes the number of page that should be read *i*-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation *P* from position *l* to position *r* inclusive, because she loves the order. For every of such sorting Vladik knows number *x* — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has *p**x* changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.
Input Specification:
First line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — permutation *P*. Note that elements in permutation are distinct.
Each of the next *m* lines contains three space-separated integers *l**i*, *r**i*, *x**i* (1<=≤<=*l**i*<=≤<=*x**i*<=≤<=*r**i*<=≤<=*n*) — left and right borders of sorted subsegment in *i*-th sorting and position that is interesting to Vladik.
Output Specification:
For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.
Demo Input:
['5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n', '6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3\n']
Demo Output:
['Yes\nNo\nYes\nYes\nNo\n', 'Yes\nNo\nYes\nNo\nYes\n']
Note:
Explanation of first test case:
1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 1. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 1. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No".
|
```python
[permutaionLenght, numberOfTestCases] = list(map(int, input().rstrip().split()))
basePermutation = list(map(int, input().rstrip().split()))
basePermutation.insert(0, 0)
basePermutationCopy = basePermutation;
permutation = [0]
def removeEquals(number):
if (number in permutation):
return False
else:
return True
def generateSequence(left, right, permutation):
rightValue = basePermutation[right]
leftValue = basePermutation[left]
if (left != 1 and right != 1 and rightValue - leftValue != (permutaionLenght - 1) and leftValue - rightValue != (permutaionLenght - 1)):
permutation.insert(1, 5)
if (rightValue < leftValue and leftValue):
tmp = list(reversed(list(range(leftValue, rightValue + 1))))
permutation.extend(tmp)
else:
tmp = list(range(leftValue, rightValue + 1))
permutation.extend(tmp)
return permutation
for i in range(numberOfTestCases+1):
[right, left, x] = list(map(int, input().rstrip().split()))
if(basePermutation[left] == basePermutation[right]):
permutation = basePermutation
else:
permutation = generateSequence(left, right, permutation)
basePermutationCopy = list(filter(removeEquals, basePermutationCopy))
permutation.extend(basePermutationCopy)
if(basePermutation[x] == permutation[x] ):
print("Yes")
else:
print("No")
permutation = [0]
basePermutationCopy = basePermutation
""""
5 1
5 4 3 2 1
1 3 1
5 1
5 4 3 2 1
4 4 4
5 5
5 4 3 2 1
1 5 3
1 3 1
2 4 3
4 4 4
2 5 3
"""
"""
should be
"""
```
| -1
|
|
37
|
A
|
Towers
|
PROGRAMMING
| 1,000
|
[
"sortings"
] |
A. Towers
|
2
|
256
|
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
|
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
|
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
|
[
"3\n1 2 3\n",
"4\n6 5 6 7\n"
] |
[
"1 3\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3\n1 2 3",
"output": "1 3"
},
{
"input": "4\n6 5 6 7",
"output": "2 3"
},
{
"input": "4\n3 2 1 1",
"output": "2 3"
},
{
"input": "4\n1 2 3 3",
"output": "2 3"
},
{
"input": "3\n20 22 36",
"output": "1 3"
},
{
"input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162",
"output": "2 23"
},
{
"input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193",
"output": "2 43"
},
{
"input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111",
"output": "2 57"
},
{
"input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642",
"output": "2 80"
},
{
"input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188",
"output": "2 92"
},
{
"input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233",
"output": "3 101"
},
{
"input": "5\n5 5 5 5 5",
"output": "5 1"
},
{
"input": "3\n1000 1000 1000",
"output": "3 1"
},
{
"input": "1\n1000",
"output": "1 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "5\n1 1000 1000 1000 1000",
"output": "4 2"
},
{
"input": "5\n1000 1000 1000 8 7",
"output": "3 3"
}
] | 1,585,137,505
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 186
| 0
|
N = int(input())
l = list(map(int,input().split()))
count = 0
a = []
for i in l:
if i not in a:
a.append(i)
for i in l:
if a.count(i) > count:
count = l.count(i)
print(count,len(a))
|
Title: Towers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
Input Specification:
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
Output Specification:
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
Demo Input:
['3\n1 2 3\n', '4\n6 5 6 7\n']
Demo Output:
['1 3\n', '2 3\n']
Note:
none
|
```python
N = int(input())
l = list(map(int,input().split()))
count = 0
a = []
for i in l:
if i not in a:
a.append(i)
for i in l:
if a.count(i) > count:
count = l.count(i)
print(count,len(a))
```
| 0
|
748
|
B
|
Santa Claus and Keyboard Check
|
PROGRAMMING
| 1,500
|
[
"implementation",
"strings"
] | null | null |
Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be.
In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard.
You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once.
|
The input consists of only two strings *s* and *t* denoting the favorite Santa's patter and the resulting string. *s* and *t* are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters.
|
If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes).
Otherwise, the first line of output should contain the only integer *k* (*k*<=≥<=0) — the number of pairs of keys that should be swapped. The following *k* lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct.
If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair.
Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes.
|
[
"helloworld\nehoolwlroz\n",
"hastalavistababy\nhastalavistababy\n",
"merrychristmas\nchristmasmerry\n"
] |
[
"3\nh e\nl o\nd z\n",
"0\n",
"-1\n"
] |
none
| 1,000
|
[
{
"input": "helloworld\nehoolwlroz",
"output": "3\nh e\nl o\nd z"
},
{
"input": "hastalavistababy\nhastalavistababy",
"output": "0"
},
{
"input": "merrychristmas\nchristmasmerry",
"output": "-1"
},
{
"input": "kusyvdgccw\nkusyvdgccw",
"output": "0"
},
{
"input": "bbbbbabbab\naaaaabaaba",
"output": "1\nb a"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzz\nqwertyuiopasdfghjklzx",
"output": "-1"
},
{
"input": "accdccdcdccacddbcacc\naccbccbcbccacbbdcacc",
"output": "1\nd b"
},
{
"input": "giiibdbebjdaihdghahccdeffjhfgidfbdhjdggajfgaidadjd\ngiiibdbebjdaihdghahccdeffjhfgidfbdhjdggajfgaidadjd",
"output": "0"
},
{
"input": "gndggadlmdefgejidmmcglbjdcmglncfmbjjndjcibnjbabfab\nfihffahlmhogfojnhmmcflkjhcmflicgmkjjihjcnkijkakgak",
"output": "5\ng f\nn i\nd h\ne o\nb k"
},
{
"input": "ijpanyhovzwjjxsvaiyhchfaulcsdgfszjnwtoqbtaqygfmxuwvynvlhqhvmkjbooklxfhmqlqvfoxlnoclfxtbhvnkmhjcmrsdc\nijpanyhovzwjjxsvaiyhchfaulcsdgfszjnwtoqbtaqygfmxuwvynvlhqhvmkjbooklxfhmqlqvfoxlnoclfxtbhvnkmhjcmrsdc",
"output": "0"
},
{
"input": "ab\naa",
"output": "-1"
},
{
"input": "a\nz",
"output": "1\na z"
},
{
"input": "zz\nzy",
"output": "-1"
},
{
"input": "as\ndf",
"output": "2\na d\ns f"
},
{
"input": "abc\nbca",
"output": "-1"
},
{
"input": "rtfg\nrftg",
"output": "1\nt f"
},
{
"input": "y\ny",
"output": "0"
},
{
"input": "qwertyuiopasdfghjklzx\nzzzzzzzzzzzzzzzzzzzzz",
"output": "-1"
},
{
"input": "qazwsxedcrfvtgbyhnujmik\nqwertyuiasdfghjkzxcvbnm",
"output": "-1"
},
{
"input": "aaaaaa\nabcdef",
"output": "-1"
},
{
"input": "qwerty\nffffff",
"output": "-1"
},
{
"input": "dofbgdppdvmwjwtdyphhmqliydxyjfxoopxiscevowleccmhwybsxitvujkfliamvqinlrpytyaqdlbywccprukoisyaseibuqbfqjcabkieimsggsakpnqliwhehnemewhychqrfiuyaecoydnromrh\ndofbgdppdvmwjwtdyphhmqliydxyjfxoopxiscevowleccmhwybsxitvujkfliamvqinlrpytyaqdlbywccprukoisyaseibuqbfqjcabkieimsggsakpnqliwhehnemewhychqrfiuyaecoydnromrh",
"output": "0"
},
{
"input": "acdbccddadbcbabbebbaebdcedbbcebeaccecdabadeabeecbacacdcbccedeadadedeccedecdaabcedccccbbcbcedcaccdede\ndcbaccbbdbacadaaeaadeabcebaaceaedccecbdadbedaeecadcdcbcaccebedbdbebeccebecbddacebccccaacacebcdccbebe",
"output": "-1"
},
{
"input": "bacccbbacabbcaacbbba\nbacccbbacabbcaacbbba",
"output": "0"
},
{
"input": "dbadbddddb\nacbacaaaac",
"output": "-1"
},
{
"input": "dacbdbbbdd\nadbdadddaa",
"output": "-1"
},
{
"input": "bbbbcbcbbc\ndaddbabddb",
"output": "-1"
},
{
"input": "dddddbcdbd\nbcbbbdacdb",
"output": "-1"
},
{
"input": "cbadcbcdaa\nabbbababbb",
"output": "-1"
},
{
"input": "dmkgadidjgdjikgkehhfkhgkeamhdkfemikkjhhkdjfaenmkdgenijinamngjgkmgmmedfdehkhdigdnnkhmdkdindhkhndnakdgdhkdefagkedndnijekdmkdfedkhekgdkhgkimfeakdhhhgkkff\nbdenailbmnbmlcnehjjkcgnehadgickhdlecmggcimkahfdeinhflmlfadfnmncdnddhbkbhgejblnbffcgdbeilfigegfifaebnijeihkanehififlmhcbdcikhieghenbejneldkhaebjggncckk",
"output": "-1"
},
{
"input": "acbbccabaa\nabbbbbabaa",
"output": "-1"
},
{
"input": "ccccaccccc\naaaabaaaac",
"output": "-1"
},
{
"input": "acbacacbbb\nacbacacbbb",
"output": "0"
},
{
"input": "abbababbcc\nccccccccbb",
"output": "-1"
},
{
"input": "jbcbbjiifdcbeajgdeabddbfcecafejddcigfcaedbgicjihifgbahjihcjefgabgbccdiibfjgacehbbdjceacdbdeaiibaicih\nhhihhhddcfihddhjfddhffhcididcdhffidjciddfhjdihdhdcjhdhhdhihdcjdhjhiifddhchjdidhhhfhiddifhfddddhddidh",
"output": "-1"
},
{
"input": "ahaeheedefeehahfefhjhhedheeeedhehhfhdejdhffhhejhhhejadhefhahhadjjhdhheeeehfdaffhhefehhhefhhhhehehjda\neiefbdfgdhffieihfhjajifgjddffgifjbhigfagjhhjicaijbdaegidhiejiegaabgjidcfcjhgehhjjchcbjjdhjbiidjdjage",
"output": "-1"
},
{
"input": "fficficbidbcbfaddifbffdbbiaccbbciiaidbcbbiadcccbccbbaibabcbbdbcibcciibiccfifbiiicadibbiaafadacdficbc\nddjhdghbgcbhadeccjdbddcbfjeiiaaigjejcaiabgechiiahibfejbeahafcfhjbihgjfgihdgdagjjhecjafjeedecehcdjhai",
"output": "-1"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "a\nz",
"output": "1\na z"
},
{
"input": "z\na",
"output": "1\nz a"
},
{
"input": "aa\nzz",
"output": "1\na z"
},
{
"input": "az\nza",
"output": "1\na z"
},
{
"input": "aa\nza",
"output": "-1"
},
{
"input": "za\nzz",
"output": "-1"
},
{
"input": "aa\nab",
"output": "-1"
},
{
"input": "hehe\nheeh",
"output": "-1"
},
{
"input": "bd\ncc",
"output": "-1"
},
{
"input": "he\nhh",
"output": "-1"
},
{
"input": "hee\nheh",
"output": "-1"
},
{
"input": "aa\nac",
"output": "-1"
},
{
"input": "ab\naa",
"output": "-1"
},
{
"input": "hello\nehlol",
"output": "-1"
},
{
"input": "ac\naa",
"output": "-1"
},
{
"input": "aaabbb\nbbbaab",
"output": "-1"
},
{
"input": "aa\nfa",
"output": "-1"
},
{
"input": "hg\nee",
"output": "-1"
},
{
"input": "helloworld\nehoolwlrow",
"output": "-1"
},
{
"input": "abb\nbab",
"output": "-1"
},
{
"input": "aaa\naae",
"output": "-1"
},
{
"input": "aba\nbaa",
"output": "-1"
},
{
"input": "aa\nba",
"output": "-1"
},
{
"input": "da\naa",
"output": "-1"
},
{
"input": "aaa\naab",
"output": "-1"
},
{
"input": "xy\nzz",
"output": "-1"
}
] | 1,634,411,402
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 13
| 77
| 0
|
def main():
a = list(input())
b = list(input())
c = []
count = 1
for i in range(len(a)):
if a[i]!=b[i]:
count += 1
if c.count(a[i])==0 and c.count(b[i])==0:
c.append(a[i])
c.append(b[i])
b[i] = a[i]
# print(count, len(a))
if a==b:
print(0)
elif len(a)==count:
print(-1)
else:
n = len(c)//2
print(n)
j = 1
while j<len(c):
print(f"{c[j-1]} {c[j]}")
j+=2
main()
|
Title: Santa Claus and Keyboard Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be.
In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard.
You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once.
Input Specification:
The input consists of only two strings *s* and *t* denoting the favorite Santa's patter and the resulting string. *s* and *t* are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters.
Output Specification:
If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes).
Otherwise, the first line of output should contain the only integer *k* (*k*<=≥<=0) — the number of pairs of keys that should be swapped. The following *k* lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct.
If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair.
Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes.
Demo Input:
['helloworld\nehoolwlroz\n', 'hastalavistababy\nhastalavistababy\n', 'merrychristmas\nchristmasmerry\n']
Demo Output:
['3\nh e\nl o\nd z\n', '0\n', '-1\n']
Note:
none
|
```python
def main():
a = list(input())
b = list(input())
c = []
count = 1
for i in range(len(a)):
if a[i]!=b[i]:
count += 1
if c.count(a[i])==0 and c.count(b[i])==0:
c.append(a[i])
c.append(b[i])
b[i] = a[i]
# print(count, len(a))
if a==b:
print(0)
elif len(a)==count:
print(-1)
else:
n = len(c)//2
print(n)
j = 1
while j<len(c):
print(f"{c[j-1]} {c[j]}")
j+=2
main()
```
| 0
|
|
466
|
C
|
Number of Ways
|
PROGRAMMING
| 1,700
|
[
"binary search",
"brute force",
"data structures",
"dp",
"two pointers"
] | null | null |
You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that .
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*.
|
Print a single integer — the number of ways to split the array into three parts with the same sum.
|
[
"5\n1 2 3 0 3\n",
"4\n0 1 -1 0\n",
"2\n4 1\n"
] |
[
"2\n",
"1\n",
"0\n"
] |
none
| 1,500
|
[
{
"input": "5\n1 2 3 0 3",
"output": "2"
},
{
"input": "4\n0 1 -1 0",
"output": "1"
},
{
"input": "2\n4 1",
"output": "0"
},
{
"input": "9\n0 0 0 0 0 0 0 0 0",
"output": "28"
},
{
"input": "10\n2 5 -2 2 -3 -2 3 5 -5 -2",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "100\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "2030"
},
{
"input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "10\n1 0 0 0 1 1 1 0 1 1",
"output": "2"
},
{
"input": "10\n-2 2 -2 0 -2 -1 1 -1 2 0",
"output": "0"
},
{
"input": "4\n0 2 -1 2",
"output": "0"
},
{
"input": "5\n3 3 -3 3 3",
"output": "3"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "8\n-1 -1 -1 -1 -1 -1 -1 -1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "4\n1 -1 0 0",
"output": "1"
},
{
"input": "3\n6 -3 6",
"output": "0"
}
] | 1,699,690,695
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 46
| 2,560,000
|
n=int(input())
a=list(map(int,input().split()))
if n<3:
print(0)
elif a.count(0)==n:
print((n-1)*(n-2)//2)
else:
tot=sum(a)
if tot%3!=0:
print(0)
else:
flag=True
cntl=0
cntr=0
for i in range(0,len(a)):
cntl+=a[i]
if cntl==tot//3:
break
elif cntl>tot//3:
flag=False
break
else:
continue
for j in range(len(a)-1,-1,-1):
cntr+=a[j]
if cntr==tot//3:
break
elif cntr>tot//3:
flag=False
break
else:
continue
if flag==False:
print(0)
else:
left=i
right=j
cnt=0
for p in range(left+1,len(a)):
if a[p]!=0:
break
else:
cnt+=1
for q in range(right-1,-1,-1):
if a[q]!=0:
break
else:
cnt+=1
print(cnt+1)
|
Title: Number of Ways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that .
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*.
Output Specification:
Print a single integer — the number of ways to split the array into three parts with the same sum.
Demo Input:
['5\n1 2 3 0 3\n', '4\n0 1 -1 0\n', '2\n4 1\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
none
|
```python
n=int(input())
a=list(map(int,input().split()))
if n<3:
print(0)
elif a.count(0)==n:
print((n-1)*(n-2)//2)
else:
tot=sum(a)
if tot%3!=0:
print(0)
else:
flag=True
cntl=0
cntr=0
for i in range(0,len(a)):
cntl+=a[i]
if cntl==tot//3:
break
elif cntl>tot//3:
flag=False
break
else:
continue
for j in range(len(a)-1,-1,-1):
cntr+=a[j]
if cntr==tot//3:
break
elif cntr>tot//3:
flag=False
break
else:
continue
if flag==False:
print(0)
else:
left=i
right=j
cnt=0
for p in range(left+1,len(a)):
if a[p]!=0:
break
else:
cnt+=1
for q in range(right-1,-1,-1):
if a[q]!=0:
break
else:
cnt+=1
print(cnt+1)
```
| 0
|
|
705
|
A
|
Hulk
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of layers of love and hate.
|
Print Dr.Banner's feeling in one line.
|
[
"1\n",
"2\n",
"3\n"
] |
[
"I hate it\n",
"I hate that I love it\n",
"I hate that I love that I hate it\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "I hate it"
},
{
"input": "2",
"output": "I hate that I love it"
},
{
"input": "3",
"output": "I hate that I love that I hate it"
},
{
"input": "4",
"output": "I hate that I love that I hate that I love it"
},
{
"input": "5",
"output": "I hate that I love that I hate that I love that I hate it"
},
{
"input": "10",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "18",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "19",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate it"
},
{
"input": "76",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "77",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "81",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "82",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "85",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "86",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "100",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "99",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "98",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "33",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate it"
},
{
"input": "34",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "57",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "100",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
}
] | 1,694,846,789
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 31
| 0
|
n = int(input())
hulkf = []
for i in range(1,n+1):
if i%2 == 1:
hulkf.append('I hate')
else:
hulkf.append('I love')
print(f'{" that ".join(hulkf)} it')
|
Title: Hulk
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of layers of love and hate.
Output Specification:
Print Dr.Banner's feeling in one line.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['I hate it\n', 'I hate that I love it\n', 'I hate that I love that I hate it\n']
Note:
none
|
```python
n = int(input())
hulkf = []
for i in range(1,n+1):
if i%2 == 1:
hulkf.append('I hate')
else:
hulkf.append('I love')
print(f'{" that ".join(hulkf)} it')
```
| 3
|
|
908
|
A
|
New Year and Counting Cards
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
|
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
|
Print a single integer, the minimum number of cards you must turn over to verify your claim.
|
[
"ee\n",
"z\n",
"0ay1\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
| 500
|
[
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01234567890123456789012345678901234567890123456789",
"output": "25"
},
{
"input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb",
"output": "10"
},
{
"input": "cjw2dwmr10pku4yxohe0wglktd",
"output": "4"
},
{
"input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6",
"output": "13"
},
{
"input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1",
"output": "13"
},
{
"input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs",
"output": "8"
},
{
"input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc",
"output": "14"
},
{
"input": "2222",
"output": "0"
},
{
"input": "13579",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "a",
"output": "1"
},
{
"input": "y",
"output": "0"
},
{
"input": "w",
"output": "0"
},
{
"input": "oo",
"output": "2"
},
{
"input": "oy",
"output": "1"
},
{
"input": "yo",
"output": "1"
},
{
"input": "yy",
"output": "0"
},
{
"input": "a0",
"output": "1"
},
{
"input": "a9",
"output": "2"
},
{
"input": "y0",
"output": "0"
},
{
"input": "y7",
"output": "1"
},
{
"input": "0a",
"output": "1"
},
{
"input": "3a",
"output": "2"
},
{
"input": "06",
"output": "0"
},
{
"input": "07",
"output": "1"
},
{
"input": "70",
"output": "1"
},
{
"input": "77",
"output": "2"
},
{
"input": "13570",
"output": "4"
},
{
"input": "0000000000011111",
"output": "5"
},
{
"input": "1357",
"output": "4"
},
{
"input": "uuuuuuuuuuuuuuuuuu",
"output": "18"
},
{
"input": "gabieurat",
"output": "5"
}
] | 1,522,671,179
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
s = input()
a = ['e', 'o', 'u', 'a', 'i', '0', '2', '4', '6', '8']
c = 0:
for i in s:
if i in a: c += 1
print(c)
|
Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
|
```python
s = input()
a = ['e', 'o', 'u', 'a', 'i', '0', '2', '4', '6', '8']
c = 0:
for i in s:
if i in a: c += 1
print(c)
```
| -1
|
|
755
|
A
|
PolandBall and Hypothesis
|
PROGRAMMING
| 800
|
[
"brute force",
"graphs",
"math",
"number theory"
] | null | null |
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
|
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
|
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
|
[
"3\n",
"4\n"
] |
[
"1",
"2"
] |
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
| 500
|
[
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "153",
"output": "1"
},
{
"input": "1000",
"output": "1"
},
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "4"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "11",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "996",
"output": "3"
},
{
"input": "36",
"output": "4"
},
{
"input": "210",
"output": "4"
},
{
"input": "270",
"output": "4"
},
{
"input": "306",
"output": "4"
},
{
"input": "330",
"output": "5"
},
{
"input": "336",
"output": "4"
},
{
"input": "600",
"output": "4"
},
{
"input": "726",
"output": "4"
},
{
"input": "988",
"output": "1"
},
{
"input": "12",
"output": "2"
},
{
"input": "987",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "986",
"output": "1"
},
{
"input": "14",
"output": "1"
},
{
"input": "985",
"output": "1"
},
{
"input": "15",
"output": "1"
},
{
"input": "984",
"output": "1"
},
{
"input": "16",
"output": "2"
},
{
"input": "983",
"output": "1"
},
{
"input": "17",
"output": "1"
},
{
"input": "982",
"output": "2"
},
{
"input": "18",
"output": "3"
},
{
"input": "981",
"output": "1"
},
{
"input": "19",
"output": "1"
},
{
"input": "980",
"output": "1"
},
{
"input": "20",
"output": "1"
},
{
"input": "979",
"output": "1"
},
{
"input": "21",
"output": "1"
},
{
"input": "978",
"output": "1"
},
{
"input": "22",
"output": "2"
},
{
"input": "977",
"output": "1"
},
{
"input": "23",
"output": "1"
},
{
"input": "976",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "975",
"output": "1"
},
{
"input": "25",
"output": "1"
},
{
"input": "2",
"output": "4"
},
{
"input": "6",
"output": "4"
},
{
"input": "996",
"output": "3"
},
{
"input": "8",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "330",
"output": "5"
},
{
"input": "18",
"output": "3"
},
{
"input": "12",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "999",
"output": "1"
},
{
"input": "726",
"output": "4"
},
{
"input": "876",
"output": "3"
}
] | 1,666,624,355
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 63
| 61
| 0
|
n = int(input())
def jn(n):
i = 2
while i*i <=n and n%i !=0:
i+=1
return (i*i>n)
m = 1
while True:
if jn((n*m+1)):
m+=1
else:
print(m)
break
|
Title: PolandBall and Hypothesis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
Input Specification:
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
Output Specification:
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
Demo Input:
['3\n', '4\n']
Demo Output:
['1', '2']
Note:
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
|
```python
n = int(input())
def jn(n):
i = 2
while i*i <=n and n%i !=0:
i+=1
return (i*i>n)
m = 1
while True:
if jn((n*m+1)):
m+=1
else:
print(m)
break
```
| 3
|
|
76
|
F
|
Tourist
|
PROGRAMMING
| 2,300
|
[
"binary search",
"data structures",
"dp"
] |
F. Tourist
|
0
|
256
|
Tourist walks along the *X* axis. He can choose either of two directions and any speed not exceeding *V*. He can also stand without moving anywhere. He knows from newspapers that at time *t*1 in the point with coordinate *x*1 an interesting event will occur, at time *t*2 in the point with coordinate *x*2 — another one, and so on up to (*x**n*,<=*t**n*). Interesting events are short so we can assume they are immediate. Event *i* counts visited if at time *t**i* tourist was at point with coordinate *x**i*.
Write program tourist that will find maximum number of events tourist if:
- at the beginning (when time is equal to 0) tourist appears at point 0, - tourist can choose initial point for himself.
Yes, you should answer on two similar but different questions.
|
The first line of input contains single integer number *N* (1<=≤<=*N*<=≤<=100000) — number of interesting events. The following *N* lines contain two integers *x**i* and *t**i* — coordinate and time of the *i*-th event. The last line of the input contains integer *V* — maximum speed of the tourist. All *x**i* will be within range <=-<=2·108<=≤<=*x**i*<=≤<=2·108, all *t**i* will be between 1 and 2·106 inclusive. *V* will be positive and will not exceed 1000. The input may contain events that happen at the same time or in the same place but not in the same place at the same time.
|
The only line of the output should contain two space-sepatated integers — maximum number of events tourist can visit in he starts moving from point 0 at time 0, and maximum number of events tourist can visit if he chooses the initial point for himself.
|
[
"3\n-1 1\n42 7\n40 8\n2\n"
] |
[
"1 2\n"
] |
none
| 0
|
[
{
"input": "3\n-1 1\n42 7\n40 8\n2",
"output": "1 2"
},
{
"input": "5\n1 5\n6 7\n17127 17\n17072 42\n17042 77\n3",
"output": "2 3"
},
{
"input": "7\n-976754 20479\n79929143 911181\n9598220 82517\n-51609349 810257\n67416551 266544\n-14170975 307690\n-15476178 491195\n352",
"output": "5 5"
},
{
"input": "12\n-6712 6\n2375 73\n4643 197\n-5660 215\n-378 223\n2228 307\n3305 340\n3449 468\n-4389 707\n-7161 729\n-1838 921\n-1655 985\n730",
"output": "11 12"
},
{
"input": "22\n-5301 602\n-1400 194\n-8512 987\n-282 962\n6484 494\n9091 523\n-7744 92\n6002 461\n5053 305\n-1339 272\n-821 165\n9308 785\n929 421\n8384 883\n1266 157\n8411 923\n-2949 885\n7725 687\n6859 197\n-7630 867\n5031 918\n5662 410\n41",
"output": "9 9"
}
] | 1,594,122,628
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 16
| 468
| 31,948,800
|
import bisect
from sys import stdin
from math import inf
def f_l(a):
n = len(a)
d = [inf for i in range(n)]
length = 0
d[0] = -inf
for i in range(0, n - 1):
j = bisect.bisect(d, a[i])
if d[j - 1] <= a[i] <= d[j]:
d[j] = a[i]
length = max(length, j)
if a[-1] > d[-1]:
length += 1
return length
def main():
n = int(stdin.readline())
x = [0 for _ in range(n)]
t = [0 for _ in range(n)]
for i in range(n):
x[i], t[i] = map(int, stdin.readline().split())
v = int(stdin.readline())
list_1 = list(map(lambda t: (-t[0] + v * t[1], t[0] + v * t[1]), zip(x, t)))
list_1.sort()
list_1_ = list(map(lambda t: t[1], list_1))
res_1 = f_l(list_1_)
list_2 = list_1
j = bisect.bisect(list_2, (0, 0))
list_2.insert(j, (0, 0))
list_2 = list(map(lambda t: t[1], list_2))
ind = list_2.index(0)
without_0 = list(filter(lambda t: (t[0] > 0 and t[1] > ind) or (t[0] < 0 and t[1] < ind), zip(list_2, range(len(list_2)))))
list_2 = list(map(lambda t: t[0], without_0))
res_2 = f_l(list_2)
print(res_2, res_1)
main()
|
Title: Tourist
Time Limit: 0 seconds
Memory Limit: 256 megabytes
Problem Description:
Tourist walks along the *X* axis. He can choose either of two directions and any speed not exceeding *V*. He can also stand without moving anywhere. He knows from newspapers that at time *t*1 in the point with coordinate *x*1 an interesting event will occur, at time *t*2 in the point with coordinate *x*2 — another one, and so on up to (*x**n*,<=*t**n*). Interesting events are short so we can assume they are immediate. Event *i* counts visited if at time *t**i* tourist was at point with coordinate *x**i*.
Write program tourist that will find maximum number of events tourist if:
- at the beginning (when time is equal to 0) tourist appears at point 0, - tourist can choose initial point for himself.
Yes, you should answer on two similar but different questions.
Input Specification:
The first line of input contains single integer number *N* (1<=≤<=*N*<=≤<=100000) — number of interesting events. The following *N* lines contain two integers *x**i* and *t**i* — coordinate and time of the *i*-th event. The last line of the input contains integer *V* — maximum speed of the tourist. All *x**i* will be within range <=-<=2·108<=≤<=*x**i*<=≤<=2·108, all *t**i* will be between 1 and 2·106 inclusive. *V* will be positive and will not exceed 1000. The input may contain events that happen at the same time or in the same place but not in the same place at the same time.
Output Specification:
The only line of the output should contain two space-sepatated integers — maximum number of events tourist can visit in he starts moving from point 0 at time 0, and maximum number of events tourist can visit if he chooses the initial point for himself.
Demo Input:
['3\n-1 1\n42 7\n40 8\n2\n']
Demo Output:
['1 2\n']
Note:
none
|
```python
import bisect
from sys import stdin
from math import inf
def f_l(a):
n = len(a)
d = [inf for i in range(n)]
length = 0
d[0] = -inf
for i in range(0, n - 1):
j = bisect.bisect(d, a[i])
if d[j - 1] <= a[i] <= d[j]:
d[j] = a[i]
length = max(length, j)
if a[-1] > d[-1]:
length += 1
return length
def main():
n = int(stdin.readline())
x = [0 for _ in range(n)]
t = [0 for _ in range(n)]
for i in range(n):
x[i], t[i] = map(int, stdin.readline().split())
v = int(stdin.readline())
list_1 = list(map(lambda t: (-t[0] + v * t[1], t[0] + v * t[1]), zip(x, t)))
list_1.sort()
list_1_ = list(map(lambda t: t[1], list_1))
res_1 = f_l(list_1_)
list_2 = list_1
j = bisect.bisect(list_2, (0, 0))
list_2.insert(j, (0, 0))
list_2 = list(map(lambda t: t[1], list_2))
ind = list_2.index(0)
without_0 = list(filter(lambda t: (t[0] > 0 and t[1] > ind) or (t[0] < 0 and t[1] < ind), zip(list_2, range(len(list_2)))))
list_2 = list(map(lambda t: t[0], without_0))
res_2 = f_l(list_2)
print(res_2, res_1)
main()
```
| 0
|
228
|
A
|
Is your horseshoe on the other hoof?
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
|
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
|
Print a single integer — the minimum number of horseshoes Valera needs to buy.
|
[
"1 7 3 3\n",
"7 7 7 7\n"
] |
[
"1\n",
"3\n"
] |
none
| 500
|
[
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,687,932,353
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
a=list(map(int,input().split()))
b=[]
for i in a:
if i in b:
continue
else:
b.append(i)
print(4-len(b))
|
Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none
|
```python
a=list(map(int,input().split()))
b=[]
for i in a:
if i in b:
continue
else:
b.append(i)
print(4-len(b))
```
| 3
|
|
742
|
A
|
Arpa’s hard exam and Mehrdad’s naive cheat
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math",
"number theory"
] | null | null |
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
|
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
|
Print single integer — the last digit of 1378*n*.
|
[
"1\n",
"2\n"
] |
[
"8",
"4"
] |
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4.
| 500
|
[
{
"input": "1",
"output": "8"
},
{
"input": "2",
"output": "4"
},
{
"input": "1000",
"output": "6"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "6"
},
{
"input": "1000000000",
"output": "6"
},
{
"input": "5",
"output": "8"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "2"
},
{
"input": "1378",
"output": "4"
},
{
"input": "13781378",
"output": "4"
},
{
"input": "51202278",
"output": "4"
},
{
"input": "999999998",
"output": "4"
},
{
"input": "999999997",
"output": "8"
},
{
"input": "12193721",
"output": "8"
},
{
"input": "0",
"output": "1"
},
{
"input": "989898989",
"output": "8"
},
{
"input": "7",
"output": "2"
},
{
"input": "8",
"output": "6"
},
{
"input": "9",
"output": "8"
},
{
"input": "10",
"output": "4"
},
{
"input": "11",
"output": "2"
},
{
"input": "12",
"output": "6"
},
{
"input": "13",
"output": "8"
},
{
"input": "14",
"output": "4"
},
{
"input": "15",
"output": "2"
},
{
"input": "16",
"output": "6"
},
{
"input": "999999996",
"output": "6"
},
{
"input": "999999995",
"output": "2"
},
{
"input": "999999994",
"output": "4"
}
] | 1,644,567,912
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 2,764,800
|
n=int(input())
l=1378**n
p=str(l)
print(p[-1])
|
Title: Arpa’s hard exam and Mehrdad’s naive cheat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input Specification:
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
Output Specification:
Print single integer — the last digit of 1378*n*.
Demo Input:
['1\n', '2\n']
Demo Output:
['8', '4']
Note:
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4.
|
```python
n=int(input())
l=1378**n
p=str(l)
print(p[-1])
```
| 0
|
|
94
|
A
|
Restoring Password
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] |
A. Restoring Password
|
2
|
256
|
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password".
Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address.
Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits.
Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
|
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
|
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
|
[
"01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n",
"10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n"
] |
[
"12345678\n",
"30234919\n"
] |
none
| 500
|
[
{
"input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110",
"output": "12345678"
},
{
"input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000",
"output": "30234919"
},
{
"input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011",
"output": "65264629"
},
{
"input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001",
"output": "98484434"
},
{
"input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001",
"output": "96071437"
},
{
"input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001",
"output": "89787267"
},
{
"input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110",
"output": "88447623"
},
{
"input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001",
"output": "80805519"
},
{
"input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110",
"output": "09250147"
},
{
"input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011",
"output": "40862358"
},
{
"input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001",
"output": "73907059"
},
{
"input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101",
"output": "57680434"
},
{
"input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010",
"output": "93678590"
},
{
"input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000",
"output": "58114879"
},
{
"input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100",
"output": "61146904"
},
{
"input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001",
"output": "23558422"
},
{
"input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111",
"output": "76827631"
},
{
"input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111",
"output": "26240666"
},
{
"input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100",
"output": "21067271"
},
{
"input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010",
"output": "77770029"
},
{
"input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011",
"output": "09448580"
},
{
"input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111",
"output": "99411277"
},
{
"input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110",
"output": "86658594"
},
{
"input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000",
"output": "04536863"
},
{
"input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110",
"output": "21066773"
},
{
"input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111",
"output": "01063858"
},
{
"input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000",
"output": "01599791"
},
{
"input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010",
"output": "40579016"
},
{
"input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001",
"output": "75424967"
},
{
"input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001",
"output": "72259657"
},
{
"input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011",
"output": "22339256"
},
{
"input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110",
"output": "70554591"
},
{
"input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011",
"output": "18124952"
},
{
"input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110",
"output": "33774052"
},
{
"input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000",
"output": "97961250"
},
{
"input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011",
"output": "21954161"
},
{
"input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100",
"output": "78740192"
},
{
"input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111",
"output": "05336882"
},
{
"input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111",
"output": "62020383"
},
{
"input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101",
"output": "24819275"
},
{
"input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001",
"output": "90010504"
},
{
"input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111",
"output": "85948866"
},
{
"input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110",
"output": "77874864"
},
{
"input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011",
"output": "10374003"
},
{
"input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001",
"output": "22955387"
},
{
"input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100",
"output": "25893541"
},
{
"input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010",
"output": "13805878"
},
{
"input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010",
"output": "06369030"
},
{
"input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110",
"output": "46909115"
},
{
"input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100",
"output": "90005541"
}
] | 1,457,102,084
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 0
| 2,000
| 512,000
|
import re
password = input()
codes = {}
for i in range(0, 10):
code = input()
codes[code]= i
password = ""
while len(password) < 8:
for code in codes:
if re.search("^"+code, password):
password += codes[code]
print(password)
|
Title: Restoring Password
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password".
Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address.
Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits.
Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
Input Specification:
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
Output Specification:
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
Demo Input:
['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n']
Demo Output:
['12345678\n', '30234919\n']
Note:
none
|
```python
import re
password = input()
codes = {}
for i in range(0, 10):
code = input()
codes[code]= i
password = ""
while len(password) < 8:
for code in codes:
if re.search("^"+code, password):
password += codes[code]
print(password)
```
| 0
|
989
|
B
|
A Tide of Riverscape
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"strings"
] | null | null |
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
|
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
|
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
|
[
"10 7\n1.0.1.0.1.\n",
"10 6\n1.0.1.1000\n",
"10 9\n1........1\n"
] |
[
"1000100010\n",
"1001101000\n",
"No\n"
] |
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
| 1,000
|
[
{
"input": "10 7\n1.0.1.0.1.",
"output": "1000100010"
},
{
"input": "10 6\n1.0.1.1000",
"output": "1001101000"
},
{
"input": "10 9\n1........1",
"output": "No"
},
{
"input": "1 1\n.",
"output": "No"
},
{
"input": "5 1\n0...1",
"output": "00001"
},
{
"input": "17 10\n..1.100..1..0.100",
"output": "00101000010000100"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "3 1\n.0.",
"output": "001"
},
{
"input": "3 1\n00.",
"output": "001"
},
{
"input": "3 2\n0..",
"output": "001"
},
{
"input": "3 2\n0.0",
"output": "No"
},
{
"input": "3 2\n1..",
"output": "100"
},
{
"input": "3 2\n.1.",
"output": "011"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "3 3\n1..",
"output": "No"
},
{
"input": "3 3\n.00",
"output": "No"
},
{
"input": "5 3\n0.000",
"output": "01000"
},
{
"input": "10 6\n10010.1001",
"output": "No"
},
{
"input": "75 38\n00.0.1.0.0110.1.00010..100.1110..110..00.0.1.0.0110.1.00010..100.1110..110.",
"output": "000001000011001000010001000111000110000000010000110010000100010001110001101"
},
{
"input": "128 108\n01100.110...000.0001.1.11.11.010010.01100.0.1.01.0.0011.11001.000101...1.0.0..100.0110.0110.0.0101.0.0.0001.01100.110...100.0001",
"output": "01100011000000000001010110110010010001100000100100000110110010000101000100000010000110001100000101000000001001100011000010000001"
},
{
"input": "5 4\n.101.",
"output": "01011"
},
{
"input": "4 2\n101.",
"output": "1011"
},
{
"input": "5 4\n.1011",
"output": "01011"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "5 3\n00.11",
"output": "00011"
},
{
"input": "10 8\n1111.00000",
"output": "1111000000"
},
{
"input": "10 3\n11111111.1",
"output": "1111111101"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "6 4\n11..10",
"output": "110010"
},
{
"input": "4 2\n.111",
"output": "0111"
},
{
"input": "3 2\n01.",
"output": "011"
},
{
"input": "5 4\n10.00",
"output": "10000"
},
{
"input": "10 9\n1........0",
"output": "1000000000"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "8 4\n111111..",
"output": "11111100"
},
{
"input": "3 2\n0.1",
"output": "001"
},
{
"input": "4 1\n111.",
"output": "1110"
},
{
"input": "3 1\n01.",
"output": "010"
},
{
"input": "10 7\n000....111",
"output": "0000000111"
}
] | 1,529,084,921
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 93
| 307,200
|
n=[int(num) for num in input().split()]
s=input()
flag=0
for i in range(0,n[0]-n[1]):
if s[i]!=s[i+n[1]] and s[i]!='.' and s[i+n[1]]!='.':
flag=1
break
elif s[i]==s[i+n[1]] and s[i]=='.':
s=s[:i]+'0'+s[i+1:i+n[1]]+'1'+s[i+n[1]+1:]
flag=1
break
elif s[i]=='.':
if s[i+n[1]]=='1':
s=s[:i]+'0'+s[i+1:]
else:
s=s[:i]+'1'+s[i+1:]
flag=1
break
elif s[i+n[1]]=='.':
if s[i]=='1':
s=s[:i+n[1]]+'0'+s[i+n[1]+1:]
else:
s=s[:i+n[1]]+'1'+s[i+n[1]+1:]
flag=1
break
if flag==1:
for i in s:
if i=='.':
i='0'
print(i,end='')
else:
print('NO')
|
Title: A Tide of Riverscape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
Input Specification:
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output Specification:
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
Demo Input:
['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n']
Demo Output:
['1000100010\n', '1001101000\n', 'No\n']
Note:
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
|
```python
n=[int(num) for num in input().split()]
s=input()
flag=0
for i in range(0,n[0]-n[1]):
if s[i]!=s[i+n[1]] and s[i]!='.' and s[i+n[1]]!='.':
flag=1
break
elif s[i]==s[i+n[1]] and s[i]=='.':
s=s[:i]+'0'+s[i+1:i+n[1]]+'1'+s[i+n[1]+1:]
flag=1
break
elif s[i]=='.':
if s[i+n[1]]=='1':
s=s[:i]+'0'+s[i+1:]
else:
s=s[:i]+'1'+s[i+1:]
flag=1
break
elif s[i+n[1]]=='.':
if s[i]=='1':
s=s[:i+n[1]]+'0'+s[i+n[1]+1:]
else:
s=s[:i+n[1]]+'1'+s[i+n[1]+1:]
flag=1
break
if flag==1:
for i in s:
if i=='.':
i='0'
print(i,end='')
else:
print('NO')
```
| 3
|
|
712
|
B
|
Memory and Trident
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
|
The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.
|
If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
|
[
"RRU\n",
"UDUR\n",
"RUUR\n"
] |
[
"-1\n",
"1\n",
"2\n"
] |
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.
| 1,000
|
[
{
"input": "RRU",
"output": "-1"
},
{
"input": "UDUR",
"output": "1"
},
{
"input": "RUUR",
"output": "2"
},
{
"input": "DDDD",
"output": "2"
},
{
"input": "RRRR",
"output": "2"
},
{
"input": "RRRUUD",
"output": "2"
},
{
"input": "UDURLRDURLRD",
"output": "1"
},
{
"input": "RLRU",
"output": "1"
},
{
"input": "RDDLLDLUUUDDRDRURLUUURLLDDLRLUURRLLRRLDRLLUDRLRULLDLRRLRLRLRUDUUDLULURLLDUURULURLLRRRURRRDRUUDLDRLRDRLRRDDLDLDLLUDRUDRLLLLDRDUULRUURRDLULLULDUDULRURRDDDLLUDRLUDDLDDDRRDDDULLLLDLDRLRRLRRDDRLULURRUDRDUUUULDURUDRDLDDUDUDRRURDULRRUDRLRRDLUURURDLDRLRDUDDDLDDDURURLUULRDUUULRURUDUDRRUDULLLUUUDRLLDRRDDLRUDRDRDLLRURURRRULURURRRLUUULRRRUURUUDURUDDLLDLDRLRDLDRLLDLDRDRRLRRRURUUUDRDLRLRUDRLULUUULUDDLULDLRLLRDUULLRLRURLRURULLLUDUDDLRULRDUURURLDLUURRRDURRLLDRUUDRDLLDUUDLURUDDUUUULRLLURLUDDRLRRDRURLRUDRLDDRLLL",
"output": "-1"
},
{
"input": "LDLDLDLDLDRULD",
"output": "5"
},
{
"input": "LULULURULLLU",
"output": "5"
},
{
"input": "DRDRDDRR",
"output": "4"
},
{
"input": "LR",
"output": "0"
},
{
"input": "UL",
"output": "1"
},
{
"input": "UD",
"output": "0"
},
{
"input": "R",
"output": "-1"
},
{
"input": "LU",
"output": "1"
},
{
"input": "RR",
"output": "1"
},
{
"input": "UDLR",
"output": "0"
},
{
"input": "RRRRRLLUUUUDD",
"output": "-1"
},
{
"input": "UUUUDLLLLR",
"output": "3"
},
{
"input": "LLRUD",
"output": "-1"
},
{
"input": "LLRDDU",
"output": "1"
}
] | 1,579,251,958
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 108
| 0
|
X = input()
RL, UD, X = 0, 0 , int(input())
for i in X:
if i in 'UD': UD += (1 if i == 'U' else -1)
if i in 'RL': RL += (1 if i == 'R' else -1)
Sum = abs(RL) + abs(UD)
print(-1 if len(X)%2==1 else Sum // 2)
# Show you deserve being the best to whom doesn't believe in you.
|
Title: Memory and Trident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
Input Specification:
The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.
Output Specification:
If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
Demo Input:
['RRU\n', 'UDUR\n', 'RUUR\n']
Demo Output:
['-1\n', '1\n', '2\n']
Note:
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.
|
```python
X = input()
RL, UD, X = 0, 0 , int(input())
for i in X:
if i in 'UD': UD += (1 if i == 'U' else -1)
if i in 'RL': RL += (1 if i == 'R' else -1)
Sum = abs(RL) + abs(UD)
print(-1 if len(X)%2==1 else Sum // 2)
# Show you deserve being the best to whom doesn't believe in you.
```
| -1
|
|
35
|
C
|
Fire Again
|
PROGRAMMING
| 1,500
|
[
"brute force",
"dfs and similar",
"shortest paths"
] |
C. Fire Again
|
2
|
64
|
After a terrifying forest fire in Berland a forest rebirth program was carried out. Due to it *N* rows with *M* trees each were planted and the rows were so neat that one could map it on a system of coordinates so that the *j*-th tree in the *i*-th row would have the coordinates of (*i*,<=*j*). However a terrible thing happened and the young forest caught fire. Now we must find the coordinates of the tree that will catch fire last to plan evacuation.
The burning began in *K* points simultaneously, which means that initially *K* trees started to burn. Every minute the fire gets from the burning trees to the ones that aren’t burning and that the distance from them to the nearest burning tree equals to 1.
Find the tree that will be the last to start burning. If there are several such trees, output any.
|
The first input line contains two integers *N*,<=*M* (1<=≤<=*N*,<=*M*<=≤<=2000) — the size of the forest. The trees were planted in all points of the (*x*,<=*y*) (1<=≤<=*x*<=≤<=*N*,<=1<=≤<=*y*<=≤<=*M*) type, *x* and *y* are integers.
The second line contains an integer *K* (1<=≤<=*K*<=≤<=10) — amount of trees, burning in the beginning.
The third line contains *K* pairs of integers: *x*1,<=*y*1,<=*x*2,<=*y*2,<=...,<=*x**k*,<=*y**k* (1<=≤<=*x**i*<=≤<=*N*,<=1<=≤<=*y**i*<=≤<=*M*) — coordinates of the points from which the fire started. It is guaranteed that no two points coincide.
|
Output a line with two space-separated integers *x* and *y* — coordinates of the tree that will be the last one to start burning. If there are several such trees, output any.
|
[
"3 3\n1\n2 2\n",
"3 3\n1\n1 1\n",
"3 3\n2\n1 1 3 3\n"
] |
[
"1 1\n",
"3 3\n",
"2 2"
] |
none
| 1,500
|
[
{
"input": "3 3\n1\n2 2",
"output": "1 1"
},
{
"input": "3 3\n1\n1 1",
"output": "3 3"
},
{
"input": "3 3\n2\n1 1 3 3",
"output": "1 3"
},
{
"input": "1 1\n1\n1 1",
"output": "1 1"
},
{
"input": "2 2\n1\n2 2",
"output": "1 1"
},
{
"input": "2 2\n2\n1 1 2 1",
"output": "1 2"
},
{
"input": "2 2\n3\n1 2 2 1 1 1",
"output": "2 2"
},
{
"input": "2 2\n4\n2 1 2 2 1 1 1 2",
"output": "1 1"
},
{
"input": "10 10\n1\n5 5",
"output": "10 10"
},
{
"input": "10 10\n2\n7 8 1 9",
"output": "3 1"
},
{
"input": "10 10\n3\n3 9 6 3 3 5",
"output": "10 7"
},
{
"input": "10 10\n4\n5 3 4 7 7 5 8 5",
"output": "10 10"
},
{
"input": "10 10\n5\n2 7 10 6 5 3 9 5 2 9",
"output": "1 1"
},
{
"input": "10 10\n6\n5 1 4 6 3 9 9 9 5 7 7 2",
"output": "1 3"
},
{
"input": "10 10\n7\n5 8 4 6 4 1 6 2 1 10 3 2 7 10",
"output": "10 5"
},
{
"input": "10 10\n8\n9 4 9 10 5 8 6 5 1 3 2 5 10 6 2 1",
"output": "1 10"
},
{
"input": "10 10\n9\n10 1 10 4 8 4 6 6 1 9 10 10 7 7 6 5 7 10",
"output": "1 1"
},
{
"input": "10 10\n10\n7 2 1 9 5 8 6 10 9 4 10 8 6 8 8 7 4 1 9 5",
"output": "1 3"
},
{
"input": "100 100\n1\n44 3",
"output": "100 100"
},
{
"input": "100 100\n2\n79 84 76 63",
"output": "1 1"
},
{
"input": "100 100\n3\n89 93 99 32 32 82",
"output": "1 1"
},
{
"input": "100 100\n4\n72 12 1 66 57 67 25 67",
"output": "100 100"
},
{
"input": "100 100\n5\n22 41 82 16 6 3 20 6 69 78",
"output": "1 100"
},
{
"input": "100 100\n6\n92 32 90 80 32 40 24 19 36 37 39 13",
"output": "1 100"
},
{
"input": "100 100\n7\n30 32 29 63 86 78 88 2 86 50 41 60 54 28",
"output": "1 100"
},
{
"input": "100 100\n8\n40 43 96 8 17 63 61 59 16 69 4 95 30 62 12 91",
"output": "100 100"
},
{
"input": "100 100\n9\n18 16 41 71 25 1 43 38 78 92 77 70 99 8 33 54 76 78",
"output": "1 100"
},
{
"input": "100 100\n10\n58 98 33 62 75 13 94 86 81 42 14 53 12 66 7 14 3 63 87 37",
"output": "40 1"
},
{
"input": "2000 2000\n1\n407 594",
"output": "2000 2000"
},
{
"input": "2000 2000\n2\n1884 43 1235 1111",
"output": "1 2000"
},
{
"input": "2000 2000\n3\n1740 1797 1279 1552 329 756",
"output": "2000 1"
},
{
"input": "2000 2000\n4\n1844 1342 171 1810 1558 1141 1917 1999",
"output": "530 1"
},
{
"input": "2000 2000\n5\n1846 327 1911 1534 134 1615 1664 682 1982 1112",
"output": "346 1"
},
{
"input": "2000 2000\n6\n1744 1102 852 723 409 179 89 1085 997 1433 1082 1680",
"output": "2000 1"
},
{
"input": "2000 2000\n7\n1890 22 288 1729 383 831 1192 1206 721 1376 969 492 510 1699",
"output": "2000 2000"
},
{
"input": "2000 2000\n8\n286 381 572 1849 1703 1574 622 1047 1507 941 871 663 1930 120 1084 1830",
"output": "1 1423"
},
{
"input": "2000 2000\n9\n226 531 56 138 722 405 1082 608 1355 1426 83 544 275 1268 683 412 1880 1049",
"output": "1701 1"
},
{
"input": "2000 2000\n10\n763 851 1182 571 1758 389 247 1907 730 881 531 1970 1430 667 169 765 1729 120 129 967",
"output": "2000 1793"
},
{
"input": "2000 2000\n10\n655 95 1640 1656 1344 79 666 1677 968 1180 522 1394 1850 1568 336 130 412 920 29 1664",
"output": "2000 570"
},
{
"input": "10 1\n10\n4 1 6 1 10 1 9 1 1 1 7 1 5 1 2 1 8 1 3 1",
"output": "1 1"
},
{
"input": "1 10\n10\n1 10 1 4 1 3 1 7 1 6 1 1 1 8 1 2 1 9 1 5",
"output": "1 1"
},
{
"input": "1 100\n10\n1 68 1 18 1 43 1 12 1 64 1 34 1 23 1 70 1 46 1 33",
"output": "1 100"
},
{
"input": "100 1\n10\n62 1 63 1 57 1 76 1 35 1 69 1 73 1 95 1 96 1 21 1",
"output": "1 1"
},
{
"input": "1 2000\n10\n1 1730 1 1374 1 831 1 1076 1 580 1 914 1 123 1 668 1 1288 1 160",
"output": "1 2000"
},
{
"input": "2000 1\n10\n1058 1 1779 1 1995 1 1398 1 96 1 1599 1 1496 1 1659 1 385 1 1485 1",
"output": "721 1"
},
{
"input": "5 5\n1\n1 1",
"output": "5 5"
},
{
"input": "2 5\n10\n1 1 1 2 1 3 1 4 1 5 2 1 2 2 2 3 2 4 2 5",
"output": "1 1"
},
{
"input": "1 10\n9\n1 1 1 2 1 3 1 4 1 6 1 7 1 8 1 9 1 10",
"output": "1 5"
},
{
"input": "3 3\n5\n1 1 1 3 2 2 3 1 3 3",
"output": "1 2"
},
{
"input": "999 999\n4\n1 499 499 1 499 999 999 499",
"output": "500 500"
}
] | 1,594,421,964
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 404
| 22,425,600
|
Matriz=[]
def valid(nx,my,x,y):
return 0<=x< nx and 0<=y< my
def parse(arreglo):
return f"{arreglo[0][1]+1} {arreglo[0][1]+1}"
def makeMat(nx,my):
global Matriz
for x in range(nx):
arreglo=[]
for y in range(my):
arreglo.append(0)
Matriz.append(arreglo)
#makeDicGraph(nx,my)
'''
def makeDicGraph(nx, my):
global d
for x in range(nx):
for y in range(my):
real_x = x+1
real_y = y+1
if valid(nx,my,real_x+1,real_y):
d[parse(nx,real_x,real_y)].append(parse(nx,real_x+1,real_y))
d[parse(nx,real_x+1,real_y)].append(parse(nx,real_x,real_y))
if valid(nx,my,real_x,real_y+1):
d[parse(nx,real_x,real_y)].append(parse(nx,real_x,real_y+1))
d[parse(nx,real_x,real_y+1)].append(parse(nx,real_x,real_y))
if valid(nx,my,real_x+1,real_y+1):
d[parse(nx,real_x,real_y)].append(parse(nx,real_x+1,real_y+1))
d[parse(nx,real_x+1,real_y+1)].append(parse(nx,real_x,real_y))
'''
def burnTree(arreglo):
global Matriz
quemados = []
for element in arreglo:
[x,y] = element
if valid(x-1,y) and Matriz[x-1][y] == 0:
quemados.append([x-1,y])
Matriz[x-1][y]=1
if valid(x+1,y) and Matriz[x+1][y] == 0:
quemados.append([x+1,y])
Matriz[x+1][y]=1
if valid(x,y-1) and Matriz[x][y-1] == 0:
quemados.append([x,y-1])
Matriz[x][y-1]=1
if valid(x,y+1) and Matriz[x][y+1] == 0:
quemados.append([x,y+1])
Matriz[x][y+1]=1
return quemados
A = open("input.txt")
linea=1
for line in A:
if linea == 1:
l = line.strip("")
elif linea == 2:
k = line.strip("")
else:
XY = line.strip("")
linea += 1
A.close()
#l = input()
[nx,my] = l.split(" ")
nx = int(nx)
my = int(my)
makeMat(nx,my)
#k = input()
k = int(k)
#XY = input()
XY = XY.split(" ")
i = 0
arreglo = []
while i < 2*k:
arreglo.append([int(XY[i])-1,int(XY[i+1])-1])
i += 2
actual_quemados = burnTree(arreglo)
while(True):
#print(actual_quemados)
aux_quemados = burnTree(actual_quemados)
if len(aux_quemados) == 0:
break
actual_quemados = aux_quemados
#print(actual_quemados[0])
#print(invparse(actual_quemados[0],nx,my))
A = open("output.txt","w")
if len(actual_quemados) == 0:
A.write(parse(arreglo[0]))
else:
A.write(parse(actual_quemados[0]))
|
Title: Fire Again
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
After a terrifying forest fire in Berland a forest rebirth program was carried out. Due to it *N* rows with *M* trees each were planted and the rows were so neat that one could map it on a system of coordinates so that the *j*-th tree in the *i*-th row would have the coordinates of (*i*,<=*j*). However a terrible thing happened and the young forest caught fire. Now we must find the coordinates of the tree that will catch fire last to plan evacuation.
The burning began in *K* points simultaneously, which means that initially *K* trees started to burn. Every minute the fire gets from the burning trees to the ones that aren’t burning and that the distance from them to the nearest burning tree equals to 1.
Find the tree that will be the last to start burning. If there are several such trees, output any.
Input Specification:
The first input line contains two integers *N*,<=*M* (1<=≤<=*N*,<=*M*<=≤<=2000) — the size of the forest. The trees were planted in all points of the (*x*,<=*y*) (1<=≤<=*x*<=≤<=*N*,<=1<=≤<=*y*<=≤<=*M*) type, *x* and *y* are integers.
The second line contains an integer *K* (1<=≤<=*K*<=≤<=10) — amount of trees, burning in the beginning.
The third line contains *K* pairs of integers: *x*1,<=*y*1,<=*x*2,<=*y*2,<=...,<=*x**k*,<=*y**k* (1<=≤<=*x**i*<=≤<=*N*,<=1<=≤<=*y**i*<=≤<=*M*) — coordinates of the points from which the fire started. It is guaranteed that no two points coincide.
Output Specification:
Output a line with two space-separated integers *x* and *y* — coordinates of the tree that will be the last one to start burning. If there are several such trees, output any.
Demo Input:
['3 3\n1\n2 2\n', '3 3\n1\n1 1\n', '3 3\n2\n1 1 3 3\n']
Demo Output:
['1 1\n', '3 3\n', '2 2']
Note:
none
|
```python
Matriz=[]
def valid(nx,my,x,y):
return 0<=x< nx and 0<=y< my
def parse(arreglo):
return f"{arreglo[0][1]+1} {arreglo[0][1]+1}"
def makeMat(nx,my):
global Matriz
for x in range(nx):
arreglo=[]
for y in range(my):
arreglo.append(0)
Matriz.append(arreglo)
#makeDicGraph(nx,my)
'''
def makeDicGraph(nx, my):
global d
for x in range(nx):
for y in range(my):
real_x = x+1
real_y = y+1
if valid(nx,my,real_x+1,real_y):
d[parse(nx,real_x,real_y)].append(parse(nx,real_x+1,real_y))
d[parse(nx,real_x+1,real_y)].append(parse(nx,real_x,real_y))
if valid(nx,my,real_x,real_y+1):
d[parse(nx,real_x,real_y)].append(parse(nx,real_x,real_y+1))
d[parse(nx,real_x,real_y+1)].append(parse(nx,real_x,real_y))
if valid(nx,my,real_x+1,real_y+1):
d[parse(nx,real_x,real_y)].append(parse(nx,real_x+1,real_y+1))
d[parse(nx,real_x+1,real_y+1)].append(parse(nx,real_x,real_y))
'''
def burnTree(arreglo):
global Matriz
quemados = []
for element in arreglo:
[x,y] = element
if valid(x-1,y) and Matriz[x-1][y] == 0:
quemados.append([x-1,y])
Matriz[x-1][y]=1
if valid(x+1,y) and Matriz[x+1][y] == 0:
quemados.append([x+1,y])
Matriz[x+1][y]=1
if valid(x,y-1) and Matriz[x][y-1] == 0:
quemados.append([x,y-1])
Matriz[x][y-1]=1
if valid(x,y+1) and Matriz[x][y+1] == 0:
quemados.append([x,y+1])
Matriz[x][y+1]=1
return quemados
A = open("input.txt")
linea=1
for line in A:
if linea == 1:
l = line.strip("")
elif linea == 2:
k = line.strip("")
else:
XY = line.strip("")
linea += 1
A.close()
#l = input()
[nx,my] = l.split(" ")
nx = int(nx)
my = int(my)
makeMat(nx,my)
#k = input()
k = int(k)
#XY = input()
XY = XY.split(" ")
i = 0
arreglo = []
while i < 2*k:
arreglo.append([int(XY[i])-1,int(XY[i+1])-1])
i += 2
actual_quemados = burnTree(arreglo)
while(True):
#print(actual_quemados)
aux_quemados = burnTree(actual_quemados)
if len(aux_quemados) == 0:
break
actual_quemados = aux_quemados
#print(actual_quemados[0])
#print(invparse(actual_quemados[0],nx,my))
A = open("output.txt","w")
if len(actual_quemados) == 0:
A.write(parse(arreglo[0]))
else:
A.write(parse(actual_quemados[0]))
```
| -1
|
362
|
B
|
Petya and Staircases
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
|
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
|
[
"10 5\n2 4 8 3 6\n",
"10 5\n2 4 5 7 9\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "10 5\n2 4 8 3 6",
"output": "NO"
},
{
"input": "10 5\n2 4 5 7 9",
"output": "YES"
},
{
"input": "10 9\n2 3 4 5 6 7 8 9 10",
"output": "NO"
},
{
"input": "5 2\n4 5",
"output": "NO"
},
{
"input": "123 13\n36 73 111 2 92 5 47 55 48 113 7 78 37",
"output": "YES"
},
{
"input": "10 10\n7 6 4 2 5 10 8 3 9 1",
"output": "NO"
},
{
"input": "12312 0",
"output": "YES"
},
{
"input": "9817239 1\n6323187",
"output": "YES"
},
{
"input": "1 1\n1",
"output": "NO"
},
{
"input": "5 4\n4 2 5 1",
"output": "NO"
},
{
"input": "5 3\n4 3 5",
"output": "NO"
},
{
"input": "500 3\n18 62 445",
"output": "YES"
},
{
"input": "500 50\n72 474 467 241 442 437 336 234 410 120 438 164 405 177 142 114 27 20 445 235 46 176 88 488 242 391 28 414 145 92 206 334 152 343 367 254 100 243 155 348 148 450 461 483 97 34 471 69 416 362",
"output": "NO"
},
{
"input": "500 8\n365 313 338 410 482 417 325 384",
"output": "YES"
},
{
"input": "1000000000 10\n2 3 5 6 8 9 123 874 1230 1000000000",
"output": "NO"
},
{
"input": "1000000000 10\n1 2 3 5 6 8 9 123 874 1230",
"output": "NO"
},
{
"input": "10 1\n1",
"output": "NO"
},
{
"input": "10 4\n1 2 4 5",
"output": "NO"
},
{
"input": "50 20\n22 33 17 23 27 5 26 31 41 20 8 24 6 3 4 29 40 25 13 16",
"output": "NO"
},
{
"input": "50 40\n14 27 19 30 31 20 28 11 37 29 23 33 7 26 22 16 1 6 18 3 47 36 38 2 48 9 41 8 5 50 4 45 44 25 39 12 43 42 40 46",
"output": "NO"
},
{
"input": "123 12\n35 95 47 99 79 122 58 94 31 57 18 10",
"output": "YES"
},
{
"input": "10 5\n1 3 5 7 9",
"output": "NO"
},
{
"input": "100 7\n2 3 5 6 8 9 100",
"output": "NO"
},
{
"input": "100 3\n98 99 100",
"output": "NO"
},
{
"input": "100 3\n97 98 99",
"output": "NO"
},
{
"input": "100 3\n96 98 99",
"output": "YES"
},
{
"input": "10 6\n2 3 5 6 8 9",
"output": "YES"
},
{
"input": "1000000000 10\n2 4 10 18 40 42 49 58 59 60",
"output": "NO"
},
{
"input": "10 3\n1 4 6",
"output": "NO"
},
{
"input": "8 3\n2 3 4",
"output": "NO"
},
{
"input": "100 3\n4 5 6",
"output": "NO"
},
{
"input": "10 2\n10 1",
"output": "NO"
},
{
"input": "10 1\n10",
"output": "NO"
},
{
"input": "4 2\n2 3",
"output": "YES"
},
{
"input": "2 1\n1",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "4 3\n2 3 4",
"output": "NO"
},
{
"input": "5 3\n4 2 3",
"output": "NO"
}
] | 1,576,705,978
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
def arr_inp():
return [int(x) for x in stdin.readline().split()]
def inp():
return map(int, stdin.readline().split())
from sys import *
n, m = inp()
a, pre, move = sorted(arr_inp()), 1, 0
for i in a:
if i - pre == 1:
if move < 2:
move += 1
else:
exit(print('NO'))
else:
move = 0
pre = i
print('YES')
|
Title: Petya and Staircases
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
Output Specification:
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
Demo Input:
['10 5\n2 4 8 3 6\n', '10 5\n2 4 5 7 9\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
def arr_inp():
return [int(x) for x in stdin.readline().split()]
def inp():
return map(int, stdin.readline().split())
from sys import *
n, m = inp()
a, pre, move = sorted(arr_inp()), 1, 0
for i in a:
if i - pre == 1:
if move < 2:
move += 1
else:
exit(print('NO'))
else:
move = 0
pre = i
print('YES')
```
| 0
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,591,343,412
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 109
| 0
|
s=input()
l=list('hello')
idx=0
for i in range(len(s)):
if idx==5:
break
if s[i]==l[idx]:
idx+=1
if idx==5:
print('YES')
else:
print('NO')
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
s=input()
l=list('hello')
idx=0
for i in range(len(s)):
if idx==5:
break
if s[i]==l[idx]:
idx+=1
if idx==5:
print('YES')
else:
print('NO')
```
| 3.9455
|
26
|
B
|
Regular Bracket Sequence
|
PROGRAMMING
| 1,400
|
[
"greedy"
] |
B. Regular Bracket Sequence
|
5
|
256
|
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
|
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
|
Output the maximum possible length of a regular bracket sequence.
|
[
"(()))(\n",
"((()())\n"
] |
[
"4\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "(()))(",
"output": "4"
},
{
"input": "((()())",
"output": "6"
},
{
"input": "(",
"output": "0"
},
{
"input": ")",
"output": "0"
},
{
"input": ")(()(",
"output": "2"
},
{
"input": "))))))(",
"output": "0"
},
{
"input": "()()(()(((",
"output": "6"
},
{
"input": "()))(()((((()(())",
"output": "10"
},
{
"input": "())))((()())())))))())",
"output": "14"
},
{
"input": ")))((((())(()((()((((()()())((",
"output": "16"
},
{
"input": "))()()((()()))())()(((((((())((((((((())()()((())(",
"output": "32"
},
{
"input": "))())))))))())))))()()))()()))))())))))()))))))))))))(()))())(()))))(()))))())))((((()()))))()))()))",
"output": "48"
}
] | 1,664,141,096
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 38
| 154
| 5,017,600
|
import sys
input = lambda: sys.stdin.readline().rstrip()
S = input()
cnt=0
ans=0
for c in S:
if c==')' and cnt>0:
cnt-=1
ans+=2
elif c=='(':
cnt+=1
print(ans)
|
Title: Regular Bracket Sequence
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input Specification:
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output Specification:
Output the maximum possible length of a regular bracket sequence.
Demo Input:
['(()))(\n', '((()())\n']
Demo Output:
['4\n', '6\n']
Note:
none
|
```python
import sys
input = lambda: sys.stdin.readline().rstrip()
S = input()
cnt=0
ans=0
for c in S:
if c==')' and cnt>0:
cnt-=1
ans+=2
elif c=='(':
cnt+=1
print(ans)
```
| 3.975254
|
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"
}
] | 1,695,099,614
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
n=input()
m=list(n)
m[0]=m[0].upper()
print("".join(m))
|
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()
m=list(n)
m[0]=m[0].upper()
print("".join(m))
```
| 3
|
|
992
|
B
|
Nastya Studies Informatics
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.
We define a pair of integers (*a*,<=*b*) good, if *GCD*(*a*,<=*b*)<==<=*x* and *LCM*(*a*,<=*b*)<==<=*y*, where *GCD*(*a*,<=*b*) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of *a* and *b*, and *LCM*(*a*,<=*b*) denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of *a* and *b*.
You are given two integers *x* and *y*. You are to find the number of good pairs of integers (*a*,<=*b*) such that *l*<=≤<=*a*,<=*b*<=≤<=*r*. Note that pairs (*a*,<=*b*) and (*b*,<=*a*) are considered different if *a*<=≠<=*b*.
|
The only line contains four integers *l*,<=*r*,<=*x*,<=*y* (1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=*y*<=≤<=109).
|
In the only line print the only integer — the answer for the problem.
|
[
"1 2 1 2\n",
"1 12 1 12\n",
"50 100 3 30\n"
] |
[
"2\n",
"4\n",
"0\n"
] |
In the first example there are two suitable good pairs of integers (*a*, *b*): (1, 2) and (2, 1).
In the second example there are four suitable good pairs of integers (*a*, *b*): (1, 12), (12, 1), (3, 4) and (4, 3).
In the third example there are good pairs of integers, for example, (3, 30), but none of them fits the condition *l* ≤ *a*, *b* ≤ *r*.
| 1,000
|
[
{
"input": "1 2 1 2",
"output": "2"
},
{
"input": "1 12 1 12",
"output": "4"
},
{
"input": "50 100 3 30",
"output": "0"
},
{
"input": "1 1000000000 1 1000000000",
"output": "4"
},
{
"input": "1 1000000000 158260522 200224287",
"output": "0"
},
{
"input": "1 1000000000 2 755829150",
"output": "8"
},
{
"input": "1 1000000000 158260522 158260522",
"output": "1"
},
{
"input": "1 1000000000 877914575 877914575",
"output": "1"
},
{
"input": "232 380232688 116 760465376",
"output": "30"
},
{
"input": "47259 3393570 267 600661890",
"output": "30"
},
{
"input": "1 1000000000 1 672672000",
"output": "64"
},
{
"input": "1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1 1000000000 1 649209600",
"output": "32"
},
{
"input": "1 1000000000 1 682290000",
"output": "32"
},
{
"input": "1 1000000000 1 228614400",
"output": "16"
},
{
"input": "1 1000000000 1 800280000",
"output": "32"
},
{
"input": "1 1000000000 1 919987200",
"output": "16"
},
{
"input": "1 1000000000 1 456537870",
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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"output": "22"
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{
"input": "22 158 2 1738",
"output": "2"
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{
"input": "1 2623 1 2623",
"output": "4"
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{
"input": "7 163677675 3 18",
"output": "0"
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{
"input": "159 20749927 1 158",
"output": "0"
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{
"input": "5252 477594071 1 5251",
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"output": "0"
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{
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"output": "2"
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{
"input": "26 46 2 598",
"output": "2"
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{
"input": "26 82 2 1066",
"output": "2"
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{
"input": "1 2993 1 2993",
"output": "4"
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{
"input": "17 17 1 289",
"output": "0"
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{
"input": "177 267 3 15753",
"output": "2"
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{
"input": "7388 22705183 1 7387",
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{
"input": "1 100 3 100",
"output": "0"
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{
"input": "1 1000 6 1024",
"output": "0"
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{
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"output": "2"
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{
"input": "1 10000 2 455",
"output": "0"
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{
"input": "1 1000000000 250000000 1000000000",
"output": "2"
},
{
"input": "3 3 1 1",
"output": "0"
},
{
"input": "1 1000000000 100000000 1000000000",
"output": "4"
},
{
"input": "5 10 3 3",
"output": "0"
},
{
"input": "1 1000 5 13",
"output": "0"
},
{
"input": "2 2 3 3",
"output": "0"
},
{
"input": "1 1000000000 499999993 999999986",
"output": "2"
},
{
"input": "1 1 1 10",
"output": "0"
},
{
"input": "1 10 10 100",
"output": "0"
},
{
"input": "1 1000 4 36",
"output": "2"
},
{
"input": "1 1000000000 10000000 20000000",
"output": "2"
},
{
"input": "100 100 5 5",
"output": "0"
},
{
"input": "3 3 3 9",
"output": "0"
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{
"input": "36 200 24 144",
"output": "2"
},
{
"input": "1 100 3 10",
"output": "0"
}
] | 1,689,443,499
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 3
| 1,000
| 1,433,600
|
from math import gcd, lcm
l, r, x, y = map(int, input().split())
count = 0
for a in range(l, r + 1):
if gcd(a, x) == x and y % a == 0:
b = y // a
if gcd(a, b) == x:
count += 1
print(count)
|
Title: Nastya Studies Informatics
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.
We define a pair of integers (*a*,<=*b*) good, if *GCD*(*a*,<=*b*)<==<=*x* and *LCM*(*a*,<=*b*)<==<=*y*, where *GCD*(*a*,<=*b*) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of *a* and *b*, and *LCM*(*a*,<=*b*) denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of *a* and *b*.
You are given two integers *x* and *y*. You are to find the number of good pairs of integers (*a*,<=*b*) such that *l*<=≤<=*a*,<=*b*<=≤<=*r*. Note that pairs (*a*,<=*b*) and (*b*,<=*a*) are considered different if *a*<=≠<=*b*.
Input Specification:
The only line contains four integers *l*,<=*r*,<=*x*,<=*y* (1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=*y*<=≤<=109).
Output Specification:
In the only line print the only integer — the answer for the problem.
Demo Input:
['1 2 1 2\n', '1 12 1 12\n', '50 100 3 30\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
In the first example there are two suitable good pairs of integers (*a*, *b*): (1, 2) and (2, 1).
In the second example there are four suitable good pairs of integers (*a*, *b*): (1, 12), (12, 1), (3, 4) and (4, 3).
In the third example there are good pairs of integers, for example, (3, 30), but none of them fits the condition *l* ≤ *a*, *b* ≤ *r*.
|
```python
from math import gcd, lcm
l, r, x, y = map(int, input().split())
count = 0
for a in range(l, r + 1):
if gcd(a, x) == x and y % a == 0:
b = y // a
if gcd(a, b) == x:
count += 1
print(count)
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,695,028,405
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
import math
m,n=map(int,input().split())
print((math.ceil(m/2))*(math.ceil(n/1)))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
import math
m,n=map(int,input().split())
print((math.ceil(m/2))*(math.ceil(n/1)))
```
| 0
|
344
|
A
|
Magnets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
|
On the single line of the output print the number of groups of magnets.
|
[
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] |
[
"3\n",
"2\n"
] |
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
| 500
|
[
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,699,527,494
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 92
| 0
|
T = int(input())
S = input()
count = 0
for _ in range(T):
PreS = S
S = input()
if S != PreS:
count += 1
print(count)
|
Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
|
```python
T = int(input())
S = input()
count = 0
for _ in range(T):
PreS = S
S = input()
if S != PreS:
count += 1
print(count)
```
| -1
|
|
575
|
D
|
Tablecity
|
PROGRAMMING
| 1,700
|
[
"constructive algorithms",
"implementation"
] | null | null |
There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves.
Tablecity can be represented as 1000<=×<=2 grid, where every cell represents one district. Each district has its own unique name “(*X*,<=*Y*)”, where *X* and *Y* are the coordinates of the district in the grid. The thief’s movement is as
Every hour the thief will leave the district (*X*,<=*Y*) he is currently hiding in, and move to one of the districts: (*X*<=-<=1,<=*Y*), (*X*<=+<=1,<=*Y*), (*X*<=-<=1,<=*Y*<=-<=1), (*X*<=-<=1,<=*Y*<=+<=1), (*X*<=+<=1,<=*Y*<=-<=1), (*X*<=+<=1,<=*Y*<=+<=1) as long as it exists in Tablecity.
Below is an example of thief’s possible movements if he is located in district (7,1):
Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that.
|
There is no input for this problem.
|
The first line of output contains integer *N* – duration of police search in hours. Each of the following *N* lines contains exactly 4 integers *X**i*1, *Y**i*1, *X**i*2, *Y**i*2 separated by spaces, that represent 2 districts (*X**i*1, *Y**i*1), (*X**i*2, *Y**i*2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement.
- *N*<=≤<=2015 - 1<=≤<=*X*<=≤<=1000 - 1<=≤<=*Y*<=≤<=2
|
[
"В этой задаче нет примеров ввода-вывода.\nThis problem doesn't have sample input and output."
] |
[
"Смотрите замечание ниже.\nSee the note below."
] |
Let's consider the following output:
2
5 1 50 2
8 1 80 2
This output is not guaranteed to catch the thief and is not correct. It is given to you only to show the expected output format. There exists a combination of an initial position and a movement strategy such that the police will not catch the thief.
Consider the following initial position and thief’s movement:
In the first hour, the thief is located in district (1,1). Police officers will search districts (5,1) and (50,2) and will not find him.
At the start of the second hour, the thief moves to district (2,2). Police officers will search districts (8,1) and (80,2) and will not find him.
Since there is no further investigation by the police, the thief escaped!
| 0
|
[
{
"input": "dummy",
"output": "2000\n1 1 1 2\n2 1 2 2\n3 1 3 2\n4 1 4 2\n5 1 5 2\n6 1 6 2\n7 1 7 2\n8 1 8 2\n9 1 9 2\n10 1 10 2\n11 1 11 2\n12 1 12 2\n13 1 13 2\n14 1 14 2\n15 1 15 2\n16 1 16 2\n17 1 17 2\n18 1 18 2\n19 1 19 2\n20 1 20 2\n21 1 21 2\n22 1 22 2\n23 1 23 2\n24 1 24 2\n25 1 25 2\n26 1 26 2\n27 1 27 2\n28 1 28 2\n29 1 29 2\n30 1 30 2\n31 1 31 2\n32 1 32 2\n33 1 33 2\n34 1 34 2\n35 1 35 2\n36 1 36 2\n37 1 37 2\n38 1 38 2\n39 1 39 2\n40 1 40 2\n41 1 41 2\n42 1 42 2\n43 1 43 2\n44 1 44 2\n45 1 45 2\n46 1 46 2\n47 1 47 2\n48 1 4..."
}
] | 1,441,527,590
| 1,190
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 512,000
|
print(1000)
for i in range(1, 1001):
print(i, 1, i, 2)
|
Title: Tablecity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves.
Tablecity can be represented as 1000<=×<=2 grid, where every cell represents one district. Each district has its own unique name “(*X*,<=*Y*)”, where *X* and *Y* are the coordinates of the district in the grid. The thief’s movement is as
Every hour the thief will leave the district (*X*,<=*Y*) he is currently hiding in, and move to one of the districts: (*X*<=-<=1,<=*Y*), (*X*<=+<=1,<=*Y*), (*X*<=-<=1,<=*Y*<=-<=1), (*X*<=-<=1,<=*Y*<=+<=1), (*X*<=+<=1,<=*Y*<=-<=1), (*X*<=+<=1,<=*Y*<=+<=1) as long as it exists in Tablecity.
Below is an example of thief’s possible movements if he is located in district (7,1):
Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that.
Input Specification:
There is no input for this problem.
Output Specification:
The first line of output contains integer *N* – duration of police search in hours. Each of the following *N* lines contains exactly 4 integers *X**i*1, *Y**i*1, *X**i*2, *Y**i*2 separated by spaces, that represent 2 districts (*X**i*1, *Y**i*1), (*X**i*2, *Y**i*2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement.
- *N*<=≤<=2015 - 1<=≤<=*X*<=≤<=1000 - 1<=≤<=*Y*<=≤<=2
Demo Input:
["В этой задаче нет примеров ввода-вывода.\nThis problem doesn't have sample input and output."]
Demo Output:
['Смотрите замечание ниже.\nSee the note below.']
Note:
Let's consider the following output:
2
5 1 50 2
8 1 80 2
This output is not guaranteed to catch the thief and is not correct. It is given to you only to show the expected output format. There exists a combination of an initial position and a movement strategy such that the police will not catch the thief.
Consider the following initial position and thief’s movement:
In the first hour, the thief is located in district (1,1). Police officers will search districts (5,1) and (50,2) and will not find him.
At the start of the second hour, the thief moves to district (2,2). Police officers will search districts (8,1) and (80,2) and will not find him.
Since there is no further investigation by the police, the thief escaped!
|
```python
print(1000)
for i in range(1, 1001):
print(i, 1, i, 2)
```
| 0
|
|
509
|
A
|
Maximum in Table
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
|
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
|
Print a single line containing a positive integer *m* — the maximum value in the table.
|
[
"1\n",
"5\n"
] |
[
"1",
"70"
] |
In the second test the rows of the table look as follows:
| 0
|
[
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
},
{
"input": "9",
"output": "12870"
},
{
"input": "10",
"output": "48620"
},
{
"input": "8",
"output": "3432"
}
] | 1,595,863,886
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 10
| 93
| 6,963,200
|
size = int(input())
matrix = []
for i in range(size):
a = []
for j in range(size):
if i == 0 or j == 0:
a.append(1)
else:
a.append(0)
matrix.append(a)
for i in range(1,size):
for j in range(1,size):
matrix[i][j] = matrix[i-1][j] + matrix[i][j-1]
print(matrix[size-1][size-1])
|
Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows:
|
```python
size = int(input())
matrix = []
for i in range(size):
a = []
for j in range(size):
if i == 0 or j == 0:
a.append(1)
else:
a.append(0)
matrix.append(a)
for i in range(1,size):
for j in range(1,size):
matrix[i][j] = matrix[i-1][j] + matrix[i][j-1]
print(matrix[size-1][size-1])
```
| 3
|
|
189
|
A
|
Cut Ribbon
|
PROGRAMMING
| 1,300
|
[
"brute force",
"dp"
] | null | null |
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
|
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
|
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
|
[
"5 5 3 2\n",
"7 5 5 2\n"
] |
[
"2\n",
"2\n"
] |
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
| 500
|
[
{
"input": "5 5 3 2",
"output": "2"
},
{
"input": "7 5 5 2",
"output": "2"
},
{
"input": "4 4 4 4",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "4000 1 2 3",
"output": "4000"
},
{
"input": "4000 3 4 5",
"output": "1333"
},
{
"input": "10 3 4 5",
"output": "3"
},
{
"input": "100 23 15 50",
"output": "2"
},
{
"input": "3119 3515 1021 7",
"output": "11"
},
{
"input": "918 102 1327 1733",
"output": "9"
},
{
"input": "3164 42 430 1309",
"output": "15"
},
{
"input": "3043 317 1141 2438",
"output": "7"
},
{
"input": "26 1 772 2683",
"output": "26"
},
{
"input": "370 2 1 15",
"output": "370"
},
{
"input": "734 12 6 2",
"output": "367"
},
{
"input": "418 18 14 17",
"output": "29"
},
{
"input": "18 16 28 9",
"output": "2"
},
{
"input": "14 6 2 17",
"output": "7"
},
{
"input": "29 27 18 2",
"output": "2"
},
{
"input": "29 12 7 10",
"output": "3"
},
{
"input": "27 23 4 3",
"output": "9"
},
{
"input": "5 14 5 2",
"output": "1"
},
{
"input": "5 17 26 5",
"output": "1"
},
{
"input": "9 1 10 3",
"output": "9"
},
{
"input": "2 19 15 1",
"output": "2"
},
{
"input": "4 6 4 9",
"output": "1"
},
{
"input": "10 6 2 9",
"output": "5"
},
{
"input": "2 2 9 6",
"output": "1"
},
{
"input": "6 2 4 1",
"output": "6"
},
{
"input": "27 24 5 27",
"output": "1"
},
{
"input": "2683 83 26 2709",
"output": "101"
},
{
"input": "728 412 789 158",
"output": "3"
},
{
"input": "3964 4 2916 176",
"output": "991"
},
{
"input": "3399 2035 2 3334",
"output": "683"
},
{
"input": "3455 244 3301 3",
"output": "991"
},
{
"input": "595 2263 3625 1",
"output": "595"
},
{
"input": "4000 1 1 1",
"output": "4000"
},
{
"input": "3999 2 2 3999",
"output": "1"
},
{
"input": "25 6 8 11",
"output": "3"
},
{
"input": "4000 500 1000 2000",
"output": "8"
},
{
"input": "53 10 11 23",
"output": "5"
},
{
"input": "100 100 1 1",
"output": "100"
},
{
"input": "17 3 4 10",
"output": "5"
},
{
"input": "413 101 102 105",
"output": "4"
},
{
"input": "490 4 49 50",
"output": "111"
},
{
"input": "3999 2 3 3",
"output": "1999"
},
{
"input": "8 3 8 4",
"output": "2"
},
{
"input": "5 1 3 3",
"output": "5"
},
{
"input": "100 3 17 22",
"output": "27"
},
{
"input": "4000 2 3 4",
"output": "2000"
},
{
"input": "4000 3 3 5",
"output": "1332"
},
{
"input": "13 4 6 7",
"output": "2"
},
{
"input": "4000 5 2 2",
"output": "2000"
},
{
"input": "3999 2 2 3",
"output": "1999"
},
{
"input": "4000 33 7 3333",
"output": "564"
},
{
"input": "60 33 20 9",
"output": "4"
},
{
"input": "100 9 11 99",
"output": "10"
},
{
"input": "2009 6 8 9",
"output": "334"
}
] | 1,682,359,997
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 77
| 0
|
n, a, b, c = map(int, input().split())
length = sorted([a, b, c])
nums = [0] * 8000
nums[a] = 1
nums[b] = 1
nums[c] = 1
for i in range(1, n):
for j in length:
if nums[i] > 0:
nums[i + j] = max(nums[i] + 1, nums[i + j])
print(nums[n])
|
Title: Cut Ribbon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input Specification:
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
Output Specification:
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Demo Input:
['5 5 3 2\n', '7 5 5 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
|
```python
n, a, b, c = map(int, input().split())
length = sorted([a, b, c])
nums = [0] * 8000
nums[a] = 1
nums[b] = 1
nums[c] = 1
for i in range(1, n):
for j in length:
if nums[i] > 0:
nums[i + j] = max(nums[i] + 1, nums[i + j])
print(nums[n])
```
| 3
|
|
886
|
B
|
Vlad and Cafes
|
PROGRAMMING
| 1,000
|
[] | null | null |
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
|
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
|
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
|
[
"5\n1 3 2 1 2\n",
"6\n2 1 2 2 4 1\n"
] |
[
"3\n",
"2\n"
] |
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
| 1,000
|
[
{
"input": "5\n1 3 2 1 2",
"output": "3"
},
{
"input": "6\n2 1 2 2 4 1",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n200000",
"output": "200000"
},
{
"input": "2\n2018 2017",
"output": "2018"
},
{
"input": "5\n100 1000 1000 1000 1000",
"output": "100"
},
{
"input": "8\n200000 1 200000 1 200000 1 200000 2",
"output": "1"
},
{
"input": "5\n20000 1 2 3 4",
"output": "20000"
},
{
"input": "2\n200000 1",
"output": "200000"
},
{
"input": "3\n2 100 2",
"output": "100"
},
{
"input": "2\n10 1",
"output": "10"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n5 5 5",
"output": "5"
},
{
"input": "2\n5 1",
"output": "5"
}
] | 1,586,334,754
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 2,000
| 21,913,600
|
n=int(input())
t=list(map(int,input().split()))
p=t[::-1]
s=list(set(t))
o=0
for j in s:
if p.index(j)>o:
o=p.index(j)
print(p[o])
|
Title: Vlad and Cafes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
Input Specification:
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Output Specification:
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
Demo Input:
['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n']
Demo Output:
['3\n', '2\n']
Note:
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
|
```python
n=int(input())
t=list(map(int,input().split()))
p=t[::-1]
s=list(set(t))
o=0
for j in s:
if p.index(j)>o:
o=p.index(j)
print(p[o])
```
| 0
|
|
863
|
B
|
Kayaking
|
PROGRAMMING
| 1,500
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
|
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
|
Print minimum possible total instability.
|
[
"2\n1 2 3 4\n",
"4\n1 3 4 6 3 4 100 200\n"
] |
[
"1\n",
"5\n"
] |
none
| 0
|
[
{
"input": "2\n1 2 3 4",
"output": "1"
},
{
"input": "4\n1 3 4 6 3 4 100 200",
"output": "5"
},
{
"input": "3\n305 139 205 406 530 206",
"output": "102"
},
{
"input": "3\n610 750 778 6 361 407",
"output": "74"
},
{
"input": "5\n97 166 126 164 154 98 221 7 51 47",
"output": "35"
},
{
"input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2",
"output": "0"
},
{
"input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4",
"output": "1"
},
{
"input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476",
"output": "368"
},
{
"input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597",
"output": "393"
},
{
"input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21",
"output": "387"
},
{
"input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637",
"output": "376"
},
{
"input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61",
"output": "351"
},
{
"input": "3\n1 2 10 11 100 100",
"output": "1"
},
{
"input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858",
"output": "318"
},
{
"input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636",
"output": "355"
},
{
"input": "2\n55 5 25 51",
"output": "4"
},
{
"input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973",
"output": "348"
},
{
"input": "4\n35 48 71 44 78 79 57 48",
"output": "10"
},
{
"input": "3\n58 89 73 15 5 47",
"output": "21"
},
{
"input": "2\n1 20 99 100",
"output": "1"
}
] | 1,601,191,822
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 108
| 307,200
|
"""Kayak"""
def process(data_in: list):
"""process"""
data = list(set(data_in))
remove = []
for i in data:
if data_in.count(i) % 2 == 0:
remove.append(i)
data = list(set(data)-set(remove))
data.sort()
total = 0
for i in range(0, len(data[:-2]), 2):
total += data[i+1]-data[i]
print(total)
def kayak():
"""Kayak"""
num = int(input())
people = input().split()
process([int(x) for x in people])
kayak()
|
Title: Kayaking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input Specification:
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
Output Specification:
Print minimum possible total instability.
Demo Input:
['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n']
Demo Output:
['1\n', '5\n']
Note:
none
|
```python
"""Kayak"""
def process(data_in: list):
"""process"""
data = list(set(data_in))
remove = []
for i in data:
if data_in.count(i) % 2 == 0:
remove.append(i)
data = list(set(data)-set(remove))
data.sort()
total = 0
for i in range(0, len(data[:-2]), 2):
total += data[i+1]-data[i]
print(total)
def kayak():
"""Kayak"""
num = int(input())
people = input().split()
process([int(x) for x in people])
kayak()
```
| 0
|
|
258
|
A
|
Little Elephant and Bits
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
|
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
|
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
|
[
"101\n",
"110010\n"
] |
[
"11\n",
"11010\n"
] |
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
| 500
|
[
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,593,527,989
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 20
| 2,000
| 12,697,600
|
n = input()
n = [x for x in n]
zero = 0
for i in range(len(n)):
if(n[i]=='0'):
n.pop(i)
zero = 1
break
if(zero==0):
n.pop(0)
s = ""
for ele in n:
s+=ele
print(s)
|
Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
|
```python
n = input()
n = [x for x in n]
zero = 0
for i in range(len(n)):
if(n[i]=='0'):
n.pop(i)
zero = 1
break
if(zero==0):
n.pop(0)
s = ""
for ele in n:
s+=ele
print(s)
```
| 0
|
|
669
|
A
|
Little Artem and Presents
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
|
Print the maximum possible number of times Artem can give presents to Masha.
|
[
"1\n",
"2\n",
"3\n",
"4\n"
] |
[
"1\n",
"1\n",
"2\n",
"3\n"
] |
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
| 500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"
},
{
"input": "1000000000",
"output": "666666667"
},
{
"input": "5",
"output": "3"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "666666666"
},
{
"input": "999999998",
"output": "666666665"
},
{
"input": "999999997",
"output": "666666665"
},
{
"input": "999999996",
"output": "666666664"
},
{
"input": "999999995",
"output": "666666663"
},
{
"input": "999999994",
"output": "666666663"
},
{
"input": "999999993",
"output": "666666662"
},
{
"input": "999999992",
"output": "666666661"
},
{
"input": "999999991",
"output": "666666661"
},
{
"input": "1000",
"output": "667"
},
{
"input": "10000",
"output": "6667"
},
{
"input": "100000",
"output": "66667"
},
{
"input": "1000000",
"output": "666667"
},
{
"input": "10000000",
"output": "6666667"
},
{
"input": "100000000",
"output": "66666667"
},
{
"input": "7",
"output": "5"
}
] | 1,627,045,611
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 77
| 6,758,400
|
n = int(input())
k = round(n/3)
print(n - k)
|
Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
|
```python
n = int(input())
k = round(n/3)
print(n - k)
```
| 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<=≤<=*k*<=≤<=*n*<=≤<=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"
}
] | 1,696,325,740
| 2,147,483,647
|
PyPy 3-64
|
MEMORY_LIMIT_EXCEEDED
|
TESTS
| 7
| 1,496
| 268,390,400
|
new_str = input().split(" ")
n = int(new_str[0])
k = int(new_str[1])
arr = []
for i in range(n):
arr.append(i + 1)
arr.sort(key=lambda x: x % 2 == 0)
print(arr[k - 1])
|
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<=≤<=*k*<=≤<=*n*<=≤<=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
new_str = input().split(" ")
n = int(new_str[0])
k = int(new_str[1])
arr = []
for i in range(n):
arr.append(i + 1)
arr.sort(key=lambda x: x % 2 == 0)
print(arr[k - 1])
```
| 0
|
|
371
|
C
|
Hamburgers
|
PROGRAMMING
| 1,600
|
[
"binary search",
"brute force"
] | null | null |
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
|
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
|
[
"BBBSSC\n6 4 1\n1 2 3\n4\n",
"BBC\n1 10 1\n1 10 1\n21\n",
"BSC\n1 1 1\n1 1 3\n1000000000000\n"
] |
[
"2\n",
"7\n",
"200000000001\n"
] |
none
| 1,500
|
[
{
"input": "BBBSSC\n6 4 1\n1 2 3\n4",
"output": "2"
},
{
"input": "BBC\n1 10 1\n1 10 1\n21",
"output": "7"
},
{
"input": "BSC\n1 1 1\n1 1 3\n1000000000000",
"output": "200000000001"
},
{
"input": "B\n1 1 1\n1 1 1\n381",
"output": "382"
},
{
"input": "BSC\n3 5 6\n7 3 9\n100",
"output": "10"
},
{
"input": "BSC\n100 1 1\n100 1 1\n100",
"output": "51"
},
{
"input": "SBBCCSBB\n1 50 100\n31 59 21\n100000",
"output": "370"
},
{
"input": "BBBBCCCCCCCCCCCCCCCCCCCCSSSSBBBBBBBBSS\n100 100 100\n1 1 1\n3628800",
"output": "95502"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n200",
"output": "0"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n2000",
"output": "1"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n300",
"output": "0"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n300000000",
"output": "42858"
},
{
"input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n914159265358",
"output": "130594181"
},
{
"input": "SSSSSSSSSSBBBBBBBBBCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSBB\n31 53 97\n13 17 31\n914159265358",
"output": "647421579"
},
{
"input": "BBBCSBSBBSSSSCCCCBBCSBBBBSSBBBCBSCCSSCSSCSBSSSCCCCBSCSSBSSSCCCBBCCCSCBCBBCCSCCCCSBBCCBBBBCCCCCCBSSCB\n91 87 17\n64 44 43\n958532915587",
"output": "191668251"
},
{
"input": "CSSCBBCCCSBSCBBBCSBBBCBSBCSCBCSCBCBSBCBCSSBBSBBCBBBBSCSBBCCBCCBCBBSBSBCSCSBBSSBBCSSBCSCSCCSSBCBBCBSB\n56 34 48\n78 6 96\n904174875419",
"output": "140968956"
},
{
"input": "CCSCCCSBBBSCBSCSCCSSBBBSSBBBSBBBCBCSSBCSCBBCCCBCBCBCCCSSBSBBCCCCCBBSCBSCBCBBCBBCSSBCSBSSCCSCCSCCBBBS\n33 73 67\n4 56 42\n886653164314",
"output": "277425898"
},
{
"input": "SBCSSCBBSSBCSSBBBSSBSCBSSSCBBSBBBBCSBCSBSCBSCBSCBSBSSCCCCBSBCCBCBSCCCBSCCBSBBCBSSCCCCSBSBBBSSSBCSCBC\n94 16 85\n14 18 91\n836590091442",
"output": "217522127"
},
{
"input": "BSCSBSCCSCSSCCCSBCSSBCBBSCCBSCCSSSSSSSSSCCSBSCCBBCBBSBSCCCCBCSBSBSSBBBBBSSBSSCBCCSSBSSSCBBCSBBSBCCCB\n67 54 8\n36 73 37\n782232051273",
"output": "154164772"
},
{
"input": "CBBCBSBCCSCBSSCCBCSBCSBBSCBBCSCCBSCCSCSBBSSBSBSCBBSBBCSSSSBBBBSBBCBCSBBCBCSSBBCSBSCCSCSBCSCBSCCBBCSC\n71 71 52\n52 88 3\n654400055575",
"output": "137826467"
},
{
"input": "CBBCBSBCCSCBSSCCBCSBCSBBSCBBCSCCBSCCSCSBBSBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBCBBCSC\n100 1 1\n1 17 23\n954400055575",
"output": "1355681897"
},
{
"input": "C\n100 100 100\n1 1 1\n1000000000000",
"output": "1000000000100"
},
{
"input": "SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n100 100 100\n100 100 100\n1000000000000",
"output": "100000001"
},
{
"input": "B\n100 100 100\n1 1 1\n1",
"output": "101"
},
{
"input": "SC\n2 1 1\n1 1 1\n100000000000",
"output": "50000000001"
},
{
"input": "B\n100 1 1\n1 1 1\n1000000000000",
"output": "1000000000100"
},
{
"input": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\n1 1 1\n100 100 100\n1000000000000",
"output": "100000000"
},
{
"input": "CC\n1 1 1\n100 100 100\n1",
"output": "0"
},
{
"input": "B\n100 100 100\n1 1 1\n1000000000000",
"output": "1000000000100"
},
{
"input": "BSC\n100 100 100\n1 1 1\n1000000000000",
"output": "333333333433"
},
{
"input": "BSC\n100 100 100\n1 1 1\n1",
"output": "100"
}
] | 1,664,257,902
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 1,000
| 1,536,000
|
buter = input()
k = 0
nb, ns, nc = map(int, input().split())
pb, ps, pc = map(int, input().split())
r = int(input())
while r >= 0:
for s in range(len(buter)):
if buter[s] == "B":
if nb > 0:
nb -= 1
else:
r -= pb
if buter[s] == "C":
if nc > 0:
nc -= 1
else:
r -= pc
if buter[s] == "S":
if ns > 0:
ns -= 1
else:
r -= ps
k += 1
print(k - 1)
|
Title: Hamburgers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
Input Specification:
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
Demo Input:
['BBBSSC\n6 4 1\n1 2 3\n4\n', 'BBC\n1 10 1\n1 10 1\n21\n', 'BSC\n1 1 1\n1 1 3\n1000000000000\n']
Demo Output:
['2\n', '7\n', '200000000001\n']
Note:
none
|
```python
buter = input()
k = 0
nb, ns, nc = map(int, input().split())
pb, ps, pc = map(int, input().split())
r = int(input())
while r >= 0:
for s in range(len(buter)):
if buter[s] == "B":
if nb > 0:
nb -= 1
else:
r -= pb
if buter[s] == "C":
if nc > 0:
nc -= 1
else:
r -= pc
if buter[s] == "S":
if ns > 0:
ns -= 1
else:
r -= ps
k += 1
print(k - 1)
```
| 0
|
|
295
|
D
|
Greg and Caves
|
PROGRAMMING
| 2,400
|
[
"combinatorics",
"dp"
] | null | null |
Greg has a pad. The pad's screen is an *n*<=×<=*m* rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to *n* from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to *m* from left to right.
Greg thinks that the pad's screen displays a cave if the following conditions hold:
- There is a segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), such that each of the rows *l*,<=*l*<=+<=1,<=...,<=*r* has exactly two black cells and all other rows have only white cells. - There is a row number *t* (*l*<=≤<=*t*<=≤<=*r*), such that for all pairs of rows with numbers *i* and *j* (*l*<=≤<=*i*<=≤<=*j*<=≤<=*t*) the set of columns between the black cells in row *i* (with the columns where is these black cells) is the subset of the set of columns between the black cells in row *j* (with the columns where is these black cells). Similarly, for all pairs of rows with numbers *i* and *j* (*t*<=≤<=*i*<=≤<=*j*<=≤<=*r*) the set of columns between the black cells in row *j* (with the columns where is these black cells) is the subset of the set of columns between the black cells in row *i* (with the columns where is these black cells).
Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.
Help Greg.
|
The first line contains two integers *n*, *m* — the pad's screen size (1<=≤<=*n*,<=*m*<=≤<=2000).
|
In the single line print the remainder after dividing the answer to the problem by 1000000007 (109<=+<=7).
|
[
"1 1\n",
"4 4\n",
"3 5\n"
] |
[
"0\n",
"485\n",
"451\n"
] |
none
| 2,000
|
[
{
"input": "1 1",
"output": "0"
},
{
"input": "4 4",
"output": "485"
},
{
"input": "3 5",
"output": "451"
},
{
"input": "5 3",
"output": "185"
},
{
"input": "5 5",
"output": "6751"
},
{
"input": "7 8",
"output": "5898445"
},
{
"input": "9 8",
"output": "72459477"
},
{
"input": "10 10",
"output": "33937168"
},
{
"input": "100 100",
"output": "631601096"
},
{
"input": "100 110",
"output": "257801865"
},
{
"input": "100 200",
"output": "852627600"
},
{
"input": "1 1000",
"output": "499500"
},
{
"input": "1000 1",
"output": "0"
},
{
"input": "1000 3",
"output": "1333331"
},
{
"input": "3 1000",
"output": "977762109"
},
{
"input": "10 1000",
"output": "298998986"
},
{
"input": "10 500",
"output": "659024105"
},
{
"input": "250 250",
"output": "331145635"
},
{
"input": "500 1000",
"output": "169229174"
},
{
"input": "1000 500",
"output": "900561408"
},
{
"input": "1000 1000",
"output": "950299696"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2000 2000",
"output": "627008355"
},
{
"input": "1500 2000",
"output": "294292096"
},
{
"input": "2000 1777",
"output": "20685302"
},
{
"input": "1999 1994",
"output": "785234759"
},
{
"input": "1994 1995",
"output": "854486105"
},
{
"input": "1 2000",
"output": "1999000"
},
{
"input": "2000 1",
"output": "0"
},
{
"input": "2 2000",
"output": "668322662"
},
{
"input": "2000 2",
"output": "2001000"
},
{
"input": "3 1999",
"output": "583178527"
},
{
"input": "1998 4",
"output": "542192517"
}
] | 1,693,827,611
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
print("_RANDOM_GUESS_1693827611.2417014")# 1693827611.241719
|
Title: Greg and Caves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has a pad. The pad's screen is an *n*<=×<=*m* rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to *n* from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to *m* from left to right.
Greg thinks that the pad's screen displays a cave if the following conditions hold:
- There is a segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), such that each of the rows *l*,<=*l*<=+<=1,<=...,<=*r* has exactly two black cells and all other rows have only white cells. - There is a row number *t* (*l*<=≤<=*t*<=≤<=*r*), such that for all pairs of rows with numbers *i* and *j* (*l*<=≤<=*i*<=≤<=*j*<=≤<=*t*) the set of columns between the black cells in row *i* (with the columns where is these black cells) is the subset of the set of columns between the black cells in row *j* (with the columns where is these black cells). Similarly, for all pairs of rows with numbers *i* and *j* (*t*<=≤<=*i*<=≤<=*j*<=≤<=*r*) the set of columns between the black cells in row *j* (with the columns where is these black cells) is the subset of the set of columns between the black cells in row *i* (with the columns where is these black cells).
Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.
Help Greg.
Input Specification:
The first line contains two integers *n*, *m* — the pad's screen size (1<=≤<=*n*,<=*m*<=≤<=2000).
Output Specification:
In the single line print the remainder after dividing the answer to the problem by 1000000007 (109<=+<=7).
Demo Input:
['1 1\n', '4 4\n', '3 5\n']
Demo Output:
['0\n', '485\n', '451\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1693827611.2417014")# 1693827611.241719
```
| 0
|
|
493
|
B
|
Vasya and Wrestling
|
PROGRAMMING
| 1,400
|
[
"implementation"
] | null | null |
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
|
The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105).
The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points.
The techniques are given in chronological order.
|
If the first wrestler wins, print string "first", otherwise print "second"
|
[
"5\n1\n2\n-3\n-4\n3\n",
"3\n-1\n-2\n3\n",
"2\n4\n-4\n"
] |
[
"second\n",
"first\n",
"second\n"
] |
Sequence *x* = *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y* = *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*| > |*y*| and *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">|*y*|</sub> = *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> > *y*<sub class="lower-index">*r* + 1</sub>.
We use notation |*a*| to denote length of sequence *a*.
| 1,000
|
[
{
"input": "5\n1\n2\n-3\n-4\n3",
"output": "second"
},
{
"input": "3\n-1\n-2\n3",
"output": "first"
},
{
"input": "2\n4\n-4",
"output": "second"
},
{
"input": "7\n1\n2\n-3\n4\n5\n-6\n7",
"output": "first"
},
{
"input": "14\n1\n2\n3\n4\n5\n6\n7\n-8\n-9\n-10\n-11\n-12\n-13\n-14",
"output": "second"
},
{
"input": "4\n16\n12\n19\n-98",
"output": "second"
},
{
"input": "5\n-6\n-1\n-1\n5\n3",
"output": "second"
},
{
"input": "11\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1",
"output": "first"
},
{
"input": "1\n-534365",
"output": "second"
},
{
"input": "1\n10253033",
"output": "first"
},
{
"input": "3\n-1\n-2\n3",
"output": "first"
},
{
"input": "8\n1\n-2\n-3\n4\n5\n-6\n-7\n8",
"output": "second"
},
{
"input": "2\n1\n-1",
"output": "second"
},
{
"input": "5\n1\n2\n3\n4\n5",
"output": "first"
},
{
"input": "5\n-1\n-2\n-3\n-4\n-5",
"output": "second"
},
{
"input": "10\n-1\n-2\n-3\n-4\n-5\n5\n4\n3\n2\n1",
"output": "first"
},
{
"input": "131\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n-1\n-1\n2",
"output": "first"
},
{
"input": "6\n-1\n-2\n-3\n1\n2\n3",
"output": "first"
},
{
"input": "3\n1000000000\n1000000000\n1000000000",
"output": "first"
},
{
"input": "12\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000",
"output": "first"
},
{
"input": "4\n1000000000\n1000000000\n1000000000\n-1000000000",
"output": "first"
},
{
"input": "20\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000",
"output": "first"
},
{
"input": "5\n1000000000\n1000000000\n-1000000000\n-1000000000\n-1000000000",
"output": "second"
},
{
"input": "4\n1\n-1000000000\n-1000000000\n-1000000000",
"output": "second"
},
{
"input": "5\n1000000000\n1000000000\n1000000000\n-1000000000\n-1000000000",
"output": "first"
},
{
"input": "4\n-1\n1000000000\n1000000000\n1000000000",
"output": "first"
},
{
"input": "11\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000",
"output": "first"
},
{
"input": "2\n-4\n4",
"output": "first"
},
{
"input": "3\n-12\n3\n9",
"output": "second"
},
{
"input": "3\n9\n1\n-10",
"output": "second"
},
{
"input": "3\n1\n2\n-3",
"output": "second"
},
{
"input": "4\n55\n5\n-5\n-55",
"output": "first"
},
{
"input": "4\n5\n-1\n1\n-5",
"output": "first"
},
{
"input": "2\n-5\n6",
"output": "first"
},
{
"input": "4\n5\n-4\n3\n-40",
"output": "second"
},
{
"input": "4\n1000000000\n1000000000\n1000000000\n-5",
"output": "first"
},
{
"input": "6\n3\n2\n1\n-3\n-1\n-2",
"output": "first"
},
{
"input": "5\n4\n1\n1\n-3\n-3",
"output": "first"
},
{
"input": "5\n208\n-52\n-52\n-52\n-52",
"output": "first"
},
{
"input": "3\n-100\n-200\n300",
"output": "first"
},
{
"input": "3\n400\n-200\n-200",
"output": "first"
},
{
"input": "3\n208\n-207\n-1",
"output": "first"
},
{
"input": "3\n98888887\n98888888\n-197777775",
"output": "second"
}
] | 1,572,255,114
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 57
| 389
| 9,728,000
|
import sys
inf = float("inf")
# sys.setrecursionlimit(10000000)
# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod, MOD = 1000000007, 998244353
# vow=['a','e','i','o','u']
# dx,dy=[-1,1,0,0],[0,0,1,-1]
# import random
# from collections import deque, Counter, OrderedDict,defaultdict
# from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# from math import ceil,floor,log,sqrt,factorial
# from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
def check_order(first, second):
len1 = len(first)
len2 = len(second)
i = j = 0
while i<len1 and j<len2:
if first[i]==second[j]:
i+=1
j+=1
elif first[i]>second[j]:
return 'first'
else:
return 'second'
return 'no'
n = int(input())
first = []
second = []
last = 0 # 1 if last turn was first else 2
for i in range(n):
x = int(input())
if x>0:
first.append(x)
last = 1
else:
second.append(abs(x))
last = 2
if sum(first)!=sum(second):
if sum(first)>sum(second):
print('first')
else:
print('second')
else:
ans = check_order(first,second)
if ans!='no':
print(ans)
else:
if last==1:
print('first')
else:
print('second')
|
Title: Vasya and Wrestling
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input Specification:
The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105).
The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points.
The techniques are given in chronological order.
Output Specification:
If the first wrestler wins, print string "first", otherwise print "second"
Demo Input:
['5\n1\n2\n-3\n-4\n3\n', '3\n-1\n-2\n3\n', '2\n4\n-4\n']
Demo Output:
['second\n', 'first\n', 'second\n']
Note:
Sequence *x* = *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y* = *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*| > |*y*| and *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">|*y*|</sub> = *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> > *y*<sub class="lower-index">*r* + 1</sub>.
We use notation |*a*| to denote length of sequence *a*.
|
```python
import sys
inf = float("inf")
# sys.setrecursionlimit(10000000)
# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod, MOD = 1000000007, 998244353
# vow=['a','e','i','o','u']
# dx,dy=[-1,1,0,0],[0,0,1,-1]
# import random
# from collections import deque, Counter, OrderedDict,defaultdict
# from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# from math import ceil,floor,log,sqrt,factorial
# from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
def check_order(first, second):
len1 = len(first)
len2 = len(second)
i = j = 0
while i<len1 and j<len2:
if first[i]==second[j]:
i+=1
j+=1
elif first[i]>second[j]:
return 'first'
else:
return 'second'
return 'no'
n = int(input())
first = []
second = []
last = 0 # 1 if last turn was first else 2
for i in range(n):
x = int(input())
if x>0:
first.append(x)
last = 1
else:
second.append(abs(x))
last = 2
if sum(first)!=sum(second):
if sum(first)>sum(second):
print('first')
else:
print('second')
else:
ans = check_order(first,second)
if ans!='no':
print(ans)
else:
if last==1:
print('first')
else:
print('second')
```
| 3
|
|
295
|
A
|
Greg and Array
|
PROGRAMMING
| 1,400
|
[
"data structures",
"implementation"
] | null | null |
Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.
|
The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces.
|
On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
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 of the %I64d specifier.
|
[
"3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n",
"1 1 1\n1\n1 1 1\n1 1\n",
"4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n"
] |
[
"9 18 17\n",
"2\n",
"5 18 31 20\n"
] |
none
| 500
|
[
{
"input": "3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3",
"output": "9 18 17"
},
{
"input": "1 1 1\n1\n1 1 1\n1 1",
"output": "2"
},
{
"input": "4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3",
"output": "5 18 31 20"
},
{
"input": "1 1 1\n0\n1 1 0\n1 1",
"output": "0"
}
] | 1,665,036,157
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,500
| 22,630,400
|
n, m, k = map(int, input().split())
operations = [None]
arr = list(map(int, input().split()))
arr.insert(0, 0)
for i in range(m):
operations.append(tuple(map(int, input().split())))
queries = []
for i in range(k):
queries.append(tuple(map(int, input().split())))
for i in range(1, m + 1):
apply_count = 0
for q in queries:
if i >= q[0] and i <= q[1]:
apply_count += 1
l, r, d = operations[i]
apply_count *= d
for j in range(l, r + 1):
arr[j] += apply_count
for x in arr[1:]:
print(str(x) + ' ', end='')
|
Title: Greg and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.
Input Specification:
The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces.
Output Specification:
On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
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 of the %I64d specifier.
Demo Input:
['3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n', '1 1 1\n1\n1 1 1\n1 1\n', '4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n']
Demo Output:
['9 18 17\n', '2\n', '5 18 31 20\n']
Note:
none
|
```python
n, m, k = map(int, input().split())
operations = [None]
arr = list(map(int, input().split()))
arr.insert(0, 0)
for i in range(m):
operations.append(tuple(map(int, input().split())))
queries = []
for i in range(k):
queries.append(tuple(map(int, input().split())))
for i in range(1, m + 1):
apply_count = 0
for q in queries:
if i >= q[0] and i <= q[1]:
apply_count += 1
l, r, d = operations[i]
apply_count *= d
for j in range(l, r + 1):
arr[j] += apply_count
for x in arr[1:]:
print(str(x) + ' ', end='')
```
| 0
|
|
988
|
B
|
Substrings Sort
|
PROGRAMMING
| 1,100
|
[
"sortings",
"strings"
] | null | null |
You are given $n$ strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.
String $a$ is a substring of string $b$ if it is possible to choose several consecutive letters in $b$ in such a way that they form $a$. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".
|
The first line contains an integer $n$ ($1 \le n \le 100$) — the number of strings.
The next $n$ lines contain the given strings. The number of letters in each string is from $1$ to $100$, inclusive. Each string consists of lowercase English letters.
Some strings might be equal.
|
If it is impossible to reorder $n$ given strings in required order, print "NO" (without quotes).
Otherwise print "YES" (without quotes) and $n$ given strings in required order.
|
[
"5\na\naba\nabacaba\nba\naba\n",
"5\na\nabacaba\nba\naba\nabab\n",
"3\nqwerty\nqwerty\nqwerty\n"
] |
[
"YES\na\nba\naba\naba\nabacaba\n",
"NO\n",
"YES\nqwerty\nqwerty\nqwerty\n"
] |
In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".
| 0
|
[
{
"input": "5\na\naba\nabacaba\nba\naba",
"output": "YES\na\nba\naba\naba\nabacaba"
},
{
"input": "5\na\nabacaba\nba\naba\nabab",
"output": "NO"
},
{
"input": "3\nqwerty\nqwerty\nqwerty",
"output": "YES\nqwerty\nqwerty\nqwerty"
},
{
"input": "1\nwronganswer",
"output": "YES\nwronganswer"
},
{
"input": "3\na\nb\nab",
"output": "NO"
},
{
"input": "2\nababaab\nabaab",
"output": "YES\nabaab\nababaab"
},
{
"input": "2\nq\nqq",
"output": "YES\nq\nqq"
},
{
"input": "5\nabab\nbab\nba\nab\na",
"output": "NO"
},
{
"input": "3\nb\nc\nd",
"output": "NO"
},
{
"input": "3\naba\nbab\nababa",
"output": "NO"
},
{
"input": "4\na\nba\nabacabac\nb",
"output": "NO"
},
{
"input": "4\nab\nba\nabab\na",
"output": "NO"
},
{
"input": "3\naaa\naab\naaab",
"output": "NO"
},
{
"input": "2\nac\nabac",
"output": "YES\nac\nabac"
},
{
"input": "2\na\nb",
"output": "NO"
},
{
"input": "3\nbaa\nbaaaaaaaab\naaaaaa",
"output": "NO"
},
{
"input": "3\naaab\naab\naaaab",
"output": "YES\naab\naaab\naaaab"
},
{
"input": "2\naaba\naba",
"output": "YES\naba\naaba"
},
{
"input": "10\na\nb\nc\nd\nab\nbc\ncd\nabc\nbcd\nabcd",
"output": "NO"
},
{
"input": "5\na\nab\nae\nabcd\nabcde",
"output": "NO"
},
{
"input": "3\nv\nab\nvab",
"output": "NO"
},
{
"input": "4\na\nb\nc\nabc",
"output": "NO"
},
{
"input": "2\nab\naab",
"output": "YES\nab\naab"
},
{
"input": "3\nabc\na\nc",
"output": "NO"
},
{
"input": "2\nabaab\nababaab",
"output": "YES\nabaab\nababaab"
},
{
"input": "3\ny\nxx\nxxy",
"output": "NO"
},
{
"input": "4\naaaa\naaaa\naaaa\nab",
"output": "NO"
},
{
"input": "3\nbad\naba\nabad",
"output": "NO"
},
{
"input": "3\nabcabc\nab\nbc",
"output": "NO"
},
{
"input": "2\naaaab\naaaaab",
"output": "YES\naaaab\naaaaab"
},
{
"input": "5\nab\naba\naba\naba\nabd",
"output": "NO"
},
{
"input": "4\nded\nd\ne\nd",
"output": "NO"
},
{
"input": "5\nekgetterherforme\ner\nter\nher\ntter",
"output": "NO"
},
{
"input": "3\naa\nbba\ncbba",
"output": "NO"
},
{
"input": "3\nab\naa\naab",
"output": "NO"
},
{
"input": "6\naaaa\naaaa\naaaa\naaaaab\nab\nab",
"output": "NO"
},
{
"input": "3\na\nbcd\nabcd",
"output": "NO"
},
{
"input": "3\naa\nab\naab",
"output": "NO"
},
{
"input": "3\nabcde\nab\ncde",
"output": "NO"
},
{
"input": "4\nanoop\np\nan\noop",
"output": "NO"
},
{
"input": "3\nab\ncd\nabcd",
"output": "NO"
},
{
"input": "3\nafaba\nafab\nfaba",
"output": "NO"
},
{
"input": "3\nababc\nabababc\nab",
"output": "YES\nab\nababc\nabababc"
},
{
"input": "3\na\nba\nbbab",
"output": "YES\na\nba\nbbab"
},
{
"input": "2\naabaa\naba",
"output": "YES\naba\naabaa"
},
{
"input": "3\nbc\nbca\na",
"output": "NO"
},
{
"input": "2\naba\nabba",
"output": "NO"
},
{
"input": "7\na\nb\nc\nab\nac\nac\nac",
"output": "NO"
},
{
"input": "2\naa\nmmmmmmmammmmmmmm",
"output": "NO"
}
] | 1,623,954,003
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 56
| 109
| 102,400
|
import sys
import os.path
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
n=int(input())
s= [input().strip() for i in range(n)]
s.sort(key=lambda x: len(x))
f=True
for i in range(1,n):
if s[i-1] not in s[i]:
f=False
break
if f: print("YES\n"+"\n".join(s))
else: print("NO")
|
Title: Substrings Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given $n$ strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.
String $a$ is a substring of string $b$ if it is possible to choose several consecutive letters in $b$ in such a way that they form $a$. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".
Input Specification:
The first line contains an integer $n$ ($1 \le n \le 100$) — the number of strings.
The next $n$ lines contain the given strings. The number of letters in each string is from $1$ to $100$, inclusive. Each string consists of lowercase English letters.
Some strings might be equal.
Output Specification:
If it is impossible to reorder $n$ given strings in required order, print "NO" (without quotes).
Otherwise print "YES" (without quotes) and $n$ given strings in required order.
Demo Input:
['5\na\naba\nabacaba\nba\naba\n', '5\na\nabacaba\nba\naba\nabab\n', '3\nqwerty\nqwerty\nqwerty\n']
Demo Output:
['YES\na\nba\naba\naba\nabacaba\n', 'NO\n', 'YES\nqwerty\nqwerty\nqwerty\n']
Note:
In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".
|
```python
import sys
import os.path
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
n=int(input())
s= [input().strip() for i in range(n)]
s.sort(key=lambda x: len(x))
f=True
for i in range(1,n):
if s[i-1] not in s[i]:
f=False
break
if f: print("YES\n"+"\n".join(s))
else: print("NO")
```
| 3
|
|
446
|
B
|
DZY Loves Modification
|
PROGRAMMING
| 2,000
|
[
"brute force",
"data structures",
"greedy"
] | null | null |
As we know, DZY loves playing games. One day DZY decided to play with a *n*<=×<=*m* matrix. To be more precise, he decided to modify the matrix with exactly *k* operations.
Each modification is one of the following:
1. Pick some row of the matrix and decrease each element of the row by *p*. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing. 1. Pick some column of the matrix and decrease each element of the column by *p*. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.
DZY wants to know: what is the largest total value of pleasure he could get after performing exactly *k* modifications? Please, help him to calculate this value.
|
The first line contains four space-separated integers *n*,<=*m*,<=*k* and *p* (1<=≤<=*n*,<=*m*<=≤<=103; 1<=≤<=*k*<=≤<=106; 1<=≤<=*p*<=≤<=100).
Then *n* lines follow. Each of them contains *m* integers representing *a**ij* (1<=≤<=*a**ij*<=≤<=103) — the elements of the current row of the matrix.
|
Output a single integer — the maximum possible total pleasure value DZY could get.
|
[
"2 2 2 2\n1 3\n2 4\n",
"2 2 5 2\n1 3\n2 4\n"
] |
[
"11\n",
"11\n"
] |
For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:
For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:
| 1,500
|
[
{
"input": "2 2 2 2\n1 3\n2 4",
"output": "11"
},
{
"input": "2 2 5 2\n1 3\n2 4",
"output": "11"
},
{
"input": "5 5 20 100\n464 757 53 708 262\n753 769 189 38 796\n394 60 381 384 935\n882 877 501 615 464\n433 798 504 301 301",
"output": "38013"
},
{
"input": "10 10 50 80\n529 349 889 455 946 983 482 179 590 907\n436 940 407 631 26 963 181 789 461 437\n367 505 888 521 449 741 900 994 342 847\n605 374 112 829 212 184 295 416 181 435\n438 647 676 144 436 629 660 282 545 470\n392 560 351 654 102 315 588 578 696 592\n837 115 349 760 324 680 584 822 909 75\n575 89 566 888 622 443 605 169 557 304\n326 797 568 251 181 466 261 158 769 192\n395 105 34 640 537 629 13 317 645 409",
"output": "175135"
}
] | 1,673,663,078
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 2
| 77
| 11,980,800
|
# O(k + (p + max(a) + log((max(n, m)))) * max(n, m)) instead of O(k*log(max(n, m))) :)
#
# Main idea:
# We can do all row operations before column operations. Furthermore, every column operation will be
# decreased by i*p from the value if no row operations were performed where i is the # of row operations.
# Thus, this implies the optimal row and column operations are independent. Thus, we can find the first k optimal
# row and column operations and then fix the row and column operations and find the decreased cost = i*(k-i)*p
#
# Time Complexity Optimization:
# The idea is to maintain a list count[i] giving the count of rows with sum i
# Then, we can just move a pointer backward to find the next count[i] >= 1 and this is the
# next best row to operate on. However, count[i] can be max(a)*max(n, m)*p in size.
# However, notice that if the new sum of the row after operating is <= the minimum,
# the optimal rows selected to be operated on will cycle :) (kind of like Project Euler 822).
# Thus, we can remove a factor of p and instead add a term of p* max(n, m)
import sys
input = sys.stdin.buffer.readline
n, m, k, p = map(int, input().split())
a = [list(map(int, input().split())) for i in range(n)]
def compute_best(b, n):
mn = min(b)
count = [0]*(10**6 + 10**5) # Extra 10**5 so values will "loop" back when we use negative indexing
for i in b:
count[i] += 1
count_i = 10**6
curr_op = 1
best = [0]*(k+1)
# Perform operations until cycle starts using count[] to maintain pleasure values
while curr_op <= k:
while not count[count_i]:
count_i -= 1
best[curr_op] = count_i + best[curr_op-1]
curr_op += 1
count[count_i] -= 1
new = count_i-p*n
count[new] += 1
if new <= mn:
# cycle starts :)
break
# Get the values of how much pleasure would be gained from each row operation
for i in range(n):
while not count[count_i]:
count_i -= 1
b[i] = count_i
count[count_i] -= 1
# Get pleasure values utilizing cycle instead of count[]
left = k - curr_op + 1
for i in range(left):
best[curr_op + i] = best[curr_op + i - 1] + b[i%n] - (i//n)*p*n
return best
best_rows = compute_best([sum(a[i]) for i in range(n)], m)
best_cols = compute_best([sum([a[i][j] for i in range(n)]) for j in range(m)], n)
ans = -1e18
for i in range(k+1):
ans = max(ans, best_rows[i] + best_cols[k-i] - i*p*(k-i))
print(ans)
|
Title: DZY Loves Modification
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As we know, DZY loves playing games. One day DZY decided to play with a *n*<=×<=*m* matrix. To be more precise, he decided to modify the matrix with exactly *k* operations.
Each modification is one of the following:
1. Pick some row of the matrix and decrease each element of the row by *p*. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing. 1. Pick some column of the matrix and decrease each element of the column by *p*. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.
DZY wants to know: what is the largest total value of pleasure he could get after performing exactly *k* modifications? Please, help him to calculate this value.
Input Specification:
The first line contains four space-separated integers *n*,<=*m*,<=*k* and *p* (1<=≤<=*n*,<=*m*<=≤<=103; 1<=≤<=*k*<=≤<=106; 1<=≤<=*p*<=≤<=100).
Then *n* lines follow. Each of them contains *m* integers representing *a**ij* (1<=≤<=*a**ij*<=≤<=103) — the elements of the current row of the matrix.
Output Specification:
Output a single integer — the maximum possible total pleasure value DZY could get.
Demo Input:
['2 2 2 2\n1 3\n2 4\n', '2 2 5 2\n1 3\n2 4\n']
Demo Output:
['11\n', '11\n']
Note:
For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:
For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:
|
```python
# O(k + (p + max(a) + log((max(n, m)))) * max(n, m)) instead of O(k*log(max(n, m))) :)
#
# Main idea:
# We can do all row operations before column operations. Furthermore, every column operation will be
# decreased by i*p from the value if no row operations were performed where i is the # of row operations.
# Thus, this implies the optimal row and column operations are independent. Thus, we can find the first k optimal
# row and column operations and then fix the row and column operations and find the decreased cost = i*(k-i)*p
#
# Time Complexity Optimization:
# The idea is to maintain a list count[i] giving the count of rows with sum i
# Then, we can just move a pointer backward to find the next count[i] >= 1 and this is the
# next best row to operate on. However, count[i] can be max(a)*max(n, m)*p in size.
# However, notice that if the new sum of the row after operating is <= the minimum,
# the optimal rows selected to be operated on will cycle :) (kind of like Project Euler 822).
# Thus, we can remove a factor of p and instead add a term of p* max(n, m)
import sys
input = sys.stdin.buffer.readline
n, m, k, p = map(int, input().split())
a = [list(map(int, input().split())) for i in range(n)]
def compute_best(b, n):
mn = min(b)
count = [0]*(10**6 + 10**5) # Extra 10**5 so values will "loop" back when we use negative indexing
for i in b:
count[i] += 1
count_i = 10**6
curr_op = 1
best = [0]*(k+1)
# Perform operations until cycle starts using count[] to maintain pleasure values
while curr_op <= k:
while not count[count_i]:
count_i -= 1
best[curr_op] = count_i + best[curr_op-1]
curr_op += 1
count[count_i] -= 1
new = count_i-p*n
count[new] += 1
if new <= mn:
# cycle starts :)
break
# Get the values of how much pleasure would be gained from each row operation
for i in range(n):
while not count[count_i]:
count_i -= 1
b[i] = count_i
count[count_i] -= 1
# Get pleasure values utilizing cycle instead of count[]
left = k - curr_op + 1
for i in range(left):
best[curr_op + i] = best[curr_op + i - 1] + b[i%n] - (i//n)*p*n
return best
best_rows = compute_best([sum(a[i]) for i in range(n)], m)
best_cols = compute_best([sum([a[i][j] for i in range(n)]) for j in range(m)], n)
ans = -1e18
for i in range(k+1):
ans = max(ans, best_rows[i] + best_cols[k-i] - i*p*(k-i))
print(ans)
```
| -1
|
|
574
|
A
|
Bear and Elections
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
|
The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
|
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
|
[
"5\n5 1 11 2 8\n",
"4\n1 8 8 8\n",
"2\n7 6\n"
] |
[
"4\n",
"6\n",
"0\n"
] |
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.
In the third sample Limak is a winner without bribing any citizen.
| 500
|
[
{
"input": "5\n5 1 11 2 8",
"output": "4"
},
{
"input": "4\n1 8 8 8",
"output": "6"
},
{
"input": "2\n7 6",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "10\n100 200 57 99 1 1000 200 200 200 500",
"output": "451"
},
{
"input": "16\n7 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "932"
},
{
"input": "100\n47 64 68 61 68 66 69 61 69 65 69 63 62 60 68 65 64 65 65 62 63 68 60 70 63 63 65 67 70 69 68 69 61 65 63 60 60 65 61 60 70 66 66 65 62 60 65 68 61 62 67 64 66 65 67 68 60 69 70 63 65 62 64 65 67 67 69 68 66 69 70 67 65 70 60 66 70 67 67 64 69 69 66 68 60 64 62 62 68 69 67 69 60 70 69 68 62 63 68 66",
"output": "23"
},
{
"input": "2\n96 97",
"output": "1"
},
{
"input": "2\n1000 1000",
"output": "1"
},
{
"input": "3\n999 1000 1000",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "7\n10 940 926 990 946 980 985",
"output": "817"
},
{
"input": "10\n5 3 4 5 5 2 1 8 4 1",
"output": "2"
},
{
"input": "15\n17 15 17 16 13 17 13 16 14 14 17 17 13 15 17",
"output": "1"
},
{
"input": "20\n90 5 62 9 50 7 14 43 44 44 56 13 71 22 43 35 52 60 73 54",
"output": "0"
},
{
"input": "30\n27 85 49 7 77 38 4 68 23 28 81 100 40 9 78 38 1 60 60 49 98 44 45 92 46 39 98 24 37 39",
"output": "58"
},
{
"input": "51\n90 47 100 12 21 96 2 68 84 60 2 9 33 8 45 13 59 50 100 93 22 97 4 81 51 2 3 78 19 16 25 63 52 34 79 32 34 87 7 42 96 93 30 33 33 43 69 8 63 58 57",
"output": "8"
},
{
"input": "77\n1000 2 2 3 1 1 1 3 3 2 1 1 3 2 2 2 3 2 3 1 3 1 1 2 2 2 3 1 1 2 2 2 3 2 1 3 3 1 2 3 3 3 2 1 3 2 1 3 3 2 3 3 2 1 3 1 1 1 2 3 2 3 1 3 1 2 1 2 2 2 1 2 2 3 2 2 2",
"output": "0"
},
{
"input": "91\n3 92 89 83 85 80 91 94 95 82 92 95 80 88 90 85 81 90 87 86 94 88 90 87 88 82 95 84 84 93 83 95 91 85 89 88 88 85 87 90 93 80 89 95 94 92 93 86 83 82 86 84 91 80 90 95 84 86 84 85 84 92 82 84 83 91 87 95 94 95 90 95 86 92 86 80 95 86 88 80 82 87 84 83 91 93 81 81 91 89 88",
"output": "89"
},
{
"input": "100\n1 3 71 47 64 82 58 61 61 35 52 36 57 62 63 54 52 21 78 100 24 94 4 80 99 62 43 72 21 70 90 4 23 14 72 4 76 49 71 96 96 99 78 7 32 11 14 61 19 69 1 68 100 77 86 54 14 86 47 53 30 88 67 66 61 70 17 63 40 5 99 53 38 31 91 18 41 5 77 61 53 30 87 21 23 54 52 17 23 75 58 99 99 63 20 1 78 72 28 11",
"output": "90"
},
{
"input": "100\n1 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99"
},
{
"input": "94\n3 100 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 100 99 100 99 100 100 100 99 100 99 100 99 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 99 99 100 99 100 99 99 99 100 100 99 100 100 99 99 100 100 100 99 100 99 99 99 99 99 100 100 100 100 100 100 100 100 100 99 99 99 99 100 99 100 99 100 100",
"output": "97"
},
{
"input": "97\n99 99 98 98 100 98 99 99 98 100 100 100 99 99 100 99 99 98 99 99 98 98 98 100 100 99 98 99 100 98 99 98 98 100 98 99 100 98 98 99 98 98 99 98 100 99 99 99 99 98 98 98 100 99 100 100 99 99 100 99 99 98 98 98 100 100 98 100 100 99 98 99 100 98 98 98 98 99 99 98 98 99 100 100 98 98 99 98 99 100 98 99 100 98 99 99 100",
"output": "2"
},
{
"input": "100\n100 55 70 81 73 51 6 75 45 85 33 61 98 63 11 59 1 8 14 28 78 74 44 80 7 69 7 5 90 73 43 78 64 64 43 92 59 70 80 19 33 39 31 70 38 85 24 23 86 79 98 56 92 63 92 4 36 8 79 74 2 81 54 13 69 44 49 63 17 76 78 99 42 36 47 71 19 90 9 58 83 53 27 2 35 51 65 59 90 51 74 87 84 48 98 44 84 100 84 93",
"output": "1"
},
{
"input": "100\n100 637 498 246 615 901 724 673 793 33 282 908 477 185 185 969 34 859 90 70 107 492 227 918 919 131 620 182 802 703 779 184 403 891 448 499 628 553 905 392 70 396 8 575 66 908 992 496 792 174 667 355 836 610 855 377 244 827 836 808 667 354 800 114 746 556 75 894 162 367 99 718 394 273 833 776 151 433 315 470 759 12 552 613 85 793 775 649 225 86 296 624 557 201 209 595 697 527 282 168",
"output": "749"
},
{
"input": "100\n107 172 549 883 564 56 399 970 173 990 224 217 601 381 948 631 159 958 512 136 61 584 633 202 652 355 26 723 663 237 410 721 688 552 699 24 748 186 461 88 34 243 872 205 471 298 654 693 244 33 359 533 471 116 386 653 654 887 531 303 335 829 319 340 827 89 602 191 422 289 361 200 593 421 592 402 256 813 606 589 741 9 148 893 3 142 50 169 219 360 642 45 810 818 507 624 561 743 303 111",
"output": "729"
},
{
"input": "90\n670 694 651 729 579 539 568 551 707 638 604 544 502 531 775 805 558 655 506 729 802 778 653 737 591 770 594 535 588 604 658 713 779 705 504 563 513 651 529 572 505 553 515 750 621 574 727 774 714 725 665 798 670 747 751 635 755 798 635 717 583 682 517 546 740 802 743 507 658 700 645 671 533 594 506 633 768 584 672 666 703 522 530 501 592 528 678 708 619 786",
"output": "111"
},
{
"input": "90\n10 265 429 431 343 305 806 746 284 313 503 221 594 351 83 653 232 431 427 610 458 88 255 215 529 205 492 549 55 694 535 104 45 327 816 432 595 549 454 141 216 557 250 415 531 494 190 749 718 380 78 447 784 347 196 814 16 780 262 462 776 315 160 307 593 694 692 41 528 725 376 777 337 44 438 630 345 502 384 184 742 429 570 361 394 267 820 778 662 377",
"output": "714"
},
{
"input": "95\n800 280 176 472 587 763 588 838 760 378 667 231 566 278 713 305 354 815 140 220 188 409 109 180 251 268 474 590 853 143 235 691 313 785 386 92 783 471 43 342 718 592 678 404 256 362 239 504 163 85 521 81 356 73 754 589 380 159 196 862 838 509 149 42 366 630 467 292 698 123 187 796 576 37 689 800 186 518 488 432 159 860 349 799 282 304 880 283 23 312 55 507 734 370 490",
"output": "52"
},
{
"input": "100\n95 88 84 85 74 97 100 52 91 94 62 66 90 56 86 66 95 73 79 68 54 67 99 52 82 62 81 71 93 85 72 72 95 52 72 63 57 90 92 89 88 77 84 78 95 59 72 86 98 64 89 64 80 70 54 93 88 86 79 78 94 64 89 66 50 90 54 82 52 96 99 54 81 66 83 79 69 80 51 73 81 69 93 82 76 52 58 87 93 92 52 67 78 63 63 87 77 95 58 78",
"output": "4"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "100\n999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "2"
},
{
"input": "100\n901 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "99"
},
{
"input": "4\n8 5 11 3",
"output": "2"
},
{
"input": "3\n3 10 1",
"output": "4"
}
] | 1,597,760,801
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 35
| 140
| 1,228,800
|
n = int(input())
a = (list(map(int,input().split())))
k = a[0]
k2 = a[0]
del a[0]
cnt = 0
while k <= max(a):
a.append(max(a) - 1)
a.remove(max(a))
k +=1
cnt +=1
#print(k - k2)
print(cnt)
|
Title: Bear and Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
Output Specification:
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
Demo Input:
['5\n5 1 11 2 8\n', '4\n1 8 8 8\n', '2\n7 6\n']
Demo Output:
['4\n', '6\n', '0\n']
Note:
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.
In the third sample Limak is a winner without bribing any citizen.
|
```python
n = int(input())
a = (list(map(int,input().split())))
k = a[0]
k2 = a[0]
del a[0]
cnt = 0
while k <= max(a):
a.append(max(a) - 1)
a.remove(max(a))
k +=1
cnt +=1
#print(k - k2)
print(cnt)
```
| 3
|
|
483
|
A
|
Counterexample
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math",
"number theory"
] | null | null |
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
|
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
|
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
|
[
"2 4\n",
"10 11\n",
"900000000000000009 900000000000000029\n"
] |
[
"2 3 4\n",
"-1\n",
"900000000000000009 900000000000000010 900000000000000021\n"
] |
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
| 500
|
[
{
"input": "2 4",
"output": "2 3 4"
},
{
"input": "10 11",
"output": "-1"
},
{
"input": "900000000000000009 900000000000000029",
"output": "900000000000000009 900000000000000010 900000000000000021"
},
{
"input": "640097987171091791 640097987171091835",
"output": "640097987171091792 640097987171091793 640097987171091794"
},
{
"input": "19534350415104721 19534350415104725",
"output": "19534350415104722 19534350415104723 19534350415104724"
},
{
"input": "933700505788726243 933700505788726280",
"output": "933700505788726244 933700505788726245 933700505788726246"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "2 3 4"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "266540997167959130 266540997167959164",
"output": "266540997167959130 266540997167959131 266540997167959132"
},
{
"input": "267367244641009850 267367244641009899",
"output": "267367244641009850 267367244641009851 267367244641009852"
},
{
"input": "268193483524125978 268193483524125993",
"output": "268193483524125978 268193483524125979 268193483524125980"
},
{
"input": "269019726702209402 269019726702209432",
"output": "269019726702209402 269019726702209403 269019726702209404"
},
{
"input": "269845965585325530 269845965585325576",
"output": "269845965585325530 269845965585325531 269845965585325532"
},
{
"input": "270672213058376250 270672213058376260",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492378",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608523",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691962",
"output": "273150934002691930 273150934002691931 273150934002691932"
},
{
"input": "996517375802030516 996517375802030524",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146694",
"output": "997343614685146644 997343614685146645 997343614685146646"
},
{
"input": "998169857863230068 998169857863230083",
"output": "998169857863230068 998169857863230069 998169857863230070"
},
{
"input": "998996101041313492 998996101041313522",
"output": "998996101041313492 998996101041313493 998996101041313494"
},
{
"input": "999822344219396916 999822344219396961",
"output": "999822344219396916 999822344219396917 999822344219396918"
},
{
"input": "648583102513043 648583102513053",
"output": "648583102513044 648583102513045 648583102513046"
},
{
"input": "266540997167959130 266540997167959131",
"output": "-1"
},
{
"input": "267367244641009850 267367244641009850",
"output": "-1"
},
{
"input": "268193483524125978 268193483524125979",
"output": "-1"
},
{
"input": "269019726702209402 269019726702209402",
"output": "-1"
},
{
"input": "269845965585325530 269845965585325530",
"output": "-1"
},
{
"input": "270672213058376250 270672213058376254",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492379",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608508",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691931",
"output": "-1"
},
{
"input": "996517375802030516 996517375802030518",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146644",
"output": "-1"
},
{
"input": "2147483647 2147483649",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "1 7",
"output": "2 3 4"
},
{
"input": "9 12",
"output": "9 11 12"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "11 13",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
}
] | 1,646,013,603
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
import math
def counter_example():
a,b=list(map(int,input().split()))
if abs(b-a)==1:print(-1)
else:
for i in range(a,b+1):
if math.gcd(i,i+1)==1 and math.gcd(i+1,i+2)==1 and math.gcd(i,i+2)!=1:print(i,i+1,i+2);break
counter_example()
|
Title: Counterexample
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
Input Specification:
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
Output Specification:
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
Demo Input:
['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n']
Demo Output:
['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n']
Note:
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
|
```python
import math
def counter_example():
a,b=list(map(int,input().split()))
if abs(b-a)==1:print(-1)
else:
for i in range(a,b+1):
if math.gcd(i,i+1)==1 and math.gcd(i+1,i+2)==1 and math.gcd(i,i+2)!=1:print(i,i+1,i+2);break
counter_example()
```
| 0
|
|
133
|
A
|
HQ9+
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
|
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
|
Output "YES", if executing the program will produce any output, and "NO" otherwise.
|
[
"Hi!\n",
"Codeforces\n"
] |
[
"YES\n",
"NO\n"
] |
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
| 500
|
[
{
"input": "Hi!",
"output": "YES"
},
{
"input": "Codeforces",
"output": "NO"
},
{
"input": "a+b=c",
"output": "NO"
},
{
"input": "hq-lowercase",
"output": "NO"
},
{
"input": "Q",
"output": "YES"
},
{
"input": "9",
"output": "YES"
},
{
"input": "H",
"output": "YES"
},
{
"input": "+",
"output": "NO"
},
{
"input": "~",
"output": "NO"
},
{
"input": "dEHsbM'gS[\\brZ_dpjXw8f?L[4E\"s4Zc9*(,j:>p$}m7HD[_9nOWQ\\uvq2mHWR",
"output": "YES"
},
{
"input": "tt6l=RHOfStm.;Qd$-}zDes*E,.F7qn5-b%HC",
"output": "YES"
},
{
"input": "@F%K2=%RyL/",
"output": "NO"
},
{
"input": "juq)k(FT.^G=G\\zcqnO\"uJIE1_]KFH9S=1c\"mJ;F9F)%>&.WOdp09+k`Yc6}\"6xw,Aos:M\\_^^:xBb[CcsHm?J",
"output": "YES"
},
{
"input": "6G_\"Fq#<AWyHG=Rci1t%#Jc#x<Fpg'N@t%F=``YO7\\Zd;6PkMe<#91YgzTC)",
"output": "YES"
},
{
"input": "Fvg_~wC>SO4lF}*c`Q;mII9E{4.QodbqN]C",
"output": "YES"
},
{
"input": "p-UXsbd&f",
"output": "NO"
},
{
"input": "<]D7NMA)yZe=`?RbP5lsa.l_Mg^V:\"-0x+$3c,q&L%18Ku<HcA\\s!^OQblk^x{35S'>yz8cKgVHWZ]kV0>_",
"output": "YES"
},
{
"input": "f.20)8b+.R}Gy!DbHU3v(.(=Q^`z[_BaQ}eO=C1IK;b2GkD\\{\\Bf\"!#qh]",
"output": "YES"
},
{
"input": "}do5RU<(w<q[\"-NR)IAH_HyiD{",
"output": "YES"
},
{
"input": "Iy^.,Aw*,5+f;l@Q;jLK'G5H-r1Pfmx?ei~`CjMmUe{K:lS9cu4ay8rqRh-W?Gqv!e-j*U)!Mzn{E8B6%~aSZ~iQ_QwlC9_cX(o8",
"output": "YES"
},
{
"input": "sKLje,:q>-D,;NvQ3,qN3-N&tPx0nL/,>Ca|z\"k2S{NF7btLa3_TyXG4XZ:`(t&\"'^M|@qObZxv",
"output": "YES"
},
{
"input": "%z:c@1ZsQ@\\6U/NQ+M9R>,$bwG`U1+C\\18^:S},;kw!&4r|z`",
"output": "YES"
},
{
"input": "OKBB5z7ud81[Tn@P\"nDUd,>@",
"output": "NO"
},
{
"input": "y{0;neX]w0IenPvPx0iXp+X|IzLZZaRzBJ>q~LhMhD$x-^GDwl;,a'<bAqH8QrFwbK@oi?I'W.bZ]MlIQ/x(0YzbTH^l.)]0Bv",
"output": "YES"
},
{
"input": "EL|xIP5_+Caon1hPpQ0[8+r@LX4;b?gMy>;/WH)pf@Ur*TiXu*e}b-*%acUA~A?>MDz#!\\Uh",
"output": "YES"
},
{
"input": "UbkW=UVb>;z6)p@Phr;^Dn.|5O{_i||:Rv|KJ_ay~V(S&Jp",
"output": "NO"
},
{
"input": "!3YPv@2JQ44@)R2O_4`GO",
"output": "YES"
},
{
"input": "Kba/Q,SL~FMd)3hOWU'Jum{9\"$Ld4:GW}D]%tr@G{hpG:PV5-c'VIZ~m/6|3I?_4*1luKnOp`%p|0H{[|Y1A~4-ZdX,Rw2[\\",
"output": "YES"
},
{
"input": "NRN*=v>;oU7[acMIJn*n^bWm!cm3#E7Efr>{g-8bl\"DN4~_=f?[T;~Fq#&)aXq%</GcTJD^e$@Extm[e\"C)q_L",
"output": "NO"
},
{
"input": "y#<fv{_=$MP!{D%I\\1OqjaqKh[pqE$KvYL<9@*V'j8uH0/gQdA'G;&y4Cv6&",
"output": "YES"
},
{
"input": "+SE_Pg<?7Fh,z&uITQut2a-mk8X8La`c2A}",
"output": "YES"
},
{
"input": "Uh3>ER](J",
"output": "NO"
},
{
"input": "!:!{~=9*\\P;Z6F?HC5GadFz)>k*=u|+\"Cm]ICTmB!`L{&oS/z6b~#Snbp/^\\Q>XWU-vY+/dP.7S=-#&whS@,",
"output": "YES"
},
{
"input": "KimtYBZp+ISeO(uH;UldoE6eAcp|9u?SzGZd6j-e}[}u#e[Cx8.qgY]$2!",
"output": "YES"
},
{
"input": "[:[SN-{r>[l+OggH3v3g{EPC*@YBATT@",
"output": "YES"
},
{
"input": "'jdL(vX",
"output": "NO"
},
{
"input": "Q;R+aay]cL?Zh*uG\"YcmO*@Dts*Gjp}D~M7Z96+<4?9I3aH~0qNdO(RmyRy=ci,s8qD_kwj;QHFzD|5,5",
"output": "YES"
},
{
"input": "{Q@#<LU_v^qdh%gGxz*pu)Y\"]k-l-N30WAxvp2IE3:jD0Wi4H/xWPH&s",
"output": "YES"
},
{
"input": "~@Gb(S&N$mBuBUMAky-z^{5VwLNTzYg|ZUZncL@ahS?K*As<$iNUARM3r43J'jJB)$ujfPAq\"G<S9flGyakZg!2Z.-NJ|2{F>]",
"output": "YES"
},
{
"input": "Jp5Aa>aP6fZ!\\6%A}<S}j{O4`C6y$8|i3IW,WHy&\"ioE&7zP\"'xHAY;:x%@SnS]Mr{R|})gU",
"output": "YES"
},
{
"input": "ZA#:U)$RI^sE\\vuAt]x\"2zipI!}YEu2<j$:H0_9/~eB?#->",
"output": "YES"
},
{
"input": "&ppw0._:\\p-PuWM@l}%%=",
"output": "NO"
},
{
"input": "P(^pix\"=oiEZu8?@d@J(I`Xp5TN^T3\\Z7P5\"ZrvZ{2Fwz3g-8`U!)(1$a<g+9Q|COhDoH;HwFY02Pa|ZGp$/WZBR=>6Jg!yr",
"output": "YES"
},
{
"input": "`WfODc\\?#ax~1xu@[ao+o_rN|L7%v,p,nDv>3+6cy.]q3)+A6b!q*Hc+#.t4f~vhUa~$^q",
"output": "YES"
},
{
"input": ",)TH9N}'6t2+0Yg?S#6/{_.,!)9d}h'wG|sY&'Ul4D0l0",
"output": "YES"
},
{
"input": "VXB&r9Z)IlKOJ:??KDA",
"output": "YES"
},
{
"input": "\")1cL>{o\\dcYJzu?CefyN^bGRviOH&P7rJS3PT4:0V3F)%\\}L=AJouYsj_>j2|7^1NWu*%NbOP>ngv-ls<;b-4Sd3Na0R",
"output": "YES"
},
{
"input": "2Y}\\A)>row{~c[g>:'.|ZC8%UTQ/jcdhK%6O)QRC.kd@%y}LJYk=V{G5pQK/yKJ%{G3C",
"output": "YES"
},
{
"input": "O.&=qt(`z(",
"output": "NO"
},
{
"input": "_^r6fyIc/~~;>l%9?aVEi7-{=,[<aMiB'-scSg$$|\"jAzY0N>QkHHGBZj2c\"=fhRlWd5;5K|GgU?7h]!;wl@",
"output": "YES"
},
{
"input": "+/`sAd&eB29E=Nu87${.u6GY@$^a$,}s^!p!F}B-z8<<wORb<S7;HM1a,gp",
"output": "YES"
},
{
"input": "U_ilyOGMT+QiW/M8/D(1=6a7)_FA,h4`8",
"output": "YES"
},
{
"input": "!0WKT:$O",
"output": "NO"
},
{
"input": "1EE*I%EQz6$~pPu7|(r7nyPQt4uGU@]~H'4uII?b1_Wn)K?ZRHrr0z&Kr;}aO3<mN=3:{}QgPxI|Ncm4#)",
"output": "YES"
},
{
"input": "[u3\"$+!:/.<Dp1M7tH}:zxjt],^kv}qP;y12\"`^'/u*h%AFmPJ>e1#Yly",
"output": "YES"
},
{
"input": "'F!_]tB<A&UO+p?7liE>(x&RFgG2~\\(",
"output": "NO"
},
{
"input": "Qv)X8",
"output": "YES"
},
{
"input": "aGv7,J@&g1(}E3g6[LuDZwZl2<v7IwQA%\"R(?ouBD>_=y\"3Kf%^>vON<a^T\\G^ootgE@whWmZo=[ex|F",
"output": "YES"
},
{
"input": "e{}2vQ+/r@p0}cLKNe4MCk",
"output": "YES"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnkH",
"output": "YES"
},
{
"input": "Qzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnky",
"output": "YES"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwb9uvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnky",
"output": "YES"
},
{
"input": "1H1",
"output": "YES"
},
{
"input": "+Q",
"output": "YES"
},
{
"input": "1ab",
"output": "NO"
},
{
"input": "!",
"output": "NO"
},
{
"input": "0+",
"output": "NO"
},
{
"input": "+H",
"output": "YES"
},
{
"input": "cH",
"output": "YES"
},
{
"input": "+8",
"output": "NO"
},
{
"input": "8+",
"output": "NO"
},
{
"input": "++++++++++++++++++++++++++",
"output": "NO"
},
{
"input": "(+)",
"output": "NO"
},
{
"input": "H+",
"output": "YES"
},
{
"input": "a!",
"output": "NO"
},
{
"input": "++++++++++++++",
"output": "NO"
},
{
"input": "+++++++++++++++++++++++++++++++++++++++++++++++++++++++++",
"output": "NO"
},
{
"input": "8",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "HQ9",
"output": "YES"
},
{
"input": "+++H",
"output": "YES"
},
{
"input": "++++++++++",
"output": "NO"
},
{
"input": "HHHHHHH",
"output": "YES"
},
{
"input": "abacabaH",
"output": "YES"
},
{
"input": "+G",
"output": "NO"
}
] | 1,695,657,175
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
s=input()
def ismatch(s):
map={'H':1,'Q':1,'9':1}
for i in s:
if i in map:
return 'Yes'
return 'No'
print(ismatch(s))
|
Title: HQ9+
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input Specification:
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output Specification:
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Demo Input:
['Hi!\n', 'Codeforces\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
|
```python
s=input()
def ismatch(s):
map={'H':1,'Q':1,'9':1}
for i in s:
if i in map:
return 'Yes'
return 'No'
print(ismatch(s))
```
| 0
|
|
404
|
A
|
Valera and X
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
|
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
|
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
|
[
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nckcccbj\nccbcbcc\ncccbccc\nccbcbcc\ncbcccbc\nbccccdt",
"output": "NO"
},
{
"input": "13\nsooooooooooos\nosoooooooooso\noosooooooosoo\nooosooooosooo\noooosooosoooo\nooooososooooo\noooooosoooooo\nooooososooooo\noooosooosoooo\nooosooooosooo\noosooooooosoo\nosoooooooooso\nsooooooooooos",
"output": "YES"
},
{
"input": "3\naaa\naaa\naaa",
"output": "NO"
},
{
"input": "3\naca\noec\nzba",
"output": "NO"
},
{
"input": "15\nrxeeeeeeeeeeeer\nereeeeeeeeeeere\needeeeeeeeeeoee\neeereeeeeeeewee\neeeereeeeebeeee\nqeeeereeejedyee\neeeeeerereeeeee\neeeeeeereeeeeee\neeeeeerereeeeze\neeeeereeereeeee\neeeereeeeegeeee\neeereeeeeeereee\neereeeeeeqeeved\ncreeeeeeceeeere\nreeerneeeeeeeer",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxxxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxoxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxxxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxoox\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxaxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\noxoxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "3\nxxx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxx\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxxx",
"output": "NO"
},
{
"input": "3\nxax\nxxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naaa\nxax",
"output": "NO"
},
{
"input": "3\naax\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxaa\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\naax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaa",
"output": "NO"
},
{
"input": "3\nxfx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\nafa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaf",
"output": "NO"
},
{
"input": "3\nxox\nxxx\nxxx",
"output": "NO"
},
{
"input": "3\naxa\naax\nxxa",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxox",
"output": "NO"
},
{
"input": "3\nxox\nooo\nxox",
"output": "NO"
},
{
"input": "3\naaa\naab\nbbb",
"output": "NO"
},
{
"input": "3\nxxx\nsxs\nxsx",
"output": "NO"
},
{
"input": "5\nabbba\nbabab\nbbbbb\nbaaab\nabbba",
"output": "NO"
},
{
"input": "5\nabaaa\nbbbbb\nbbabb\nbabab\nabbba",
"output": "NO"
},
{
"input": "5\nxoxox\noxoxo\nooxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxxx",
"output": "NO"
},
{
"input": "5\nxoooo\noxooo\nooxoo\noooxo\noooox",
"output": "NO"
},
{
"input": "5\nxoooo\noxoxx\nooxoo\noxoxo\noxoox",
"output": "NO"
},
{
"input": "3\naaa\nbab\naba",
"output": "NO"
}
] | 1,666,185,904
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 0
|
n=int(input())
count=0
for i in range(n):
s=input()
for i in range(len(s)):
if s[0]==s[n-1]:
count=count+1
break
if (count==n):
print("YES")
else :
print("NO")
|
Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n=int(input())
count=0
for i in range(n):
s=input()
for i in range(len(s)):
if s[0]==s[n-1]:
count=count+1
break
if (count==n):
print("YES")
else :
print("NO")
```
| 0
|
|
453
|
A
|
Little Pony and Expected Maximum
|
PROGRAMMING
| 1,600
|
[
"probabilities"
] | null | null |
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
|
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
|
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
|
[
"6 1\n",
"6 3\n",
"2 2\n"
] |
[
"3.500000000000\n",
"4.958333333333\n",
"1.750000000000\n"
] |
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
| 500
|
[
{
"input": "6 1",
"output": "3.500000000000"
},
{
"input": "6 3",
"output": "4.958333333333"
},
{
"input": "2 2",
"output": "1.750000000000"
},
{
"input": "5 4",
"output": "4.433600000000"
},
{
"input": "5 8",
"output": "4.814773760000"
},
{
"input": "3 10",
"output": "2.982641534996"
},
{
"input": "3 6",
"output": "2.910836762689"
},
{
"input": "1 8",
"output": "1.000000000000"
},
{
"input": "24438 9",
"output": "21994.699969310015"
},
{
"input": "94444 9",
"output": "85000.099992058866"
},
{
"input": "8 66716",
"output": "8.000000000000"
},
{
"input": "4 25132",
"output": "4.000000000000"
},
{
"input": "51520 73331",
"output": "51519.682650242677"
},
{
"input": "54230 31747",
"output": "54228.743352775018"
},
{
"input": "24236 90163",
"output": "24235.975171545670"
},
{
"input": "26946 99523",
"output": "26945.974480086279"
},
{
"input": "50323 7",
"output": "44033.124988408454"
},
{
"input": "53033 3",
"output": "39775.249995286234"
},
{
"input": "55743 5",
"output": "46452.999992525307"
},
{
"input": "59964 79",
"output": "59214.949890211828"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "3 1",
"output": "2.000000000000"
},
{
"input": "1 2",
"output": "1.000000000000"
},
{
"input": "53513 34040",
"output": "53511.875329020870"
},
{
"input": "100000 100000",
"output": "99999.418033254507"
},
{
"input": "1 100000",
"output": "1.000000000000"
},
{
"input": "100000 1",
"output": "50000.499999999935"
},
{
"input": "2 100000",
"output": "2.000000000000"
},
{
"input": "100000 2",
"output": "66667.166665000332"
},
{
"input": "50000 100000",
"output": "49999.843487110789"
},
{
"input": "99999 1111",
"output": "99909.571915885972"
},
{
"input": "99999 99999",
"output": "99998.418033254609"
},
{
"input": "1000 1000",
"output": "999.419018443269"
},
{
"input": "50000 50000",
"output": "49999.418043215679"
},
{
"input": "88888 88888",
"output": "88887.418034499773"
},
{
"input": "99999 100000",
"output": "99998.418042461126"
},
{
"input": "100000 555",
"output": "99820.643422392372"
},
{
"input": "10000 10000",
"output": "9999.418122897887"
},
{
"input": "100000 5001",
"output": "99980.503829474910"
},
{
"input": "100000 1000",
"output": "99900.599066768002"
}
] | 1,593,004,373
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 202
| 24,166,400
|
import sys,math
from collections import deque
#input = sys.stdin.buffer.readline
def solve():
return;
def poo(x,p):
if p==0:
return 1
a = poo(x,p//2)
if p%2==0:
return a*a
else:
return a*a*x
# for _ in range(int(input())):
# solve()
m,n = map(int,input().split())
ans = 0
for i in range(1,m+1):
ans+=i*(poo(i/m,n)-poo((i-1)/m,n))
#print(ans)
# ans/=poo(m,n)
print(ans)
|
Title: Little Pony and Expected Maximum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
Input Specification:
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output Specification:
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
Demo Input:
['6 1\n', '6 3\n', '2 2\n']
Demo Output:
['3.500000000000\n', '4.958333333333\n', '1.750000000000\n']
Note:
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
|
```python
import sys,math
from collections import deque
#input = sys.stdin.buffer.readline
def solve():
return;
def poo(x,p):
if p==0:
return 1
a = poo(x,p//2)
if p%2==0:
return a*a
else:
return a*a*x
# for _ in range(int(input())):
# solve()
m,n = map(int,input().split())
ans = 0
for i in range(1,m+1):
ans+=i*(poo(i/m,n)-poo((i-1)/m,n))
#print(ans)
# ans/=poo(m,n)
print(ans)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
In the year of $30XX$ participants of some world programming championship live in a single large hotel. The hotel has $n$ floors. Each floor has $m$ sections with a single corridor connecting all of them. The sections are enumerated from $1$ to $m$ along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height $n$ and width $m$. We can denote sections with pairs of integers $(i, j)$, where $i$ is the floor, and $j$ is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections $(1, x)$, $(2, x)$, $\ldots$, $(n, x)$ for some $x$ between $1$ and $m$. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to $v$ floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process $q$ queries. Each query is a question "what is the minimum time needed to go from a room in section $(x_1, y_1)$ to a room in section $(x_2, y_2)$?"
|
The first line contains five integers $n, m, c_l, c_e, v$ ($2 \leq n, m \leq 10^8$, $0 \leq c_l, c_e \leq 10^5$, $1 \leq c_l + c_e \leq m - 1$, $1 \leq v \leq n - 1$) — the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains $c_l$ integers $l_1, \ldots, l_{c_l}$ in increasing order ($1 \leq l_i \leq m$), denoting the positions of the stairs. If $c_l = 0$, the second line is empty.
The third line contains $c_e$ integers $e_1, \ldots, e_{c_e}$ in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers $l_i$ and $e_i$ are distinct.
The fourth line contains a single integer $q$ ($1 \leq q \leq 10^5$) — the number of queries.
The next $q$ lines describe queries. Each of these lines contains four integers $x_1, y_1, x_2, y_2$ ($1 \leq x_1, x_2 \leq n$, $1 \leq y_1, y_2 \leq m$) — the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. $y_1$ and $y_2$ are not among $l_i$ and $e_i$.
|
Print $q$ integers, one per line — the answers for the queries.
|
[
"5 6 1 1 3\n2\n5\n3\n1 1 5 6\n1 3 5 4\n3 3 5 3\n"
] |
[
"7\n5\n4\n"
] |
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
| 0
|
[
{
"input": "5 6 1 1 3\n2\n5\n3\n1 1 5 6\n1 3 5 4\n3 3 5 3",
"output": "7\n5\n4"
},
{
"input": "2 2 0 1 1\n\n1\n1\n1 2 2 2",
"output": "3"
},
{
"input": "4 4 1 0 1\n4\n\n5\n1 1 2 2\n1 3 2 2\n3 3 4 3\n3 2 2 2\n1 2 2 3",
"output": "6\n4\n3\n5\n4"
},
{
"input": "10 10 1 8 4\n10\n2 3 4 5 6 7 8 9\n10\n1 1 3 1\n2 1 7 1\n1 1 9 1\n7 1 4 1\n10 1 7 1\n2 1 7 1\n3 1 2 1\n5 1 2 1\n10 1 5 1\n6 1 9 1",
"output": "3\n4\n4\n3\n3\n4\n3\n3\n4\n3"
},
{
"input": "2 5 1 0 1\n2\n\n1\n1 4 1 5",
"output": "1"
},
{
"input": "2 10 1 1 1\n1\n10\n1\n1 5 1 8",
"output": "3"
},
{
"input": "4 4 1 0 1\n1\n\n1\n1 2 1 4",
"output": "2"
},
{
"input": "2 4 1 1 1\n1\n2\n1\n2 3 2 4",
"output": "1"
},
{
"input": "1000 1000 1 1 10\n1\n2\n1\n1 900 1 1000",
"output": "100"
},
{
"input": "2 4 1 1 1\n1\n4\n1\n1 2 1 3",
"output": "1"
},
{
"input": "5 5 1 1 1\n3\n2\n1\n1 5 1 1",
"output": "4"
}
] | 1,663,753,549
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
a=int(input())
b=0
while a > 0:
if a >= 100:
a=a-100
b=b+1
elif a >= 50:
a=a-50
b=b+1
elif a >= 20:
a=a-20
b=b+1
elif a>=5:
a=a-5
b=b+1
elif a>=1:
a=a-1
b=b+1
print(b)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the year of $30XX$ participants of some world programming championship live in a single large hotel. The hotel has $n$ floors. Each floor has $m$ sections with a single corridor connecting all of them. The sections are enumerated from $1$ to $m$ along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height $n$ and width $m$. We can denote sections with pairs of integers $(i, j)$, where $i$ is the floor, and $j$ is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections $(1, x)$, $(2, x)$, $\ldots$, $(n, x)$ for some $x$ between $1$ and $m$. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to $v$ floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process $q$ queries. Each query is a question "what is the minimum time needed to go from a room in section $(x_1, y_1)$ to a room in section $(x_2, y_2)$?"
Input Specification:
The first line contains five integers $n, m, c_l, c_e, v$ ($2 \leq n, m \leq 10^8$, $0 \leq c_l, c_e \leq 10^5$, $1 \leq c_l + c_e \leq m - 1$, $1 \leq v \leq n - 1$) — the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains $c_l$ integers $l_1, \ldots, l_{c_l}$ in increasing order ($1 \leq l_i \leq m$), denoting the positions of the stairs. If $c_l = 0$, the second line is empty.
The third line contains $c_e$ integers $e_1, \ldots, e_{c_e}$ in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers $l_i$ and $e_i$ are distinct.
The fourth line contains a single integer $q$ ($1 \leq q \leq 10^5$) — the number of queries.
The next $q$ lines describe queries. Each of these lines contains four integers $x_1, y_1, x_2, y_2$ ($1 \leq x_1, x_2 \leq n$, $1 \leq y_1, y_2 \leq m$) — the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. $y_1$ and $y_2$ are not among $l_i$ and $e_i$.
Output Specification:
Print $q$ integers, one per line — the answers for the queries.
Demo Input:
['5 6 1 1 3\n2\n5\n3\n1 1 5 6\n1 3 5 4\n3 3 5 3\n']
Demo Output:
['7\n5\n4\n']
Note:
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
|
```python
a=int(input())
b=0
while a > 0:
if a >= 100:
a=a-100
b=b+1
elif a >= 50:
a=a-50
b=b+1
elif a >= 20:
a=a-20
b=b+1
elif a>=5:
a=a-5
b=b+1
elif a>=1:
a=a-1
b=b+1
print(b)
```
| -1
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,685,450,378
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 46
| 0
|
def countDistinctLetters(s):
s = s[1:-1]
s = s.replace(" ", "")
if len(s) == 0:
return 0
letters = s.split(",")
distinct_letters = set()
for letter in letters:
distinct_letters.add(letter)
return len(distinct_letters)
letters_set = input()
distinct_count = countDistinctLetters(letters_set)
print(distinct_count)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
def countDistinctLetters(s):
s = s[1:-1]
s = s.replace(" ", "")
if len(s) == 0:
return 0
letters = s.split(",")
distinct_letters = set()
for letter in letters:
distinct_letters.add(letter)
return len(distinct_letters)
letters_set = input()
distinct_count = countDistinctLetters(letters_set)
print(distinct_count)
```
| 3
|
|
625
|
A
|
Guest From the Past
|
PROGRAMMING
| 1,700
|
[
"implementation",
"math"
] | null | null |
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
|
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
|
[
"10\n11\n9\n8\n",
"10\n5\n6\n1\n"
] |
[
"2\n",
"2\n"
] |
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
| 750
|
[
{
"input": "10\n11\n9\n8",
"output": "2"
},
{
"input": "10\n5\n6\n1",
"output": "2"
},
{
"input": "2\n2\n2\n1",
"output": "1"
},
{
"input": "10\n3\n3\n1",
"output": "4"
},
{
"input": "10\n1\n2\n1",
"output": "10"
},
{
"input": "10\n2\n3\n1",
"output": "5"
},
{
"input": "9\n2\n4\n1",
"output": "4"
},
{
"input": "9\n2\n2\n1",
"output": "8"
},
{
"input": "9\n10\n10\n1",
"output": "0"
},
{
"input": "10\n2\n2\n1",
"output": "9"
},
{
"input": "1000000000000000000\n2\n10\n9",
"output": "999999999999999995"
},
{
"input": "501000000000000000\n300000000000000000\n301000000000000000\n100000000000000000",
"output": "2"
},
{
"input": "10\n1\n9\n8",
"output": "10"
},
{
"input": "10\n8\n8\n7",
"output": "3"
},
{
"input": "10\n5\n5\n1",
"output": "2"
},
{
"input": "29\n3\n3\n1",
"output": "14"
},
{
"input": "45\n9\n9\n8",
"output": "37"
},
{
"input": "45\n9\n9\n1",
"output": "5"
},
{
"input": "100\n10\n10\n9",
"output": "91"
},
{
"input": "179\n10\n9\n1",
"output": "22"
},
{
"input": "179\n2\n2\n1",
"output": "178"
},
{
"input": "179\n179\n179\n1",
"output": "1"
},
{
"input": "179\n59\n59\n58",
"output": "121"
},
{
"input": "500\n250\n250\n1",
"output": "2"
},
{
"input": "500\n1\n250\n1",
"output": "500"
},
{
"input": "501\n500\n500\n499",
"output": "2"
},
{
"input": "501\n450\n52\n1",
"output": "9"
},
{
"input": "501\n300\n301\n100",
"output": "2"
},
{
"input": "500\n179\n10\n1",
"output": "55"
},
{
"input": "1000\n500\n10\n9",
"output": "991"
},
{
"input": "1000\n2\n10\n9",
"output": "995"
},
{
"input": "1001\n1000\n1000\n999",
"output": "2"
},
{
"input": "10000\n10000\n10000\n1",
"output": "1"
},
{
"input": "10000\n10\n5000\n4999",
"output": "5500"
},
{
"input": "1000000000\n999999998\n999999999\n999999998",
"output": "3"
},
{
"input": "1000000000\n50\n50\n49",
"output": "999999951"
},
{
"input": "1000000000\n500\n5000\n4999",
"output": "999995010"
},
{
"input": "1000000000\n51\n100\n98",
"output": "499999952"
},
{
"input": "1000000000\n100\n51\n50",
"output": "999999950"
},
{
"input": "1000000000\n2\n5\n4",
"output": "999999998"
},
{
"input": "1000000000000000000\n999999998000000000\n999999999000000000\n999999998000000000",
"output": "3"
},
{
"input": "1000000000\n2\n2\n1",
"output": "999999999"
},
{
"input": "999999999\n2\n999999998\n1",
"output": "499999999"
},
{
"input": "999999999999999999\n2\n2\n1",
"output": "999999999999999998"
},
{
"input": "999999999999999999\n10\n10\n9",
"output": "999999999999999990"
},
{
"input": "999999999999999999\n999999999999999998\n999999999999999998\n999999999999999997",
"output": "2"
},
{
"input": "999999999999999999\n501\n501\n1",
"output": "1999999999999999"
},
{
"input": "999999999999999999\n2\n50000000000000000\n49999999999999999",
"output": "974999999999999999"
},
{
"input": "999999999999999999\n180\n180\n1",
"output": "5586592178770949"
},
{
"input": "1000000000000000000\n42\n41\n1",
"output": "24999999999999999"
},
{
"input": "1000000000000000000\n41\n40\n1",
"output": "25641025641025641"
},
{
"input": "100000000000000000\n79\n100\n25",
"output": "1333333333333333"
},
{
"input": "1\n100\n5\n4",
"output": "0"
},
{
"input": "1000000000000000000\n1000000000000000000\n10000000\n9999999",
"output": "999999999990000001"
},
{
"input": "999999999999999999\n999999999000000000\n900000000000000000\n899999999999999999",
"output": "100000000000000000"
},
{
"input": "13\n10\n15\n11",
"output": "1"
},
{
"input": "1\n1000\n5\n4",
"output": "0"
},
{
"input": "10\n100\n10\n1",
"output": "1"
},
{
"input": "3\n2\n100000\n99999",
"output": "1"
},
{
"input": "4\n2\n4\n2",
"output": "2"
},
{
"input": "5\n3\n6\n4",
"output": "1"
},
{
"input": "1\n7\n65\n49",
"output": "0"
},
{
"input": "10\n20\n100\n99",
"output": "0"
},
{
"input": "10000000000\n10000000000\n9000000000\n8999999999",
"output": "1000000001"
},
{
"input": "90\n30\n101\n100",
"output": "3"
},
{
"input": "999999999999999\n5\n500000000000000\n499999999999999",
"output": "599999999999999"
},
{
"input": "1000000000000000000\n1000000000000000000\n1000000000\n999999999",
"output": "999999999000000001"
},
{
"input": "1\n1000000000000000000\n1000000000\n999999999",
"output": "0"
},
{
"input": "100000000000000000\n100000000000000000\n1000000000\n999999999",
"output": "99999999000000001"
},
{
"input": "100000000000000009\n100\n1000000000000000\n999999999999999",
"output": "99010000000000009"
},
{
"input": "10\n20\n10\n9",
"output": "1"
},
{
"input": "10\n4\n14\n13",
"output": "2"
},
{
"input": "11\n3\n9\n7",
"output": "4"
},
{
"input": "1000000000\n5\n7\n4",
"output": "333333332"
},
{
"input": "12155\n1943\n28717\n24074",
"output": "6"
},
{
"input": "1000000000000000000\n10\n20\n5",
"output": "100000000000000000"
},
{
"input": "98\n33\n440\n314",
"output": "2"
},
{
"input": "1070252292\n57449678\n237309920\n221182550",
"output": "56"
},
{
"input": "100\n3\n102\n101",
"output": "33"
},
{
"input": "100000000000000000\n100000000000000001\n1000000000000000\n999999999999999",
"output": "99000000000000001"
},
{
"input": "66249876257975628\n302307316\n406102416\n182373516",
"output": "296116756"
},
{
"input": "10\n5\n10\n1",
"output": "2"
},
{
"input": "1000000000000000000\n10\n1000000000\n999999998",
"output": "499999999600000000"
}
] | 1,455,017,409
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 62
| 0
|
n=int(input())
a=int(input())
b=int(input())
c=int(input())
ans=0
extra=0
if( b-c < a ):
ans += int((n-b+1)/(b-c))
n = b-1 + int( (n-b+1)%(b-c) )
ans += int(n/a)
else:
ans = int(n/a)
print(ans)
|
Title: Guest From the Past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output Specification:
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Demo Input:
['10\n11\n9\n8\n', '10\n5\n6\n1\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
|
```python
n=int(input())
a=int(input())
b=int(input())
c=int(input())
ans=0
extra=0
if( b-c < a ):
ans += int((n-b+1)/(b-c))
n = b-1 + int( (n-b+1)%(b-c) )
ans += int(n/a)
else:
ans = int(n/a)
print(ans)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
|
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
|
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
|
[
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] |
[
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] |
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.
| 0
|
[
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
"input": "15 42\n8 1\n3 14\n7 14\n12 3\n7 9\n6 7\n6 12\n14 12\n3 10\n10 14\n6 3\n3 13\n13 10\n7 12\n7 2\n6 10\n11 4\n9 3\n8 4\n7 3\n2 3\n2 10\n9 13\n2 14\n6 14\n13 2\n1 4\n13 6\n7 10\n13 14\n12 10\n13 7\n12 2\n9 10\n13 12\n2 6\n9 14\n6 9\n12 9\n11 1\n2 9\n11 8",
"output": "YES"
},
{
"input": "20 80\n17 4\n10 1\n11 10\n17 7\n15 10\n14 15\n13 1\n18 13\n3 13\n12 7\n9 13\n10 12\n14 12\n18 11\n4 7\n10 13\n11 3\n19 8\n14 7\n10 17\n14 3\n7 11\n11 14\n19 5\n10 14\n15 17\n3 1\n9 10\n11 1\n4 1\n11 4\n9 1\n12 3\n13 7\n1 14\n11 12\n7 1\n9 12\n18 15\n17 3\n7 15\n4 10\n7 18\n7 9\n12 17\n14 18\n3 18\n18 17\n9 15\n14 4\n14 9\n9 18\n12 4\n7 10\n15 4\n4 18\n15 13\n1 12\n7 3\n13 11\n4 13\n5 8\n12 18\n12 15\n17 9\n11 15\n3 10\n18 10\n4 3\n15 3\n13 12\n9 4\n9 11\n14 17\n13 17\n3 9\n13 14\n1 17\n15 1\n17 11",
"output": "NO"
},
{
"input": "99 26\n64 17\n48 70\n71 50\n3 50\n9 60\n61 64\n53 50\n25 12\n3 71\n71 53\n3 53\n65 70\n9 25\n9 12\n59 56\n39 60\n64 69\n65 94\n70 94\n25 60\n60 12\n94 48\n17 69\n61 17\n65 48\n61 69",
"output": "NO"
},
{
"input": "3 1\n1 2",
"output": "YES"
},
{
"input": "3 2\n3 2\n1 3",
"output": "NO"
},
{
"input": "3 3\n2 3\n1 2\n1 3",
"output": "YES"
},
{
"input": "4 2\n4 1\n2 1",
"output": "NO"
},
{
"input": "4 3\n3 1\n2 1\n3 2",
"output": "YES"
},
{
"input": "5 9\n1 2\n5 1\n3 1\n1 4\n2 4\n5 3\n5 4\n2 3\n5 2",
"output": "NO"
},
{
"input": "10 5\n9 5\n1 2\n6 8\n6 3\n10 6",
"output": "NO"
},
{
"input": "10 8\n10 7\n9 7\n5 7\n6 8\n3 5\n8 10\n3 4\n7 8",
"output": "NO"
},
{
"input": "10 20\n8 2\n8 3\n1 8\n9 5\n2 4\n10 1\n10 5\n7 5\n7 8\n10 7\n6 5\n3 7\n1 9\n9 8\n7 2\n2 10\n2 1\n6 4\n9 7\n4 3",
"output": "NO"
},
{
"input": "150000 10\n62562 50190\n48849 60549\n139470 18456\n21436 25159\n66845 120884\n99972 114453\n11631 99153\n62951 134848\n78114 146050\n136760 131762",
"output": "YES"
},
{
"input": "150000 0",
"output": "YES"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n1 4",
"output": "NO"
},
{
"input": "30 73\n25 2\n2 16\n20 12\n16 20\n7 18\n11 15\n13 11\n30 29\n16 12\n12 25\n2 1\n18 14\n9 8\n28 16\n2 9\n22 21\n1 25\n12 28\n14 7\n4 9\n26 7\n14 27\n12 2\n29 22\n1 9\n13 15\n3 10\n1 12\n8 20\n30 24\n25 20\n4 1\n4 12\n20 1\n8 4\n2 28\n25 16\n16 8\n20 4\n9 12\n21 30\n23 11\n19 6\n28 4\n29 21\n9 28\n30 10\n22 24\n25 8\n27 26\n25 4\n28 20\n9 25\n24 29\n20 9\n18 26\n1 28\n30 22\n23 15\n28 27\n8 2\n23 13\n12 8\n14 26\n16 4\n28 25\n8 1\n4 2\n9 16\n20 2\n18 27\n28 8\n27 7",
"output": "NO"
},
{
"input": "5 4\n1 2\n2 5\n3 4\n4 5",
"output": "NO"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n4 1",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 4\n4 3\n1 5\n5 6\n6 3",
"output": "NO"
},
{
"input": "3 2\n1 2\n1 3",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n1 6",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n2 4\n3 4",
"output": "NO"
},
{
"input": "6 9\n1 4\n1 5\n1 6\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n3 4",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n2 4",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 1",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "NO"
},
{
"input": "150000 3\n150000 149999\n149998 149999\n149998 150000",
"output": "YES"
},
{
"input": "10 15\n1 2\n1 3\n2 4\n2 5\n3 4\n3 5\n4 5\n1 6\n6 7\n6 8\n7 9\n7 10\n8 9\n8 10\n9 10",
"output": "NO"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "NO"
},
{
"input": "5 4\n1 2\n1 3\n1 4\n1 5",
"output": "NO"
},
{
"input": "9 9\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 1",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n3 4",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n4 2\n4 3",
"output": "NO"
},
{
"input": "5 5\n1 4\n5 3\n5 1\n2 3\n4 5",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n2 3\n1 4\n2 4",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n2 4",
"output": "NO"
},
{
"input": "6 12\n1 2\n1 4\n1 5\n1 6\n2 3\n2 5\n2 6\n3 4\n3 5\n3 6\n4 5\n4 6",
"output": "NO"
},
{
"input": "1000 4\n100 132\n100 164\n196 132\n196 164",
"output": "NO"
}
] | 1,694,808,816
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
def es_racional(n, conexiones):
amigos = {}
for i in range(1, n + 1):
amigos[i] = set()
for x, y in conexiones:
amigos[x].add(y)
amigos[y].add(x)
for x in range(1, n + 1):
for y in amigos[x]:
for z in amigos[y]:
if z != x and z not in amigos[x]:
print("NO")
return
print("SÍ")
n, m = map(int, input().split())
conexiones = [tuple(map(int, input().split())) for _ in range(m)]
es_racional(n, conexiones)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.
|
```python
def es_racional(n, conexiones):
amigos = {}
for i in range(1, n + 1):
amigos[i] = set()
for x, y in conexiones:
amigos[x].add(y)
amigos[y].add(x)
for x in range(1, n + 1):
for y in amigos[x]:
for z in amigos[y]:
if z != x and z not in amigos[x]:
print("NO")
return
print("SÍ")
n, m = map(int, input().split())
conexiones = [tuple(map(int, input().split())) for _ in range(m)]
es_racional(n, conexiones)
```
| -1
|
|
488
|
B
|
Candy Boxes
|
PROGRAMMING
| 1,900
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {*x*1,<=*x*2,<=*x*3,<=*x*4} (*x*1<=≤<=*x*2<=≤<=*x*3<=≤<=*x*4) arithmetic mean is , median is and range is *x*4<=-<=*x*1. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs.
For example, 1,<=1,<=3,<=3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.
Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only *n* (0<=≤<=*n*<=≤<=4) boxes remaining. The *i*-th remaining box contains *a**i* candies.
Now Jeff wants to know: is there a possible way to find the number of candies of the 4<=-<=*n* missing boxes, meeting the condition above (the mean, median and range are equal)?
|
The first line of input contains an only integer *n* (0<=≤<=*n*<=≤<=4).
The next *n* lines contain integers *a**i*, denoting the number of candies in the *i*-th box (1<=≤<=*a**i*<=≤<=500).
|
In the first output line, print "YES" if a solution exists, or print "NO" if there is no solution.
If a solution exists, you should output 4<=-<=*n* more lines, each line containing an integer *b*, denoting the number of candies in a missing box.
All your numbers *b* must satisfy inequality 1<=≤<=*b*<=≤<=106. It is guaranteed that if there exists a positive integer solution, you can always find such *b*'s meeting the condition. If there are multiple answers, you are allowed to print any of them.
Given numbers *a**i* may follow in any order in the input, not necessary in non-decreasing.
*a**i* may have stood at any positions in the original set, not necessary on lowest *n* first positions.
|
[
"2\n1\n1\n",
"3\n1\n1\n1\n",
"4\n1\n2\n2\n3\n"
] |
[
"YES\n3\n3\n",
"NO\n",
"YES\n"
] |
For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.
For the second sample, it's impossible to find the missing number of candies.
In the third example no box has been lost and numbers satisfy the condition.
You may output *b* in any order.
| 1,500
|
[
{
"input": "2\n1\n1",
"output": "YES\n3\n3"
},
{
"input": "3\n1\n1\n1",
"output": "NO"
},
{
"input": "4\n1\n2\n2\n3",
"output": "YES"
},
{
"input": "0",
"output": "YES\n1\n1\n3\n3"
},
{
"input": "1\n125",
"output": "YES\n125\n375\n375"
},
{
"input": "2\n472\n107",
"output": "NO"
},
{
"input": "3\n215\n137\n256",
"output": "NO"
},
{
"input": "4\n49\n464\n28\n118",
"output": "NO"
},
{
"input": "4\n172\n84\n252\n163",
"output": "NO"
},
{
"input": "2\n66\n135",
"output": "YES\n198\n129"
},
{
"input": "1\n190",
"output": "YES\n190\n570\n570"
},
{
"input": "3\n184\n100\n71",
"output": "YES\n213"
},
{
"input": "3\n361\n387\n130",
"output": "NO"
},
{
"input": "3\n146\n247\n182",
"output": "NO"
},
{
"input": "3\n132\n44\n126",
"output": "YES\n50"
},
{
"input": "2\n172\n148",
"output": "YES\n444\n420"
},
{
"input": "3\n276\n311\n442",
"output": "NO"
},
{
"input": "3\n324\n301\n131",
"output": "YES\n108"
},
{
"input": "4\n186\n129\n119\n62",
"output": "YES"
},
{
"input": "3\n31\n72\n65",
"output": "YES\n24"
},
{
"input": "1\n318",
"output": "YES\n318\n954\n954"
},
{
"input": "2\n68\n151",
"output": "YES\n204\n121"
},
{
"input": "1\n67",
"output": "YES\n67\n201\n201"
},
{
"input": "3\n63\n28\n56",
"output": "YES\n21"
},
{
"input": "2\n288\n399",
"output": "YES\n864\n753"
},
{
"input": "3\n257\n86\n258",
"output": "YES\n87"
},
{
"input": "1\n71",
"output": "YES\n71\n213\n213"
},
{
"input": "4\n104\n84\n47\n141",
"output": "YES"
},
{
"input": "2\n2\n2",
"output": "YES\n6\n6"
},
{
"input": "4\n258\n312\n158\n104",
"output": "YES"
},
{
"input": "1\n121",
"output": "YES\n121\n363\n363"
},
{
"input": "1\n500",
"output": "YES\n500\n1500\n1500"
},
{
"input": "2\n3\n13",
"output": "NO"
},
{
"input": "2\n200\n200",
"output": "YES\n600\n600"
},
{
"input": "3\n1\n1\n3",
"output": "YES\n3"
},
{
"input": "2\n500\n497",
"output": "YES\n1491\n1488"
},
{
"input": "3\n2\n2\n3",
"output": "YES\n1"
}
] | 1,416,601,086
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 77
| 0
|
n = int(input())
a = []
for i in range(n):
a.append(int(input()))
a = sorted(a)
#print(a)
if n == 0:
print('YES')
print(1)
print(1)
print(3)
print(3)
elif n == 1:
print('YES')
print(a[0])
print(3 * a[0])
print(3 * a[0])
elif n == 2:
if a[0] * 3 >= a[1]:
print('YES')
print(a[0] * 3)
print(a[0] * 4 - a[1])
elif a[1] % 3 == 0 and a[1] // 3 <= a[0]:
print('YES')
print(a[1] // 3)
print(a[1] + a[1] // 3 - a[0])
elif (a[0] + a[1]) % 4 == 0 and (a[0] + a[1]) // 4 <= a[0]:
print('YES')
print((a[0] + a[1]) // 4)
print(3 * ((a[0] + a[1]) // 4))
else:
print('NO')
elif n == 3:
if a[0] * 3 >= a[2] and 4 * a[0] == a[1] + a[2]:
print('YES')
print(a[0] * 3)
elif a[2] % 3 == 0 and a[2] // 3 <= a[0] and a[2] + a[2] // 3 == a[0] + a[1]:
print('YES')
print(a[2] // 3)
elif a[2] == a[0] * 3:
print('YES')
print(4 * a[0] - a[1])
else:
print('NO')
elif n == 4:
if a[3] == 3 * a[0] and a[0] + a[3] == a[1] + a[2]:
print('YES')
else:
print('NO')
|
Title: Candy Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {*x*1,<=*x*2,<=*x*3,<=*x*4} (*x*1<=≤<=*x*2<=≤<=*x*3<=≤<=*x*4) arithmetic mean is , median is and range is *x*4<=-<=*x*1. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs.
For example, 1,<=1,<=3,<=3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.
Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only *n* (0<=≤<=*n*<=≤<=4) boxes remaining. The *i*-th remaining box contains *a**i* candies.
Now Jeff wants to know: is there a possible way to find the number of candies of the 4<=-<=*n* missing boxes, meeting the condition above (the mean, median and range are equal)?
Input Specification:
The first line of input contains an only integer *n* (0<=≤<=*n*<=≤<=4).
The next *n* lines contain integers *a**i*, denoting the number of candies in the *i*-th box (1<=≤<=*a**i*<=≤<=500).
Output Specification:
In the first output line, print "YES" if a solution exists, or print "NO" if there is no solution.
If a solution exists, you should output 4<=-<=*n* more lines, each line containing an integer *b*, denoting the number of candies in a missing box.
All your numbers *b* must satisfy inequality 1<=≤<=*b*<=≤<=106. It is guaranteed that if there exists a positive integer solution, you can always find such *b*'s meeting the condition. If there are multiple answers, you are allowed to print any of them.
Given numbers *a**i* may follow in any order in the input, not necessary in non-decreasing.
*a**i* may have stood at any positions in the original set, not necessary on lowest *n* first positions.
Demo Input:
['2\n1\n1\n', '3\n1\n1\n1\n', '4\n1\n2\n2\n3\n']
Demo Output:
['YES\n3\n3\n', 'NO\n', 'YES\n']
Note:
For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.
For the second sample, it's impossible to find the missing number of candies.
In the third example no box has been lost and numbers satisfy the condition.
You may output *b* in any order.
|
```python
n = int(input())
a = []
for i in range(n):
a.append(int(input()))
a = sorted(a)
#print(a)
if n == 0:
print('YES')
print(1)
print(1)
print(3)
print(3)
elif n == 1:
print('YES')
print(a[0])
print(3 * a[0])
print(3 * a[0])
elif n == 2:
if a[0] * 3 >= a[1]:
print('YES')
print(a[0] * 3)
print(a[0] * 4 - a[1])
elif a[1] % 3 == 0 and a[1] // 3 <= a[0]:
print('YES')
print(a[1] // 3)
print(a[1] + a[1] // 3 - a[0])
elif (a[0] + a[1]) % 4 == 0 and (a[0] + a[1]) // 4 <= a[0]:
print('YES')
print((a[0] + a[1]) // 4)
print(3 * ((a[0] + a[1]) // 4))
else:
print('NO')
elif n == 3:
if a[0] * 3 >= a[2] and 4 * a[0] == a[1] + a[2]:
print('YES')
print(a[0] * 3)
elif a[2] % 3 == 0 and a[2] // 3 <= a[0] and a[2] + a[2] // 3 == a[0] + a[1]:
print('YES')
print(a[2] // 3)
elif a[2] == a[0] * 3:
print('YES')
print(4 * a[0] - a[1])
else:
print('NO')
elif n == 4:
if a[3] == 3 * a[0] and a[0] + a[3] == a[1] + a[2]:
print('YES')
else:
print('NO')
```
| 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<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
|
Print a single number — 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"
}
] | 1,696,410,479
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 62
| 1,638,400
|
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def MI(): return map(int, sys.stdin.readline().rstrip().split())
def SI(): return sys.stdin.readline().rstrip()
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def main():
n,k = MI()
a = LI()
a.sort()
count = 0
for i in range(n):
if a[i] + k <= 5:
count += 1
print(count//3)
if __name__ == "__main__":
main()
|
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<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=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 — 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
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def MI(): return map(int, sys.stdin.readline().rstrip().split())
def SI(): return sys.stdin.readline().rstrip()
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def main():
n,k = MI()
a = LI()
a.sort()
count = 0
for i in range(n):
if a[i] + k <= 5:
count += 1
print(count//3)
if __name__ == "__main__":
main()
```
| 3
|
|
734
|
A
|
Anton and Danik
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.
|
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
|
[
"6\nADAAAA\n",
"7\nDDDAADA\n",
"6\nDADADA\n"
] |
[
"Anton\n",
"Danik\n",
"Friendship\n"
] |
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
| 500
|
[
{
"input": "6\nADAAAA",
"output": "Anton"
},
{
"input": "7\nDDDAADA",
"output": "Danik"
},
{
"input": "6\nDADADA",
"output": "Friendship"
},
{
"input": "10\nDDDDADDADD",
"output": "Danik"
},
{
"input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA",
"output": "Anton"
},
{
"input": "200\nDDDDDDDADDDDDDAADADAADAAADAADADAAADDDADDDDDDADDDAADDDAADADDDDDADDDAAAADAAADDDDDAAADAADDDAAAADDADADDDAADDAADAAADAADAAAADDAADDADAAAADADDDAAAAAADDAADAADAADADDDAAADAAAADADDADAAAAAADADADDDADDDAADDADDDAAAAD",
"output": "Friendship"
},
{
"input": "1\nA",
"output": "Anton"
},
{
"input": "1\nD",
"output": "Danik"
},
{
"input": "2\nDA",
"output": "Friendship"
},
{
"input": "4\nDADA",
"output": "Friendship"
},
{
"input": "4\nDAAD",
"output": "Friendship"
},
{
"input": "3\nADD",
"output": "Danik"
},
{
"input": "3\nDAD",
"output": "Danik"
},
{
"input": "2\nDA",
"output": "Friendship"
},
{
"input": "379\nAADAAAAAADDAAAAAADAADADADDAAAAADADDAADAAAADDDADAAAAAAADAADAAAAAAADAAAAAAAAADAAAAAAADAAAAAAAAAAADDDADAAAAAAAADAADADAAAADAAAAAAAAAAAAAAAAADAAAADDDAADAAAAAAADAAADAAADAADDDADDAAADAAAAAADDDADDDAAADAAAADAAAAAAAAADAAADAAAAAAAAADAAAAAAAAAAAAAAAAAADADAAAAAAAAAAADAAAAADAAAADAAAAAAAAAAAAADADAADAAAAAAAADAADAAAAAAAADAAAAAAAADDDAAAAAADAAADAAAAAADAADAAAAAADAAAADADAADAAAAAADAAAADAADDAADAADAAA",
"output": "Anton"
}
] | 1,697,446,539
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 46
| 921,600
|
num_games=int(input())
game_winner=list(input())
Anton=0
Danik=0
for n in range(num_games):
if game_winner[n] =='A':
Anton+=1
else:
Danik+=1
if Anton > Danik:
print("Anton")
elif Anton < Danik:
print("Danik")
else:
print("Friendship")
|
Title: Anton and Danik
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.
Output Specification:
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Demo Input:
['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n']
Demo Output:
['Anton\n', 'Danik\n', 'Friendship\n']
Note:
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
|
```python
num_games=int(input())
game_winner=list(input())
Anton=0
Danik=0
for n in range(num_games):
if game_winner[n] =='A':
Anton+=1
else:
Danik+=1
if Anton > Danik:
print("Anton")
elif Anton < Danik:
print("Danik")
else:
print("Friendship")
```
| 3
|
|
698
|
A
|
Vacations
|
PROGRAMMING
| 1,400
|
[
"dp"
] | null | null |
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
|
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
|
[
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] |
[
"2\n",
"0\n",
"1\n"
] |
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
| 500
|
[
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 3 2 2 3 2 3 3 3 3 3 3 1 2 2 3 1 3 3 2 2 2 3 1 0 3 3 3 2 3 3 1 1 3 1 3 3 3 1 3 1 3 0 1 3 2 3 2 1 1 3 2 3 3 3 2 3 1 3 3 3 3 2 2 2 1 3 1 3 3 3 3 1 3 2 3 3 0 3 3 3 3 3 1 0 2 1 3 3 0 2 3 3",
"output": "16"
},
{
"input": "10\n2 3 0 1 3 1 2 2 1 0",
"output": "3"
},
{
"input": "45\n3 3 2 3 2 3 3 3 0 3 3 3 3 3 3 3 1 3 2 3 2 3 2 2 2 3 2 3 3 3 3 3 1 2 3 3 2 2 2 3 3 3 3 1 3",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "0"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "2\n3 3",
"output": "0"
},
{
"input": "3\n3 3 3",
"output": "0"
},
{
"input": "2\n3 2",
"output": "0"
},
{
"input": "2\n0 2",
"output": "1"
},
{
"input": "10\n2 2 3 3 3 3 2 1 3 2",
"output": "2"
},
{
"input": "15\n0 1 0 0 0 2 0 1 0 0 0 2 0 0 0",
"output": "11"
},
{
"input": "15\n1 3 2 2 2 3 3 3 3 2 3 2 2 1 1",
"output": "4"
},
{
"input": "15\n3 1 3 2 3 2 2 2 3 3 3 3 2 3 2",
"output": "3"
},
{
"input": "20\n0 2 0 1 0 0 0 1 2 0 1 1 1 0 1 1 0 1 1 0",
"output": "12"
},
{
"input": "20\n2 3 2 3 3 3 3 2 0 3 1 1 2 3 0 3 2 3 0 3",
"output": "5"
},
{
"input": "20\n3 3 3 3 2 3 3 2 1 3 3 2 2 2 3 2 2 2 2 2",
"output": "4"
},
{
"input": "25\n0 0 1 0 0 1 0 0 1 0 0 1 0 2 0 0 2 0 0 1 0 2 0 1 1",
"output": "16"
},
{
"input": "25\n1 3 3 2 2 3 3 3 3 3 1 2 2 3 2 0 2 1 0 1 3 2 2 3 3",
"output": "5"
},
{
"input": "25\n2 3 1 3 3 2 1 3 3 3 1 3 3 1 3 2 3 3 1 3 3 3 2 3 3",
"output": "3"
},
{
"input": "30\n0 0 1 0 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 2 0 0 1 1 2 0 0 0",
"output": "22"
},
{
"input": "30\n1 1 3 2 2 0 3 2 3 3 1 2 0 1 1 2 3 3 2 3 1 3 2 3 0 2 0 3 3 2",
"output": "9"
},
{
"input": "30\n1 2 3 2 2 3 3 3 3 3 3 3 3 3 3 1 2 2 3 2 3 3 3 2 1 3 3 3 1 3",
"output": "2"
},
{
"input": "35\n0 1 1 0 0 2 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 2 1 0 2 2 1 0 1 0 1 1 1 0 0",
"output": "21"
},
{
"input": "35\n2 2 0 3 2 2 0 3 3 1 1 3 3 1 2 2 0 2 2 2 2 3 1 0 2 1 3 2 2 3 2 3 3 1 2",
"output": "11"
},
{
"input": "35\n1 2 2 3 3 3 3 3 2 2 3 3 2 3 3 2 3 2 3 3 2 2 2 3 3 2 3 3 3 1 3 3 2 2 2",
"output": "7"
},
{
"input": "40\n2 0 1 1 0 0 0 0 2 0 1 1 1 0 0 1 0 0 0 0 0 2 0 0 0 2 1 1 1 3 0 0 0 0 0 0 0 1 1 0",
"output": "28"
},
{
"input": "40\n2 2 3 2 0 2 3 2 1 2 3 0 2 3 2 1 1 3 1 1 0 2 3 1 3 3 1 1 3 3 2 2 1 3 3 3 2 3 3 1",
"output": "10"
},
{
"input": "40\n1 3 2 3 3 2 3 3 2 2 3 1 2 1 2 2 3 1 2 2 1 2 2 2 1 2 2 3 2 3 2 3 2 3 3 3 1 3 2 3",
"output": "8"
},
{
"input": "45\n2 1 0 0 0 2 1 0 1 0 0 2 2 1 1 0 0 2 0 0 0 0 0 0 1 0 0 2 0 0 1 1 0 0 1 0 0 1 1 2 0 0 2 0 2",
"output": "29"
},
{
"input": "45\n3 3 2 3 3 3 2 2 3 2 3 1 3 2 3 2 2 1 1 3 2 3 2 1 3 1 2 3 2 2 0 3 3 2 3 2 3 2 3 2 0 3 1 1 3",
"output": "8"
},
{
"input": "50\n3 0 0 0 2 0 0 0 0 0 0 0 2 1 0 2 0 1 0 1 3 0 2 1 1 0 0 1 1 0 0 1 2 1 1 2 1 1 0 0 0 0 0 0 0 1 2 2 0 0",
"output": "32"
},
{
"input": "50\n3 3 3 3 1 0 3 3 0 2 3 1 1 1 3 2 3 3 3 3 3 1 0 1 2 2 3 3 2 3 0 0 0 2 1 0 1 2 2 2 2 0 2 2 2 1 2 3 3 2",
"output": "16"
},
{
"input": "50\n3 2 3 1 2 1 2 3 3 2 3 3 2 1 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 2 3 3 3 3 2 3 1 2 3 3 2 3 3 1 2 2 1 1 3 3",
"output": "7"
},
{
"input": "55\n0 0 1 1 0 1 0 0 1 0 1 0 0 0 2 0 0 1 0 0 0 1 0 0 0 0 3 1 0 0 0 1 0 0 0 0 2 0 0 0 2 0 2 1 0 0 0 0 0 0 0 0 2 0 0",
"output": "40"
},
{
"input": "55\n3 0 3 3 3 2 0 2 3 0 3 2 3 3 0 3 3 1 3 3 1 2 3 2 0 3 3 2 1 2 3 2 3 0 3 2 2 1 2 3 2 2 1 3 2 2 3 1 3 2 2 3 3 2 2",
"output": "13"
},
{
"input": "55\n3 3 1 3 2 3 2 3 2 2 3 3 3 3 3 1 1 3 3 2 3 2 3 2 0 1 3 3 3 3 2 3 2 3 1 1 2 2 2 3 3 3 3 3 2 2 2 3 2 3 3 3 3 1 3",
"output": "7"
},
{
"input": "60\n0 1 0 0 0 0 0 0 0 2 1 1 3 0 0 0 0 0 1 0 1 1 0 0 0 3 0 1 0 1 0 2 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0",
"output": "44"
},
{
"input": "60\n3 2 1 3 2 2 3 3 3 1 1 3 2 2 3 3 1 3 2 2 3 3 2 2 2 2 0 2 2 3 2 3 0 3 3 3 2 3 3 0 1 3 2 1 3 1 1 2 1 3 1 1 2 2 1 3 3 3 2 2",
"output": "15"
},
{
"input": "60\n3 2 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 1 2 3 3 3 2 1 3 3 1 3 1 3 0 3 3 3 2 3 2 3 2 3 3 1 1 2 3 3 3 3 2 1 3 2 3",
"output": "8"
},
{
"input": "65\n1 0 2 1 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 2 0 2 1 0 2 1 0 1 0 1 1 0 1 1 1 2 1 0 1 0 0 0 0 1 2 2 1 0 0 1 2 1 2 0 2 0 0 0 1 1",
"output": "35"
},
{
"input": "65\n2 2 2 3 0 2 1 2 3 3 1 3 1 2 1 3 2 3 2 2 2 1 2 0 3 1 3 1 1 3 1 3 3 3 3 3 1 3 0 3 1 3 1 2 2 3 2 0 3 1 3 2 1 2 2 2 3 3 2 3 3 3 2 2 3",
"output": "13"
},
{
"input": "65\n3 2 3 3 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 3 3 2 2 2 3 3 2 3 3 2 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 3 2 2 3 3 2 3 3 1 3 3 3 3",
"output": "6"
},
{
"input": "70\n1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 3 1 1 0 1 2 0 2 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 1 3 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1",
"output": "43"
},
{
"input": "70\n2 3 3 3 1 3 3 1 2 1 1 2 2 3 0 2 3 3 1 3 3 2 2 3 3 3 2 2 2 2 1 3 3 0 2 1 1 3 2 3 3 2 2 3 1 3 1 2 3 2 3 3 2 2 2 3 1 1 2 1 3 3 2 2 3 3 3 1 1 1",
"output": "16"
},
{
"input": "70\n3 3 2 2 1 2 1 2 2 2 2 2 3 3 2 3 3 3 3 2 2 2 2 3 3 3 1 3 3 3 2 3 3 3 3 2 3 3 1 3 1 3 2 3 3 2 3 3 3 2 3 2 3 3 1 2 3 3 2 2 2 3 2 3 3 3 3 3 3 1",
"output": "10"
},
{
"input": "75\n1 0 0 1 1 0 0 1 0 1 2 0 0 2 1 1 0 0 0 0 0 0 2 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 2 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0",
"output": "51"
},
{
"input": "75\n1 3 3 3 1 1 3 2 3 3 1 3 3 3 2 1 3 2 2 3 1 1 1 1 1 1 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 2 2 2 1 2 3 3 2 2 3 0 1 1 3 3 0 0 1 1 3 2 3 3 3 3 1 2 2 3 3 3 3 1",
"output": "16"
},
{
"input": "75\n3 3 3 3 2 2 3 2 2 3 2 2 1 2 3 3 2 2 3 3 1 2 2 2 1 3 3 3 1 2 2 3 3 3 2 3 2 2 2 3 3 1 3 2 2 3 3 3 0 3 2 1 3 3 2 3 3 3 3 1 2 3 3 3 2 2 3 3 3 3 2 2 3 3 1",
"output": "11"
},
{
"input": "80\n0 0 0 0 2 0 1 1 1 1 1 0 0 0 0 2 0 0 1 0 0 0 0 1 1 0 2 2 1 1 0 1 0 1 0 1 1 1 0 1 2 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 2 2 0 1 1 0 0 0 0 0 0 0 0 1",
"output": "56"
},
{
"input": "80\n2 2 3 3 2 1 0 1 0 3 2 2 3 2 1 3 1 3 3 2 3 3 3 2 3 3 3 2 1 3 3 1 3 3 3 3 3 3 2 2 2 1 3 2 1 3 2 1 1 0 1 1 2 1 3 0 1 2 3 2 2 3 2 3 1 3 3 2 1 1 0 3 3 3 3 1 2 1 2 0",
"output": "17"
},
{
"input": "80\n2 3 3 2 2 2 3 3 2 3 3 3 3 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 1 3 2 3 3 0 3 1 2 3 3 1 2 3 2 3 3 2 3 3 3 3 3 2 2 3 0 3 3 3 3 3 2 2 3 2 3 3 3 3 3 2 3 2 3 3 3 3 2 3",
"output": "9"
},
{
"input": "85\n0 1 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 2 0 1 0 0 2 0 1 1 0 0 0 0 2 2 0 0 0 1 0 0 0 1 2 0 1 0 0 0 2 1 1 2 0 3 1 0 2 2 1 0 0 1 1 0 0 0 0 1 0 2 1 1 2 1 0 0 1 2 1 2 0 0 1 0 1 0",
"output": "54"
},
{
"input": "85\n2 3 1 3 2 3 1 3 3 2 1 2 1 2 2 3 2 2 3 2 0 3 3 2 1 2 2 2 3 3 2 3 3 3 2 1 1 3 1 3 2 2 2 3 3 2 3 2 3 1 1 3 2 3 1 3 3 2 3 3 2 2 3 0 1 1 2 2 2 2 1 2 3 1 3 3 1 3 2 2 3 2 3 3 3",
"output": "19"
},
{
"input": "85\n1 2 1 2 3 2 3 3 3 3 3 3 3 2 1 3 2 3 3 3 3 2 3 3 3 1 3 3 3 3 2 3 3 3 3 3 3 2 2 1 3 3 3 3 2 2 3 1 1 2 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 3 3 1 1 1 3 3 3 3 1 3 3 3 1 3 3 1 3 2 3",
"output": "9"
},
{
"input": "90\n2 0 1 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 0 2 2 0 2 0 0 2 1 0 2 0 1 0 1 0 0 1 2 2 0 0 1 0 0 1 0 1 0 2 0 1 1 1 0 1 1 0 1 0 2 0 1 0 1 0 0 0 1 0 0 1 2 0 0 0 1 0 0 2 2 0 0 0 0 0 1 3 1 1 0 1",
"output": "57"
},
{
"input": "90\n2 3 3 3 2 3 2 1 3 0 3 2 3 3 2 1 3 3 2 3 2 3 3 2 1 3 1 3 3 1 2 2 3 3 2 1 2 3 2 3 0 3 3 2 2 3 1 0 3 3 1 3 3 3 3 2 1 2 2 1 3 2 1 3 3 1 2 0 2 2 3 2 2 3 3 3 1 3 2 1 2 3 3 2 3 2 3 3 2 1",
"output": "17"
},
{
"input": "90\n2 3 2 3 2 2 3 3 2 3 2 1 2 3 3 3 2 3 2 3 3 2 3 3 3 1 3 3 1 3 2 3 2 2 1 3 3 3 3 3 3 3 3 3 3 2 3 2 3 2 1 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 1 3 2 3 3 3 2 2 3 2 3 2 1 3 2",
"output": "9"
},
{
"input": "95\n0 0 3 0 2 0 1 0 0 2 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 1 0 0 2 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1 2 0 1 2 2 0 0 1 0 2 0 0 0 1 0 2 1 2 1 0 1 0 0 0 1 0 0 1 1 2 1 1 1 1 2 0 0 0 0 0 1 1 0 1",
"output": "61"
},
{
"input": "95\n2 3 3 2 1 1 3 3 3 2 3 3 3 2 3 2 3 3 3 2 3 2 2 3 3 2 1 2 3 3 3 1 3 0 3 3 1 3 3 1 0 1 3 3 3 0 2 1 3 3 3 3 0 1 3 2 3 3 2 1 3 1 2 1 1 2 3 0 3 3 2 1 3 2 1 3 3 3 2 2 3 2 3 3 3 2 1 3 3 3 2 3 3 1 2",
"output": "15"
},
{
"input": "95\n2 3 3 2 3 2 2 1 3 1 2 1 2 3 1 2 3 3 1 3 3 3 1 2 3 2 2 2 2 3 3 3 2 2 3 3 3 3 3 1 2 2 3 3 3 3 2 3 2 2 2 3 3 2 3 3 3 3 3 3 3 0 3 2 0 3 3 1 3 3 3 2 3 2 3 2 3 3 3 3 2 2 1 1 3 3 3 3 3 1 3 3 3 3 2",
"output": "14"
},
{
"input": "100\n1 0 2 0 0 0 0 2 0 0 0 1 0 1 0 0 1 0 1 2 0 1 1 0 0 1 0 1 1 0 0 0 2 0 1 0 0 2 0 0 0 0 0 1 1 1 0 0 1 0 2 0 0 0 0 1 0 1 0 1 0 1 0 1 2 2 0 0 2 0 1 0 1 0 1 0 0 0 1 0 0 2 1 1 1 0 0 1 0 0 0 2 0 0 2 1 1 0 0 2",
"output": "63"
},
{
"input": "100\n3 2 1 3 2 3 2 3 2 2 3 1 3 3 3 3 3 2 2 3 2 2 3 2 3 3 3 2 3 1 2 1 3 3 3 3 1 3 3 3 3 3 2 3 2 1 3 3 1 2 2 3 1 3 3 1 2 2 1 3 1 3 2 2 3 3 1 3 2 3 1 2 1 2 3 3 2 2 1 2 3 3 3 3 3 1 3 3 3 3 2 1 3 0 3 3 3 2 3 3",
"output": "15"
},
{
"input": "100\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"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",
"output": "0"
},
{
"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",
"output": "50"
},
{
"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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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": "50"
},
{
"input": "99\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",
"output": "49"
},
{
"input": "100\n2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "100"
},
{
"input": "2\n0 3",
"output": "1"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n2 0",
"output": "1"
},
{
"input": "2\n2 1",
"output": "0"
},
{
"input": "2\n2 3",
"output": "0"
},
{
"input": "2\n3 0",
"output": "1"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "100\n3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0",
"output": "50"
}
] | 1,684,045,925
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 88
| 78
| 0
|
input()
p = 3
c = 0
for i in list(map(int, input().split())):
if i == p and p != 3:
i = 0
elif i==3 and p!=3:
i -= p
if i == 0:
c += 1
p = i
print(c)
quit()
|
Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
|
```python
input()
p = 3
c = 0
for i in list(map(int, input().split())):
if i == p and p != 3:
i = 0
elif i==3 and p!=3:
i -= p
if i == 0:
c += 1
p = i
print(c)
quit()
```
| 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) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — 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 — 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"
}
] | 1,665,473,808
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
possible_degrees = [60,90,108,120,135,140,144,150,156,160,162,165,168,170,171,172,174,175,176,177,178,179]
for _ in range(int(input()))
degree = int(input())
print(['NO','YES'][degree in possible_degrees])
|
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) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — 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 — a square.
|
```python
possible_degrees = [60,90,108,120,135,140,144,150,156,160,162,165,168,170,171,172,174,175,176,177,178,179]
for _ in range(int(input()))
degree = int(input())
print(['NO','YES'][degree in possible_degrees])
```
| -1
|
|
854
|
A
|
Fraction
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
|
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
|
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
|
[
"3\n",
"4\n",
"12\n"
] |
[
"1 2\n",
"1 3\n",
"5 7\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,541,597,393
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 109
| 0
|
IN,mx = int(input()),0
for i in range(1,(IN+1)//2) :
if i/(IN-i) >= mx :
mx = i/(IN-i)
a,b = i,(IN-i)
for i in range((IN+1)//2,IN-1) :
if i/(IN-i) >= mx :
x,y = i,(IN-i)
for d in range(2,max(x,y)+1) :
if all([x/d%1 != 0.0,y/d%1 != 0.0]) or all([max(x,y)/d%1 != 0.0,min(x,y)/d%1 == 0.0,d==min(x,y)]):
print(x,y)
sd = 1
break
else :
sd = 0
if sd == 1 :
mx = i/(IN-i)
a,b = i,(IN-i)
print(a,b)
|
Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none
|
```python
IN,mx = int(input()),0
for i in range(1,(IN+1)//2) :
if i/(IN-i) >= mx :
mx = i/(IN-i)
a,b = i,(IN-i)
for i in range((IN+1)//2,IN-1) :
if i/(IN-i) >= mx :
x,y = i,(IN-i)
for d in range(2,max(x,y)+1) :
if all([x/d%1 != 0.0,y/d%1 != 0.0]) or all([max(x,y)/d%1 != 0.0,min(x,y)/d%1 == 0.0,d==min(x,y)]):
print(x,y)
sd = 1
break
else :
sd = 0
if sd == 1 :
mx = i/(IN-i)
a,b = i,(IN-i)
print(a,b)
```
| 0
|
|
884
|
A
|
Book Reading
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
|
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
|
[
"2 2\n86400 86398\n",
"2 86400\n0 86400\n"
] |
[
"2\n",
"1\n"
] |
none
| 0
|
[
{
"input": "2 2\n86400 86398",
"output": "2"
},
{
"input": "2 86400\n0 86400",
"output": "1"
},
{
"input": "2 86400\n1 86399",
"output": "2"
},
{
"input": "100 1000000\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "12"
},
{
"input": "1 1\n86399",
"output": "1"
},
{
"input": "6 1200\n86400 86400 86000 86000 86000 86400",
"output": "5"
},
{
"input": "6 1200\n86400 86400 86000 86000 86001 86399",
"output": "6"
},
{
"input": "4 172799\n1 1 86400 0",
"output": "4"
},
{
"input": "4 172799\n0 86400 86399 0",
"output": "4"
},
{
"input": "6 1\n1 1 86400 1 86399 1",
"output": "1"
},
{
"input": "4 1\n86400 86399 86400 86400",
"output": "2"
},
{
"input": "4 1\n86400 86400 0 86400",
"output": "3"
}
] | 1,670,688,953
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
def count_days(number_of_days, seconds_needed, input2):
counter = 0
for i in range(number_of_days):
if seconds_needed + input2[i] <= 86400:
counter = i + 1
return counter
def main():
input1 = list(map(int, input().split()))
number_of_days = input1[0]
seconds_needed = input1[1]
input2 = list(map(int, input().split()))
count = count_days(number_of_days, seconds_needed, input2)
print(count)
main()
|
Title: Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
Output Specification:
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
Demo Input:
['2 2\n86400 86398\n', '2 86400\n0 86400\n']
Demo Output:
['2\n', '1\n']
Note:
none
|
```python
def count_days(number_of_days, seconds_needed, input2):
counter = 0
for i in range(number_of_days):
if seconds_needed + input2[i] <= 86400:
counter = i + 1
return counter
def main():
input1 = list(map(int, input().split()))
number_of_days = input1[0]
seconds_needed = input1[1]
input2 = list(map(int, input().split()))
count = count_days(number_of_days, seconds_needed, input2)
print(count)
main()
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,692,371,576
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 92
| 0
|
# A. Укладка доминошками
mn = [int(i) for i in input().split()]
print(round(mn[0] * mn[1] / 2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
# A. Укладка доминошками
mn = [int(i) for i in input().split()]
print(round(mn[0] * mn[1] / 2))
```
| 0
|
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) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — 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 — 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"
}
] | 1,633,260,212
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
t=int(input())
n=int()
for i in range(t):
a=int(input())
if a*n==(n-2)*180:
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) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — 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 — a square.
|
```python
t=int(input())
n=int()
for i in range(t):
a=int(input())
if a*n==(n-2)*180:
print("YES")
else:
print("NO")
```
| 0
|
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