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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
177
|
A1
|
Good Matrix Elements
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
|
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
|
Print a single integer — the sum of good matrix elements.
|
[
"3\n1 2 3\n4 5 6\n7 8 9\n",
"5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n"
] |
[
"45\n",
"17\n"
] |
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
| 30
|
[
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "45"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "17"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33",
"output": "756"
},
{
"input": "3\n19 7 16\n12 15 5\n15 15 5",
"output": "109"
},
{
"input": "3\n36 4 33\n11 46 32\n20 49 34",
"output": "265"
},
{
"input": "3\n79 91 74\n33 82 22\n18 28 54",
"output": "481"
},
{
"input": "5\n7 0 8 1 7\n5 1 1 0 4\n4 2 8 1 6\n1 2 3 2 7\n6 0 1 9 6",
"output": "65"
},
{
"input": "5\n27 20 28 11 17\n25 21 1 20 14\n14 22 28 1 6\n1 2 23 2 7\n6 0 1 29 6",
"output": "225"
},
{
"input": "5\n57 50 58 41 17\n25 21 1 50 44\n44 22 28 31 36\n31 32 23 32 37\n6 0 31 59 6",
"output": "495"
},
{
"input": "5\n57 80 28 41 47\n85 51 61 50 74\n44 82 28 31 36\n31 32 23 32 37\n66 60 31 59 6",
"output": "705"
},
{
"input": "5\n13 58 10 17 43\n61 73 100 0 9\n52 38 16 22 96\n11 4 14 67 62\n70 89 7 98 83",
"output": "708"
},
{
"input": "5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 1 0 1 0\n0 0 0 0 1\n0 0 5 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "5"
},
{
"input": "5\n0 0 0 0 1\n0 1 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "3\n0 0 0\n0 0 0\n0 0 0",
"output": "0"
}
] | 1,547,477,523
| 2,147,483,647
|
Python 3
|
OK
|
TESTS1
| 17
| 248
| 0
|
a=int(input())
b=[]
c=0
d=0
for i in range(a):
c=list(map(int,input().split()))
b.append(c)
for i in range(len(b)):
d+=b[i][i]
d+=b[i][0-i-1]
d+=b[(a-1)//2][i]
d+=b[i][(a-1)//2]
print(d-(b[(a-1)//2][(a-1)//2]*3))
|
Title: Good Matrix Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
Input Specification:
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
Output Specification:
Print a single integer — the sum of good matrix elements.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n']
Demo Output:
['45\n', '17\n']
Note:
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
|
```python
a=int(input())
b=[]
c=0
d=0
for i in range(a):
c=list(map(int,input().split()))
b.append(c)
for i in range(len(b)):
d+=b[i][i]
d+=b[i][0-i-1]
d+=b[(a-1)//2][i]
d+=b[i][(a-1)//2]
print(d-(b[(a-1)//2][(a-1)//2]*3))
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
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},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,634,391,068
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
a = str(input())
b = str(input())
ans = []
n = int(len(a))
for i in range(0,n) :
if a[i] != b[i] :
ans.append(1)
if a[i] == b[i] :
ans.append(0)
for i in ans :
print(i,end="")
print("\n")
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
a = str(input())
b = str(input())
ans = []
n = int(len(a))
for i in range(0,n) :
if a[i] != b[i] :
ans.append(1)
if a[i] == b[i] :
ans.append(0)
for i in ans :
print(i,end="")
print("\n")
```
| 3.9885
|
446
|
A
|
DZY Loves Sequences
|
PROGRAMMING
| 1,600
|
[
"dp",
"implementation",
"two pointers"
] | null | null |
DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
In a single line print the answer to the problem — the maximum length of the required subsegment.
|
[
"6\n7 2 3 1 5 6\n"
] |
[
"5\n"
] |
You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4.
| 500
|
[
{
"input": "6\n7 2 3 1 5 6",
"output": "5"
},
{
"input": "10\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422",
"output": "9"
},
{
"input": "50\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 189641422 25202363 350490028 783368691 102520060 44897764 967513927 365180541 540383427 304089173 303455737 35005212 521595369 294702568 726956430 336465783 861021531 59961394 89018457 101513930 125898168 131176230 145174068 233665124 278722863 315634023 369133070 468703136 628175012 635723059 653377374 656478043 801979803 859484422 914544920 608413785 756898538 734575199 973594325 149798316 38664371",
"output": "19"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "2"
},
{
"input": "5\n1 2 3 4 1",
"output": "5"
},
{
"input": "10\n1 2 3 4 5 5 6 7 8 9",
"output": "6"
},
{
"input": "5\n1 1 1 1 1",
"output": "2"
},
{
"input": "5\n1 1 2 3 4",
"output": "5"
},
{
"input": "5\n1 2 3 1 6",
"output": "5"
},
{
"input": "1\n42",
"output": "1"
},
{
"input": "5\n1 2 42 3 4",
"output": "4"
},
{
"input": "5\n1 5 9 6 10",
"output": "4"
},
{
"input": "5\n5 2 3 4 5",
"output": "5"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "5\n1 2 3 3 4",
"output": "4"
},
{
"input": "8\n1 2 3 4 1 5 6 7",
"output": "5"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "3\n5 1 2",
"output": "3"
},
{
"input": "4\n1 4 3 4",
"output": "4"
},
{
"input": "6\n7 2 12 4 5 6",
"output": "5"
},
{
"input": "6\n7 2 3 1 4 5",
"output": "4"
},
{
"input": "6\n2 3 5 5 6 7",
"output": "6"
},
{
"input": "5\n2 4 7 6 8",
"output": "5"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "5\n4 1 2 3 4",
"output": "5"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6",
"output": "7"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "4\n4 3 1 2",
"output": "3"
},
{
"input": "6\n1 2 2 3 4 5",
"output": "5"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n5 1 2 3",
"output": "4"
},
{
"input": "5\n9 1 2 3 4",
"output": "5"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "5\n1 3 2 4 5",
"output": "4"
},
{
"input": "6\n1 2 1 2 4 5",
"output": "5"
},
{
"input": "10\n1 1 5 3 2 9 9 7 7 6",
"output": "3"
},
{
"input": "6\n1 2 3 100000 100 101",
"output": "6"
},
{
"input": "4\n3 3 3 4",
"output": "3"
},
{
"input": "3\n4 3 5",
"output": "3"
},
{
"input": "5\n1 3 2 3 4",
"output": "4"
},
{
"input": "10\n1 2 3 4 5 10 10 11 12 13",
"output": "10"
},
{
"input": "7\n11 2 1 2 13 4 14",
"output": "5"
},
{
"input": "3\n5 1 3",
"output": "3"
},
{
"input": "4\n1 5 3 4",
"output": "4"
},
{
"input": "10\n1 2 3 4 100 6 7 8 9 10",
"output": "10"
},
{
"input": "3\n5 3 5",
"output": "3"
},
{
"input": "5\n100 100 7 8 9",
"output": "4"
},
{
"input": "5\n1 2 3 4 5",
"output": "5"
},
{
"input": "5\n1 2 4 4 5",
"output": "5"
},
{
"input": "6\n7 4 5 6 7 8",
"output": "6"
},
{
"input": "9\n3 4 1 6 3 4 5 6 7",
"output": "7"
},
{
"input": "3\n1000 1 2",
"output": "3"
},
{
"input": "3\n20 1 9",
"output": "3"
},
{
"input": "6\n7 2 3 1 4 6",
"output": "4"
},
{
"input": "3\n100 5 10",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "6\n4 2 8 1 2 5",
"output": "4"
},
{
"input": "3\n25 1 6",
"output": "3"
},
{
"input": "10\n17 99 23 72 78 36 5 43 95 9",
"output": "5"
},
{
"input": "7\n21 16 22 21 11 13 19",
"output": "4"
},
{
"input": "5\n1 2 5 3 4",
"output": "4"
},
{
"input": "6\n2 2 2 3 4 5",
"output": "5"
},
{
"input": "5\n1 3 1 2 3",
"output": "4"
},
{
"input": "3\n81 33 64",
"output": "3"
},
{
"input": "7\n14 3 3 19 13 19 15",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 42 7 8 9",
"output": "9"
},
{
"input": "5\n2 3 7 5 6",
"output": "5"
},
{
"input": "5\n1 3 3 4 5",
"output": "5"
},
{
"input": "6\n1 5 4 3 4 5",
"output": "4"
}
] | 1,687,800,540
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 61
| 0
|
n = int(input())
arr = [int(x) for x in input().split()]
right = [1] * n
left = [1] * n
for i in range(1,n):
if arr[i] > arr[i-1]:
left[i] += left[i-1]
for i in range(n-2, -1 , -1):
if arr[i+1] > arr[i]:
right[i] += right[i+1]
ans = max(max(right), max(left))
for i in range(1,n-1):
if arr[i-1] + 1< arr[i+1]:
ans = max(ans, left[i-1]+ right[i+1] + 1)
print(ans)
|
Title: DZY Loves Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
In a single line print the answer to the problem — the maximum length of the required subsegment.
Demo Input:
['6\n7 2 3 1 5 6\n']
Demo Output:
['5\n']
Note:
You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4.
|
```python
n = int(input())
arr = [int(x) for x in input().split()]
right = [1] * n
left = [1] * n
for i in range(1,n):
if arr[i] > arr[i-1]:
left[i] += left[i-1]
for i in range(n-2, -1 , -1):
if arr[i+1] > arr[i]:
right[i] += right[i+1]
ans = max(max(right), max(left))
for i in range(1,n-1):
if arr[i-1] + 1< arr[i+1]:
ans = max(ans, left[i-1]+ right[i+1] + 1)
print(ans)
```
| 0
|
|
296
|
A
|
Yaroslav and Permutations
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
|
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
|
[
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
| 500
|
[
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"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": "NO"
},
{
"input": "99\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",
"output": "NO"
},
{
"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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "YES"
},
{
"input": "99\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,579,280,978
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 216
| 307,200
|
#80
n=int(input())
l=list(map(int,input().split()))
s=0
c=0
t=0
for a in range(len(l)-1):
for b in range(a+1,len(l)):
if l[a]!=l[b]:
s=l[a+1]
l[a+1]=l[b]
l[b]=s
c=1
break
if c==0:
print('No')
t=1
break
else:
c=0
if t==0:
print('YES')
|
Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
|
```python
#80
n=int(input())
l=list(map(int,input().split()))
s=0
c=0
t=0
for a in range(len(l)-1):
for b in range(a+1,len(l)):
if l[a]!=l[b]:
s=l[a+1]
l[a+1]=l[b]
l[b]=s
c=1
break
if c==0:
print('No')
t=1
break
else:
c=0
if t==0:
print('YES')
```
| 0
|
|
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"
}
] | 1,670,176,606
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 46
| 0
|
numbers = [int(x) for x in input().split()]
a, b, c, d = numbers[0], numbers[1], numbers[2], numbers[3]
print((a / b) / (1 - ((1 - a / b) * (1 - c / d))))
|
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
numbers = [int(x) for x in input().split()]
a, b, c, d = numbers[0], numbers[1], numbers[2], numbers[3]
print((a / b) / (1 - ((1 - a / b) * (1 - c / d))))
```
| 3
|
|
220
|
A
|
Little Elephant and Problem
|
PROGRAMMING
| 1,300
|
[
"implementation",
"sortings"
] | null | null |
The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array *a* of length *n* and possibly swapped some elements of the array.
The Little Elephant doesn't want to call the police until he understands if he could have accidentally changed the array himself. He thinks that he could have accidentally changed array *a*, only if array *a* can be sorted in no more than one operation of swapping elements (not necessarily adjacent). That is, the Little Elephant could have accidentally swapped some two elements.
Help the Little Elephant, determine if he could have accidentally changed the array *a*, sorted by non-decreasing, himself.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the size of array *a*. The next line contains *n* positive integers, separated by single spaces and not exceeding 109, — array *a*.
Note that the elements of the array are not necessarily distinct numbers.
|
In a single line print "YES" (without the quotes) if the Little Elephant could have accidentally changed the array himself, and "NO" (without the quotes) otherwise.
|
[
"2\n1 2\n",
"3\n3 2 1\n",
"4\n4 3 2 1\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the array has already been sorted, so to sort it, we need 0 swap operations, that is not more than 1. Thus, the answer is "YES".
In the second sample we can sort the array if we swap elements 1 and 3, so we need 1 swap operation to sort the array. Thus, the answer is "YES".
In the third sample we can't sort the array in more than one swap operation, so the answer is "NO".
| 500
|
[
{
"input": "2\n1 2",
"output": "YES"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "4\n4 3 2 1",
"output": "NO"
},
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "YES"
},
{
"input": "9\n7 7 8 8 10 10 10 10 1000000000",
"output": "YES"
},
{
"input": "10\n1 2 9 4 5 6 7 8 3 10",
"output": "YES"
},
{
"input": "4\n2 2 2 1",
"output": "YES"
},
{
"input": "10\n1 2 4 4 4 5 5 7 7 10",
"output": "YES"
},
{
"input": "10\n4 5 11 12 13 14 16 16 16 18",
"output": "YES"
},
{
"input": "20\n38205814 119727790 127848638 189351562 742927936 284688399 318826601 326499046 387938139 395996609 494453625 551393005 561264192 573569187 600766727 606718722 730549586 261502770 751513115 943272321",
"output": "YES"
},
{
"input": "47\n6 277 329 393 410 432 434 505 529 545 650 896 949 1053 1543 1554 1599 1648 1927 1976 1998 2141 2248 2384 2542 2638 2995 3155 3216 3355 3409 3597 3851 3940 4169 4176 4378 4378 4425 4490 4627 4986 5025 5033 5374 5453 5644",
"output": "YES"
},
{
"input": "50\n6 7 8 4 10 3 2 7 1 3 10 3 4 7 2 3 7 4 10 6 8 10 9 6 5 10 9 6 1 8 9 4 3 7 3 10 5 3 10 1 6 10 6 7 10 7 1 5 9 5",
"output": "NO"
},
{
"input": "100\n3 7 7 8 15 25 26 31 37 41 43 43 46 64 65 82 94 102 102 103 107 124 125 131 140 145 146 150 151 160 160 161 162 165 169 175 182 191 201 211 214 216 218 304 224 229 236 241 244 249 252 269 270 271 273 289 285 295 222 307 312 317 319 319 320 321 325 330 340 341 345 347 354 356 366 366 375 376 380 383 386 398 401 407 414 417 423 426 431 438 440 444 446 454 457 458 458 466 466 472",
"output": "NO"
},
{
"input": "128\n1 2 4 6 8 17 20 20 23 33 43 49 49 49 52 73 74 75 82 84 85 87 90 91 102 103 104 105 111 111 401 142 142 152 155 160 175 176 178 181 183 184 187 188 191 193 326 202 202 214 224 225 236 239 240 243 246 247 249 249 257 257 261 264 265 271 277 281 284 284 286 289 290 296 297 303 305 307 307 317 318 320 322 200 332 342 393 349 350 350 369 375 381 381 385 385 387 393 347 397 398 115 402 407 407 408 410 411 411 416 423 426 429 429 430 440 447 449 463 464 466 471 473 480 480 483 497 503",
"output": "NO"
},
{
"input": "4\n5 12 12 6",
"output": "YES"
},
{
"input": "5\n1 3 3 3 2",
"output": "YES"
},
{
"input": "4\n2 1 1 1",
"output": "YES"
},
{
"input": "2\n1 1",
"output": "YES"
},
{
"input": "4\n1000000000 1 1000000000 1",
"output": "YES"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 1",
"output": "YES"
},
{
"input": "6\n1 2 3 4 5 3",
"output": "NO"
},
{
"input": "9\n3 3 3 2 2 2 1 1 1",
"output": "NO"
},
{
"input": "4\n4 1 2 3",
"output": "NO"
},
{
"input": "6\n3 4 5 6 7 2",
"output": "NO"
},
{
"input": "4\n4 2 1 3",
"output": "NO"
},
{
"input": "4\n3 3 2 2",
"output": "NO"
},
{
"input": "4\n3 2 1 1",
"output": "NO"
},
{
"input": "4\n4 5 1 1",
"output": "NO"
},
{
"input": "6\n1 6 2 4 3 5",
"output": "NO"
},
{
"input": "5\n1 4 5 2 3",
"output": "NO"
},
{
"input": "4\n2 2 1 1",
"output": "NO"
},
{
"input": "5\n1 4 3 2 1",
"output": "NO"
},
{
"input": "5\n1 4 2 2 3",
"output": "NO"
},
{
"input": "6\n1 2 3 1 2 3",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "NO"
},
{
"input": "5\n5 1 2 3 4",
"output": "NO"
},
{
"input": "5\n3 3 3 2 2",
"output": "NO"
},
{
"input": "5\n100 5 6 10 7",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "NO"
},
{
"input": "5\n4 4 1 1 1",
"output": "NO"
},
{
"input": "5\n1 2 5 3 4",
"output": "NO"
},
{
"input": "4\n3 4 1 2",
"output": "NO"
},
{
"input": "4\n2 4 1 5",
"output": "NO"
},
{
"input": "5\n1 3 3 2 2",
"output": "NO"
},
{
"input": "5\n1 5 4 4 4",
"output": "YES"
},
{
"input": "7\n3 2 1 2 3 5 4",
"output": "NO"
},
{
"input": "5\n1 1 3 2 2",
"output": "YES"
},
{
"input": "9\n1 8 7 7 7 7 7 8 3",
"output": "YES"
},
{
"input": "5\n1 3 2 3 3",
"output": "YES"
},
{
"input": "10\n4 4 4 4 10 4 4 4 4 4",
"output": "YES"
},
{
"input": "8\n3 6 6 6 6 6 4 9",
"output": "YES"
},
{
"input": "4\n4 4 3 3",
"output": "NO"
},
{
"input": "4\n3 2 2 4",
"output": "YES"
},
{
"input": "5\n2 2 1 3 3",
"output": "YES"
},
{
"input": "5\n1 2 7 3 5",
"output": "NO"
},
{
"input": "5\n2 3 4 5 1",
"output": "NO"
},
{
"input": "6\n1 4 3 6 2 5",
"output": "NO"
},
{
"input": "5\n3 3 1 5 4",
"output": "NO"
},
{
"input": "4\n1 2 1 2",
"output": "YES"
},
{
"input": "6\n4 5 3 4 2 6",
"output": "NO"
},
{
"input": "11\n1 2 3 4 5 1 2 3 4 5 1",
"output": "NO"
},
{
"input": "6\n6 1 2 3 4 5",
"output": "NO"
},
{
"input": "5\n4 1 1 1 1",
"output": "YES"
},
{
"input": "9\n1 2 3 5 4 6 7 8 9",
"output": "YES"
},
{
"input": "6\n6 1 2 3 4 2",
"output": "NO"
},
{
"input": "6\n2 2 2 2 3 2",
"output": "YES"
},
{
"input": "3\n2 1 1",
"output": "YES"
},
{
"input": "5\n1 2 1 1 2",
"output": "YES"
},
{
"input": "5\n1 2 2 1 2",
"output": "YES"
},
{
"input": "8\n5 5 5 5 5 5 1 1",
"output": "NO"
}
] | 1,591,219,239
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 96
| 249
| 11,059,200
|
BS="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def to_base(s, b):
res = ""
while s:
res+=BS[s%b]
s//= b
return res[::-1] or "0"
alpha = "abcdefghijklmnopqrstuvwxyz"
from collections import Counter
t = int(input())
nums = list(map(int, input().split()))
so = sorted(nums)
counter = 0
for i in range(len(so)):
if so[i]!=nums[i]:
counter += 1
if counter==2 or counter==0:
print("YES")
else:
print("NO")
|
Title: Little Elephant and Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array *a* of length *n* and possibly swapped some elements of the array.
The Little Elephant doesn't want to call the police until he understands if he could have accidentally changed the array himself. He thinks that he could have accidentally changed array *a*, only if array *a* can be sorted in no more than one operation of swapping elements (not necessarily adjacent). That is, the Little Elephant could have accidentally swapped some two elements.
Help the Little Elephant, determine if he could have accidentally changed the array *a*, sorted by non-decreasing, himself.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the size of array *a*. The next line contains *n* positive integers, separated by single spaces and not exceeding 109, — array *a*.
Note that the elements of the array are not necessarily distinct numbers.
Output Specification:
In a single line print "YES" (without the quotes) if the Little Elephant could have accidentally changed the array himself, and "NO" (without the quotes) otherwise.
Demo Input:
['2\n1 2\n', '3\n3 2 1\n', '4\n4 3 2 1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the array has already been sorted, so to sort it, we need 0 swap operations, that is not more than 1. Thus, the answer is "YES".
In the second sample we can sort the array if we swap elements 1 and 3, so we need 1 swap operation to sort the array. Thus, the answer is "YES".
In the third sample we can't sort the array in more than one swap operation, so the answer is "NO".
|
```python
BS="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def to_base(s, b):
res = ""
while s:
res+=BS[s%b]
s//= b
return res[::-1] or "0"
alpha = "abcdefghijklmnopqrstuvwxyz"
from collections import Counter
t = int(input())
nums = list(map(int, input().split()))
so = sorted(nums)
counter = 0
for i in range(len(so)):
if so[i]!=nums[i]:
counter += 1
if counter==2 or counter==0:
print("YES")
else:
print("NO")
```
| 3
|
|
330
|
A
|
Cakeminator
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
|
The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
|
Output the maximum number of cake cells that the cakeminator can eat.
|
[
"3 4\nS...\n....\n..S.\n"
] |
[
"8\n"
] |
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).
| 500
|
[
{
"input": "3 4\nS...\n....\n..S.",
"output": "8"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "2 2\nSS\nSS",
"output": "0"
},
{
"input": "7 3\nS..\nS..\nS..\nS..\nS..\nS..\nS..",
"output": "14"
},
{
"input": "3 5\n..S..\nSSSSS\n..S..",
"output": "0"
},
{
"input": "10 10\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS",
"output": "0"
},
{
"input": "10 10\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS",
"output": "30"
},
{
"input": "10 10\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..",
"output": "80"
},
{
"input": "9 5\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS",
"output": "0"
},
{
"input": "9 9\n...S.....\nS.S.....S\n.S....S..\n.S.....SS\n.........\n..S.S..S.\n.SS......\n....S....\n..S...S..",
"output": "17"
},
{
"input": "5 6\nSSSSSS\nSSSSSS\nSSSSSS\nSS.S..\nS.S.SS",
"output": "0"
},
{
"input": "9 8\n........\n.......S\n........\nS.......\n........\n........\nS.......\n........\n.......S",
"output": "64"
},
{
"input": "9 7\n......S\n......S\nS.S.S..\n.......\n.......\n.S.....\n.S....S\n..S....\n.S....S",
"output": "28"
},
{
"input": "10 10\n.....S....\n....SS..S.\n.S...S....\n........SS\n.S.......S\nSS..S.....\n.SS.....SS\nS..S......\n.......SSS\nSSSSS....S",
"output": "10"
},
{
"input": "6 7\n..S.SS.\n......S\n....S.S\nSS..S..\nS..SS.S\n.....S.",
"output": "0"
},
{
"input": "10 6\n.SSSSS\nSSS.SS\nSSSSSS\nS.SSSS\nSSSSS.\nS.SSSS\nSS.SSS\n.SSS.S\n.SSS..\nSS..SS",
"output": "0"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "4 3\n.S.\nS.S\n.S.\nS.S",
"output": "0"
},
{
"input": "2 3\n...\nSSS",
"output": "3"
},
{
"input": "2 4\nS.SS\nS.SS",
"output": "2"
},
{
"input": "2 2\n..\n.S",
"output": "3"
},
{
"input": "3 2\n.S\n.S\nSS",
"output": "0"
},
{
"input": "2 4\nSS.S\n..S.",
"output": "0"
},
{
"input": "2 3\n...\nS..",
"output": "5"
}
] | 1,622,518,070
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 218
| 0
|
def f(i):
for j in range(n):
if l[j][i]=='S':
return False
return True
n,m=map(int,input().split())
r=0
c=0
l=[]
for i in range(n):
s=input()
if "S" not in s:
r+=1
c+=m
l.append(list(s))
for i in range(m):
if f(i):
c+=n-r
print(c)
|
Title: Cakeminator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
Input Specification:
The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
Output Specification:
Output the maximum number of cake cells that the cakeminator can eat.
Demo Input:
['3 4\nS...\n....\n..S.\n']
Demo Output:
['8\n']
Note:
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).
|
```python
def f(i):
for j in range(n):
if l[j][i]=='S':
return False
return True
n,m=map(int,input().split())
r=0
c=0
l=[]
for i in range(n):
s=input()
if "S" not in s:
r+=1
c+=m
l.append(list(s))
for i in range(m):
if f(i):
c+=n-r
print(c)
```
| 3
|
|
712
|
C
|
Memory and De-Evolution
|
PROGRAMMING
| 1,600
|
[
"greedy",
"math"
] | null | null |
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*?
|
The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=<<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively.
|
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*.
|
[
"6 3\n",
"8 5\n",
"22 4\n"
] |
[
"4\n",
"3\n",
"6\n"
] |
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 1,500
|
[
{
"input": "6 3",
"output": "4"
},
{
"input": "8 5",
"output": "3"
},
{
"input": "22 4",
"output": "6"
},
{
"input": "4 3",
"output": "3"
},
{
"input": "57 27",
"output": "4"
},
{
"input": "61 3",
"output": "9"
},
{
"input": "5 4",
"output": "3"
},
{
"input": "10 6",
"output": "3"
},
{
"input": "20 10",
"output": "4"
},
{
"input": "30 5",
"output": "6"
},
{
"input": "25 24",
"output": "3"
},
{
"input": "25 3",
"output": "7"
},
{
"input": "12 7",
"output": "3"
},
{
"input": "18 6",
"output": "5"
},
{
"input": "100000 3",
"output": "25"
},
{
"input": "100000 9999",
"output": "7"
},
{
"input": "9999 3",
"output": "20"
},
{
"input": "5323 32",
"output": "13"
},
{
"input": "6666 66",
"output": "12"
},
{
"input": "38578 32201",
"output": "3"
},
{
"input": "49449 5291",
"output": "7"
},
{
"input": "65310 32879",
"output": "3"
},
{
"input": "41183 4453",
"output": "7"
},
{
"input": "49127 9714",
"output": "6"
},
{
"input": "19684 12784",
"output": "3"
},
{
"input": "15332 5489",
"output": "4"
},
{
"input": "33904 32701",
"output": "3"
},
{
"input": "9258 2966",
"output": "5"
},
{
"input": "21648 11231",
"output": "3"
},
{
"input": "90952 47239",
"output": "3"
},
{
"input": "49298 23199",
"output": "4"
},
{
"input": "33643 24915",
"output": "3"
},
{
"input": "40651 5137",
"output": "6"
},
{
"input": "52991 15644",
"output": "5"
},
{
"input": "97075 62157",
"output": "3"
},
{
"input": "82767 53725",
"output": "3"
},
{
"input": "58915 26212",
"output": "4"
},
{
"input": "86516 16353",
"output": "6"
},
{
"input": "14746 7504",
"output": "3"
},
{
"input": "20404 7529",
"output": "4"
},
{
"input": "52614 8572",
"output": "6"
},
{
"input": "50561 50123",
"output": "3"
},
{
"input": "37509 7908",
"output": "5"
},
{
"input": "36575 23933",
"output": "3"
},
{
"input": "75842 8002",
"output": "7"
},
{
"input": "47357 2692",
"output": "8"
},
{
"input": "23214 4255",
"output": "6"
},
{
"input": "9474 46",
"output": "13"
},
{
"input": "79874 76143",
"output": "3"
},
{
"input": "63784 31333",
"output": "4"
},
{
"input": "70689 29493",
"output": "4"
},
{
"input": "43575 4086",
"output": "7"
},
{
"input": "87099 7410",
"output": "7"
},
{
"input": "75749 55910",
"output": "3"
},
{
"input": "87827 20996",
"output": "5"
},
{
"input": "31162 4580",
"output": "6"
},
{
"input": "63175 33696",
"output": "3"
},
{
"input": "15108 10033",
"output": "3"
},
{
"input": "82991 29195",
"output": "4"
},
{
"input": "48258 12837",
"output": "5"
},
{
"input": "59859 33779",
"output": "3"
},
{
"input": "93698 23890",
"output": "5"
},
{
"input": "42724 379",
"output": "12"
},
{
"input": "70434 39286",
"output": "3"
},
{
"input": "69826 18300",
"output": "5"
},
{
"input": "57825 17636",
"output": "5"
},
{
"input": "64898 2076",
"output": "9"
},
{
"input": "76375 67152",
"output": "3"
},
{
"input": "30698 3778",
"output": "7"
},
{
"input": "100 3",
"output": "10"
},
{
"input": "41 3",
"output": "8"
},
{
"input": "28 4",
"output": "7"
},
{
"input": "2487 19",
"output": "12"
},
{
"input": "100000 25000",
"output": "5"
},
{
"input": "10000 3",
"output": "20"
},
{
"input": "16 3",
"output": "6"
}
] | 1,621,205,004
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 76
| 109
| 0
|
x, y = map(int, input().split())
a, b, c = y, y, y
cnt = 0
while True:
if a >= x and b >= x and c >= x:
break
cnt += 1
if cnt % 3 == 0:
a = b+c - 1
elif cnt % 3 == 1:
b = c + a - 1
elif cnt % 3 == 2:
c = b+a - 1
print(cnt)
|
Title: Memory and De-Evolution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*?
Input Specification:
The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=<<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively.
Output Specification:
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*.
Demo Input:
['6 3\n', '8 5\n', '22 4\n']
Demo Output:
['4\n', '3\n', '6\n']
Note:
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
x, y = map(int, input().split())
a, b, c = y, y, y
cnt = 0
while True:
if a >= x and b >= x and c >= x:
break
cnt += 1
if cnt % 3 == 0:
a = b+c - 1
elif cnt % 3 == 1:
b = c + a - 1
elif cnt % 3 == 2:
c = b+a - 1
print(cnt)
```
| 3
|
|
355
|
B
|
Vasya and Public Transport
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Vasya often uses public transport. The transport in the city is of two types: trolleys and buses. The city has *n* buses and *m* trolleys, the buses are numbered by integers from 1 to *n*, the trolleys are numbered by integers from 1 to *m*.
Public transport is not free. There are 4 types of tickets:
1. A ticket for one ride on some bus or trolley. It costs *c*1 burles; 1. A ticket for an unlimited number of rides on some bus or on some trolley. It costs *c*2 burles; 1. A ticket for an unlimited number of rides on all buses or all trolleys. It costs *c*3 burles; 1. A ticket for an unlimited number of rides on all buses and trolleys. It costs *c*4 burles.
Vasya knows for sure the number of rides he is going to make and the transport he is going to use. He asked you for help to find the minimum sum of burles he will have to spend on the tickets.
|
The first line contains four integers *c*1,<=*c*2,<=*c*3,<=*c*4 (1<=≤<=*c*1,<=*c*2,<=*c*3,<=*c*4<=≤<=1000) — the costs of the tickets.
The second line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of buses and trolleys Vasya is going to use.
The third line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1000) — the number of times Vasya is going to use the bus number *i*.
The fourth line contains *m* integers *b**i* (0<=≤<=*b**i*<=≤<=1000) — the number of times Vasya is going to use the trolley number *i*.
|
Print a single number — the minimum sum of burles Vasya will have to spend on the tickets.
|
[
"1 3 7 19\n2 3\n2 5\n4 4 4\n",
"4 3 2 1\n1 3\n798\n1 2 3\n",
"100 100 8 100\n3 5\n7 94 12\n100 1 47 0 42\n"
] |
[
"12\n",
"1\n",
"16\n"
] |
In the first sample the profitable strategy is to buy two tickets of the first type (for the first bus), one ticket of the second type (for the second bus) and one ticket of the third type (for all trolleys). It totals to (2·1) + 3 + 7 = 12 burles.
In the second sample the profitable strategy is to buy one ticket of the fourth type.
In the third sample the profitable strategy is to buy two tickets of the third type: for all buses and for all trolleys.
| 1,000
|
[
{
"input": "1 3 7 19\n2 3\n2 5\n4 4 4",
"output": "12"
},
{
"input": "4 3 2 1\n1 3\n798\n1 2 3",
"output": "1"
},
{
"input": "100 100 8 100\n3 5\n7 94 12\n100 1 47 0 42",
"output": "16"
},
{
"input": "3 103 945 1000\n7 9\n34 35 34 35 34 35 34\n0 0 0 0 0 0 0 0 0",
"output": "717"
},
{
"input": "7 11 597 948\n4 1\n5 1 0 11\n7",
"output": "40"
},
{
"input": "7 32 109 645\n1 3\n0\n0 0 0",
"output": "0"
},
{
"input": "680 871 347 800\n10 100\n872 156 571 136 703 201 832 213 15 333\n465 435 870 95 660 237 694 594 423 405 27 866 325 490 255 989 128 345 278 125 708 210 771 848 961 448 871 190 745 343 532 174 103 999 874 221 252 500 886 129 185 208 137 425 800 34 696 39 198 981 91 50 545 885 194 583 475 415 162 712 116 911 313 488 646 189 429 756 728 30 985 114 823 111 106 447 296 430 307 388 345 458 84 156 169 859 274 934 634 62 12 839 323 831 24 907 703 754 251 938",
"output": "694"
},
{
"input": "671 644 748 783\n100 10\n520 363 816 957 635 753 314 210 763 819 27 970 520 164 195 230 708 587 568 707 343 30 217 227 755 277 773 497 900 589 826 666 115 784 494 467 217 892 658 388 764 812 248 447 876 581 94 915 675 967 508 754 768 79 261 934 603 712 20 199 997 501 465 91 897 257 820 645 217 105 564 8 668 171 168 18 565 840 418 42 808 918 409 617 132 268 13 161 194 628 213 199 545 448 113 410 794 261 211 539\n147 3 178 680 701 193 697 666 846 389",
"output": "783"
},
{
"input": "2 7 291 972\n63 92\n7 0 0 6 0 13 0 20 2 8 0 17 7 0 0 0 0 2 2 0 0 8 20 0 0 0 3 0 0 0 4 20 0 0 0 12 0 8 17 9 0 0 0 0 4 0 0 0 17 11 3 0 2 15 0 18 11 19 14 0 0 20 13\n0 0 0 3 7 0 0 0 0 8 13 6 15 0 7 0 0 20 0 0 12 0 12 0 15 0 0 1 11 14 0 11 12 0 0 0 0 0 16 16 0 17 20 0 11 0 0 20 14 0 16 0 3 6 12 0 0 0 0 0 15 3 0 9 17 12 20 17 0 0 0 0 15 9 0 14 10 10 1 20 16 17 20 6 6 0 0 16 4 6 0 7",
"output": "494"
},
{
"input": "4 43 490 945\n63 92\n0 0 0 0 0 0 6 5 18 0 6 4 0 17 0 19 0 19 7 16 0 0 0 9 10 13 7 0 10 16 0 0 0 0 0 14 0 14 9 15 0 0 2 0 0 0 0 5 0 0 0 11 11 0 0 0 0 0 10 12 3 0 0\n0 12 0 18 7 7 0 0 9 0 0 13 17 0 18 12 4 0 0 14 18 20 0 0 12 9 17 1 19 0 11 0 5 0 0 14 0 0 16 0 19 15 9 14 7 10 0 19 19 0 0 1 0 0 0 6 0 0 0 6 0 20 1 9 0 0 10 17 5 2 5 4 16 6 0 11 0 8 13 4 0 2 0 0 13 10 0 13 0 0 8 4",
"output": "945"
},
{
"input": "2 50 258 922\n42 17\n0 2 0 1 0 1 0 11 18 9 0 0 0 0 10 15 17 4 20 0 5 0 0 13 13 0 0 2 0 7 0 20 4 0 19 3 7 0 0 0 0 0\n8 4 19 0 0 19 14 17 6 0 18 0 0 0 0 9 0",
"output": "486"
},
{
"input": "1 1 3 4\n2 3\n1 1\n1 1 1",
"output": "4"
},
{
"input": "4 4 4 1\n1 1\n0\n0",
"output": "0"
},
{
"input": "100 100 1 100\n10 10\n100 100 100 100 100 100 100 100 100 100\n100 100 100 100 100 100 100 100 100 100",
"output": "2"
}
] | 1,653,299,864
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 46
| 0
|
import sys
input = sys.stdin.readline
c1, c2, c3, c4 = map(int, input().split())
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
z = c2/c1
s1, s2 = 0, 0
for i in a:
if i > z:
s1 += c2
else:
s1 += c1*i
for i in b:
if i > z:
s2 += c2
else:
s2 += c1*i
print(min(c4, min(s1, c3) + min(s2, c3)))
|
Title: Vasya and Public Transport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya often uses public transport. The transport in the city is of two types: trolleys and buses. The city has *n* buses and *m* trolleys, the buses are numbered by integers from 1 to *n*, the trolleys are numbered by integers from 1 to *m*.
Public transport is not free. There are 4 types of tickets:
1. A ticket for one ride on some bus or trolley. It costs *c*1 burles; 1. A ticket for an unlimited number of rides on some bus or on some trolley. It costs *c*2 burles; 1. A ticket for an unlimited number of rides on all buses or all trolleys. It costs *c*3 burles; 1. A ticket for an unlimited number of rides on all buses and trolleys. It costs *c*4 burles.
Vasya knows for sure the number of rides he is going to make and the transport he is going to use. He asked you for help to find the minimum sum of burles he will have to spend on the tickets.
Input Specification:
The first line contains four integers *c*1,<=*c*2,<=*c*3,<=*c*4 (1<=≤<=*c*1,<=*c*2,<=*c*3,<=*c*4<=≤<=1000) — the costs of the tickets.
The second line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of buses and trolleys Vasya is going to use.
The third line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1000) — the number of times Vasya is going to use the bus number *i*.
The fourth line contains *m* integers *b**i* (0<=≤<=*b**i*<=≤<=1000) — the number of times Vasya is going to use the trolley number *i*.
Output Specification:
Print a single number — the minimum sum of burles Vasya will have to spend on the tickets.
Demo Input:
['1 3 7 19\n2 3\n2 5\n4 4 4\n', '4 3 2 1\n1 3\n798\n1 2 3\n', '100 100 8 100\n3 5\n7 94 12\n100 1 47 0 42\n']
Demo Output:
['12\n', '1\n', '16\n']
Note:
In the first sample the profitable strategy is to buy two tickets of the first type (for the first bus), one ticket of the second type (for the second bus) and one ticket of the third type (for all trolleys). It totals to (2·1) + 3 + 7 = 12 burles.
In the second sample the profitable strategy is to buy one ticket of the fourth type.
In the third sample the profitable strategy is to buy two tickets of the third type: for all buses and for all trolleys.
|
```python
import sys
input = sys.stdin.readline
c1, c2, c3, c4 = map(int, input().split())
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
z = c2/c1
s1, s2 = 0, 0
for i in a:
if i > z:
s1 += c2
else:
s1 += c1*i
for i in b:
if i > z:
s2 += c2
else:
s2 += c1*i
print(min(c4, min(s1, c3) + min(s2, c3)))
```
| 3
|
|
764
|
A
|
Taymyr is calling you
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
|
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
|
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
|
[
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] |
[
"10\n",
"2\n",
"1\n"
] |
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
| 500
|
[
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"input": "24 22 9235",
"output": "34"
},
{
"input": "74 8 417",
"output": "1"
},
{
"input": "972 1 203",
"output": "0"
},
{
"input": "550 1 754",
"output": "1"
},
{
"input": "860 1 884",
"output": "1"
},
{
"input": "358 2 809",
"output": "2"
},
{
"input": "33 27 216",
"output": "0"
},
{
"input": "2940 1 9311",
"output": "3"
},
{
"input": "4624 1 1953",
"output": "0"
},
{
"input": "2696 2 7345",
"output": "2"
},
{
"input": "3443 2 6701",
"output": "0"
},
{
"input": "3 613 2275",
"output": "1"
},
{
"input": "1 10000 10000",
"output": "1"
},
{
"input": "10000 1 10000",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1 10000",
"output": "10000"
},
{
"input": "34 27 10000",
"output": "10"
},
{
"input": "2 2 9999",
"output": "4999"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "6 4 36",
"output": "3"
},
{
"input": "33 6 3005",
"output": "45"
},
{
"input": "5 1 20",
"output": "4"
},
{
"input": "1 2 10",
"output": "5"
},
{
"input": "2 1 100",
"output": "50"
},
{
"input": "10 20 10000",
"output": "500"
},
{
"input": "8 12 12",
"output": "0"
}
] | 1,695,106,926
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
import math
n,m,z=map(int,input().split())
u=m*n/math.gcd(m,n)
print(z//u)
|
Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
|
```python
import math
n,m,z=map(int,input().split())
u=m*n/math.gcd(m,n)
print(z//u)
```
| 0
|
|
302
|
A
|
Eugeny and Array
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
|
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
|
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
|
[
"2 3\n1 -1\n1 1\n1 2\n2 2\n",
"5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n"
] |
[
"0\n1\n0\n",
"0\n1\n0\n1\n0\n"
] |
none
| 500
|
[
{
"input": "2 3\n1 -1\n1 1\n1 2\n2 2",
"output": "0\n1\n0"
},
{
"input": "5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5",
"output": "0\n1\n0\n1\n0"
},
{
"input": "3 3\n1 1 1\n2 2\n1 1\n1 1",
"output": "0\n0\n0"
},
{
"input": "4 4\n-1 -1 -1 -1\n1 3\n1 2\n1 2\n1 1",
"output": "0\n0\n0\n0"
},
{
"input": "5 5\n-1 -1 -1 -1 -1\n1 1\n1 1\n3 4\n1 1\n1 4",
"output": "0\n0\n0\n0\n0"
},
{
"input": "6 6\n-1 -1 1 -1 -1 1\n1 1\n3 4\n1 1\n1 1\n1 3\n1 4",
"output": "0\n1\n0\n0\n0\n1"
},
{
"input": "7 7\n-1 -1 -1 1 -1 -1 -1\n1 1\n2 7\n1 3\n1 5\n4 7\n1 7\n6 7",
"output": "0\n0\n0\n0\n0\n0\n1"
},
{
"input": "8 8\n1 1 1 1 1 1 1 1\n5 8\n2 6\n2 3\n1 7\n7 7\n1 6\n1 8\n1 3",
"output": "0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "9 9\n-1 1 1 1 1 1 1 1 1\n1 7\n5 6\n1 4\n1 1\n1 1\n6 8\n1 1\n6 7\n3 5",
"output": "0\n1\n0\n0\n0\n0\n0\n1\n0"
},
{
"input": "10 10\n-1 1 -1 1 -1 -1 -1 -1 -1 -1\n6 7\n2 5\n3 6\n1 3\n3 5\n4 5\n3 4\n1 6\n1 1\n1 1",
"output": "1\n1\n1\n0\n0\n1\n1\n0\n0\n0"
},
{
"input": "1 1\n-1\n1 1",
"output": "0"
},
{
"input": "1 1\n1\n1 1",
"output": "0"
}
] | 1,668,078,540
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 3,276,800
|
n, m = map(int, input().split())
a = list(map(int, input().split()))
for _ in range(m):
l, r = map(int, input().split())
l -= 1
r -= 1
k_pol = a.count(1)
k_neg = a.count(-1)
if (r - l + 1) % 2 == 0:
if k_pol >= (r - l + 1)//2 and k_neg >= (r - l + 1)//2:
print('1')
else:
print('0')
else:
print('0')
|
Title: Eugeny and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
Demo Input:
['2 3\n1 -1\n1 1\n1 2\n2 2\n', '5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n']
Demo Output:
['0\n1\n0\n', '0\n1\n0\n1\n0\n']
Note:
none
|
```python
n, m = map(int, input().split())
a = list(map(int, input().split()))
for _ in range(m):
l, r = map(int, input().split())
l -= 1
r -= 1
k_pol = a.count(1)
k_neg = a.count(-1)
if (r - l + 1) % 2 == 0:
if k_pol >= (r - l + 1)//2 and k_neg >= (r - l + 1)//2:
print('1')
else:
print('0')
else:
print('0')
```
| 0
|
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,693,920,696
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 124
| 0
|
k=int(input());print(["YES","NO"][all(k%a for a in[4,7,47,744,477])])
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
k=int(input());print(["YES","NO"][all(k%a for a in[4,7,47,744,477])])
```
| 3
|
|
839
|
A
|
Arya and Bran
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integer — the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
|
[
"2 3\n1 2\n",
"3 17\n10 10 10\n",
"1 9\n10\n"
] |
[
"2",
"3",
"-1"
] |
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
| 500
|
[
{
"input": "2 3\n1 2",
"output": "2"
},
{
"input": "3 17\n10 10 10",
"output": "3"
},
{
"input": "1 9\n10",
"output": "-1"
},
{
"input": "10 70\n6 5 2 3 3 2 1 4 3 2",
"output": "-1"
},
{
"input": "20 140\n40 4 81 40 10 54 34 50 84 60 16 1 90 78 38 93 99 60 81 99",
"output": "18"
},
{
"input": "30 133\n3 2 3 4 3 7 4 5 5 6 7 2 1 3 4 6 7 4 6 4 7 5 7 1 3 4 1 6 8 5",
"output": "30"
},
{
"input": "40 320\n70 79 21 64 95 36 63 29 66 89 30 34 100 76 42 12 4 56 80 78 83 1 39 9 34 45 6 71 27 31 55 52 72 71 38 21 43 83 48 47",
"output": "40"
},
{
"input": "50 300\n5 3 11 8 7 4 9 5 5 1 6 3 5 7 4 2 2 10 8 1 7 10 4 4 11 5 2 4 9 1 5 4 11 9 11 2 7 4 4 8 10 9 1 11 10 2 4 11 6 9",
"output": "-1"
},
{
"input": "37 30\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",
"output": "30"
},
{
"input": "100 456\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": "57"
},
{
"input": "90 298\n94 90 98 94 93 90 99 98 90 96 93 96 92 92 97 98 94 94 96 100 93 96 95 98 94 91 95 95 94 90 93 96 93 100 99 98 94 95 98 91 91 98 97 100 98 93 92 93 91 100 92 97 95 95 97 94 98 97 99 100 90 96 93 100 95 99 92 100 99 91 97 99 98 93 90 93 97 95 94 96 90 100 94 93 91 92 97 97 97 100",
"output": "38"
},
{
"input": "7 43\n4 3 7 9 3 8 10",
"output": "-1"
},
{
"input": "99 585\n8 2 3 3 10 7 9 4 7 4 6 8 7 11 5 8 7 4 7 7 6 7 11 8 1 7 3 2 10 1 6 10 10 5 10 2 5 5 11 6 4 1 5 10 5 8 1 3 7 10 6 1 1 3 8 11 5 8 2 2 5 4 7 6 7 5 8 7 10 9 6 11 4 8 2 7 1 7 1 4 11 1 9 6 1 10 6 10 1 5 6 5 2 5 11 5 1 10 8",
"output": "-1"
},
{
"input": "30 177\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 8",
"output": "-1"
},
{
"input": "19 129\n3 3 10 11 4 7 3 8 10 2 11 6 11 9 4 2 11 10 5",
"output": "-1"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "13 104\n94 55 20 96 86 76 13 71 13 1 32 76 69",
"output": "13"
},
{
"input": "85 680\n61 44 55 6 30 74 27 26 17 45 73 1 67 71 39 32 13 25 79 66 4 59 49 28 29 22 10 17 98 80 36 99 52 24 59 44 27 79 29 46 29 12 47 72 82 25 6 30 81 72 95 65 30 71 72 45 39 16 16 89 48 42 59 71 50 58 31 65 91 70 48 56 28 34 53 89 94 98 49 55 94 65 91 11 53",
"output": "85"
},
{
"input": "100 458\n3 6 4 1 8 4 1 5 4 4 5 8 4 4 6 6 5 1 2 2 2 1 7 1 1 2 6 5 7 8 3 3 8 3 7 5 7 6 6 2 4 2 2 1 1 8 6 1 5 3 3 4 1 4 6 8 5 4 8 5 4 5 5 1 3 1 6 7 6 2 7 3 4 8 1 8 6 7 1 2 4 6 7 4 8 8 8 4 8 7 5 2 8 4 2 5 6 8 8 5",
"output": "100"
},
{
"input": "98 430\n4 7 6 3 4 1 7 1 1 6 6 1 5 4 6 1 5 4 6 6 1 5 1 1 8 1 6 6 2 6 8 4 4 6 6 8 8 7 4 1 2 4 1 5 4 3 7 3 2 5 7 7 7 2 2 2 7 2 8 7 3 4 5 7 8 3 7 6 7 3 2 4 7 1 4 4 7 1 1 8 4 5 8 3 1 5 3 5 2 1 3 3 8 1 3 5 8 6",
"output": "98"
},
{
"input": "90 80\n6 1 7 1 1 8 6 6 6 1 5 4 2 2 8 4 8 7 7 2 5 7 7 8 5 5 6 3 3 8 3 5 6 3 4 2 6 5 5 3 3 3 8 6 6 1 8 3 6 5 4 8 5 4 3 7 1 3 2 3 3 7 7 7 3 5 2 6 2 3 6 4 6 5 5 3 2 1 1 7 3 3 4 3 4 2 1 2 3 1",
"output": "18"
},
{
"input": "89 99\n7 7 3 5 2 7 8 8 1 1 5 7 7 4 1 5 3 4 4 8 8 3 3 2 6 3 8 2 7 5 8 1 3 5 3 6 4 3 6 2 3 3 4 5 1 6 1 7 7 7 6 7 7 7 8 8 8 2 1 7 5 8 6 7 7 4 7 5 7 8 1 3 5 8 7 1 4 2 5 8 3 4 4 5 5 6 2 4 2",
"output": "21"
},
{
"input": "50 700\n4 3 2 8 8 5 5 3 3 4 7 2 6 6 3 3 8 4 2 4 8 6 5 4 5 4 5 8 6 5 4 7 2 4 1 6 2 6 8 6 2 5 8 1 3 8 3 8 4 1",
"output": "-1"
},
{
"input": "82 359\n95 98 95 90 90 96 91 94 93 99 100 100 92 99 96 94 99 90 94 96 91 91 90 93 97 96 90 94 97 99 93 90 99 98 96 100 93 97 100 91 100 92 93 100 92 90 90 94 99 95 100 98 99 96 94 96 96 99 99 91 97 100 95 100 99 91 94 91 98 98 100 97 93 93 96 97 94 94 92 100 91 91",
"output": "45"
},
{
"input": "60 500\n93 93 100 99 91 92 95 93 95 99 93 91 97 98 90 91 98 100 95 100 94 93 92 91 91 98 98 90 93 91 90 96 92 93 92 94 94 91 96 94 98 100 97 96 96 97 91 99 97 95 96 94 91 92 99 95 97 92 98 90",
"output": "-1"
},
{
"input": "98 776\n48 63 26 3 88 81 27 33 37 10 2 89 41 84 98 93 25 44 42 90 41 65 97 1 28 69 42 14 86 18 96 28 28 94 78 8 44 31 96 45 26 52 93 25 48 39 3 75 94 93 63 59 67 86 18 74 27 38 68 7 31 60 69 67 20 11 19 34 47 43 86 96 3 49 56 60 35 49 89 28 92 69 48 15 17 73 99 69 2 73 27 35 28 53 11 1 96 50",
"output": "97"
},
{
"input": "100 189\n15 14 32 65 28 96 33 93 48 28 57 20 32 20 90 42 57 53 18 58 94 21 27 29 37 22 94 45 67 60 83 23 20 23 35 93 3 42 6 46 68 46 34 25 17 16 50 5 49 91 23 76 69 100 58 68 81 32 88 41 64 29 37 13 95 25 6 59 74 58 31 35 16 80 13 80 10 59 85 18 16 70 51 40 44 28 8 76 8 87 53 86 28 100 2 73 14 100 52 9",
"output": "24"
},
{
"input": "99 167\n72 4 79 73 49 58 15 13 92 92 42 36 35 21 13 10 51 94 64 35 86 50 6 80 93 77 59 71 2 88 22 10 27 30 87 12 77 6 34 56 31 67 78 84 36 27 15 15 12 56 80 7 56 14 10 9 14 59 15 20 34 81 8 49 51 72 4 58 38 77 31 86 18 61 27 86 95 36 46 36 39 18 78 39 48 37 71 12 51 92 65 48 39 22 16 87 4 5 42",
"output": "21"
},
{
"input": "90 4\n48 4 4 78 39 3 85 29 69 52 70 39 11 98 42 56 65 98 77 24 61 31 6 59 60 62 84 46 67 59 15 44 99 23 12 74 2 48 84 60 51 28 17 90 10 82 3 43 50 100 45 57 57 95 53 71 20 74 52 46 64 59 72 33 74 16 44 44 80 71 83 1 70 59 61 6 82 69 81 45 88 28 17 24 22 25 53 97 1 100",
"output": "1"
},
{
"input": "30 102\n55 94 3 96 3 47 92 85 25 78 27 70 97 83 40 2 55 12 74 84 91 37 31 85 7 40 33 54 72 5",
"output": "13"
},
{
"input": "81 108\n61 59 40 100 8 75 5 74 87 12 6 23 98 26 59 68 27 4 98 79 14 44 4 11 89 77 29 90 33 3 43 1 87 91 28 24 4 84 75 7 37 46 15 46 8 87 68 66 5 21 36 62 77 74 91 95 88 28 12 48 18 93 14 51 33 5 99 62 99 38 49 15 56 87 52 64 69 46 41 12 92",
"output": "14"
},
{
"input": "2 16\n10 6",
"output": "2"
},
{
"input": "2 8\n7 8",
"output": "2"
},
{
"input": "2 9\n4 8",
"output": "2"
},
{
"input": "3 19\n9 9 1",
"output": "3"
},
{
"input": "4 32\n9 9 9 5",
"output": "4"
},
{
"input": "2 15\n14 1",
"output": "2"
},
{
"input": "2 3\n3 3",
"output": "1"
},
{
"input": "3 10\n10 1 1",
"output": "2"
},
{
"input": "12 20\n3 16 19 10 1 6 17 8 6 20 1 4",
"output": "4"
},
{
"input": "4 15\n14 3 3 3",
"output": "2"
},
{
"input": "5 40\n10 10 10 10 1",
"output": "5"
},
{
"input": "4 31\n9 9 8 5",
"output": "4"
},
{
"input": "4 31\n20 7 1 1",
"output": "-1"
},
{
"input": "2 10\n9 1",
"output": "2"
},
{
"input": "10 50\n100 10 1 1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2 11\n10 2",
"output": "2"
},
{
"input": "3 21\n10 10 1",
"output": "3"
},
{
"input": "2 2\n1 2",
"output": "2"
},
{
"input": "3 2\n1 8 8",
"output": "2"
},
{
"input": "2 11\n10 1",
"output": "2"
},
{
"input": "2 16\n12 4",
"output": "2"
},
{
"input": "3 11\n9 2 2",
"output": "2"
},
{
"input": "3 11\n4 3 4",
"output": "3"
},
{
"input": "2 13\n7 6",
"output": "2"
},
{
"input": "3 24\n14 3 4",
"output": "-1"
},
{
"input": "2 13\n10 3",
"output": "2"
},
{
"input": "3 11\n9 2 1",
"output": "2"
},
{
"input": "2 15\n12 3",
"output": "2"
},
{
"input": "2 14\n11 4",
"output": "2"
}
] | 1,650,838,940
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 61
| 0
|
import sys
Input = sys.stdin.readlines()
Lin1 = str(Input[0])
Lin2 = str(Input[1])
Lin1 = Lin1.split(" ")
Lin2 = Lin2.split(" ")
N = int(Lin1[0])
K = int(Lin1[1])
GivenCandy = 0
if N * 8 < K:
print("-1")
else:
for i in range(len(Lin2)):
if int(Lin2[i]) >= 8:
GivenCandy += 8
else:
GivenCandy += int(Lin2[i])
if GivenCandy >= K:
print(i+1)
break
|
Title: Arya and Bran
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integer — the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
Demo Input:
['2 3\n1 2\n', '3 17\n10 10 10\n', '1 9\n10\n']
Demo Output:
['2', '3', '-1']
Note:
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
|
```python
import sys
Input = sys.stdin.readlines()
Lin1 = str(Input[0])
Lin2 = str(Input[1])
Lin1 = Lin1.split(" ")
Lin2 = Lin2.split(" ")
N = int(Lin1[0])
K = int(Lin1[1])
GivenCandy = 0
if N * 8 < K:
print("-1")
else:
for i in range(len(Lin2)):
if int(Lin2[i]) >= 8:
GivenCandy += 8
else:
GivenCandy += int(Lin2[i])
if GivenCandy >= K:
print(i+1)
break
```
| 0
|
|
92
|
A
|
Chips
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] |
A. Chips
|
2
|
256
|
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
|
Print the number of chips the presenter ended up with.
|
[
"4 11\n",
"17 107\n",
"3 8\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
| 500
|
[
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,512,106,649
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 124
| 5,632,000
|
import math
arr = input().split()
n = int(arr[0])
c = int(arr[1])
w = n*(n+1)/2
#print(w)
c %= w
#print(c)
m = int((-1 + math.sqrt(1 + 8*c))/2)
#print(m)
ans = c - m*(m+1)/2
print(int(ans))
|
Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
|
```python
import math
arr = input().split()
n = int(arr[0])
c = int(arr[1])
w = n*(n+1)/2
#print(w)
c %= w
#print(c)
m = int((-1 + math.sqrt(1 + 8*c))/2)
#print(m)
ans = c - m*(m+1)/2
print(int(ans))
```
| 3.95851
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,671,631,230
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 30
| 0
|
n=int(input())
li=list()
for i in range(n):
w=(input())
li.append(w)
for i in range(n):
x=li[i]
l=len(x)
if(l<10):
print(x)
else:
s,e=x[0],x[l-1]
print(s,l-2,e,sep="")
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n=int(input())
li=list()
for i in range(n):
w=(input())
li.append(w)
for i in range(n):
x=li[i]
l=len(x)
if(l<10):
print(x)
else:
s,e=x[0],x[l-1]
print(s,l-2,e,sep="")
```
| 0
|
614
|
B
|
Gena's Code
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
|
The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000.
|
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
|
[
"3\n5 10 1\n",
"4\n1 1 10 11\n",
"5\n0 3 1 100 1\n"
] |
[
"50",
"110",
"0"
] |
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
| 1,000
|
[
{
"input": "3\n5 10 1",
"output": "50"
},
{
"input": "4\n1 1 10 11",
"output": "110"
},
{
"input": "5\n0 3 1 100 1",
"output": "0"
},
{
"input": "40\n10 100 10 1 10 10 100 10 10 100 10 100 100 10 1824868942 100 100 1 10 100 100 10 100 100 10 100 10 1 10 100 100 100 10 1 10 1 10 10 100 100",
"output": "1824868942000000000000000000000000000000000000000000000000000"
},
{
"input": "6\n1000000000000000000000000000000000000 6643573784 1000000000000000000000000000000000000 1000000000000000000000000000000000000 1000000000000000000000000000000000000 1000000000000000000000000000000000000",
"output": "6643573784000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n9",
"output": "9"
},
{
"input": "2\n10 50",
"output": "500"
},
{
"input": "3\n500 1 10",
"output": "5000"
}
] | 1,687,192,056
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 249
| 3,379,200
|
n = input()
s = input()
lst = list(s.split(" "))
l = len(lst)
out = 1
for i in range(0,l):
out *= int(lst[i])
print(out)
|
Title: Gena's Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input Specification:
The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000.
Output Specification:
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Demo Input:
['3\n5 10 1\n', '4\n1 1 10 11\n', '5\n0 3 1 100 1\n']
Demo Output:
['50', '110', '0']
Note:
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
|
```python
n = input()
s = input()
lst = list(s.split(" "))
l = len(lst)
out = 1
for i in range(0,l):
out *= int(lst[i])
print(out)
```
| 3
|
|
719
|
B
|
Anatoly and Cockroaches
|
PROGRAMMING
| 1,400
|
[
"greedy"
] | null | null |
Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are *n* cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cockroaches.
The second line contains a string of length *n*, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.
|
Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.
|
[
"5\nrbbrr\n",
"5\nbbbbb\n",
"3\nrbr\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0.
| 1,000
|
[
{
"input": "5\nrbbrr",
"output": "1"
},
{
"input": "5\nbbbbb",
"output": "2"
},
{
"input": "3\nrbr",
"output": "0"
},
{
"input": "13\nrbbbrbrrbrrbb",
"output": "3"
},
{
"input": "18\nrrrrrrrrrrrrrrrrrb",
"output": "8"
},
{
"input": "100\nbrbbbrrrbbrbrbbrbbrbbbbrbbrrbbbrrbbbbrbrbbbbbbbbbbbbbbbbrrrrbbbbrrrbbbbbbbrbrrbrbbbbrrrbbbbrbbrbbbrb",
"output": "34"
},
{
"input": "166\nrbbbbbbbbbbbbrbrrbbrbbbrbbbbbbbbbbrbbbbbbrbbbrbbbbbrbbbbbbbrbbbbbbbrbbrbbbbbbbbrbbbbbbbbbbbbbbrrbbbrbbbbbbbbbbbbbbrbrbbbbbbbbbbbrbbbbbbbbbbbbbbrbbbbbbbbbbbbbbbbbbbbbb",
"output": "70"
},
{
"input": "1\nr",
"output": "0"
},
{
"input": "1\nb",
"output": "0"
},
{
"input": "2\nrb",
"output": "0"
},
{
"input": "2\nbr",
"output": "0"
},
{
"input": "2\nrr",
"output": "1"
},
{
"input": "2\nbb",
"output": "1"
},
{
"input": "8\nrbbrbrbr",
"output": "1"
},
{
"input": "7\nrrbrbrb",
"output": "1"
}
] | 1,599,820,204
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 46
| 187
| 7,270,400
|
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
N = int(input())
s = [*input()]
x = "rb"*((N//2)+1)
y = "br"*((N//2)+1)
mini = N
if N==1:
print(0)
exit(0)
def fun(t):
P = [];Q= []
for i in range(N):
if s[i] != t[i]:
if s[i] == 'b':
P.append(i)
else:
Q.append(i)
#mm = min(len(P),len(Q))
nn = max(len(P),len(Q))
return nn
print(min(fun(x),fun(y)))
|
Title: Anatoly and Cockroaches
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are *n* cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cockroaches.
The second line contains a string of length *n*, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.
Output Specification:
Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.
Demo Input:
['5\nrbbrr\n', '5\nbbbbb\n', '3\nrbr\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0.
|
```python
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
N = int(input())
s = [*input()]
x = "rb"*((N//2)+1)
y = "br"*((N//2)+1)
mini = N
if N==1:
print(0)
exit(0)
def fun(t):
P = [];Q= []
for i in range(N):
if s[i] != t[i]:
if s[i] == 'b':
P.append(i)
else:
Q.append(i)
#mm = min(len(P),len(Q))
nn = max(len(P),len(Q))
return nn
print(min(fun(x),fun(y)))
```
| 3
|
|
609
|
A
|
USB Flash Drives
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
|
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
|
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
|
[
"3\n5\n2\n1\n3\n",
"3\n6\n2\n3\n2\n",
"2\n5\n5\n10\n"
] |
[
"2\n",
"3\n",
"1\n"
] |
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
| 0
|
[
{
"input": "3\n5\n2\n1\n3",
"output": "2"
},
{
"input": "3\n6\n2\n3\n2",
"output": "3"
},
{
"input": "2\n5\n5\n10",
"output": "1"
},
{
"input": "5\n16\n8\n1\n3\n4\n9",
"output": "2"
},
{
"input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100",
"output": "2"
},
{
"input": "12\n4773\n325\n377\n192\n780\n881\n816\n839\n223\n215\n125\n952\n8",
"output": "7"
},
{
"input": "15\n7758\n182\n272\n763\n910\n24\n359\n583\n890\n735\n819\n66\n992\n440\n496\n227",
"output": "15"
},
{
"input": "30\n70\n6\n2\n10\n4\n7\n10\n5\n1\n8\n10\n4\n3\n5\n9\n3\n6\n6\n4\n2\n6\n5\n10\n1\n9\n7\n2\n1\n10\n7\n5",
"output": "8"
},
{
"input": "40\n15705\n702\n722\n105\n873\n417\n477\n794\n300\n869\n496\n572\n232\n456\n298\n473\n584\n486\n713\n934\n121\n303\n956\n934\n840\n358\n201\n861\n497\n131\n312\n957\n96\n914\n509\n60\n300\n722\n658\n820\n103",
"output": "21"
},
{
"input": "50\n18239\n300\n151\n770\n9\n200\n52\n247\n753\n523\n263\n744\n463\n540\n244\n608\n569\n771\n32\n425\n777\n624\n761\n628\n124\n405\n396\n726\n626\n679\n237\n229\n49\n512\n18\n671\n290\n768\n632\n739\n18\n136\n413\n117\n83\n413\n452\n767\n664\n203\n404",
"output": "31"
},
{
"input": "70\n149\n5\n3\n3\n4\n6\n1\n2\n9\n8\n3\n1\n8\n4\n4\n3\n6\n10\n7\n1\n10\n8\n4\n9\n3\n8\n3\n2\n5\n1\n8\n6\n9\n10\n4\n8\n6\n9\n9\n9\n3\n4\n2\n2\n5\n8\n9\n1\n10\n3\n4\n3\n1\n9\n3\n5\n1\n3\n7\n6\n9\n8\n9\n1\n7\n4\n4\n2\n3\n5\n7",
"output": "17"
},
{
"input": "70\n2731\n26\n75\n86\n94\n37\n25\n32\n35\n92\n1\n51\n73\n53\n66\n16\n80\n15\n81\n100\n87\n55\n48\n30\n71\n39\n87\n77\n25\n70\n22\n75\n23\n97\n16\n75\n95\n61\n61\n28\n10\n78\n54\n80\n51\n25\n24\n90\n58\n4\n77\n40\n54\n53\n47\n62\n30\n38\n71\n97\n71\n60\n58\n1\n21\n15\n55\n99\n34\n88\n99",
"output": "35"
},
{
"input": "70\n28625\n34\n132\n181\n232\n593\n413\n862\n887\n808\n18\n35\n89\n356\n640\n339\n280\n975\n82\n345\n398\n948\n372\n91\n755\n75\n153\n948\n603\n35\n694\n722\n293\n363\n884\n264\n813\n175\n169\n646\n138\n449\n488\n828\n417\n134\n84\n763\n288\n845\n801\n556\n972\n332\n564\n934\n699\n842\n942\n644\n203\n406\n140\n37\n9\n423\n546\n675\n491\n113\n587",
"output": "45"
},
{
"input": "80\n248\n3\n9\n4\n5\n10\n7\n2\n6\n2\n2\n8\n2\n1\n3\n7\n9\n2\n8\n4\n4\n8\n5\n4\n4\n10\n2\n1\n4\n8\n4\n10\n1\n2\n10\n2\n3\n3\n1\n1\n8\n9\n5\n10\n2\n8\n10\n5\n3\n6\n1\n7\n8\n9\n10\n5\n10\n10\n2\n10\n1\n2\n4\n1\n9\n4\n7\n10\n8\n5\n8\n1\n4\n2\n2\n3\n9\n9\n9\n10\n6",
"output": "27"
},
{
"input": "80\n2993\n18\n14\n73\n38\n14\n73\n77\n18\n81\n6\n96\n65\n77\n86\n76\n8\n16\n81\n83\n83\n34\n69\n58\n15\n19\n1\n16\n57\n95\n35\n5\n49\n8\n15\n47\n84\n99\n94\n93\n55\n43\n47\n51\n61\n57\n13\n7\n92\n14\n4\n83\n100\n60\n75\n41\n95\n74\n40\n1\n4\n95\n68\n59\n65\n15\n15\n75\n85\n46\n77\n26\n30\n51\n64\n75\n40\n22\n88\n68\n24",
"output": "38"
},
{
"input": "80\n37947\n117\n569\n702\n272\n573\n629\n90\n337\n673\n589\n576\n205\n11\n284\n645\n719\n777\n271\n567\n466\n251\n402\n3\n97\n288\n699\n208\n173\n530\n782\n266\n395\n957\n159\n463\n43\n316\n603\n197\n386\n132\n799\n778\n905\n784\n71\n851\n963\n883\n705\n454\n275\n425\n727\n223\n4\n870\n833\n431\n463\n85\n505\n800\n41\n954\n981\n242\n578\n336\n48\n858\n702\n349\n929\n646\n528\n993\n506\n274\n227",
"output": "70"
},
{
"input": "90\n413\n5\n8\n10\n7\n5\n7\n5\n7\n1\n7\n8\n4\n3\n9\n4\n1\n10\n3\n1\n10\n9\n3\n1\n8\n4\n7\n5\n2\n9\n3\n10\n10\n3\n6\n3\n3\n10\n7\n5\n1\n1\n2\n4\n8\n2\n5\n5\n3\n9\n5\n5\n3\n10\n2\n3\n8\n5\n9\n1\n3\n6\n5\n9\n2\n3\n7\n10\n3\n4\n4\n1\n5\n9\n2\n6\n9\n1\n1\n9\n9\n7\n7\n7\n8\n4\n5\n3\n4\n6\n9",
"output": "59"
},
{
"input": "90\n4226\n33\n43\n83\n46\n75\n14\n88\n36\n8\n25\n47\n4\n96\n19\n33\n49\n65\n17\n59\n72\n1\n55\n94\n92\n27\n33\n39\n14\n62\n79\n12\n89\n22\n86\n13\n19\n77\n53\n96\n74\n24\n25\n17\n64\n71\n81\n87\n52\n72\n55\n49\n74\n36\n65\n86\n91\n33\n61\n97\n38\n87\n61\n14\n73\n95\n43\n67\n42\n67\n22\n12\n62\n32\n96\n24\n49\n82\n46\n89\n36\n75\n91\n11\n10\n9\n33\n86\n28\n75\n39",
"output": "64"
},
{
"input": "90\n40579\n448\n977\n607\n745\n268\n826\n479\n59\n330\n609\n43\n301\n970\n726\n172\n632\n600\n181\n712\n195\n491\n312\n849\n722\n679\n682\n780\n131\n404\n293\n387\n567\n660\n54\n339\n111\n833\n612\n911\n869\n356\n884\n635\n126\n639\n712\n473\n663\n773\n435\n32\n973\n484\n662\n464\n699\n274\n919\n95\n904\n253\n589\n543\n454\n250\n349\n237\n829\n511\n536\n36\n45\n152\n626\n384\n199\n877\n941\n84\n781\n115\n20\n52\n726\n751\n920\n291\n571\n6\n199",
"output": "64"
},
{
"input": "100\n66\n7\n9\n10\n5\n2\n8\n6\n5\n4\n10\n10\n6\n5\n2\n2\n1\n1\n5\n8\n7\n8\n10\n5\n6\n6\n5\n9\n9\n6\n3\n8\n7\n10\n5\n9\n6\n7\n3\n5\n8\n6\n8\n9\n1\n1\n1\n2\n4\n5\n5\n1\n1\n2\n6\n7\n1\n5\n8\n7\n2\n1\n7\n10\n9\n10\n2\n4\n10\n4\n10\n10\n5\n3\n9\n1\n2\n1\n10\n5\n1\n7\n4\n4\n5\n7\n6\n10\n4\n7\n3\n4\n3\n6\n2\n5\n2\n4\n9\n5\n3",
"output": "7"
},
{
"input": "100\n4862\n20\n47\n85\n47\n76\n38\n48\n93\n91\n81\n31\n51\n23\n60\n59\n3\n73\n72\n57\n67\n54\n9\n42\n5\n32\n46\n72\n79\n95\n61\n79\n88\n33\n52\n97\n10\n3\n20\n79\n82\n93\n90\n38\n80\n18\n21\n43\n60\n73\n34\n75\n65\n10\n84\n100\n29\n94\n56\n22\n59\n95\n46\n22\n57\n69\n67\n90\n11\n10\n61\n27\n2\n48\n69\n86\n91\n69\n76\n36\n71\n18\n54\n90\n74\n69\n50\n46\n8\n5\n41\n96\n5\n14\n55\n85\n39\n6\n79\n75\n87",
"output": "70"
},
{
"input": "100\n45570\n14\n881\n678\n687\n993\n413\n760\n451\n426\n787\n503\n343\n234\n530\n294\n725\n941\n524\n574\n441\n798\n399\n360\n609\n376\n525\n229\n995\n478\n347\n47\n23\n468\n525\n749\n601\n235\n89\n995\n489\n1\n239\n415\n122\n671\n128\n357\n886\n401\n964\n212\n968\n210\n130\n871\n360\n661\n844\n414\n187\n21\n824\n266\n713\n126\n496\n916\n37\n193\n755\n894\n641\n300\n170\n176\n383\n488\n627\n61\n897\n33\n242\n419\n881\n698\n107\n391\n418\n774\n905\n87\n5\n896\n835\n318\n373\n916\n393\n91\n460",
"output": "78"
},
{
"input": "100\n522\n1\n5\n2\n4\n2\n6\n3\n4\n2\n10\n10\n6\n7\n9\n7\n1\n7\n2\n5\n3\n1\n5\n2\n3\n5\n1\n7\n10\n10\n4\n4\n10\n9\n10\n6\n2\n8\n2\n6\n10\n9\n2\n7\n5\n9\n4\n6\n10\n7\n3\n1\n1\n9\n5\n10\n9\n2\n8\n3\n7\n5\n4\n7\n5\n9\n10\n6\n2\n9\n2\n5\n10\n1\n7\n7\n10\n5\n6\n2\n9\n4\n7\n10\n10\n8\n3\n4\n9\n3\n6\n9\n10\n2\n9\n9\n3\n4\n1\n10\n2",
"output": "74"
},
{
"input": "100\n32294\n414\n116\n131\n649\n130\n476\n630\n605\n213\n117\n757\n42\n109\n85\n127\n635\n629\n994\n410\n764\n204\n161\n231\n577\n116\n936\n537\n565\n571\n317\n722\n819\n229\n284\n487\n649\n304\n628\n727\n816\n854\n91\n111\n549\n87\n374\n417\n3\n868\n882\n168\n743\n77\n534\n781\n75\n956\n910\n734\n507\n568\n802\n946\n891\n659\n116\n678\n375\n380\n430\n627\n873\n350\n930\n285\n6\n183\n96\n517\n81\n794\n235\n360\n551\n6\n28\n799\n226\n996\n894\n981\n551\n60\n40\n460\n479\n161\n318\n952\n433",
"output": "42"
},
{
"input": "100\n178\n71\n23\n84\n98\n8\n14\n4\n42\n56\n83\n87\n28\n22\n32\n50\n5\n96\n90\n1\n59\n74\n56\n96\n77\n88\n71\n38\n62\n36\n85\n1\n97\n98\n98\n32\n99\n42\n6\n81\n20\n49\n57\n71\n66\n9\n45\n41\n29\n28\n32\n68\n38\n29\n35\n29\n19\n27\n76\n85\n68\n68\n41\n32\n78\n72\n38\n19\n55\n83\n83\n25\n46\n62\n48\n26\n53\n14\n39\n31\n94\n84\n22\n39\n34\n96\n63\n37\n42\n6\n78\n76\n64\n16\n26\n6\n79\n53\n24\n29\n63",
"output": "2"
},
{
"input": "100\n885\n226\n266\n321\n72\n719\n29\n121\n533\n85\n672\n225\n830\n783\n822\n30\n791\n618\n166\n487\n922\n434\n814\n473\n5\n741\n947\n910\n305\n998\n49\n945\n588\n868\n809\n803\n168\n280\n614\n434\n634\n538\n591\n437\n540\n445\n313\n177\n171\n799\n778\n55\n617\n554\n583\n611\n12\n94\n599\n182\n765\n556\n965\n542\n35\n460\n177\n313\n485\n744\n384\n21\n52\n879\n792\n411\n614\n811\n565\n695\n428\n587\n631\n794\n461\n258\n193\n696\n936\n646\n756\n267\n55\n690\n730\n742\n734\n988\n235\n762\n440",
"output": "1"
},
{
"input": "100\n29\n9\n2\n10\n8\n6\n7\n7\n3\n3\n10\n4\n5\n2\n5\n1\n6\n3\n2\n5\n10\n10\n9\n1\n4\n5\n2\n2\n3\n1\n2\n2\n9\n6\n9\n7\n8\n8\n1\n5\n5\n3\n1\n5\n6\n1\n9\n2\n3\n8\n10\n8\n3\n2\n7\n1\n2\n1\n2\n8\n10\n5\n2\n3\n1\n10\n7\n1\n7\n4\n9\n6\n6\n4\n7\n1\n2\n7\n7\n9\n9\n7\n10\n4\n10\n8\n2\n1\n5\n5\n10\n5\n8\n1\n5\n6\n5\n1\n5\n6\n8",
"output": "3"
},
{
"input": "100\n644\n94\n69\n43\n36\n54\n93\n30\n74\n56\n95\n70\n49\n11\n36\n57\n30\n59\n3\n52\n59\n90\n82\n39\n67\n32\n8\n80\n64\n8\n65\n51\n48\n89\n90\n35\n4\n54\n66\n96\n68\n90\n30\n4\n13\n97\n41\n90\n85\n17\n45\n94\n31\n58\n4\n39\n76\n95\n92\n59\n67\n46\n96\n55\n82\n64\n20\n20\n83\n46\n37\n15\n60\n37\n79\n45\n47\n63\n73\n76\n31\n52\n36\n32\n49\n26\n61\n91\n31\n25\n62\n90\n65\n65\n5\n94\n7\n15\n97\n88\n68",
"output": "7"
},
{
"input": "100\n1756\n98\n229\n158\n281\n16\n169\n149\n239\n235\n182\n147\n215\n49\n270\n194\n242\n295\n289\n249\n19\n12\n144\n157\n92\n270\n122\n212\n97\n152\n14\n42\n12\n198\n98\n295\n154\n229\n191\n294\n5\n156\n43\n185\n184\n20\n125\n23\n10\n257\n244\n264\n79\n46\n277\n13\n22\n97\n212\n77\n293\n20\n51\n17\n109\n37\n68\n117\n51\n248\n10\n149\n179\n192\n239\n161\n13\n173\n297\n73\n43\n109\n288\n198\n81\n70\n254\n187\n277\n1\n295\n113\n95\n291\n293\n119\n205\n191\n37\n34\n116",
"output": "6"
},
{
"input": "100\n20562\n721\n452\n11\n703\n376\n183\n197\n203\n406\n642\n346\n446\n256\n760\n201\n360\n702\n707\n388\n779\n653\n610\n497\n768\n670\n134\n780\n306\n661\n180\n259\n256\n362\n6\n121\n415\n747\n170\n67\n439\n728\n193\n622\n481\n38\n225\n343\n303\n253\n436\n305\n68\n794\n247\n291\n600\n750\n188\n199\n757\n28\n776\n749\n253\n351\n53\n629\n129\n578\n209\n89\n651\n262\n638\n353\n469\n31\n144\n460\n176\n535\n562\n366\n639\n234\n577\n364\n761\n617\n303\n450\n778\n311\n289\n221\n274\n239\n626\n194\n36",
"output": "30"
},
{
"input": "100\n32630\n548\n21\n756\n138\n56\n719\n97\n86\n599\n531\n464\n137\n737\n239\n626\n179\n170\n271\n335\n178\n298\n597\n616\n104\n190\n216\n333\n606\n621\n310\n70\n286\n593\n655\n472\n556\n77\n625\n744\n395\n487\n762\n316\n469\n258\n475\n596\n651\n180\n737\n101\n511\n708\n163\n713\n475\n557\n649\n705\n583\n97\n447\n119\n91\n372\n194\n394\n667\n384\n635\n611\n746\n758\n613\n54\n508\n405\n188\n570\n254\n28\n599\n767\n115\n737\n194\n148\n686\n705\n236\n762\n613\n770\n633\n401\n773\n448\n274\n141\n718",
"output": "52"
},
{
"input": "1\n1\n1",
"output": "1"
}
] | 1,568,415,688
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 202
| 2,150,400
|
number_usbs = input()
size_file = input()
sizes_usb = []
for i in range (0,number_usbs):
add = input()
sizes_usb.append(add)
sizes_usb.sort(reverse = True)
usb_needed = 0
size_remaining = size_file
for element in sizes_usb:
size_remaining -= element
usb_needed += 1
if size_remaining <= 0:
break
print(usb_needed)
|
Title: USB Flash Drives
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Output Specification:
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
Demo Input:
['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
|
```python
number_usbs = input()
size_file = input()
sizes_usb = []
for i in range (0,number_usbs):
add = input()
sizes_usb.append(add)
sizes_usb.sort(reverse = True)
usb_needed = 0
size_remaining = size_file
for element in sizes_usb:
size_remaining -= element
usb_needed += 1
if size_remaining <= 0:
break
print(usb_needed)
```
| -1
|
|
749
|
A
|
Bachgold Problem
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
|
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
|
[
"5\n",
"6\n"
] |
[
"2\n2 3\n",
"3\n2 2 2\n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "100000",
"output": "50000\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99996",
"output": "49998\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99998",
"output": "49999\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99994",
"output": "49997\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\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 3"
},
{
"input": "57",
"output": "28\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 3"
},
{
"input": "61",
"output": "30\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 3"
},
{
"input": "774",
"output": "387\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "202",
"output": "101\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 2"
},
{
"input": "530",
"output": "265\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7166",
"output": "3583\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "9294",
"output": "4647\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "2422",
"output": "1211\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "15326",
"output": "7663\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "11454",
"output": "5727\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "14878",
"output": "7439\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "90672",
"output": "45336\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99544",
"output": "49772\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "90472",
"output": "45236\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\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 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
}
] | 1,621,682,059
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 124
| 102,400
|
a=int(input())
if a%2!=0:
a=a-3
x=a//2
print(x+1)
for i in range (x):
print (2, end=' ')
print (3, end=' ')
else:
x=a//2
print (a//2)
for i in range (x):
print (a//x, end=' ')
|
Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none
|
```python
a=int(input())
if a%2!=0:
a=a-3
x=a//2
print(x+1)
for i in range (x):
print (2, end=' ')
print (3, end=' ')
else:
x=a//2
print (a//2)
for i in range (x):
print (a//x, end=' ')
```
| 3
|
|
294
|
A
|
Shaass and Oskols
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
|
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
|
On the *i*-th line of the output print the number of birds on the *i*-th wire.
|
[
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] |
[
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] |
none
| 500
|
[
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,670,251,795
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
lines = input()
lines = int(lines)
lstOfBirds = list(map(int, input().split()))
shots= input()
shots = int(shots)
for i in range (shots):
shot, numOfBird=map(int,input().split())
lstOfBirds[shot] = lstOfBirds[shot]+lstOfBirds[shot-1]-numOfBird
if(shot >= 2):
lstOfBirds[shot-2]=lstOfBirds[shot-2]+numOfBird-1
lstOfBirds[shot - 1] = 0
print(lstOfBirds)
|
Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none
|
```python
lines = input()
lines = int(lines)
lstOfBirds = list(map(int, input().split()))
shots= input()
shots = int(shots)
for i in range (shots):
shot, numOfBird=map(int,input().split())
lstOfBirds[shot] = lstOfBirds[shot]+lstOfBirds[shot-1]-numOfBird
if(shot >= 2):
lstOfBirds[shot-2]=lstOfBirds[shot-2]+numOfBird-1
lstOfBirds[shot - 1] = 0
print(lstOfBirds)
```
| 0
|
|
197
|
B
|
Limit
|
PROGRAMMING
| 1,400
|
[
"math"
] | null | null |
You are given two polynomials:
- *P*(*x*)<==<=*a*0·*x**n*<=+<=*a*1·*x**n*<=-<=1<=+<=...<=+<=*a**n*<=-<=1·*x*<=+<=*a**n* and - *Q*(*x*)<==<=*b*0·*x**m*<=+<=*b*1·*x**m*<=-<=1<=+<=...<=+<=*b**m*<=-<=1·*x*<=+<=*b**m*.
Calculate limit .
|
The first line contains two space-separated integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=100) — degrees of polynomials *P*(*x*) and *Q*(*x*) correspondingly.
The second line contains *n*<=+<=1 space-separated integers — the factors of polynomial *P*(*x*): *a*0, *a*1, ..., *a**n*<=-<=1, *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100,<=*a*0<=≠<=0).
The third line contains *m*<=+<=1 space-separated integers — the factors of polynomial *Q*(*x*): *b*0, *b*1, ..., *b**m*<=-<=1, *b**m* (<=-<=100<=≤<=*b**i*<=≤<=100,<=*b*0<=≠<=0).
|
If the limit equals <=+<=∞, print "Infinity" (without quotes). If the limit equals <=-<=∞, print "-Infinity" (without the quotes).
If the value of the limit equals zero, print "0/1" (without the quotes).
Otherwise, print an irreducible fraction — the value of limit , in the format "p/q" (without the quotes), where *p* is the — numerator, *q* (*q*<=><=0) is the denominator of the fraction.
|
[
"2 1\n1 1 1\n2 5\n",
"1 0\n-1 3\n2\n",
"0 1\n1\n1 0\n",
"2 2\n2 1 6\n4 5 -7\n",
"1 1\n9 0\n-5 2\n"
] |
[
"Infinity\n",
"-Infinity\n",
"0/1\n",
"1/2\n",
"-9/5\n"
] |
Let's consider all samples:
1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c28febca257452afdfcbd6984ba8623911f9bdbc.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1e55ecd04e54a45e5e0092ec9a5c1ea03bb29255.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/2c95fb684d373fcc1a481cfabeda4d5c2f3673ee.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4dc40cb8b3cd6375c42445366e50369649a2801a.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c6455aba35cfb3c4397505121d1f77afcd17c98e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
You can learn more about the definition and properties of limits if you follow the link: http://en.wikipedia.org/wiki/Limit_of_a_function
| 500
|
[
{
"input": "2 1\n1 1 1\n2 5",
"output": "Infinity"
},
{
"input": "1 0\n-1 3\n2",
"output": "-Infinity"
},
{
"input": "0 1\n1\n1 0",
"output": "0/1"
},
{
"input": "2 2\n2 1 6\n4 5 -7",
"output": "1/2"
},
{
"input": "1 1\n9 0\n-5 2",
"output": "-9/5"
},
{
"input": "1 2\n5 3\n-3 2 -1",
"output": "0/1"
},
{
"input": "1 2\n-4 8\n-2 5 -3",
"output": "0/1"
},
{
"input": "3 2\n4 3 1 2\n-5 7 0",
"output": "-Infinity"
},
{
"input": "2 1\n-3 5 1\n-8 0",
"output": "Infinity"
},
{
"input": "1 1\n-5 7\n3 1",
"output": "-5/3"
},
{
"input": "2 2\n-4 2 1\n-5 8 -19",
"output": "4/5"
},
{
"input": "0 100\n1\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0/1"
},
{
"input": "100 0\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100\n1",
"output": "Infinity"
},
{
"input": "0 0\n36\n-54",
"output": "-2/3"
},
{
"input": "0 0\n36\n-8",
"output": "-9/2"
},
{
"input": "0 0\n-6\n-8",
"output": "3/4"
},
{
"input": "0 2\n-3\n1 4 6",
"output": "0/1"
},
{
"input": "0 0\n-21\n13",
"output": "-21/13"
},
{
"input": "0 0\n-34\n21",
"output": "-34/21"
},
{
"input": "0 0\n-55\n34",
"output": "-55/34"
},
{
"input": "33 100\n-15 -90 -84 57 67 60 -40 -82 83 -80 43 -15 -36 -14 -37 -49 42 -79 49 -7 -12 53 -44 -21 87 -91 -73 -27 13 65 5 74 -21 -52\n-67 -17 36 -46 -5 31 -45 -35 -49 13 -7 -82 92 -55 -67 -96 31 -70 76 24 -29 26 96 19 -40 99 -26 74 -17 -56 -72 24 -71 -62 10 -56 -74 75 -48 -98 -67 -26 47 7 63 -38 99 66 -25 -31 -24 -42 -49 -27 -45 -2 -37 -16 5 -21 97 33 85 -33 93 30 84 73 -48 18 -36 71 -38 -41 28 1 -7 -15 60 59 -20 -38 -86 90 2 -12 72 -43 26 76 97 7 -2 -47 -4 100 -40 -48 53 -54 0",
"output": "0/1"
},
{
"input": "39 87\n78 -50 18 -32 -12 -65 83 41 -6 53 -26 64 -19 -53 -61 91 -49 -66 67 69 100 -39 95 99 86 -67 -66 63 48 26 -4 95 -54 -71 26 -74 -93 79 -91 -45\n-18 23 48 59 76 82 95 2 -26 18 -39 -74 44 -92 40 -44 1 -97 -100 -63 -54 -3 -86 85 28 -50 41 -53 -74 -29 -91 87 27 -42 -90 -15 -26 -15 -100 -70 -10 -41 16 85 71 -39 -31 -65 80 98 9 23 -40 14 -88 15 -34 10 -67 -94 -58 -24 75 48 -42 56 -77 -13 -25 -79 -100 -57 89 45 22 85 78 -93 -79 69 63 44 74 94 35 -65 -12 -88",
"output": "0/1"
},
{
"input": "47 56\n31 -99 -97 6 -45 -5 89 35 -77 69 57 91 -32 -66 -36 16 30 61 -36 32 48 67 5 -85 65 -11 -51 -63 -51 -16 39 -26 -60 -28 91 43 -90 32 44 83 70 -53 51 56 68 -81 76 79\n61 -21 -75 -36 -24 -19 80 26 -28 93 27 72 -39 -46 -38 68 -29 -16 -63 84 -13 64 55 63 77 5 68 70 15 99 12 -69 50 -48 -82 -3 52 -54 68 91 -37 -100 -5 74 24 91 -1 74 28 29 -87 -13 -88 82 -13 58 23",
"output": "0/1"
},
{
"input": "9 100\n-34 88 33 -80 87 31 -53 -3 8 -70\n31 -25 46 78 8 82 -92 -36 -30 85 -93 86 -87 75 8 -71 44 -41 -83 19 89 -28 81 42 79 86 41 -23 64 -31 46 24 -79 23 71 63 99 90 -16 -70 -1 88 10 65 3 -99 95 52 -80 53 -24 -43 -30 -7 51 40 -47 44 -10 -18 -61 -67 -84 37 45 93 -5 68 32 3 -61 -100 38 -21 -91 90 83 -45 75 89 17 -44 75 14 -28 1 -84 -100 -36 84 -40 88 -84 -54 2 -32 92 -49 77 85 91",
"output": "0/1"
},
{
"input": "28 87\n-77 49 37 46 -92 65 89 100 53 76 -43 47 -80 -46 -94 -4 20 46 81 -41 86 25 69 60 15 -78 -98 -7 -42\n-85 96 59 -40 90 -72 41 -17 -40 -15 -98 66 47 9 -33 -63 59 -25 -31 25 -94 35 28 -36 -41 -38 -38 -54 -40 90 7 -10 98 -19 54 -10 46 -58 -88 -21 90 82 37 -70 -98 -63 41 75 -50 -59 -69 79 -93 -3 -45 14 76 28 -28 -98 -44 -39 71 44 90 91 0 45 7 65 68 39 -27 58 68 -47 -41 100 14 -95 -80 69 -88 -51 -89 -70 -23 95",
"output": "0/1"
},
{
"input": "100 4\n-5 -93 89 -26 -79 14 -28 13 -45 69 50 -84 21 -68 62 30 -26 99 -12 39 20 -74 -39 -41 -28 -72 -55 28 20 31 -92 -20 76 -65 57 72 -36 4 33 -28 -19 -41 -40 40 84 -36 -83 75 -74 -80 32 -50 -56 72 16 75 57 90 -19 -10 67 -71 69 -48 -48 23 37 -31 -64 -86 20 67 97 14 82 -41 2 87 65 -81 -27 9 -79 -1 -5 84 -8 29 -34 31 82 40 21 -53 -31 -45 17 -33 79 50 -94\n56 -4 -90 36 84",
"output": "-Infinity"
},
{
"input": "77 51\n89 45 -33 -87 33 -61 -79 40 -76 16 -17 31 27 25 99 82 51 -40 85 -66 19 89 -62 24 -61 -53 -77 17 21 83 53 -18 -56 75 9 -78 33 -11 -6 96 -33 -2 -57 97 30 20 -41 42 -13 45 -99 67 37 -20 51 -33 88 -62 2 40 17 36 45 71 4 -44 24 20 -2 29 -12 -84 -7 -84 -38 48 -73 79\n60 -43 60 1 90 -1 19 -18 -21 31 -76 51 79 91 12 39 -33 -14 71 -90 -65 -93 -58 93 49 17 77 19 32 -8 14 58 -9 85 -95 -73 0 85 -91 -99 -30 -43 61 20 -89 93 53 20 -33 -38 79 54",
"output": "Infinity"
},
{
"input": "84 54\n82 -54 28 68 74 -61 54 98 59 67 -65 -1 16 65 -78 -16 61 -79 2 14 44 96 -62 77 51 87 37 66 65 28 88 -99 -21 -83 24 80 39 64 -65 45 86 -53 -49 94 -75 -31 -42 -1 -35 -18 74 30 31 -40 30 -6 47 58 -71 -21 20 13 75 -79 15 -98 -26 76 99 -77 -9 85 48 51 -87 56 -53 37 47 -3 94 64 -7 74 86\n72 51 -74 20 41 -76 98 58 24 -61 -97 -73 62 29 6 42 -92 -6 -65 89 -32 -9 82 -13 -88 -70 -97 25 -48 12 -54 33 -92 -29 48 60 -21 86 -17 -86 45 -34 -3 -9 -62 12 25 -74 -76 -89 48 55 -30 86 51",
"output": "Infinity"
},
{
"input": "73 15\n-70 78 51 -33 -95 46 87 -33 16 62 67 -85 -57 75 -93 -59 98 -45 -90 -88 9 53 35 37 28 3 40 -87 28 5 18 11 9 1 72 69 -65 -62 1 73 -3 3 35 17 -28 -31 -45 60 64 18 60 38 -47 12 2 -90 -4 33 -51 -55 -54 90 38 -65 39 32 -70 0 -5 3 -12 100 78 55\n46 33 41 52 -89 -9 53 -81 34 -45 -11 -41 14 -28 95 -50",
"output": "-Infinity"
},
{
"input": "33 1\n-75 -83 87 -27 -48 47 -90 -84 -18 -4 14 -1 -83 -98 -68 -85 -86 28 2 45 96 -59 86 -25 -2 -64 -92 65 69 72 72 -58 -99 90\n-1 72",
"output": "Infinity"
},
{
"input": "58 58\n-25 40 -34 23 -52 94 -30 -99 -71 -90 -44 -71 69 48 -45 -59 0 66 -70 -96 95 91 82 90 -95 87 3 -77 -77 -26 15 87 -82 5 -24 82 -11 99 35 49 22 44 18 -60 -26 79 67 71 -13 29 -23 9 58 -90 88 18 77 5 -7\n-30 -11 -13 -50 61 -78 11 -74 -73 13 -66 -65 -82 38 58 25 -64 -24 78 -87 6 6 -80 -96 47 -25 -54 10 -41 -22 -50 -1 -6 -22 27 54 -32 30 93 88 -70 -100 -69 -47 -20 -92 -24 70 -93 42 78 42 -35 41 31 75 -67 -62 -83",
"output": "5/6"
},
{
"input": "20 20\n5 4 91 -66 -57 55 -79 -2 -54 -72 -49 21 -23 -5 57 -48 70 -16 -86 -26 -19\n51 -60 64 -8 89 27 -96 4 95 -24 -2 -27 -41 -14 -88 -19 24 68 -31 34 -62",
"output": "5/51"
},
{
"input": "69 69\n-90 -63 -21 23 23 -14 -82 65 42 -60 -42 -39 67 34 96 93 -42 -24 21 -80 44 -81 45 -74 -19 -88 39 58 90 87 16 48 -19 -2 36 87 4 -66 -82 -49 -32 -43 -65 12 34 -29 -58 46 -67 -20 -30 91 21 65 15 2 3 -92 -67 -68 39 -24 77 76 -17 -34 5 63 88 83\n-55 98 -79 18 -100 -67 -79 -85 -75 -44 -6 -73 -11 -12 -24 -78 47 -51 25 -29 -34 25 27 11 -87 15 -44 41 -44 46 -67 70 -35 41 62 -36 27 -41 -42 -50 96 31 26 -66 9 74 34 31 25 6 -84 41 74 -7 49 5 35 -5 -71 -37 28 58 -8 -40 -19 -83 -34 64 7 15",
"output": "18/11"
},
{
"input": "0 0\n46\n-33",
"output": "-46/33"
},
{
"input": "67 67\n-8 11 55 80 -26 -38 58 73 -48 -10 35 75 16 -84 55 -51 98 58 -28 98 77 81 51 -86 -46 68 -87 -80 -49 81 96 -97 -42 25 6 -8 -55 -25 93 -29 -33 -6 -26 -85 73 97 63 57 51 92 -6 -8 4 86 46 -45 36 -19 -71 1 71 39 97 -44 -34 -1 2 -46\n91 -32 -76 11 -40 91 -8 -100 73 80 47 82 24 0 -71 82 -93 38 -54 1 -55 -53 90 -86 0 10 -35 49 90 56 25 17 46 -43 13 16 -82 -33 64 -83 -56 22 12 -74 4 -68 85 -27 60 -28 -47 73 -93 69 -37 54 -3 90 -56 56 78 61 7 -79 48 -42 -10 -48",
"output": "-8/91"
},
{
"input": "69 69\n-7 38 -3 -22 65 -78 -65 -99 -76 63 0 -4 -78 -51 54 -61 -53 60 80 34 -96 99 -78 -96 21 -10 -86 33 -9 -81 -19 -2 -76 -3 -66 -80 -55 -21 -50 37 -86 -37 47 44 76 -39 54 -25 41 -86 -3 -25 -67 94 18 67 27 -5 -30 -69 2 -76 7 -97 -52 -35 -55 -20 92 2\n90 -94 37 41 -27 -54 96 -15 -60 -29 -75 -93 -57 62 48 -88 -99 -62 4 -9 85 33 65 -95 -30 16 -29 -89 -33 -83 -35 -21 53 -52 80 -40 76 -33 86 47 18 43 -67 -36 -99 -42 1 -94 -78 34 -41 73 96 2 -60 29 68 -96 -21 -61 -98 -67 1 40 85 55 66 -25 -50 -83",
"output": "-7/90"
},
{
"input": "17 17\n-54 59 -95 87 3 -27 -30 49 -87 74 45 78 36 60 -95 41 -53 -70\n-27 16 -67 -24 10 -73 -41 12 -52 53 -73 -17 -56 -74 -33 -8 100 -39",
"output": "2/1"
},
{
"input": "1 1\n36 -49\n-32 -40",
"output": "-9/8"
},
{
"input": "1 1\n1 1\n1 1",
"output": "1/1"
},
{
"input": "1 1\n-2 1\n4 1",
"output": "-1/2"
},
{
"input": "0 0\n2\n1",
"output": "2/1"
},
{
"input": "0 0\n4\n-3",
"output": "-4/3"
},
{
"input": "0 0\n2\n2",
"output": "1/1"
},
{
"input": "0 0\n17\n-10",
"output": "-17/10"
},
{
"input": "0 0\n-1\n2",
"output": "-1/2"
},
{
"input": "0 0\n1\n1",
"output": "1/1"
},
{
"input": "0 0\n50\n20",
"output": "5/2"
},
{
"input": "0 0\n20\n20",
"output": "1/1"
},
{
"input": "0 0\n4\n-2",
"output": "-2/1"
},
{
"input": "0 0\n4\n-6",
"output": "-2/3"
},
{
"input": "0 0\n1\n-2",
"output": "-1/2"
},
{
"input": "0 0\n4\n2",
"output": "2/1"
},
{
"input": "0 0\n2\n-4",
"output": "-1/2"
},
{
"input": "1 1\n4 1\n2 1",
"output": "2/1"
},
{
"input": "2 2\n-13 1 3\n6 3 2",
"output": "-13/6"
},
{
"input": "99 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 1\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": "1/2"
},
{
"input": "0 0\n5\n5",
"output": "1/1"
},
{
"input": "0 0\n2\n-1",
"output": "-2/1"
}
] | 1,585,765,849
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 278
| 0
|
[n,m]=list(map(int,input().strip().split()))
l1=list(map(int,input().strip().split()))
l2=list(map(int,input().strip().split()))
def computeGCD(x, y):
if x > y:
small = y
else:
small = x
for i in range(1, small+1):
if((x % i == 0) and (y % i == 0)):
gcd = i
return gcd
if n>m:
if abs(l1[0])//l1[0]+abs(l2[0])//l2[0]==0:
print("-Infinity")
else:
print("Infinity")
elif m>n:
print("0/1")
else:
divide=computeGCD(abs(l1[0]),abs(l2[0]))
print(divide)
che=abs(l1[0])//l1[0]+abs(l2[0])//l2[0]
s=""
if che==0:
s="-"
print(s+str(abs(l1[0]//divide))+"/"+str(abs(l2[0]//divide)))
|
Title: Limit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two polynomials:
- *P*(*x*)<==<=*a*0·*x**n*<=+<=*a*1·*x**n*<=-<=1<=+<=...<=+<=*a**n*<=-<=1·*x*<=+<=*a**n* and - *Q*(*x*)<==<=*b*0·*x**m*<=+<=*b*1·*x**m*<=-<=1<=+<=...<=+<=*b**m*<=-<=1·*x*<=+<=*b**m*.
Calculate limit .
Input Specification:
The first line contains two space-separated integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=100) — degrees of polynomials *P*(*x*) and *Q*(*x*) correspondingly.
The second line contains *n*<=+<=1 space-separated integers — the factors of polynomial *P*(*x*): *a*0, *a*1, ..., *a**n*<=-<=1, *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100,<=*a*0<=≠<=0).
The third line contains *m*<=+<=1 space-separated integers — the factors of polynomial *Q*(*x*): *b*0, *b*1, ..., *b**m*<=-<=1, *b**m* (<=-<=100<=≤<=*b**i*<=≤<=100,<=*b*0<=≠<=0).
Output Specification:
If the limit equals <=+<=∞, print "Infinity" (without quotes). If the limit equals <=-<=∞, print "-Infinity" (without the quotes).
If the value of the limit equals zero, print "0/1" (without the quotes).
Otherwise, print an irreducible fraction — the value of limit , in the format "p/q" (without the quotes), where *p* is the — numerator, *q* (*q*<=><=0) is the denominator of the fraction.
Demo Input:
['2 1\n1 1 1\n2 5\n', '1 0\n-1 3\n2\n', '0 1\n1\n1 0\n', '2 2\n2 1 6\n4 5 -7\n', '1 1\n9 0\n-5 2\n']
Demo Output:
['Infinity\n', '-Infinity\n', '0/1\n', '1/2\n', '-9/5\n']
Note:
Let's consider all samples:
1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c28febca257452afdfcbd6984ba8623911f9bdbc.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1e55ecd04e54a45e5e0092ec9a5c1ea03bb29255.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/2c95fb684d373fcc1a481cfabeda4d5c2f3673ee.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4dc40cb8b3cd6375c42445366e50369649a2801a.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c6455aba35cfb3c4397505121d1f77afcd17c98e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
You can learn more about the definition and properties of limits if you follow the link: http://en.wikipedia.org/wiki/Limit_of_a_function
|
```python
[n,m]=list(map(int,input().strip().split()))
l1=list(map(int,input().strip().split()))
l2=list(map(int,input().strip().split()))
def computeGCD(x, y):
if x > y:
small = y
else:
small = x
for i in range(1, small+1):
if((x % i == 0) and (y % i == 0)):
gcd = i
return gcd
if n>m:
if abs(l1[0])//l1[0]+abs(l2[0])//l2[0]==0:
print("-Infinity")
else:
print("Infinity")
elif m>n:
print("0/1")
else:
divide=computeGCD(abs(l1[0]),abs(l2[0]))
print(divide)
che=abs(l1[0])//l1[0]+abs(l2[0])//l2[0]
s=""
if che==0:
s="-"
print(s+str(abs(l1[0]//divide))+"/"+str(abs(l2[0]//divide)))
```
| 0
|
|
579
|
A
|
Raising Bacteria
|
PROGRAMMING
| 1,000
|
[
"bitmasks"
] | null | null |
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
|
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
|
The only line containing one integer: the answer.
|
[
"5\n",
"8\n"
] |
[
"2\n",
"1\n"
] |
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
| 250
|
[
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697681824",
"output": "14"
},
{
"input": "41313494",
"output": "14"
},
{
"input": "673935585",
"output": "13"
},
{
"input": "117422204",
"output": "19"
},
{
"input": "954746654",
"output": "15"
},
{
"input": "536838144",
"output": "14"
},
{
"input": "1000000000",
"output": "13"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "6",
"output": "2"
},
{
"input": "7",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "999999999",
"output": "21"
},
{
"input": "999999998",
"output": "20"
},
{
"input": "999999997",
"output": "20"
},
{
"input": "999999996",
"output": "19"
},
{
"input": "999999995",
"output": "20"
},
{
"input": "999999994",
"output": "19"
},
{
"input": "999999993",
"output": "19"
},
{
"input": "999999992",
"output": "18"
},
{
"input": "999999991",
"output": "20"
},
{
"input": "999999990",
"output": "19"
},
{
"input": "536870910",
"output": "28"
},
{
"input": "536870912",
"output": "1"
},
{
"input": "536870913",
"output": "2"
},
{
"input": "536870914",
"output": "2"
},
{
"input": "22",
"output": "3"
},
{
"input": "14",
"output": "3"
}
] | 1,695,901,830
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 61
| 0
|
#Coder_1_neel
a=int(input())
k=a%2
if k==0:
print(1)
else:
print(k+1)
|
Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
|
```python
#Coder_1_neel
a=int(input())
k=a%2
if k==0:
print(1)
else:
print(k+1)
```
| 0
|
|
129
|
A
|
Cookies
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
|
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
|
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
|
[
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] |
[
"1\n",
"8\n",
"1\n"
] |
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
| 500
|
[
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,696,183,165
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 92
| 0
|
def count_ways_to_steal_bag(n, bags):
even_count = 0
odd_count = 0
for bag in bags:
if bag % 2 == 0:
even_count += 1
else:
odd_count += 1
# If there is exactly one bag with an odd number of cookies, return the count
if odd_count == 1:
return 1
else:
return even_count
# Read input
n = int(input())
bags = list(map(int, input().split()))
# Calculate and print the result
result = count_ways_to_steal_bag(n, bags)
print(result)
|
Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
|
```python
def count_ways_to_steal_bag(n, bags):
even_count = 0
odd_count = 0
for bag in bags:
if bag % 2 == 0:
even_count += 1
else:
odd_count += 1
# If there is exactly one bag with an odd number of cookies, return the count
if odd_count == 1:
return 1
else:
return even_count
# Read input
n = int(input())
bags = list(map(int, input().split()))
# Calculate and print the result
result = count_ways_to_steal_bag(n, bags)
print(result)
```
| 0
|
|
622
|
C
|
Not Equal on a Segment
|
PROGRAMMING
| 1,700
|
[
"data structures",
"implementation"
] | null | null |
You are given array *a* with *n* integers and *m* queries. The *i*-th query is given with three integers *l**i*,<=*r**i*,<=*x**i*.
For the *i*-th query find any position *p**i* (*l**i*<=≤<=*p**i*<=≤<=*r**i*) so that *a**p**i*<=≠<=*x**i*.
|
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the number of elements in *a* and the number of queries.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=106) — the elements of the array *a*.
Each of the next *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*,<=1<=≤<=*x**i*<=≤<=106) — the parameters of the *i*-th query.
|
Print *m* lines. On the *i*-th line print integer *p**i* — the position of any number not equal to *x**i* in segment [*l**i*,<=*r**i*] or the value <=-<=1 if there is no such number.
|
[
"6 4\n1 2 1 1 3 5\n1 4 1\n2 6 2\n3 4 1\n3 4 2\n"
] |
[
"2\n6\n-1\n4\n"
] |
none
| 0
|
[
{
"input": "6 4\n1 2 1 1 3 5\n1 4 1\n2 6 2\n3 4 1\n3 4 2",
"output": "2\n6\n-1\n4"
},
{
"input": "1 1\n1\n1 1 1",
"output": "-1"
},
{
"input": "1 1\n2\n1 1 2",
"output": "-1"
},
{
"input": "1 1\n569888\n1 1 967368",
"output": "1"
},
{
"input": "10 10\n1 1 1 1 1 1 1 1 1 1\n3 10 1\n3 6 1\n1 8 1\n1 7 1\n1 5 1\n3 7 1\n4 7 1\n9 9 1\n6 7 1\n3 4 1",
"output": "-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "10 10\n1 2 2 2 2 1 1 2 1 1\n3 3 1\n4 9 1\n4 8 1\n2 7 2\n2 8 2\n3 10 1\n7 7 2\n10 10 2\n1 5 1\n1 2 1",
"output": "3\n8\n8\n7\n7\n8\n7\n10\n5\n2"
},
{
"input": "10 10\n318890 307761 832732 700511 820583 522866 130891 914566 128429 739710\n4 9 178864\n6 9 741003\n4 9 172997\n4 6 314469\n1 4 694802\n8 8 401658\n7 10 376243\n7 8 508771\n3 5 30038\n2 10 591490",
"output": "9\n9\n9\n6\n4\n8\n10\n8\n5\n10"
},
{
"input": "1 1\n2\n1 1 1",
"output": "1"
},
{
"input": "10 10\n1 1 1 1 1 2 1 1 1 1\n1 9 1\n6 7 1\n2 4 1\n7 8 1\n1 3 1\n10 10 1\n3 5 1\n6 7 1\n1 10 1\n6 6 1",
"output": "6\n6\n-1\n-1\n-1\n-1\n-1\n6\n6\n6"
},
{
"input": "7 1\n2 1 3 2 2 2 2\n1 7 2",
"output": "3"
},
{
"input": "4 1\n3 1 2 2\n1 4 2",
"output": "2"
},
{
"input": "6 1\n3 2 4 3 3 3\n1 6 3",
"output": "3"
},
{
"input": "4 1\n1 3 2 2\n1 4 2",
"output": "2"
},
{
"input": "5 1\n2 3 1 2 2\n1 5 2",
"output": "3"
},
{
"input": "3 1\n1 9 5\n1 3 5",
"output": "2"
},
{
"input": "4 1\n4 2 6 4\n1 4 4",
"output": "3"
},
{
"input": "2 1\n1 3\n1 2 2",
"output": "2"
},
{
"input": "10 1\n2 2 1 3 2 2 2 2 2 2\n2 5 2",
"output": "4"
},
{
"input": "7 1\n6 5 7 6 6 6 6\n1 7 6",
"output": "3"
},
{
"input": "3 1\n2 4 3\n1 3 3",
"output": "2"
},
{
"input": "4 1\n4 2 3 3\n1 4 3",
"output": "2"
},
{
"input": "5 1\n3 2 4 5 5\n1 3 3",
"output": "3"
},
{
"input": "2 6\n1 1\n1 1 1\n1 1 2\n1 2 1\n1 2 2\n2 2 1\n2 2 2",
"output": "-1\n1\n-1\n2\n-1\n2"
}
] | 1,463,498,487
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 13
| 1,000
| 5,222,400
|
n,m = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
ans = []
for i in range(m):
l,r,x = [int(j) for j in input().split()]
t = 0
for k in a[l - 1:r]:
t += 1
if k != x:
ans.append(l - 1 + t)
break
if t == r - l + 1:
ans.append(-1)
print(*ans, sep = '\n')
|
Title: Not Equal on a Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given array *a* with *n* integers and *m* queries. The *i*-th query is given with three integers *l**i*,<=*r**i*,<=*x**i*.
For the *i*-th query find any position *p**i* (*l**i*<=≤<=*p**i*<=≤<=*r**i*) so that *a**p**i*<=≠<=*x**i*.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the number of elements in *a* and the number of queries.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=106) — the elements of the array *a*.
Each of the next *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*,<=1<=≤<=*x**i*<=≤<=106) — the parameters of the *i*-th query.
Output Specification:
Print *m* lines. On the *i*-th line print integer *p**i* — the position of any number not equal to *x**i* in segment [*l**i*,<=*r**i*] or the value <=-<=1 if there is no such number.
Demo Input:
['6 4\n1 2 1 1 3 5\n1 4 1\n2 6 2\n3 4 1\n3 4 2\n']
Demo Output:
['2\n6\n-1\n4\n']
Note:
none
|
```python
n,m = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
ans = []
for i in range(m):
l,r,x = [int(j) for j in input().split()]
t = 0
for k in a[l - 1:r]:
t += 1
if k != x:
ans.append(l - 1 + t)
break
if t == r - l + 1:
ans.append(-1)
print(*ans, sep = '\n')
```
| 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"
}
] | 1,447,958,477
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 25
| 1,000
| 0
|
def pw(a, n):
res = 1
while(n):
if (n & 1):
res *= a
n -= 1
else:
a *= a
n >>= 1
return res
m, n = map(int, input().split())
ans = 0
for i in range(1, m + 1):
ans += (pw(i / m, n) - pw((i - 1) / m, n)) * i
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
def pw(a, n):
res = 1
while(n):
if (n & 1):
res *= a
n -= 1
else:
a *= a
n >>= 1
return res
m, n = map(int, input().split())
ans = 0
for i in range(1, m + 1):
ans += (pw(i / m, n) - pw((i - 1) / m, n)) * i
print(ans)
```
| 0
|
|
876
|
A
|
Trip For Meal
|
PROGRAMMING
| 900
|
[
"math"
] | null | null |
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
|
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
|
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
|
[
"3\n2\n3\n1\n",
"1\n2\n3\n5\n"
] |
[
"3\n",
"0\n"
] |
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
| 500
|
[
{
"input": "3\n2\n3\n1",
"output": "3"
},
{
"input": "1\n2\n3\n5",
"output": "0"
},
{
"input": "10\n1\n8\n3",
"output": "9"
},
{
"input": "7\n10\n5\n6",
"output": "30"
},
{
"input": "9\n9\n7\n5",
"output": "42"
},
{
"input": "9\n37\n85\n76",
"output": "296"
},
{
"input": "76\n46\n77\n11",
"output": "860"
},
{
"input": "80\n42\n1\n37",
"output": "79"
},
{
"input": "8\n80\n55\n1",
"output": "61"
},
{
"input": "10\n13\n72\n17",
"output": "117"
},
{
"input": "9\n24\n1\n63",
"output": "8"
},
{
"input": "65\n5\n8\n7",
"output": "320"
},
{
"input": "56\n8\n9\n3",
"output": "170"
},
{
"input": "59\n8\n1\n2",
"output": "58"
},
{
"input": "75\n50\n50\n5",
"output": "415"
},
{
"input": "75\n54\n76\n66",
"output": "3996"
},
{
"input": "73\n71\n69\n66",
"output": "4755"
},
{
"input": "83\n58\n88\n16",
"output": "1354"
},
{
"input": "74\n31\n11\n79",
"output": "803"
},
{
"input": "62\n27\n16\n72",
"output": "976"
},
{
"input": "72\n95\n27\n9",
"output": "657"
},
{
"input": "1\n2\n2\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n99",
"output": "0"
},
{
"input": "100\n100\n100\n100",
"output": "9900"
},
{
"input": "2\n1\n1\n3",
"output": "1"
},
{
"input": "1\n3\n2\n1",
"output": "0"
},
{
"input": "1\n5\n6\n1",
"output": "0"
},
{
"input": "1\n2\n6\n1",
"output": "0"
},
{
"input": "1\n30\n20\n1",
"output": "0"
},
{
"input": "1\n15\n13\n11",
"output": "0"
},
{
"input": "1\n100\n50\n1",
"output": "0"
},
{
"input": "1\n2\n3\n1",
"output": "0"
},
{
"input": "5\n5\n5\n5",
"output": "20"
},
{
"input": "1\n100\n50\n5",
"output": "0"
},
{
"input": "1\n5\n7\n3",
"output": "0"
},
{
"input": "1\n3\n4\n2",
"output": "0"
},
{
"input": "3\n3\n3\n3",
"output": "6"
},
{
"input": "1\n99\n98\n1",
"output": "0"
},
{
"input": "5\n3\n3\n1",
"output": "6"
},
{
"input": "2\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1\n7\n8",
"output": "1"
},
{
"input": "2\n2\n2\n1",
"output": "2"
},
{
"input": "2\n2\n3\n1",
"output": "2"
},
{
"input": "1\n5\n6\n2",
"output": "0"
},
{
"input": "100\n1\n1\n100",
"output": "99"
},
{
"input": "2\n3\n2\n1",
"output": "2"
},
{
"input": "1\n99\n98\n97",
"output": "0"
}
] | 1,647,672,872
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 48
| 109
| 0
|
n = int(input())
a = int(input())
b = int(input())
c = int(input())
dist = 0
cur_pos = "R"
cur_n = 1
d = {"RO": a, "RE": b, "OE": c}
while cur_n < n:
j = [i for i in d.keys() if cur_pos in i]
if d[j[0]] > d[j[1]]:
k = j[1]
else:
k = j[0]
cur_pos = k.replace(cur_pos, "")
cur_n += 1
dist += d[k]
print(dist)
|
Title: Trip For Meal
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
Input Specification:
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
Output Specification:
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
Demo Input:
['3\n2\n3\n1\n', '1\n2\n3\n5\n']
Demo Output:
['3\n', '0\n']
Note:
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
|
```python
n = int(input())
a = int(input())
b = int(input())
c = int(input())
dist = 0
cur_pos = "R"
cur_n = 1
d = {"RO": a, "RE": b, "OE": c}
while cur_n < n:
j = [i for i in d.keys() if cur_pos in i]
if d[j[0]] > d[j[1]]:
k = j[1]
else:
k = j[0]
cur_pos = k.replace(cur_pos, "")
cur_n += 1
dist += d[k]
print(dist)
```
| 3
|
|
817
|
E
|
Choosing The Commander
|
PROGRAMMING
| 2,000
|
[
"bitmasks",
"data structures",
"trees"
] | null | null |
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality — an integer number *p**i*. Each commander has two characteristics — his personality *p**j* and leadership *l**j* (both are integer numbers). Warrior *i* respects commander *j* only if ( is the bitwise excluding OR of *x* and *y*).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
- 1 *p**i* — one warrior with personality *p**i* joins Vova's army; - 2 *p**i* — one warrior with personality *p**i* leaves Vova's army; - 3 *p**i* *l**i* — Vova tries to hire a commander with personality *p**i* and leadership *l**i*.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
|
The first line contains one integer *q* (1<=≤<=*q*<=≤<=100000) — the number of events.
Then *q* lines follow. Each line describes the event:
- 1 *p**i* (1<=≤<=*p**i*<=≤<=108) — one warrior with personality *p**i* joins Vova's army; - 2 *p**i* (1<=≤<=*p**i*<=≤<=108) — one warrior with personality *p**i* leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment); - 3 *p**i* *l**i* (1<=≤<=*p**i*,<=*l**i*<=≤<=108) — Vova tries to hire a commander with personality *p**i* and leadership *l**i*. There is at least one event of this type.
|
For each event of the third type print one integer — the number of warriors who respect the commander Vova tries to hire in the event.
|
[
"5\n1 3\n1 4\n3 6 3\n2 4\n3 6 3\n"
] |
[
"1\n0\n"
] |
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3d8e4cd0a5c3c0cdc8b35097f3dc7317604240a6.png" style="max-width: 100.0%;max-height: 100.0%;"/>, and 2 < 3, but <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4e3eeab99fa495ecdcd103c68de47dd72943016f.png" style="max-width: 100.0%;max-height: 100.0%;"/>, and 5 ≥ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him.
| 0
|
[
{
"input": "5\n1 3\n1 4\n3 6 3\n2 4\n3 6 3",
"output": "1\n0"
}
] | 1,689,637,514
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1689637513.913444")# 1689637513.9134648
|
Title: Choosing The Commander
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality — an integer number *p**i*. Each commander has two characteristics — his personality *p**j* and leadership *l**j* (both are integer numbers). Warrior *i* respects commander *j* only if ( is the bitwise excluding OR of *x* and *y*).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
- 1 *p**i* — one warrior with personality *p**i* joins Vova's army; - 2 *p**i* — one warrior with personality *p**i* leaves Vova's army; - 3 *p**i* *l**i* — Vova tries to hire a commander with personality *p**i* and leadership *l**i*.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
Input Specification:
The first line contains one integer *q* (1<=≤<=*q*<=≤<=100000) — the number of events.
Then *q* lines follow. Each line describes the event:
- 1 *p**i* (1<=≤<=*p**i*<=≤<=108) — one warrior with personality *p**i* joins Vova's army; - 2 *p**i* (1<=≤<=*p**i*<=≤<=108) — one warrior with personality *p**i* leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment); - 3 *p**i* *l**i* (1<=≤<=*p**i*,<=*l**i*<=≤<=108) — Vova tries to hire a commander with personality *p**i* and leadership *l**i*. There is at least one event of this type.
Output Specification:
For each event of the third type print one integer — the number of warriors who respect the commander Vova tries to hire in the event.
Demo Input:
['5\n1 3\n1 4\n3 6 3\n2 4\n3 6 3\n']
Demo Output:
['1\n0\n']
Note:
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3d8e4cd0a5c3c0cdc8b35097f3dc7317604240a6.png" style="max-width: 100.0%;max-height: 100.0%;"/>, and 2 < 3, but <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4e3eeab99fa495ecdcd103c68de47dd72943016f.png" style="max-width: 100.0%;max-height: 100.0%;"/>, and 5 ≥ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him.
|
```python
print("_RANDOM_GUESS_1689637513.913444")# 1689637513.9134648
```
| 0
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,611,054,108
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
t=int(input())
x=[]
y=[]
z=[]
li=[]
for i in range(t):
li.append([int(x) for x in input().split()]
for i in li:
x.append(i[0])
y.append(i[1])
z.append(i[2])
if(sum(x)==0 and sum(y)==0 and sum(z)==0):
print("Yes")
else:
print("No")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
t=int(input())
x=[]
y=[]
z=[]
li=[]
for i in range(t):
li.append([int(x) for x in input().split()]
for i in li:
x.append(i[0])
y.append(i[1])
z.append(i[2])
if(sum(x)==0 and sum(y)==0 and sum(z)==0):
print("Yes")
else:
print("No")
```
| -1
|
126
|
B
|
Password
|
PROGRAMMING
| 1,700
|
[
"binary search",
"dp",
"hashing",
"string suffix structures",
"strings"
] | null | null |
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
|
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
|
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
|
[
"fixprefixsuffix\n",
"abcdabc\n"
] |
[
"fix",
"Just a legend"
] |
none
| 1,000
|
[
{
"input": "fixprefixsuffix",
"output": "fix"
},
{
"input": "abcdabc",
"output": "Just a legend"
},
{
"input": "qwertyqwertyqwerty",
"output": "qwerty"
},
{
"input": "papapapap",
"output": "papap"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaa"
},
{
"input": "ghbdtn",
"output": "Just a legend"
},
{
"input": "a",
"output": "Just a legend"
},
{
"input": "aa",
"output": "Just a legend"
},
{
"input": "ab",
"output": "Just a legend"
},
{
"input": "aaa",
"output": "a"
},
{
"input": "aba",
"output": "Just a legend"
},
{
"input": "aab",
"output": "Just a legend"
},
{
"input": "abb",
"output": "Just a legend"
},
{
"input": "abc",
"output": "Just a legend"
},
{
"input": "aaabaabaaaaab",
"output": "Just a legend"
},
{
"input": "aabaaabaaaaab",
"output": "aab"
},
{
"input": "aaabaaaabab",
"output": "Just a legend"
},
{
"input": "abcabcabcabcabc",
"output": "abcabcabc"
},
{
"input": "aaaaabaaaa",
"output": "aaaa"
},
{
"input": "aaaabaaaaaaa",
"output": "aaaa"
},
{
"input": "ghghghgxghghghg",
"output": "ghghg"
},
{
"input": "kincenvizh",
"output": "Just a legend"
},
{
"input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw",
"output": "Just a legend"
},
{
"input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx",
"output": "Just a legend"
},
{
"input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb",
"output": "nfbdzgdlbjhrlvfryyjb"
},
{
"input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp",
"output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp"
},
{
"input": "abcazabca",
"output": "a"
}
] | 1,698,435,377
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
def computeLPSArray(P, M):
lps = [0 for _ in range(M)]
len = 0
lps[0] = 0
i = 1
while i < M:
if P[i] == P[len]:
len += 1
lps[i] = len
i += 1
else:
if len != 0:
len = lps[len - 1]
else:
lps[i] = 0
i += 1
return lps
def KMPSearch(P, T):
M = len(P)
N = len(T)
lps = computeLPSArray(P, M)
i = 0
j = 0
while i < N:
if P[j] == T[i]:
j += 1
i += 1
if j == M:
return True
if i < N and P[j] != T[i]:
if j != 0:
j = lps[j - 1]
else:
i += 1
return False
def main():
s = input()
n = len(s)
ans = "Just a legend"
i = 1
while i <= (n // 2 + 1):
while s[:i] != s[n - i:]:
i += 1
if KMPSearch(s[:i], s[1:-1]):
ans = s[:i]
i += 1
print(ans)
main()
|
Title: Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
Input Specification:
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Output Specification:
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
Demo Input:
['fixprefixsuffix\n', 'abcdabc\n']
Demo Output:
['fix', 'Just a legend']
Note:
none
|
```python
def computeLPSArray(P, M):
lps = [0 for _ in range(M)]
len = 0
lps[0] = 0
i = 1
while i < M:
if P[i] == P[len]:
len += 1
lps[i] = len
i += 1
else:
if len != 0:
len = lps[len - 1]
else:
lps[i] = 0
i += 1
return lps
def KMPSearch(P, T):
M = len(P)
N = len(T)
lps = computeLPSArray(P, M)
i = 0
j = 0
while i < N:
if P[j] == T[i]:
j += 1
i += 1
if j == M:
return True
if i < N and P[j] != T[i]:
if j != 0:
j = lps[j - 1]
else:
i += 1
return False
def main():
s = input()
n = len(s)
ans = "Just a legend"
i = 1
while i <= (n // 2 + 1):
while s[:i] != s[n - i:]:
i += 1
if KMPSearch(s[:i], s[1:-1]):
ans = s[:i]
i += 1
print(ans)
main()
```
| 0
|
|
958
|
C2
|
Encryption (medium)
|
PROGRAMMING
| 2,000
|
[
"dp"
] | null | null |
Heidi has now broken the first level of encryption of the Death Star plans, and is staring at the screen presenting her with the description of the next code she has to enter. It looks surprisingly similar to the first one – seems like the Empire engineers were quite lazy...
Heidi is once again given a sequence *A*, but now she is also given two integers *k* and *p*. She needs to find out what the encryption key *S* is.
Let *X* be a sequence of integers, and *p* a positive integer. We define the score of *X* to be the sum of the elements of *X* modulo *p*.
Heidi is given a sequence *A* that consists of *N* integers, and also given integers *k* and *p*. Her goal is to split *A* into *k* part such that:
- Each part contains at least 1 element of *A*, and each part consists of contiguous elements of *A*. - No two parts overlap. - The total sum *S* of the scores of those parts is maximized.
Output the sum *S* – the encryption code.
|
The first line of the input contains three space-separated integer *N*, *k* and *p* (*k*<=≤<=*N*<=≤<=20<=000, 2<=≤<=*k*<=≤<=50, 2<=≤<=*p*<=≤<=100) – the number of elements in *A*, the number of parts *A* should be split into, and the modulo for computing scores, respectively.
The second line contains *N* space-separated integers that are the elements of *A*. Each integer is from the interval [1,<=1<=000<=000].
|
Output the number *S* as described in the problem statement.
|
[
"4 3 10\n3 4 7 2\n",
"10 5 12\n16 3 24 13 9 8 7 5 12 12\n"
] |
[
"16\n",
"37\n"
] |
In the first example, if the input sequence is split as (3, 4), (7), (2), the total score would be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5a7944578a08885d55a30192c8f0b817cad2ac96.png" style="max-width: 100.0%;max-height: 100.0%;"/>. It is easy to see that this score is maximum.
In the second example, one possible way to obtain score 37 is to make the following split: (16, 3, 24), (13, 9), (8), (7), (5, 12, 12).
| 0
|
[] | 1,523,723,818
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
//
// main.cpp
// cf3
//
// Created by joe on 2018/4/11.
// Copyright © 2018 joe. All rights reserved.
//
#include <iostream>
#include <algorithm>
using namespace std;
int n,k,p;
int *ns;
int **dp;
int **nextt;
void pri(){
for(int i=0;i<k;i++){
for(int j=0;j<p;j++)
cout<<dp[i][j]<<' ';
cout<<endl;
}
}
int main() {
std::ios::sync_with_stdio(false);
cin>>n>>k>>p;
ns=new int[n];
for(int i=0;i<n;i++){
int t;
cin>>t;
ns[i]=t%p;
}
dp=(int**)malloc(k*sizeof(int*));
for(int i=0;i<k;i++){
dp[i]=(int*)malloc(p*sizeof(int));
for(int j=0;j<p;j++)
dp[i][j]=-1;
}
nextt=(int**)malloc(k*sizeof(int*));
for(int i=0;i<k;i++){
nextt[i]=(int*)malloc(p*sizeof(int));
for(int j=0;j<p;j++)
nextt[i][j]=-1;
}
dp[0][ns[0]]=ns[0];
for(int i=1;i<n;i++){
//pri();
for(int i=0;i<k;i++){
nextt[i]=(int*)malloc(p*sizeof(int));
for(int j=0;j<p;j++)
nextt[i][j]=-1;
}
for(int k1=0;k1<k;k1++){
for(int k2=0;k2<p;k2++){
if(dp[k1][k2]==-1)
continue;
int k2_loc=(k2+ns[i])%p;
int change=k2_loc-k2;
nextt[k1][k2_loc] = dp[k1][k2] + change;
}
}
for(int k1=0;k1<k-1;k1++){
int t=-1;
for(int i=0;i<p;i++){
t=max(t,dp[k1][i]);
}
if(t!=-1){
if(nextt[k1+1][ns[i]]==-1)
nextt[k1+1][ns[i]]=t+ns[i];
else
nextt[k1+1][ns[i]]=max(t+ns[i],nextt[k1+1][ns[i]]);
}
}
int **tt;
tt=dp;
dp=nextt;
nextt=tt;
}
int t=-1;
for(int i=0;i<p;i++){
t=max(t,dp[k-1][i]);
}
cout<<t<<endl;
return 0;
}
|
Title: Encryption (medium)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Heidi has now broken the first level of encryption of the Death Star plans, and is staring at the screen presenting her with the description of the next code she has to enter. It looks surprisingly similar to the first one – seems like the Empire engineers were quite lazy...
Heidi is once again given a sequence *A*, but now she is also given two integers *k* and *p*. She needs to find out what the encryption key *S* is.
Let *X* be a sequence of integers, and *p* a positive integer. We define the score of *X* to be the sum of the elements of *X* modulo *p*.
Heidi is given a sequence *A* that consists of *N* integers, and also given integers *k* and *p*. Her goal is to split *A* into *k* part such that:
- Each part contains at least 1 element of *A*, and each part consists of contiguous elements of *A*. - No two parts overlap. - The total sum *S* of the scores of those parts is maximized.
Output the sum *S* – the encryption code.
Input Specification:
The first line of the input contains three space-separated integer *N*, *k* and *p* (*k*<=≤<=*N*<=≤<=20<=000, 2<=≤<=*k*<=≤<=50, 2<=≤<=*p*<=≤<=100) – the number of elements in *A*, the number of parts *A* should be split into, and the modulo for computing scores, respectively.
The second line contains *N* space-separated integers that are the elements of *A*. Each integer is from the interval [1,<=1<=000<=000].
Output Specification:
Output the number *S* as described in the problem statement.
Demo Input:
['4 3 10\n3 4 7 2\n', '10 5 12\n16 3 24 13 9 8 7 5 12 12\n']
Demo Output:
['16\n', '37\n']
Note:
In the first example, if the input sequence is split as (3, 4), (7), (2), the total score would be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5a7944578a08885d55a30192c8f0b817cad2ac96.png" style="max-width: 100.0%;max-height: 100.0%;"/>. It is easy to see that this score is maximum.
In the second example, one possible way to obtain score 37 is to make the following split: (16, 3, 24), (13, 9), (8), (7), (5, 12, 12).
|
```python
//
// main.cpp
// cf3
//
// Created by joe on 2018/4/11.
// Copyright © 2018 joe. All rights reserved.
//
#include <iostream>
#include <algorithm>
using namespace std;
int n,k,p;
int *ns;
int **dp;
int **nextt;
void pri(){
for(int i=0;i<k;i++){
for(int j=0;j<p;j++)
cout<<dp[i][j]<<' ';
cout<<endl;
}
}
int main() {
std::ios::sync_with_stdio(false);
cin>>n>>k>>p;
ns=new int[n];
for(int i=0;i<n;i++){
int t;
cin>>t;
ns[i]=t%p;
}
dp=(int**)malloc(k*sizeof(int*));
for(int i=0;i<k;i++){
dp[i]=(int*)malloc(p*sizeof(int));
for(int j=0;j<p;j++)
dp[i][j]=-1;
}
nextt=(int**)malloc(k*sizeof(int*));
for(int i=0;i<k;i++){
nextt[i]=(int*)malloc(p*sizeof(int));
for(int j=0;j<p;j++)
nextt[i][j]=-1;
}
dp[0][ns[0]]=ns[0];
for(int i=1;i<n;i++){
//pri();
for(int i=0;i<k;i++){
nextt[i]=(int*)malloc(p*sizeof(int));
for(int j=0;j<p;j++)
nextt[i][j]=-1;
}
for(int k1=0;k1<k;k1++){
for(int k2=0;k2<p;k2++){
if(dp[k1][k2]==-1)
continue;
int k2_loc=(k2+ns[i])%p;
int change=k2_loc-k2;
nextt[k1][k2_loc] = dp[k1][k2] + change;
}
}
for(int k1=0;k1<k-1;k1++){
int t=-1;
for(int i=0;i<p;i++){
t=max(t,dp[k1][i]);
}
if(t!=-1){
if(nextt[k1+1][ns[i]]==-1)
nextt[k1+1][ns[i]]=t+ns[i];
else
nextt[k1+1][ns[i]]=max(t+ns[i],nextt[k1+1][ns[i]]);
}
}
int **tt;
tt=dp;
dp=nextt;
nextt=tt;
}
int t=-1;
for(int i=0;i<p;i++){
t=max(t,dp[k-1][i]);
}
cout<<t<<endl;
return 0;
}
```
| -1
|
|
937
|
B
|
Vile Grasshoppers
|
PROGRAMMING
| 1,400
|
[
"brute force",
"math",
"number theory"
] | null | null |
The weather is fine today and hence it's high time to climb the nearby pine and enjoy the landscape.
The pine's trunk includes several branches, located one above another and numbered from 2 to *y*. Some of them (more precise, from 2 to *p*) are occupied by tiny vile grasshoppers which you're at war with. These grasshoppers are known for their awesome jumping skills: the grasshopper at branch *x* can jump to branches .
Keeping this in mind, you wisely decided to choose such a branch that none of the grasshoppers could interrupt you. At the same time you wanna settle as high as possible since the view from up there is simply breathtaking.
In other words, your goal is to find the highest branch that cannot be reached by any of the grasshoppers or report that it's impossible.
|
The only line contains two integers *p* and *y* (2<=≤<=*p*<=≤<=*y*<=≤<=109).
|
Output the number of the highest suitable branch. If there are none, print -1 instead.
|
[
"3 6\n",
"3 4\n"
] |
[
"5\n",
"-1\n"
] |
In the first sample case grasshopper from branch 2 reaches branches 2, 4 and 6 while branch 3 is initially settled by another grasshopper. Therefore the answer is 5.
It immediately follows that there are no valid branches in second sample case.
| 1,000
|
[
{
"input": "3 6",
"output": "5"
},
{
"input": "3 4",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "5 50",
"output": "49"
},
{
"input": "944192806 944193066",
"output": "944192807"
},
{
"input": "1000000000 1000000000",
"output": "-1"
},
{
"input": "2 1000000000",
"output": "999999999"
},
{
"input": "28788 944193066",
"output": "944192833"
},
{
"input": "49 52",
"output": "-1"
},
{
"input": "698964997 734575900",
"output": "734575871"
},
{
"input": "287894773 723316271",
"output": "723316207"
},
{
"input": "171837140 733094070",
"output": "733094069"
},
{
"input": "37839169 350746807",
"output": "350746727"
},
{
"input": "125764821 234689174",
"output": "234689137"
},
{
"input": "413598841 430509920",
"output": "430509917"
},
{
"input": "145320418 592508508",
"output": "592508479"
},
{
"input": "155098216 476450875",
"output": "476450861"
},
{
"input": "459843315 950327842",
"output": "950327831"
},
{
"input": "469621113 834270209",
"output": "834270209"
},
{
"input": "13179877 557546766",
"output": "557546753"
},
{
"input": "541748242 723508350",
"output": "723508301"
},
{
"input": "607450717 924641194",
"output": "924641189"
},
{
"input": "786360384 934418993",
"output": "934418981"
},
{
"input": "649229491 965270051",
"output": "965270051"
},
{
"input": "144179719 953974590",
"output": "953974583"
},
{
"input": "28122086 963752388",
"output": "963752347"
},
{
"input": "268497487 501999053",
"output": "501999053"
},
{
"input": "356423140 385941420",
"output": "385941419"
},
{
"input": "71233638 269883787",
"output": "269883787"
},
{
"input": "2601 698964997",
"output": "698964983"
},
{
"input": "4096 287894773",
"output": "287894771"
},
{
"input": "5675 171837140",
"output": "171837131"
},
{
"input": "13067 350746807",
"output": "350746727"
},
{
"input": "8699 234689174",
"output": "234689137"
},
{
"input": "12190 413598841",
"output": "413598817"
},
{
"input": "20555 592508508",
"output": "592508479"
},
{
"input": "19137 476450875",
"output": "476450861"
},
{
"input": "8793 950327842",
"output": "950327831"
},
{
"input": "1541 834270209",
"output": "834270209"
},
{
"input": "1082 13179877",
"output": "13179871"
},
{
"input": "3888 723508350",
"output": "723508301"
},
{
"input": "14078 607450717",
"output": "607450703"
},
{
"input": "20869 786360384",
"output": "786360373"
},
{
"input": "13689 965270051",
"output": "965270051"
},
{
"input": "782 144179719",
"output": "144179719"
},
{
"input": "404 28122086",
"output": "28122079"
},
{
"input": "21992 501999053",
"output": "501999053"
},
{
"input": "13745 385941420",
"output": "385941419"
},
{
"input": "8711 269883787",
"output": "269883787"
},
{
"input": "31333 981756889",
"output": "981756871"
},
{
"input": "944192808 944193061",
"output": "-1"
},
{
"input": "3 9",
"output": "7"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "2 13",
"output": "13"
},
{
"input": "7 53",
"output": "53"
},
{
"input": "10 1000000000",
"output": "999999997"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "4 9",
"output": "7"
}
] | 1,574,484,510
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 124
| 0
|
def SieveOfEratosthenes(n):
c = []
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
# Print all prime numbers
for p in range(2, n):
if prime[p]:
c.append(p)
return c
n,m = map(int,input().split())
k = 0
c = SieveOfEratosthenes(n+1)
if m%2==0:
m-=1
for i in range(len(c)):
if m%c[i]==0:
m-=2
print(m)
|
Title: Vile Grasshoppers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The weather is fine today and hence it's high time to climb the nearby pine and enjoy the landscape.
The pine's trunk includes several branches, located one above another and numbered from 2 to *y*. Some of them (more precise, from 2 to *p*) are occupied by tiny vile grasshoppers which you're at war with. These grasshoppers are known for their awesome jumping skills: the grasshopper at branch *x* can jump to branches .
Keeping this in mind, you wisely decided to choose such a branch that none of the grasshoppers could interrupt you. At the same time you wanna settle as high as possible since the view from up there is simply breathtaking.
In other words, your goal is to find the highest branch that cannot be reached by any of the grasshoppers or report that it's impossible.
Input Specification:
The only line contains two integers *p* and *y* (2<=≤<=*p*<=≤<=*y*<=≤<=109).
Output Specification:
Output the number of the highest suitable branch. If there are none, print -1 instead.
Demo Input:
['3 6\n', '3 4\n']
Demo Output:
['5\n', '-1\n']
Note:
In the first sample case grasshopper from branch 2 reaches branches 2, 4 and 6 while branch 3 is initially settled by another grasshopper. Therefore the answer is 5.
It immediately follows that there are no valid branches in second sample case.
|
```python
def SieveOfEratosthenes(n):
c = []
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
# Print all prime numbers
for p in range(2, n):
if prime[p]:
c.append(p)
return c
n,m = map(int,input().split())
k = 0
c = SieveOfEratosthenes(n+1)
if m%2==0:
m-=1
for i in range(len(c)):
if m%c[i]==0:
m-=2
print(m)
```
| 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"
}
] | 1,576,953,533
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 108
| 0
|
s=input()
def k(x,y):
return s.index(x,s.index(y))
if 'h' in s:
s=s[s.index('h'):len(s)]
if 'e' in s and 'l' in s and 'o' in s:
print('YES' if k('h','h')<k('e','h')<k('l','e')<k('o','l') and
s.count('l',k('e','h'))>=2 else 'NO')
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()
def k(x,y):
return s.index(x,s.index(y))
if 'h' in s:
s=s[s.index('h'):len(s)]
if 'e' in s and 'l' in s and 'o' in s:
print('YES' if k('h','h')<k('e','h')<k('l','e')<k('o','l') and
s.count('l',k('e','h'))>=2 else 'NO')
else:
print('NO')
```
| 0
|
484
|
A
|
Bits
|
PROGRAMMING
| 1,700
|
[
"bitmasks",
"constructive algorithms"
] | null | null |
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
|
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
|
For each query print the answer in a separate line.
|
[
"3\n1 2\n2 4\n1 10\n"
] |
[
"1\n3\n7\n"
] |
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>
| 500
|
[
{
"input": "3\n1 2\n2 4\n1 10",
"output": "1\n3\n7"
},
{
"input": "55\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n2 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n3 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n4 4\n4 5\n4 6\n4 7\n4 8\n4 9\n4 10\n5 5\n5 6\n5 7\n5 8\n5 9\n5 10\n6 6\n6 7\n6 8\n6 9\n6 10\n7 7\n7 8\n7 9\n7 10\n8 8\n8 9\n8 10\n9 9\n9 10\n10 10",
"output": "1\n1\n3\n3\n3\n3\n7\n7\n7\n7\n2\n3\n3\n3\n3\n7\n7\n7\n7\n3\n3\n3\n3\n7\n7\n7\n7\n4\n5\n5\n7\n7\n7\n7\n5\n5\n7\n7\n7\n7\n6\n7\n7\n7\n7\n7\n7\n7\n7\n8\n9\n9\n9\n9\n10"
},
{
"input": "18\n1 10\n1 100\n1 1000\n1 10000\n1 100000\n1 1000000\n1 10000000\n1 100000000\n1 1000000000\n1 10000000000\n1 100000000000\n1 1000000000000\n1 10000000000000\n1 100000000000000\n1 1000000000000000\n1 10000000000000000\n1 100000000000000000\n1 1000000000000000000",
"output": "7\n63\n511\n8191\n65535\n524287\n8388607\n67108863\n536870911\n8589934591\n68719476735\n549755813887\n8796093022207\n70368744177663\n562949953421311\n9007199254740991\n72057594037927935\n576460752303423487"
},
{
"input": "3\n0 0\n1 3\n2 4",
"output": "0\n3\n3"
},
{
"input": "17\n0 0\n0 8\n1 8\n36 39\n3 4\n3 7\n2 17\n8 12\n9 12\n10 12\n10 15\n6 14\n8 15\n9 15\n15 15\n100000000000000000 1000000000000000000\n99999999999999999 1000000000000000000",
"output": "0\n7\n7\n39\n3\n7\n15\n11\n11\n11\n15\n7\n15\n15\n15\n576460752303423487\n576460752303423487"
}
] | 1,519,505,446
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 5,529,600
|
import sys
f = sys.stdin.readline
n = int(f().strip())
for i in range(n):
l, r = map(int, (f().strip().split()))
cur = 1
j = 1
while cur<=r:
cur += (1<<j)
j += 1
print(cur>>1)
|
Title: Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
Input Specification:
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
Output Specification:
For each query print the answer in a separate line.
Demo Input:
['3\n1 2\n2 4\n1 10\n']
Demo Output:
['1\n3\n7\n']
Note:
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>
|
```python
import sys
f = sys.stdin.readline
n = int(f().strip())
for i in range(n):
l, r = map(int, (f().strip().split()))
cur = 1
j = 1
while cur<=r:
cur += (1<<j)
j += 1
print(cur>>1)
```
| 0
|
|
391
|
A
|
Genetic Engineering
|
PROGRAMMING
| 0
|
[
"implementation",
"two pointers"
] | null | null |
You will receive 3 points for solving this problem.
Manao is designing the genetic code for a new type of algae to efficiently produce fuel. Specifically, Manao is focusing on a stretch of DNA that encodes one protein. The stretch of DNA is represented by a string containing only the characters 'A', 'T', 'G' and 'C'.
Manao has determined that if the stretch of DNA contains a maximal sequence of consecutive identical nucleotides that is of even length, then the protein will be nonfunctional. For example, consider a protein described by DNA string "GTTAAAG". It contains four maximal sequences of consecutive identical nucleotides: "G", "TT", "AAA", and "G". The protein is nonfunctional because sequence "TT" has even length.
Manao is trying to obtain a functional protein from the protein he currently has. Manao can insert additional nucleotides into the DNA stretch. Each additional nucleotide is a character from the set {'A', 'T', 'G', 'C'}. Manao wants to determine the minimum number of insertions necessary to make the DNA encode a functional protein.
|
The input consists of a single line, containing a string *s* of length *n* (1<=≤<=*n*<=≤<=100). Each character of *s* will be from the set {'A', 'T', 'G', 'C'}.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
|
The program should print on one line a single integer representing the minimum number of 'A', 'T', 'G', 'C' characters that are required to be inserted into the input string in order to make all runs of identical characters have odd length.
|
[
"GTTAAAG\n",
"AACCAACCAAAAC\n"
] |
[
"1\n",
"5\n"
] |
In the first example, it is sufficient to insert a single nucleotide of any type between the two 'T's in the sequence to restore the functionality of the protein.
| 3
|
[
{
"input": "GTTAAAG",
"output": "1"
},
{
"input": "AACCAACCAAAAC",
"output": "5"
},
{
"input": "GTGAATTTCC",
"output": "2"
},
{
"input": "CAGGGGGCCGCCCATGAAAAAAACCCGGCCCCTTGGGAAAACTTGGGTTA",
"output": "7"
},
{
"input": "CCCTTCACCCGGATCCAAATCCCTTAGAAATAATCCCCGACGGCGTTGTATCACCTCTGCACTTGTTAGTAAGGTCAGGCGTCCATTACGGAAGAACGTA",
"output": "19"
},
{
"input": "GCATTACATGGGGGGGTCCTACGAGCCCGGCATCCCGGAAACTAGCCGGTTAATTTGGTTTAAACCCTCCCACCCCGGATTGTAACCCCCCTCATTGGTT",
"output": "17"
},
{
"input": "TTCCCAGAGAAAAAAAGGGGCCCAAATGCCCTAAAAACCCCCTTTGCCCCCCAACCCCTTTTTAAAATAAAAAGGGGCCCATTCCCTTAAAAATTTTTTG",
"output": "10"
},
{
"input": "AGCCGCCCCCCCAAAAAAGGGGGAAAAAAAAAAAAAAAAAAAAACTTTTGGAAACCCCCCCCTTTTTTTTTTTTTTTTTTTTTTTTTGGGGAAGGGGGGG",
"output": "7"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "1"
},
{
"input": "AAAAAAAAAAAAAAAAAATTTTTTTTTTTTTTTTGGGGGGGGGGGGGGGGGGGGGGGTTTTTTTTTTTTTTGGGGGGGGGGGGGGGGGGGGAAAAATTTT",
"output": "5"
},
{
"input": "AACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTAACCGGTTCCGG",
"output": "50"
},
{
"input": "A",
"output": "0"
},
{
"input": "TTT",
"output": "0"
},
{
"input": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "0"
},
{
"input": "G",
"output": "0"
},
{
"input": "T",
"output": "0"
},
{
"input": "C",
"output": "0"
},
{
"input": "AA",
"output": "1"
},
{
"input": "GGG",
"output": "0"
},
{
"input": "AAG",
"output": "1"
}
] | 1,586,114,467
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 307,200
|
import math
def split(word):
return [char for char in word]
fin = open("segment.in" , 'r')
fout = open("segment.out" , "w" )
gene = split(fin.readline().strip('\n'))
n = len(gene)
max = []
bool = True
counter = 1
for i in range(1, n):
if gene[i] == gene[i-1]:
counter += 1
else:
max.append(counter)
counter =1
max.append(counter)
answer = 0
for i in range(0, len(max) ):
if max[i] %2 == 0:
answer += 1
else:
answer += 0
fout.write(str(answer))
fout.close()
|
Title: Genetic Engineering
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You will receive 3 points for solving this problem.
Manao is designing the genetic code for a new type of algae to efficiently produce fuel. Specifically, Manao is focusing on a stretch of DNA that encodes one protein. The stretch of DNA is represented by a string containing only the characters 'A', 'T', 'G' and 'C'.
Manao has determined that if the stretch of DNA contains a maximal sequence of consecutive identical nucleotides that is of even length, then the protein will be nonfunctional. For example, consider a protein described by DNA string "GTTAAAG". It contains four maximal sequences of consecutive identical nucleotides: "G", "TT", "AAA", and "G". The protein is nonfunctional because sequence "TT" has even length.
Manao is trying to obtain a functional protein from the protein he currently has. Manao can insert additional nucleotides into the DNA stretch. Each additional nucleotide is a character from the set {'A', 'T', 'G', 'C'}. Manao wants to determine the minimum number of insertions necessary to make the DNA encode a functional protein.
Input Specification:
The input consists of a single line, containing a string *s* of length *n* (1<=≤<=*n*<=≤<=100). Each character of *s* will be from the set {'A', 'T', 'G', 'C'}.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
The program should print on one line a single integer representing the minimum number of 'A', 'T', 'G', 'C' characters that are required to be inserted into the input string in order to make all runs of identical characters have odd length.
Demo Input:
['GTTAAAG\n', 'AACCAACCAAAAC\n']
Demo Output:
['1\n', '5\n']
Note:
In the first example, it is sufficient to insert a single nucleotide of any type between the two 'T's in the sequence to restore the functionality of the protein.
|
```python
import math
def split(word):
return [char for char in word]
fin = open("segment.in" , 'r')
fout = open("segment.out" , "w" )
gene = split(fin.readline().strip('\n'))
n = len(gene)
max = []
bool = True
counter = 1
for i in range(1, n):
if gene[i] == gene[i-1]:
counter += 1
else:
max.append(counter)
counter =1
max.append(counter)
answer = 0
for i in range(0, len(max) ):
if max[i] %2 == 0:
answer += 1
else:
answer += 0
fout.write(str(answer))
fout.close()
```
| -1
|
|
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"
}
] | 1,607,262,250
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 342
| 2,048,000
|
n=int(input())
a=[]
c=0
d=0
e=-1
o=-1
a=list(map(int,input().split()))
for i in range(n):
if(a[i]%2==0):
e=a[i]
c=c+1
if(c==2 and o!=-1):
print(o)
break
else:
o=e[i]
d=d+1
if(d==2 and e!=-1):
print(e)
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())
a=[]
c=0
d=0
e=-1
o=-1
a=list(map(int,input().split()))
for i in range(n):
if(a[i]%2==0):
e=a[i]
c=c+1
if(c==2 and o!=-1):
print(o)
break
else:
o=e[i]
d=d+1
if(d==2 and e!=-1):
print(e)
break
```
| -1
|
832
|
A
|
Sasha and Sticks
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
|
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
|
[
"1 1\n",
"10 4\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
| 500
|
[
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output": "NO"
},
{
"input": "253308697183523656 25332878317796706",
"output": "YES"
},
{
"input": "669038685745448997 501718093668307460",
"output": "YES"
},
{
"input": "116453141993601660 87060381463547965",
"output": "YES"
},
{
"input": "766959657 370931668",
"output": "NO"
},
{
"input": "255787422422806632 146884995820359999",
"output": "YES"
},
{
"input": "502007866464507926 71266379084204128",
"output": "YES"
},
{
"input": "257439908778973480 64157133126869976",
"output": "NO"
},
{
"input": "232709385 91708542",
"output": "NO"
},
{
"input": "252482458300407528 89907711721009125",
"output": "NO"
},
{
"input": "6 2",
"output": "YES"
},
{
"input": "6 3",
"output": "NO"
},
{
"input": "6 4",
"output": "YES"
},
{
"input": "6 5",
"output": "YES"
},
{
"input": "6 6",
"output": "YES"
},
{
"input": "258266151957056904 30153168463725364",
"output": "NO"
},
{
"input": "83504367885565783 52285355047292458",
"output": "YES"
},
{
"input": "545668929424440387 508692735816921376",
"output": "YES"
},
{
"input": "547321411485639939 36665750286082900",
"output": "NO"
},
{
"input": "548973893546839491 183137237979822911",
"output": "NO"
},
{
"input": "544068082 193116851",
"output": "NO"
},
{
"input": "871412474 749817171",
"output": "YES"
},
{
"input": "999999999 1247",
"output": "NO"
},
{
"input": "851941088 712987048",
"output": "YES"
},
{
"input": "559922900 418944886",
"output": "YES"
},
{
"input": "293908937 37520518",
"output": "YES"
},
{
"input": "650075786 130049650",
"output": "NO"
},
{
"input": "1000000000 1000000000",
"output": "YES"
},
{
"input": "548147654663723363 107422751713800746",
"output": "YES"
},
{
"input": "828159210 131819483",
"output": "NO"
},
{
"input": "6242634 4110365",
"output": "YES"
},
{
"input": "458601973 245084155",
"output": "YES"
},
{
"input": "349593257 18089089",
"output": "YES"
},
{
"input": "814768821 312514745",
"output": "NO"
},
{
"input": "697884949 626323363",
"output": "YES"
},
{
"input": "667011589 54866795",
"output": "NO"
},
{
"input": "1000000000000000000 2",
"output": "NO"
},
{
"input": "1000000000000000000 3",
"output": "YES"
},
{
"input": "1000000000000000000 4",
"output": "NO"
},
{
"input": "999999999999999 1",
"output": "YES"
},
{
"input": "17 4",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
},
{
"input": "1000000000000000 2",
"output": "NO"
},
{
"input": "12 4",
"output": "YES"
},
{
"input": "6 1",
"output": "NO"
},
{
"input": "2 1",
"output": "NO"
},
{
"input": "10000000005 1",
"output": "YES"
},
{
"input": "10000000000000009 2",
"output": "NO"
},
{
"input": "12457895452123 1",
"output": "YES"
},
{
"input": "999999999999999999 9",
"output": "YES"
},
{
"input": "1000000000000 3",
"output": "YES"
},
{
"input": "13099714659575475 6549849616514894",
"output": "NO"
},
{
"input": "100000000000000001 1",
"output": "YES"
},
{
"input": "825175814723458 324",
"output": "YES"
},
{
"input": "20 4",
"output": "YES"
},
{
"input": "100000176877 4",
"output": "YES"
},
{
"input": "100000 3",
"output": "YES"
},
{
"input": "946744073709551614 10",
"output": "YES"
}
] | 1,587,367,174
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 41
| 124
| 0
|
n, k = input().split(' ')
n, k = int(n), int(k)
if int(n/k)%2 == 0:
print('NO')
else:
print('YES')
|
Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
|
```python
n, k = input().split(' ')
n, k = int(n), int(k)
if int(n/k)%2 == 0:
print('NO')
else:
print('YES')
```
| 0
|
|
709
|
A
|
Juicer
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
|
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
|
Print one integer — the number of times Kolya will have to empty the waste section.
|
[
"2 7 10\n5 6\n",
"1 5 10\n7\n",
"3 10 10\n5 7 7\n",
"1 1 1\n1\n"
] |
[
"1\n",
"0\n",
"1\n",
"0\n"
] |
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
| 500
|
[
{
"input": "2 7 10\n5 6",
"output": "1"
},
{
"input": "1 5 10\n7",
"output": "0"
},
{
"input": "3 10 10\n5 7 7",
"output": "1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 951637 951638\n44069 951637",
"output": "1"
},
{
"input": "50 100 129\n55 130 91 19 116 3 63 52 104 76 75 27 151 99 149 147 39 148 84 9 132 49 40 112 124 141 144 93 36 32 146 74 48 38 150 55 94 32 107 69 77 81 33 57 62 98 78 127 154 126",
"output": "12"
},
{
"input": "100 1000 1083\n992 616 818 359 609 783 263 989 501 929 362 394 919 1081 870 830 1097 975 62 346 531 367 323 457 707 360 949 334 867 116 478 417 961 963 1029 114 867 1008 988 916 983 1077 959 942 572 961 579 318 721 337 488 717 111 70 416 685 987 130 353 107 61 191 827 849 106 815 211 953 111 398 889 860 801 71 375 320 395 1059 116 222 931 444 582 74 677 655 88 173 686 491 661 186 114 832 615 814 791 464 517 850",
"output": "36"
},
{
"input": "2 6 8\n2 1",
"output": "0"
},
{
"input": "5 15 16\n7 11 5 12 8",
"output": "2"
},
{
"input": "15 759966 759967\n890397 182209 878577 548548 759966 812923 759966 860479 200595 381358 299175 339368 759966 907668 69574",
"output": "4"
},
{
"input": "5 234613 716125\n642626 494941 234613 234613 234613",
"output": "0"
},
{
"input": "50 48547 567054\n529808 597004 242355 559114 78865 537318 631455 733020 655072 645093 309010 855034 306058 625046 524574 834944 27330 664392 443637 821584 338013 490702 289520 675471 885846 258814 134220 571301 84875 94132 200425 928833 375166 521232 317961 175315 947093 89971 322071 174033 48547 998535 954205 704114 943163 438900 48547 538422 48547 48547",
"output": "0"
},
{
"input": "5 10 20\n10 10 10 10 1",
"output": "1"
},
{
"input": "5 10 11\n10 10 10 10 1",
"output": "2"
},
{
"input": "3 10 10\n4 3 3",
"output": "0"
},
{
"input": "3 5 5\n5 5 5",
"output": "1"
},
{
"input": "3 4 14\n5 5 5",
"output": "0"
},
{
"input": "2 7 10\n1234 1234",
"output": "0"
},
{
"input": "1 5 6\n10",
"output": "0"
},
{
"input": "3 4 6\n1 2 3",
"output": "0"
},
{
"input": "5 10 12\n13 13 13 13 13",
"output": "0"
},
{
"input": "3 4 5\n5 7 9",
"output": "0"
},
{
"input": "3 10 100\n5 5 5",
"output": "0"
},
{
"input": "5 1 2\n2 2 2 2 2",
"output": "0"
},
{
"input": "5 5 5\n5 5 5 5 5",
"output": "2"
},
{
"input": "2 2 5\n5 5",
"output": "0"
},
{
"input": "3 1 4\n2 2 2",
"output": "0"
}
] | 1,632,036,273
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 124
| 14,438,400
|
a , b ,c =map(int, input().split())
l =list (map(int, input().split()) )
p=0
r=0
for i in range(len(l)):
if(l[i] <= b):
p+=l[i]
if p > c :
r+=1
p=0
print(r)
|
Title: Juicer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
Input Specification:
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Output Specification:
Print one integer — the number of times Kolya will have to empty the waste section.
Demo Input:
['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n']
Demo Output:
['1\n', '0\n', '1\n', '0\n']
Note:
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
|
```python
a , b ,c =map(int, input().split())
l =list (map(int, input().split()) )
p=0
r=0
for i in range(len(l)):
if(l[i] <= b):
p+=l[i]
if p > c :
r+=1
p=0
print(r)
```
| 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"
}
] | 1,629,298,939
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 6,963,200
|
def gm(a, i):
if i == len(a):
return [0, 0, 0]
f0, f1,f2 = gm(a, i+1)
if not a[i]:
return [f0 + 1, f1 + 1, f2 + 1]
if a[i] == 1:
return [f0, f1 + 1, f2]
if a[i] == 2:
return [f0, f1, f2 + 1]
return [f0, f1, f2]
lst = []
n = int(input())
for i in range(0, n):
ele = int(input())
lst.append(ele)
print(min(gm(lst, 0)));
|
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
def gm(a, i):
if i == len(a):
return [0, 0, 0]
f0, f1,f2 = gm(a, i+1)
if not a[i]:
return [f0 + 1, f1 + 1, f2 + 1]
if a[i] == 1:
return [f0, f1 + 1, f2]
if a[i] == 2:
return [f0, f1, f2 + 1]
return [f0, f1, f2]
lst = []
n = int(input())
for i in range(0, n):
ele = int(input())
lst.append(ele)
print(min(gm(lst, 0)));
```
| -1
|
|
784
|
E
|
Twisted Circuit
|
PROGRAMMING
| 2,000
|
[
"*special",
"brute force",
"implementation"
] | null | null |
The input consists of four lines, each line containing a single digit 0 or 1.
|
Output a single digit, 0 or 1.
|
[
"0\n1\n1\n0\n"
] |
[
"0\n"
] |
none
| 0
|
[
{
"input": "0\n1\n1\n0",
"output": "0"
},
{
"input": "0\n0\n0\n0",
"output": "0"
},
{
"input": "1\n0\n0\n0",
"output": "1"
},
{
"input": "0\n1\n0\n0",
"output": "0"
},
{
"input": "1\n1\n0\n0",
"output": "1"
},
{
"input": "0\n0\n1\n0",
"output": "0"
},
{
"input": "1\n0\n1\n0",
"output": "0"
},
{
"input": "1\n1\n1\n0",
"output": "1"
},
{
"input": "0\n0\n0\n1",
"output": "1"
},
{
"input": "1\n0\n0\n1",
"output": "1"
},
{
"input": "0\n1\n0\n1",
"output": "0"
},
{
"input": "1\n1\n0\n1",
"output": "0"
},
{
"input": "0\n0\n1\n1",
"output": "1"
},
{
"input": "1\n0\n1\n1",
"output": "1"
},
{
"input": "0\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "1"
}
] | 1,491,824,799
| 6,099
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 5,529,600
|
a = int(input())
b = int(input())
c = int(input())
d = int(input())
if (not a) & b & c & (not d):
print(0)
else:
print((((a | b) & (c ^ d)) | ((b & c) ^ (a | d))))
|
Title: Twisted Circuit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input consists of four lines, each line containing a single digit 0 or 1.
Output Specification:
Output a single digit, 0 or 1.
Demo Input:
['0\n1\n1\n0\n']
Demo Output:
['0\n']
Note:
none
|
```python
a = int(input())
b = int(input())
c = int(input())
d = int(input())
if (not a) & b & c & (not d):
print(0)
else:
print((((a | b) & (c ^ d)) | ((b & c) ^ (a | d))))
```
| 0
|
||
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,668,432,165
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
n = int(input())
l = []
x, count1, count2 = 0, 0, 0
for _ in range(n):
s = str(input())
if s not in l:
l.append(s)
if s == l[0]:
count1 += 1
elif s == l[1]:
count2 += 1
if count1 > count2:
print(l[0])
elif count2 > count1:
print(l[1])
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
n = int(input())
l = []
x, count1, count2 = 0, 0, 0
for _ in range(n):
s = str(input())
if s not in l:
l.append(s)
if s == l[0]:
count1 += 1
elif s == l[1]:
count2 += 1
if count1 > count2:
print(l[0])
elif count2 > count1:
print(l[1])
```
| 3.977
|
379
|
A
|
New Year Candles
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
|
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
|
Print a single integer — the number of hours Vasily can light up the room for.
|
[
"4 2\n",
"6 3\n"
] |
[
"7\n",
"8\n"
] |
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
| 500
|
[
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,679,662,039
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
x,y=map(int,input().split())
nx1=int(x/y)
if nx1>=y:
nx2=int(nx1/y)
print(x+nx1+nx2)
else:
print(x+nx1)
|
Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
|
```python
x,y=map(int,input().split())
nx1=int(x/y)
if nx1>=y:
nx2=int(nx1/y)
print(x+nx1+nx2)
else:
print(x+nx1)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
There is a rectangular grid of *n* rows of *m* initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the *i*-th operation, a non-empty subset of rows *R**i* and a non-empty subset of columns *C**i* are chosen. For each row *r* in *R**i* and each column *c* in *C**i*, the intersection of row *r* and column *c* is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (*i*,<=*j*) (*i*<=<<=*j*) exists such that or , where denotes intersection of sets, and denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and columns of the grid, respectively.
Each of the following *n* lines contains a string of *m* characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
|
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
|
[
"5 8\n.#.#..#.\n.....#..\n.#.#..#.\n#.#....#\n.....#..\n",
"5 5\n..#..\n..#..\n#####\n..#..\n..#..\n",
"5 9\n........#\n#........\n..##.#...\n.......#.\n....#.#.#\n"
] |
[
"Yes\n",
"No\n",
"No\n"
] |
For the first example, the desired setup can be produced by 3 operations, as is shown below.
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
| 0
|
[
{
"input": "5 8\n.#.#..#.\n.....#..\n.#.#..#.\n#.#....#\n.....#..",
"output": "Yes"
},
{
"input": "5 5\n..#..\n..#..\n#####\n..#..\n..#..",
"output": "No"
},
{
"input": "5 9\n........#\n#........\n..##.#...\n.......#.\n....#.#.#",
"output": "No"
},
{
"input": "1 1\n#",
"output": "Yes"
},
{
"input": "2 1\n.\n#",
"output": "Yes"
},
{
"input": "2 5\n.####\n#..##",
"output": "No"
},
{
"input": "5 2\n##\n##\n..\n##\n..",
"output": "Yes"
},
{
"input": "5 2\n#.\n##\n##\n#.\n..",
"output": "No"
},
{
"input": "4 10\n###..#..##\n...##..#..\n.##..#..#.\n.........#",
"output": "No"
},
{
"input": "4 10\n..#......#\n.....##...\n#.........\n.#.......#",
"output": "No"
},
{
"input": "10 15\n.......#.......\n.....#.........\n....#..........\n....#..........\n.....#.........\n.....#.........\n#.............#\n...#..#........\n...............\n.............#.",
"output": "Yes"
},
{
"input": "50 1\n.\n.\n#\n.\n#\n.\n#\n.\n.\n#\n#\n#\n.\n#\n#\n#\n#\n.\n.\n.\n.\n.\n.\n.\n.\n.\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": "Yes"
},
{
"input": "2 50\n...#.##.###...#.#..##....##..........#.#..#.#####.\n...#.##.###...#.#..##....##..........#.#..#.#####.",
"output": "Yes"
},
{
"input": "50 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..",
"output": "No"
},
{
"input": "1 1\n.",
"output": "Yes"
},
{
"input": "2 3\n#.#\n###",
"output": "No"
},
{
"input": "2 3\n#.#\n##.",
"output": "No"
},
{
"input": "4 4\n###.\n##.#\n#.##\n.###",
"output": "No"
},
{
"input": "3 3\n.##\n#.#\n##.",
"output": "No"
},
{
"input": "2 2\n##\n#.",
"output": "No"
}
] | 1,521,908,070
| 2,370
|
Python 3
|
OK
|
TESTS
| 50
| 93
| 7,065,600
|
import sys
n,m = [int(x) for x in input().split(' ')]
a=[]
for i in range(n):
b=[]
s = input().strip()
for j in range(m):
if s[j]=='#':
b.append(j)
if b not in a:
a.append(b)
c=[0]*m
ans=True
#print(a)
for i in a:
for j in i:
c[j]+=1
if c[j]>1:
ans=False
break
if ans:
print("Yes")
else:
print("No")
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a rectangular grid of *n* rows of *m* initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the *i*-th operation, a non-empty subset of rows *R**i* and a non-empty subset of columns *C**i* are chosen. For each row *r* in *R**i* and each column *c* in *C**i*, the intersection of row *r* and column *c* is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (*i*,<=*j*) (*i*<=<<=*j*) exists such that or , where denotes intersection of sets, and denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and columns of the grid, respectively.
Each of the following *n* lines contains a string of *m* characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output Specification:
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Demo Input:
['5 8\n.#.#..#.\n.....#..\n.#.#..#.\n#.#....#\n.....#..\n', '5 5\n..#..\n..#..\n#####\n..#..\n..#..\n', '5 9\n........#\n#........\n..##.#...\n.......#.\n....#.#.#\n']
Demo Output:
['Yes\n', 'No\n', 'No\n']
Note:
For the first example, the desired setup can be produced by 3 operations, as is shown below.
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
|
```python
import sys
n,m = [int(x) for x in input().split(' ')]
a=[]
for i in range(n):
b=[]
s = input().strip()
for j in range(m):
if s[j]=='#':
b.append(j)
if b not in a:
a.append(b)
c=[0]*m
ans=True
#print(a)
for i in a:
for j in i:
c[j]+=1
if c[j]>1:
ans=False
break
if ans:
print("Yes")
else:
print("No")
```
| 3
|
|
770
|
A
|
New Password
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
|
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
|
Print any password which satisfies all conditions given by Innokentiy.
|
[
"4 3\n",
"6 6\n",
"5 2\n"
] |
[
"java\n",
"python\n",
"phphp\n"
] |
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
| 500
|
[
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,589,975,730
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 47
| 140
| 0
|
n1=input().split(" ")
n=int(n1[0])
k=int(n1[1])
alp="abcdefghijklmnopqrstuvwxyz"
s=""
s1=""
s2=""
for i in range(k):
s+=alp[i]
s2=s*(n-k)
for i in range(n-k):
s1+=s2[i]
print(s+s1)
|
Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
|
```python
n1=input().split(" ")
n=int(n1[0])
k=int(n1[1])
alp="abcdefghijklmnopqrstuvwxyz"
s=""
s1=""
s2=""
for i in range(k):
s+=alp[i]
s2=s*(n-k)
for i in range(n-k):
s1+=s2[i]
print(s+s1)
```
| 3
|
|
357
|
A
|
Group of Students
|
PROGRAMMING
| 1,000
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
At the beginning of the school year Berland State University starts two city school programming groups, for beginners and for intermediate coders. The children were tested in order to sort them into groups. According to the results, each student got some score from 1 to *m* points. We know that *c*1 schoolchildren got 1 point, *c*2 children got 2 points, ..., *c**m* children got *m* points. Now you need to set the passing rate *k* (integer from 1 to *m*): all schoolchildren who got less than *k* points go to the beginner group and those who get at strictly least *k* points go to the intermediate group. We know that if the size of a group is more than *y*, then the university won't find a room for them. We also know that if a group has less than *x* schoolchildren, then it is too small and there's no point in having classes with it. So, you need to split all schoolchildren into two groups so that the size of each group was from *x* to *y*, inclusive.
Help the university pick the passing rate in a way that meets these requirements.
|
The first line contains integer *m* (2<=≤<=*m*<=≤<=100). The second line contains *m* integers *c*1, *c*2, ..., *c**m*, separated by single spaces (0<=≤<=*c**i*<=≤<=100). The third line contains two space-separated integers *x* and *y* (1<=≤<=*x*<=≤<=*y*<=≤<=10000). At least one *c**i* is greater than 0.
|
If it is impossible to pick a passing rate in a way that makes the size of each resulting groups at least *x* and at most *y*, print 0. Otherwise, print an integer from 1 to *m* — the passing rate you'd like to suggest. If there are multiple possible answers, print any of them.
|
[
"5\n3 4 3 2 1\n6 8\n",
"5\n0 3 3 4 2\n3 10\n",
"2\n2 5\n3 6\n"
] |
[
"3\n",
"4\n",
"0\n"
] |
In the first sample the beginner group has 7 students, the intermediate group has 6 of them.
In the second sample another correct answer is 3.
| 500
|
[
{
"input": "5\n3 4 3 2 1\n6 8",
"output": "3"
},
{
"input": "5\n0 3 3 4 2\n3 10",
"output": "4"
},
{
"input": "2\n2 5\n3 6",
"output": "0"
},
{
"input": "3\n0 1 0\n2 10",
"output": "0"
},
{
"input": "5\n2 2 2 2 2\n5 5",
"output": "0"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1\n1 10",
"output": "10"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1\n5 5",
"output": "6"
},
{
"input": "6\n0 0 1 1 0 0\n1 6",
"output": "4"
},
{
"input": "7\n3 2 3 3 2 1 1\n5 10",
"output": "4"
},
{
"input": "4\n1 0 0 100\n1 100",
"output": "4"
},
{
"input": "100\n46 6 71 27 94 59 99 82 5 41 18 89 86 2 31 35 52 18 1 14 54 11 28 83 42 15 13 77 22 70 87 65 79 35 44 71 79 9 95 57 5 59 42 62 66 26 33 66 67 45 39 17 97 28 36 100 52 23 68 29 83 6 61 85 71 2 85 98 85 65 95 53 35 96 29 28 82 80 52 60 61 46 46 80 11 3 35 6 12 10 64 7 7 7 65 93 58 85 20 12\n2422 2429",
"output": "52"
},
{
"input": "10\n3 6 1 5 3 7 0 1 0 8\n16 18",
"output": "6"
},
{
"input": "10\n3 3 0 4 0 5 2 10 7 0\n10 24",
"output": "8"
},
{
"input": "10\n9 4 7 7 1 3 7 3 8 5\n23 31",
"output": "7"
},
{
"input": "10\n9 6 9 5 5 4 3 3 9 10\n9 54",
"output": "10"
},
{
"input": "10\n2 4 8 5 2 2 2 5 6 2\n14 24",
"output": "7"
},
{
"input": "10\n10 58 86 17 61 12 75 93 37 30\n10 469",
"output": "10"
},
{
"input": "10\n56 36 0 28 68 54 34 48 28 92\n92 352",
"output": "10"
},
{
"input": "10\n2 81 94 40 74 62 39 70 87 86\n217 418",
"output": "8"
},
{
"input": "10\n48 93 9 96 70 14 100 93 44 79\n150 496",
"output": "8"
},
{
"input": "10\n94 85 4 9 30 45 90 76 0 65\n183 315",
"output": "7"
},
{
"input": "100\n1 0 7 9 0 4 3 10 9 4 9 7 4 4 7 7 6 1 3 3 8 1 4 3 5 8 0 0 6 2 3 5 0 1 5 8 6 3 2 4 9 5 8 6 0 2 5 1 9 5 9 0 6 0 4 5 9 7 1 4 7 5 4 5 6 8 2 3 3 2 8 2 9 5 9 2 4 7 7 8 10 1 3 0 8 0 9 1 1 7 7 8 9 3 5 9 9 8 0 8\n200 279",
"output": "63"
},
{
"input": "100\n5 4 9 7 8 10 7 8 10 0 10 9 7 1 0 7 8 5 5 8 7 7 7 2 5 8 0 7 5 7 1 7 6 5 4 10 6 1 4 4 8 7 0 3 2 10 8 6 1 3 2 6 8 1 9 3 9 5 2 0 3 6 7 5 10 0 2 8 3 10 1 3 8 8 0 2 10 3 4 4 0 7 4 0 9 7 10 2 7 10 9 9 6 6 8 1 10 1 2 0\n52 477",
"output": "91"
},
{
"input": "100\n5 1 6 6 5 4 5 8 0 2 10 1 10 0 6 6 0 1 5 7 10 5 8 4 4 5 10 4 10 3 0 10 10 1 2 6 2 6 3 9 4 4 5 5 7 7 7 4 3 2 1 4 5 0 2 1 8 5 4 5 10 7 0 3 5 4 10 4 10 7 10 1 8 3 9 8 6 9 5 7 3 4 7 8 4 0 3 4 4 1 6 6 2 0 1 5 3 3 9 10\n22 470",
"output": "98"
},
{
"input": "100\n73 75 17 93 35 7 71 88 11 58 78 33 7 38 14 83 30 25 75 23 60 10 100 7 90 51 82 0 78 54 61 32 20 90 54 45 100 62 40 99 43 86 87 64 10 41 29 51 38 22 5 63 10 64 90 20 100 33 95 72 40 82 92 30 38 3 71 85 99 66 4 26 33 41 85 14 26 61 21 96 29 40 25 14 48 4 30 44 6 41 71 71 4 66 13 50 30 78 64 36\n2069 2800",
"output": "57"
},
{
"input": "100\n86 19 100 37 9 49 97 9 70 51 14 31 47 53 76 65 10 40 4 92 2 79 22 70 85 58 73 96 89 91 41 88 70 31 53 33 22 51 10 56 90 39 70 38 86 15 94 63 82 19 7 65 22 83 83 71 53 6 95 89 53 41 95 11 32 0 7 84 39 11 37 73 20 46 18 28 72 23 17 78 37 49 43 62 60 45 30 69 38 41 71 43 47 80 64 40 77 99 36 63\n1348 3780",
"output": "74"
},
{
"input": "100\n65 64 26 48 16 90 68 32 95 11 27 29 87 46 61 35 24 99 34 17 79 79 11 66 14 75 31 47 43 61 100 32 75 5 76 11 46 74 81 81 1 25 87 45 16 57 24 76 58 37 42 0 46 23 75 66 75 11 50 5 10 11 43 26 38 42 88 15 70 57 2 74 7 72 52 8 72 19 37 38 66 24 51 42 40 98 19 25 37 7 4 92 47 72 26 76 66 88 53 79\n1687 2986",
"output": "65"
},
{
"input": "100\n78 43 41 93 12 76 62 54 85 5 42 61 93 37 22 6 50 80 63 53 66 47 0 60 43 93 90 8 97 64 80 22 23 47 30 100 80 75 84 95 35 69 36 20 58 99 78 88 1 100 10 69 57 77 68 61 62 85 4 45 24 4 24 74 65 73 91 47 100 35 25 53 27 66 62 55 38 83 56 20 62 10 71 90 41 5 75 83 36 75 15 97 79 52 88 32 55 42 59 39\n873 4637",
"output": "85"
},
{
"input": "100\n12 25 47 84 72 40 85 37 8 92 85 90 12 7 45 14 98 62 31 62 10 89 37 65 77 29 5 3 21 21 10 98 44 37 37 37 50 15 69 27 19 99 98 91 63 42 32 68 77 88 78 35 13 44 4 82 42 76 28 50 65 64 88 46 94 37 40 7 10 58 21 31 17 91 75 86 3 9 9 14 72 20 40 57 11 75 91 48 79 66 53 24 93 16 58 4 10 89 75 51\n666 4149",
"output": "88"
},
{
"input": "10\n8 0 2 2 5 1 3 5 2 2\n13 17",
"output": "6"
},
{
"input": "10\n10 4 4 6 2 2 0 5 3 7\n19 24",
"output": "5"
},
{
"input": "10\n96 19 75 32 94 16 81 2 93 58\n250 316",
"output": "6"
},
{
"input": "10\n75 65 68 43 89 57 7 58 51 85\n258 340",
"output": "6"
},
{
"input": "100\n59 51 86 38 90 10 36 3 97 35 32 20 25 96 49 39 66 44 64 50 97 68 50 79 3 33 72 96 32 74 67 9 17 77 67 15 1 100 99 81 18 1 15 36 7 34 30 78 10 97 7 19 87 47 62 61 40 29 1 34 6 77 76 21 66 11 65 96 82 54 49 65 56 90 29 75 48 77 48 53 91 21 98 26 80 44 57 97 11 78 98 45 40 88 27 27 47 5 26 6\n2479 2517",
"output": "53"
},
{
"input": "100\n5 11 92 53 49 42 15 86 31 10 30 49 21 66 14 13 80 25 21 25 86 20 86 83 36 81 21 23 0 30 64 85 15 33 74 96 83 51 84 4 35 65 10 7 11 11 41 80 51 51 74 52 43 83 88 38 77 20 14 40 37 25 27 93 27 77 48 56 93 65 79 33 91 14 9 95 13 36 24 2 66 31 56 28 49 58 74 17 88 36 46 73 54 18 63 22 2 41 8 50\n2229 2279",
"output": "52"
},
{
"input": "2\n0 1\n1 1",
"output": "0"
},
{
"input": "4\n1 0 0 4\n1 3",
"output": "0"
},
{
"input": "4\n1 0 0 0\n1 10",
"output": "0"
},
{
"input": "3\n2 1 4\n3 3",
"output": "0"
},
{
"input": "5\n2 0 2 0 0\n2 2",
"output": "3"
},
{
"input": "4\n1 2 3 4\n1 7",
"output": "4"
},
{
"input": "2\n7 1\n1 6",
"output": "0"
},
{
"input": "5\n1 3 7 8 9\n4 6",
"output": "0"
},
{
"input": "2\n5 2\n5 6",
"output": "0"
},
{
"input": "2\n1 0\n1 2",
"output": "0"
},
{
"input": "4\n2 3 9 10\n5 14",
"output": "4"
},
{
"input": "3\n1 2 1\n1 1",
"output": "0"
},
{
"input": "4\n2 3 9 50\n5 30",
"output": "0"
},
{
"input": "3\n7 1 1\n6 8",
"output": "0"
},
{
"input": "6\n1 1 2 3 4 5\n3 9",
"output": "5"
},
{
"input": "3\n4 5 5\n4 9",
"output": "3"
},
{
"input": "6\n1 2 3 4 5 6\n1 3",
"output": "0"
},
{
"input": "5\n3 4 3 2 10\n6 8",
"output": "0"
},
{
"input": "5\n1 1 3 4 6\n2 2",
"output": "0"
},
{
"input": "5\n5 3 5 8 10\n2 20",
"output": "4"
},
{
"input": "4\n0 0 5 0\n3 6",
"output": "0"
},
{
"input": "8\n1 1 1 1 2 2 2 1\n3 7",
"output": "6"
},
{
"input": "3\n1 100 100\n101 200",
"output": "0"
}
] | 1,465,738,869
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 62
| 5,120,000
|
from collections import defaultdict, deque, Counter, OrderedDict
#import sys, threading
def main():
m = int(input())
sum = [int(i) for i in input().split()]
for i in range(1,m): sum[i] += sum[i-1]
x,y = map(int,input().split())
ans = 0
for i in range(m):
s = sum[i]
r = sum[m-1] - s
if x <= s <= y and x <= r <= y:
ans = i+2
break
print(ans)
if __name__ == "__main__":
"""sys.setrecursionlimit(200000)
threading.stack_size(10240000)
thread = threading.Thread(target=main)
thread.start()"""
main()
|
Title: Group of Students
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At the beginning of the school year Berland State University starts two city school programming groups, for beginners and for intermediate coders. The children were tested in order to sort them into groups. According to the results, each student got some score from 1 to *m* points. We know that *c*1 schoolchildren got 1 point, *c*2 children got 2 points, ..., *c**m* children got *m* points. Now you need to set the passing rate *k* (integer from 1 to *m*): all schoolchildren who got less than *k* points go to the beginner group and those who get at strictly least *k* points go to the intermediate group. We know that if the size of a group is more than *y*, then the university won't find a room for them. We also know that if a group has less than *x* schoolchildren, then it is too small and there's no point in having classes with it. So, you need to split all schoolchildren into two groups so that the size of each group was from *x* to *y*, inclusive.
Help the university pick the passing rate in a way that meets these requirements.
Input Specification:
The first line contains integer *m* (2<=≤<=*m*<=≤<=100). The second line contains *m* integers *c*1, *c*2, ..., *c**m*, separated by single spaces (0<=≤<=*c**i*<=≤<=100). The third line contains two space-separated integers *x* and *y* (1<=≤<=*x*<=≤<=*y*<=≤<=10000). At least one *c**i* is greater than 0.
Output Specification:
If it is impossible to pick a passing rate in a way that makes the size of each resulting groups at least *x* and at most *y*, print 0. Otherwise, print an integer from 1 to *m* — the passing rate you'd like to suggest. If there are multiple possible answers, print any of them.
Demo Input:
['5\n3 4 3 2 1\n6 8\n', '5\n0 3 3 4 2\n3 10\n', '2\n2 5\n3 6\n']
Demo Output:
['3\n', '4\n', '0\n']
Note:
In the first sample the beginner group has 7 students, the intermediate group has 6 of them.
In the second sample another correct answer is 3.
|
```python
from collections import defaultdict, deque, Counter, OrderedDict
#import sys, threading
def main():
m = int(input())
sum = [int(i) for i in input().split()]
for i in range(1,m): sum[i] += sum[i-1]
x,y = map(int,input().split())
ans = 0
for i in range(m):
s = sum[i]
r = sum[m-1] - s
if x <= s <= y and x <= r <= y:
ans = i+2
break
print(ans)
if __name__ == "__main__":
"""sys.setrecursionlimit(200000)
threading.stack_size(10240000)
thread = threading.Thread(target=main)
thread.start()"""
main()
```
| 3
|
|
607
|
B
|
Zuma
|
PROGRAMMING
| 1,900
|
[
"dp"
] | null | null |
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of *n* gemstones, the *i*-th of which has color *c**i*. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of gemstones.
The second line contains *n* space-separated integers, the *i*-th of which is *c**i* (1<=≤<=*c**i*<=≤<=*n*) — the color of the *i*-th gemstone in a line.
|
Print a single integer — the minimum number of seconds needed to destroy the entire line.
|
[
"3\n1 2 1\n",
"3\n1 2 3\n",
"7\n1 4 4 2 3 2 1\n"
] |
[
"1\n",
"3\n",
"2\n"
] |
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
| 1,250
|
[
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "7\n1 4 4 2 3 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "8\n1 2 1 3 4 1 2 1",
"output": "2"
},
{
"input": "50\n5 7 5 10 7 9 1 9 10 2 8 3 5 7 3 10 2 3 7 6 2 7 1 2 2 2 4 7 3 5 8 3 4 4 1 6 7 10 5 4 8 1 9 5 5 3 4 4 8 3",
"output": "21"
},
{
"input": "50\n13 17 20 5 14 19 4 17 9 13 10 19 16 13 17 2 18 3 1 9 19 4 19 10 17 12 16 20 10 11 15 10 3 19 8 6 2 8 9 15 13 7 8 8 5 8 15 18 9 4",
"output": "28"
},
{
"input": "50\n22 19 14 22 20 11 16 28 23 15 3 23 6 16 30 15 15 10 24 28 19 19 22 30 28 1 27 12 12 14 17 30 17 26 21 26 27 1 11 23 9 30 18 19 17 29 11 20 29 24",
"output": "25"
},
{
"input": "50\n30 17 31 15 10 3 39 36 5 29 16 11 31 2 38 1 32 40 7 15 39 34 24 11 4 23 9 35 39 32 4 5 14 37 10 34 11 33 30 14 4 34 23 10 34 34 26 34 26 16",
"output": "36"
},
{
"input": "50\n19 25 46 17 1 41 50 19 7 1 43 8 19 38 42 32 38 22 8 5 5 31 29 35 43 12 23 48 40 29 30 9 46 3 39 24 36 36 32 22 21 29 43 33 36 49 48 22 47 37",
"output": "36"
},
{
"input": "6\n1 2 1 1 3 1",
"output": "2"
}
] | 1,553,961,038
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 2,000
| 3,379,200
|
d = {}
def solve(s, i, j):
if j < i:
return 0
if i == j:
return 1
tmp = d.get((i + 1, j))
if tmp is None:
tmp = solve(s, i + 1, j)
d[(i + 1, j)] = tmp
r = 1 + tmp
for k in range(i + 2, j + 1):
if s[i] == s[k]:
left = d.get((i + 1, k - 1))
if left is None:
left = solve(s, i + 1, k - 1)
d[(i + 1, k - 1)] = left
right = d.get((k + 1, j))
if right is None:
right = solve(s, k + 1, j)
d[(k + 1, j)] = right
r = min(r, left + right)
if s[i] == s[i + 1]:
tmp = d.get((i + 2, j))
if tmp is None:
tmp = solve(s, i + 2, j)
d[(i + 2, j)] = tmp
r = min(r, 1 + tmp)
return r
if __name__ == "__main__":
n = int(input())
c = list(map(int, input().split()))
print(solve(c, 0, len(c) - 1))
|
Title: Zuma
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of *n* gemstones, the *i*-th of which has color *c**i*. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of gemstones.
The second line contains *n* space-separated integers, the *i*-th of which is *c**i* (1<=≤<=*c**i*<=≤<=*n*) — the color of the *i*-th gemstone in a line.
Output Specification:
Print a single integer — the minimum number of seconds needed to destroy the entire line.
Demo Input:
['3\n1 2 1\n', '3\n1 2 3\n', '7\n1 4 4 2 3 2 1\n']
Demo Output:
['1\n', '3\n', '2\n']
Note:
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
|
```python
d = {}
def solve(s, i, j):
if j < i:
return 0
if i == j:
return 1
tmp = d.get((i + 1, j))
if tmp is None:
tmp = solve(s, i + 1, j)
d[(i + 1, j)] = tmp
r = 1 + tmp
for k in range(i + 2, j + 1):
if s[i] == s[k]:
left = d.get((i + 1, k - 1))
if left is None:
left = solve(s, i + 1, k - 1)
d[(i + 1, k - 1)] = left
right = d.get((k + 1, j))
if right is None:
right = solve(s, k + 1, j)
d[(k + 1, j)] = right
r = min(r, left + right)
if s[i] == s[i + 1]:
tmp = d.get((i + 2, j))
if tmp is None:
tmp = solve(s, i + 2, j)
d[(i + 2, j)] = tmp
r = min(r, 1 + tmp)
return r
if __name__ == "__main__":
n = int(input())
c = list(map(int, input().split()))
print(solve(c, 0, len(c) - 1))
```
| 0
|
|
327
|
B
|
Hungry Sequence
|
PROGRAMMING
| 1,200
|
[
"math"
] | null | null |
Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of *n* integers.
A sequence *a*1, *a*2, ..., *a**n*, consisting of *n* integers, is Hungry if and only if:
- Its elements are in increasing order. That is an inequality *a**i*<=<<=*a**j* holds for any two indices *i*,<=*j* (*i*<=<<=*j*). - For any two indices *i* and *j* (*i*<=<<=*j*), *a**j* must not be divisible by *a**i*.
Iahub is in trouble, so he asks you for help. Find a Hungry sequence with *n* elements.
|
The input contains a single integer: *n* (1<=≤<=*n*<=≤<=105).
|
Output a line that contains *n* space-separated integers *a*1 *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=107), representing a possible Hungry sequence. Note, that each *a**i* must not be greater than 10000000 (107) and less than 1.
If there are multiple solutions you can output any one.
|
[
"3\n",
"5\n"
] |
[
"2 9 15\n",
"11 14 20 27 31\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "2 9 15"
},
{
"input": "5",
"output": "11 14 20 27 31"
},
{
"input": "1",
"output": "3"
},
{
"input": "1000",
"output": "3000 3001 3002 3003 3004 3005 3006 3007 3008 3009 3010 3011 3012 3013 3014 3015 3016 3017 3018 3019 3020 3021 3022 3023 3024 3025 3026 3027 3028 3029 3030 3031 3032 3033 3034 3035 3036 3037 3038 3039 3040 3041 3042 3043 3044 3045 3046 3047 3048 3049 3050 3051 3052 3053 3054 3055 3056 3057 3058 3059 3060 3061 3062 3063 3064 3065 3066 3067 3068 3069 3070 3071 3072 3073 3074 3075 3076 3077 3078 3079 3080 3081 3082 3083 3084 3085 3086 3087 3088 3089 3090 3091 3092 3093 3094 3095 3096 3097 3098 3099 3100 3101 3..."
},
{
"input": "100000",
"output": "300000 300001 300002 300003 300004 300005 300006 300007 300008 300009 300010 300011 300012 300013 300014 300015 300016 300017 300018 300019 300020 300021 300022 300023 300024 300025 300026 300027 300028 300029 300030 300031 300032 300033 300034 300035 300036 300037 300038 300039 300040 300041 300042 300043 300044 300045 300046 300047 300048 300049 300050 300051 300052 300053 300054 300055 300056 300057 300058 300059 300060 300061 300062 300063 300064 300065 300066 300067 300068 300069 300070 300071 300072 ..."
},
{
"input": "46550",
"output": "139650 139651 139652 139653 139654 139655 139656 139657 139658 139659 139660 139661 139662 139663 139664 139665 139666 139667 139668 139669 139670 139671 139672 139673 139674 139675 139676 139677 139678 139679 139680 139681 139682 139683 139684 139685 139686 139687 139688 139689 139690 139691 139692 139693 139694 139695 139696 139697 139698 139699 139700 139701 139702 139703 139704 139705 139706 139707 139708 139709 139710 139711 139712 139713 139714 139715 139716 139717 139718 139719 139720 139721 139722 ..."
},
{
"input": "61324",
"output": "183972 183973 183974 183975 183976 183977 183978 183979 183980 183981 183982 183983 183984 183985 183986 183987 183988 183989 183990 183991 183992 183993 183994 183995 183996 183997 183998 183999 184000 184001 184002 184003 184004 184005 184006 184007 184008 184009 184010 184011 184012 184013 184014 184015 184016 184017 184018 184019 184020 184021 184022 184023 184024 184025 184026 184027 184028 184029 184030 184031 184032 184033 184034 184035 184036 184037 184038 184039 184040 184041 184042 184043 184044 ..."
},
{
"input": "13176",
"output": "39528 39529 39530 39531 39532 39533 39534 39535 39536 39537 39538 39539 39540 39541 39542 39543 39544 39545 39546 39547 39548 39549 39550 39551 39552 39553 39554 39555 39556 39557 39558 39559 39560 39561 39562 39563 39564 39565 39566 39567 39568 39569 39570 39571 39572 39573 39574 39575 39576 39577 39578 39579 39580 39581 39582 39583 39584 39585 39586 39587 39588 39589 39590 39591 39592 39593 39594 39595 39596 39597 39598 39599 39600 39601 39602 39603 39604 39605 39606 39607 39608 39609 39610 39611 39612 3..."
},
{
"input": "73274",
"output": "219822 219823 219824 219825 219826 219827 219828 219829 219830 219831 219832 219833 219834 219835 219836 219837 219838 219839 219840 219841 219842 219843 219844 219845 219846 219847 219848 219849 219850 219851 219852 219853 219854 219855 219856 219857 219858 219859 219860 219861 219862 219863 219864 219865 219866 219867 219868 219869 219870 219871 219872 219873 219874 219875 219876 219877 219878 219879 219880 219881 219882 219883 219884 219885 219886 219887 219888 219889 219890 219891 219892 219893 219894 ..."
},
{
"input": "86947",
"output": "260841 260842 260843 260844 260845 260846 260847 260848 260849 260850 260851 260852 260853 260854 260855 260856 260857 260858 260859 260860 260861 260862 260863 260864 260865 260866 260867 260868 260869 260870 260871 260872 260873 260874 260875 260876 260877 260878 260879 260880 260881 260882 260883 260884 260885 260886 260887 260888 260889 260890 260891 260892 260893 260894 260895 260896 260897 260898 260899 260900 260901 260902 260903 260904 260905 260906 260907 260908 260909 260910 260911 260912 260913 ..."
},
{
"input": "26342",
"output": "79026 79027 79028 79029 79030 79031 79032 79033 79034 79035 79036 79037 79038 79039 79040 79041 79042 79043 79044 79045 79046 79047 79048 79049 79050 79051 79052 79053 79054 79055 79056 79057 79058 79059 79060 79061 79062 79063 79064 79065 79066 79067 79068 79069 79070 79071 79072 79073 79074 79075 79076 79077 79078 79079 79080 79081 79082 79083 79084 79085 79086 79087 79088 79089 79090 79091 79092 79093 79094 79095 79096 79097 79098 79099 79100 79101 79102 79103 79104 79105 79106 79107 79108 79109 79110 7..."
},
{
"input": "22345",
"output": "67035 67036 67037 67038 67039 67040 67041 67042 67043 67044 67045 67046 67047 67048 67049 67050 67051 67052 67053 67054 67055 67056 67057 67058 67059 67060 67061 67062 67063 67064 67065 67066 67067 67068 67069 67070 67071 67072 67073 67074 67075 67076 67077 67078 67079 67080 67081 67082 67083 67084 67085 67086 67087 67088 67089 67090 67091 67092 67093 67094 67095 67096 67097 67098 67099 67100 67101 67102 67103 67104 67105 67106 67107 67108 67109 67110 67111 67112 67113 67114 67115 67116 67117 67118 67119 6..."
},
{
"input": "19639",
"output": "58917 58918 58919 58920 58921 58922 58923 58924 58925 58926 58927 58928 58929 58930 58931 58932 58933 58934 58935 58936 58937 58938 58939 58940 58941 58942 58943 58944 58945 58946 58947 58948 58949 58950 58951 58952 58953 58954 58955 58956 58957 58958 58959 58960 58961 58962 58963 58964 58965 58966 58967 58968 58969 58970 58971 58972 58973 58974 58975 58976 58977 58978 58979 58980 58981 58982 58983 58984 58985 58986 58987 58988 58989 58990 58991 58992 58993 58994 58995 58996 58997 58998 58999 59000 59001 5..."
},
{
"input": "12337",
"output": "37011 37012 37013 37014 37015 37016 37017 37018 37019 37020 37021 37022 37023 37024 37025 37026 37027 37028 37029 37030 37031 37032 37033 37034 37035 37036 37037 37038 37039 37040 37041 37042 37043 37044 37045 37046 37047 37048 37049 37050 37051 37052 37053 37054 37055 37056 37057 37058 37059 37060 37061 37062 37063 37064 37065 37066 37067 37068 37069 37070 37071 37072 37073 37074 37075 37076 37077 37078 37079 37080 37081 37082 37083 37084 37085 37086 37087 37088 37089 37090 37091 37092 37093 37094 37095 3..."
},
{
"input": "67989",
"output": "203967 203968 203969 203970 203971 203972 203973 203974 203975 203976 203977 203978 203979 203980 203981 203982 203983 203984 203985 203986 203987 203988 203989 203990 203991 203992 203993 203994 203995 203996 203997 203998 203999 204000 204001 204002 204003 204004 204005 204006 204007 204008 204009 204010 204011 204012 204013 204014 204015 204016 204017 204018 204019 204020 204021 204022 204023 204024 204025 204026 204027 204028 204029 204030 204031 204032 204033 204034 204035 204036 204037 204038 204039 ..."
},
{
"input": "57610",
"output": "172830 172831 172832 172833 172834 172835 172836 172837 172838 172839 172840 172841 172842 172843 172844 172845 172846 172847 172848 172849 172850 172851 172852 172853 172854 172855 172856 172857 172858 172859 172860 172861 172862 172863 172864 172865 172866 172867 172868 172869 172870 172871 172872 172873 172874 172875 172876 172877 172878 172879 172880 172881 172882 172883 172884 172885 172886 172887 172888 172889 172890 172891 172892 172893 172894 172895 172896 172897 172898 172899 172900 172901 172902 ..."
},
{
"input": "63287",
"output": "189861 189862 189863 189864 189865 189866 189867 189868 189869 189870 189871 189872 189873 189874 189875 189876 189877 189878 189879 189880 189881 189882 189883 189884 189885 189886 189887 189888 189889 189890 189891 189892 189893 189894 189895 189896 189897 189898 189899 189900 189901 189902 189903 189904 189905 189906 189907 189908 189909 189910 189911 189912 189913 189914 189915 189916 189917 189918 189919 189920 189921 189922 189923 189924 189925 189926 189927 189928 189929 189930 189931 189932 189933 ..."
},
{
"input": "952",
"output": "2856 2857 2858 2859 2860 2861 2862 2863 2864 2865 2866 2867 2868 2869 2870 2871 2872 2873 2874 2875 2876 2877 2878 2879 2880 2881 2882 2883 2884 2885 2886 2887 2888 2889 2890 2891 2892 2893 2894 2895 2896 2897 2898 2899 2900 2901 2902 2903 2904 2905 2906 2907 2908 2909 2910 2911 2912 2913 2914 2915 2916 2917 2918 2919 2920 2921 2922 2923 2924 2925 2926 2927 2928 2929 2930 2931 2932 2933 2934 2935 2936 2937 2938 2939 2940 2941 2942 2943 2944 2945 2946 2947 2948 2949 2950 2951 2952 2953 2954 2955 2956 2957 2..."
},
{
"input": "77840",
"output": "233520 233521 233522 233523 233524 233525 233526 233527 233528 233529 233530 233531 233532 233533 233534 233535 233536 233537 233538 233539 233540 233541 233542 233543 233544 233545 233546 233547 233548 233549 233550 233551 233552 233553 233554 233555 233556 233557 233558 233559 233560 233561 233562 233563 233564 233565 233566 233567 233568 233569 233570 233571 233572 233573 233574 233575 233576 233577 233578 233579 233580 233581 233582 233583 233584 233585 233586 233587 233588 233589 233590 233591 233592 ..."
},
{
"input": "42157",
"output": "126471 126472 126473 126474 126475 126476 126477 126478 126479 126480 126481 126482 126483 126484 126485 126486 126487 126488 126489 126490 126491 126492 126493 126494 126495 126496 126497 126498 126499 126500 126501 126502 126503 126504 126505 126506 126507 126508 126509 126510 126511 126512 126513 126514 126515 126516 126517 126518 126519 126520 126521 126522 126523 126524 126525 126526 126527 126528 126529 126530 126531 126532 126533 126534 126535 126536 126537 126538 126539 126540 126541 126542 126543 ..."
},
{
"input": "46375",
"output": "139125 139126 139127 139128 139129 139130 139131 139132 139133 139134 139135 139136 139137 139138 139139 139140 139141 139142 139143 139144 139145 139146 139147 139148 139149 139150 139151 139152 139153 139154 139155 139156 139157 139158 139159 139160 139161 139162 139163 139164 139165 139166 139167 139168 139169 139170 139171 139172 139173 139174 139175 139176 139177 139178 139179 139180 139181 139182 139183 139184 139185 139186 139187 139188 139189 139190 139191 139192 139193 139194 139195 139196 139197 ..."
},
{
"input": "55142",
"output": "165426 165427 165428 165429 165430 165431 165432 165433 165434 165435 165436 165437 165438 165439 165440 165441 165442 165443 165444 165445 165446 165447 165448 165449 165450 165451 165452 165453 165454 165455 165456 165457 165458 165459 165460 165461 165462 165463 165464 165465 165466 165467 165468 165469 165470 165471 165472 165473 165474 165475 165476 165477 165478 165479 165480 165481 165482 165483 165484 165485 165486 165487 165488 165489 165490 165491 165492 165493 165494 165495 165496 165497 165498 ..."
},
{
"input": "60299",
"output": "180897 180898 180899 180900 180901 180902 180903 180904 180905 180906 180907 180908 180909 180910 180911 180912 180913 180914 180915 180916 180917 180918 180919 180920 180921 180922 180923 180924 180925 180926 180927 180928 180929 180930 180931 180932 180933 180934 180935 180936 180937 180938 180939 180940 180941 180942 180943 180944 180945 180946 180947 180948 180949 180950 180951 180952 180953 180954 180955 180956 180957 180958 180959 180960 180961 180962 180963 180964 180965 180966 180967 180968 180969 ..."
},
{
"input": "28862",
"output": "86586 86587 86588 86589 86590 86591 86592 86593 86594 86595 86596 86597 86598 86599 86600 86601 86602 86603 86604 86605 86606 86607 86608 86609 86610 86611 86612 86613 86614 86615 86616 86617 86618 86619 86620 86621 86622 86623 86624 86625 86626 86627 86628 86629 86630 86631 86632 86633 86634 86635 86636 86637 86638 86639 86640 86641 86642 86643 86644 86645 86646 86647 86648 86649 86650 86651 86652 86653 86654 86655 86656 86657 86658 86659 86660 86661 86662 86663 86664 86665 86666 86667 86668 86669 86670 8..."
},
{
"input": "48251",
"output": "144753 144754 144755 144756 144757 144758 144759 144760 144761 144762 144763 144764 144765 144766 144767 144768 144769 144770 144771 144772 144773 144774 144775 144776 144777 144778 144779 144780 144781 144782 144783 144784 144785 144786 144787 144788 144789 144790 144791 144792 144793 144794 144795 144796 144797 144798 144799 144800 144801 144802 144803 144804 144805 144806 144807 144808 144809 144810 144811 144812 144813 144814 144815 144816 144817 144818 144819 144820 144821 144822 144823 144824 144825 ..."
},
{
"input": "2",
"output": "6 7"
},
{
"input": "4",
"output": "12 13 14 15"
}
] | 1,670,335,479
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 0
|
n=int(input())
k=[i for i in range(2,n+2)]
for i in k:
print(i,end=' ')
|
Title: Hungry Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of *n* integers.
A sequence *a*1, *a*2, ..., *a**n*, consisting of *n* integers, is Hungry if and only if:
- Its elements are in increasing order. That is an inequality *a**i*<=<<=*a**j* holds for any two indices *i*,<=*j* (*i*<=<<=*j*). - For any two indices *i* and *j* (*i*<=<<=*j*), *a**j* must not be divisible by *a**i*.
Iahub is in trouble, so he asks you for help. Find a Hungry sequence with *n* elements.
Input Specification:
The input contains a single integer: *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
Output a line that contains *n* space-separated integers *a*1 *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=107), representing a possible Hungry sequence. Note, that each *a**i* must not be greater than 10000000 (107) and less than 1.
If there are multiple solutions you can output any one.
Demo Input:
['3\n', '5\n']
Demo Output:
['2 9 15\n', '11 14 20 27 31\n']
Note:
none
|
```python
n=int(input())
k=[i for i in range(2,n+2)]
for i in k:
print(i,end=' ')
```
| 0
|
|
152
|
A
|
Marks
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook.
Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject.
Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at.
Your task is to find the number of successful students in the group.
|
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.
|
Print the single number — the number of successful students in the given group.
|
[
"3 3\n223\n232\n112\n",
"3 5\n91728\n11828\n11111\n"
] |
[
"2\n",
"3\n"
] |
In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject.
In the second sample test each student is the best at at least one subject.
| 500
|
[
{
"input": "3 3\n223\n232\n112",
"output": "2"
},
{
"input": "3 5\n91728\n11828\n11111",
"output": "3"
},
{
"input": "2 2\n48\n27",
"output": "1"
},
{
"input": "2 1\n4\n6",
"output": "1"
},
{
"input": "1 2\n57",
"output": "1"
},
{
"input": "1 1\n5",
"output": "1"
},
{
"input": "3 4\n2553\n6856\n5133",
"output": "2"
},
{
"input": "8 7\n6264676\n7854895\n3244128\n2465944\n8958761\n1378945\n3859353\n6615285",
"output": "6"
},
{
"input": "9 8\n61531121\n43529859\n18841327\n88683622\n98995641\n62741632\n57441743\n49396792\n63381994",
"output": "4"
},
{
"input": "10 20\n26855662887514171367\n48525577498621511535\n47683778377545341138\n47331616748732562762\n44876938191354974293\n24577238399664382695\n42724955594463126746\n79187344479926159359\n48349683283914388185\n82157191115518781898",
"output": "9"
},
{
"input": "20 15\n471187383859588\n652657222494199\n245695867594992\n726154672861295\n614617827782772\n862889444974692\n373977167653235\n645434268565473\n785993468314573\n722176861496755\n518276853323939\n723712762593348\n728935312568886\n373898548522463\n769777587165681\n247592995114377\n182375946483965\n497496542536127\n988239919677856\n859844339819143",
"output": "18"
},
{
"input": "13 9\n514562255\n322655246\n135162979\n733845982\n473117129\n513967187\n965649829\n799122777\n661249521\n298618978\n659352422\n747778378\n723261619",
"output": "11"
},
{
"input": "75 1\n2\n3\n8\n3\n2\n1\n3\n1\n5\n1\n5\n4\n8\n8\n4\n2\n5\n1\n7\n6\n3\n2\n2\n3\n5\n5\n2\n4\n7\n7\n9\n2\n9\n5\n1\n4\n9\n5\n2\n4\n6\n6\n3\n3\n9\n3\n3\n2\n3\n4\n2\n6\n9\n1\n1\n1\n1\n7\n2\n3\n2\n9\n7\n4\n9\n1\n7\n5\n6\n8\n3\n4\n3\n4\n6",
"output": "7"
},
{
"input": "92 3\n418\n665\n861\n766\n529\n416\n476\n676\n561\n995\n415\n185\n291\n176\n776\n631\n556\n488\n118\n188\n437\n496\n466\n131\n914\n118\n766\n365\n113\n897\n386\n639\n276\n946\n759\n169\n494\n837\n338\n351\n783\n311\n261\n862\n598\n132\n246\n982\n575\n364\n615\n347\n374\n368\n523\n132\n774\n161\n552\n492\n598\n474\n639\n681\n635\n342\n516\n483\n141\n197\n571\n336\n175\n596\n481\n327\n841\n133\n142\n146\n246\n396\n287\n582\n556\n996\n479\n814\n497\n363\n963\n162",
"output": "23"
},
{
"input": "100 1\n1\n6\n9\n1\n1\n5\n5\n4\n6\n9\n6\n1\n7\n8\n7\n3\n8\n8\n7\n6\n2\n1\n5\n8\n7\n3\n5\n4\n9\n7\n1\n2\n4\n1\n6\n5\n1\n3\n9\n4\n5\n8\n1\n2\n1\n9\n7\n3\n7\n1\n2\n2\n2\n2\n3\n9\n7\n2\n4\n7\n1\n6\n8\n1\n5\n6\n1\n1\n2\n9\n7\n4\n9\n1\n9\n4\n1\n3\n5\n2\n4\n4\n6\n5\n1\n4\n5\n8\n4\n7\n6\n5\n6\n9\n5\n8\n1\n5\n1\n6",
"output": "10"
},
{
"input": "100 2\n71\n87\n99\n47\n22\n87\n49\n73\n21\n12\n77\n43\n18\n41\n78\n62\n61\n16\n64\n89\n81\n54\n53\n92\n93\n94\n68\n93\n15\n68\n42\n93\n28\n19\n86\n16\n97\n17\n11\n43\n72\n76\n54\n95\n58\n53\n48\n45\n85\n85\n74\n21\n44\n51\n89\n75\n76\n17\n38\n62\n81\n22\n66\n59\n89\n85\n91\n87\n12\n97\n52\n87\n43\n89\n51\n58\n57\n98\n78\n68\n82\n41\n87\n29\n75\n72\n48\n14\n35\n71\n74\n91\n66\n67\n42\n98\n52\n54\n22\n41",
"output": "21"
},
{
"input": "5 20\n11111111111111111111\n11111111111111111111\n11111111111111111111\n99999999999999999999\n11111111111111111119",
"output": "2"
},
{
"input": "3 3\n111\n111\n999",
"output": "1"
},
{
"input": "3 3\n119\n181\n711",
"output": "3"
},
{
"input": "15 5\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111",
"output": "15"
},
{
"input": "2 20\n22222222222222222222\n11111111111111111111",
"output": "1"
},
{
"input": "3 3\n233\n222\n111",
"output": "2"
},
{
"input": "4 15\n222222222222222\n111111111111119\n111111111111119\n111111111111111",
"output": "3"
},
{
"input": "4 1\n1\n9\n9\n9",
"output": "3"
},
{
"input": "3 3\n123\n321\n132",
"output": "3"
},
{
"input": "3 3\n113\n332\n322",
"output": "3"
},
{
"input": "2 100\n2222222222222222222222222222222222222222222222222222222222222222222222221222222222222222222222222222\n1111111111111111111111111111111111111111111111111111111111111111111111119111111111111111111111111111",
"output": "2"
},
{
"input": "3 3\n321\n231\n123",
"output": "3"
},
{
"input": "2 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222",
"output": "1"
},
{
"input": "3 3\n221\n111\n111",
"output": "3"
}
] | 1,596,208,789
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
r=raw_input;z=map(r," "*int(r().split()[0]))
print sum(any(i==max(j)for i,j in zip(s,zip(*z)))for s in z)
|
Title: Marks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook.
Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject.
Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at.
Your task is to find the number of successful students in the group.
Input Specification:
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.
Output Specification:
Print the single number — the number of successful students in the given group.
Demo Input:
['3 3\n223\n232\n112\n', '3 5\n91728\n11828\n11111\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject.
In the second sample test each student is the best at at least one subject.
|
```python
r=raw_input;z=map(r," "*int(r().split()[0]))
print sum(any(i==max(j)for i,j in zip(s,zip(*z)))for s in z)
```
| -1
|
|
40
|
A
|
Find Color
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"geometry",
"implementation",
"math"
] |
A. Find Color
|
2
|
256
|
Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture:
The picture shows only the central part of the clock. This coloring naturally extends to infinity.
The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball.
All the points located on the border of one of the areas have to be considered painted black.
|
The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000.
|
Find the required color.
All the points between which and the origin of coordinates the distance is integral-value are painted black.
|
[
"-2 1\n",
"2 1\n",
"4 3\n"
] |
[
"white\n",
"black\n",
"black\n"
] |
none
| 500
|
[
{
"input": "-2 1",
"output": "white"
},
{
"input": "2 1",
"output": "black"
},
{
"input": "4 3",
"output": "black"
},
{
"input": "3 3",
"output": "black"
},
{
"input": "4 4",
"output": "white"
},
{
"input": "-4 4",
"output": "black"
},
{
"input": "4 -4",
"output": "black"
},
{
"input": "-4 -4",
"output": "white"
},
{
"input": "0 0",
"output": "black"
},
{
"input": "0 1",
"output": "black"
},
{
"input": "0 2",
"output": "black"
},
{
"input": "0 1000",
"output": "black"
},
{
"input": "1000 0",
"output": "black"
},
{
"input": "-1000 0",
"output": "black"
},
{
"input": "0 -1000",
"output": "black"
},
{
"input": "1000 -1000",
"output": "white"
},
{
"input": "12 5",
"output": "black"
},
{
"input": "12 -5",
"output": "black"
},
{
"input": "-12 -35",
"output": "black"
},
{
"input": "20 -21",
"output": "black"
},
{
"input": "-677 492",
"output": "white"
},
{
"input": "-673 -270",
"output": "white"
},
{
"input": "-668 970",
"output": "black"
},
{
"input": "-220 208",
"output": "white"
},
{
"input": "-215 -996",
"output": "black"
},
{
"input": "-211 243",
"output": "black"
},
{
"input": "-206 -518",
"output": "white"
},
{
"input": "-201 278",
"output": "black"
},
{
"input": "-196 -484",
"output": "black"
},
{
"input": "902 479",
"output": "white"
},
{
"input": "-441 572",
"output": "white"
},
{
"input": "217 221",
"output": "white"
},
{
"input": "875 -129",
"output": "white"
},
{
"input": "-469 -36",
"output": "black"
},
{
"input": "189 -387",
"output": "white"
},
{
"input": "847 -294",
"output": "white"
},
{
"input": "-496 -644",
"output": "black"
},
{
"input": "-281 -552",
"output": "white"
},
{
"input": "377 -902",
"output": "black"
},
{
"input": "165 -738",
"output": "white"
},
{
"input": "61 -175",
"output": "black"
},
{
"input": "-42 389",
"output": "black"
},
{
"input": "-589 952",
"output": "black"
},
{
"input": "-693 -929",
"output": "white"
},
{
"input": "-796 -365",
"output": "white"
},
{
"input": "658 198",
"output": "white"
},
{
"input": "555 319",
"output": "black"
},
{
"input": "8 882",
"output": "black"
},
{
"input": "-96 -556",
"output": "black"
},
{
"input": "-129 489",
"output": "black"
},
{
"input": "207 -224",
"output": "black"
},
{
"input": "64 0",
"output": "black"
},
{
"input": "17 144",
"output": "black"
},
{
"input": "60 -448",
"output": "black"
},
{
"input": "-399 -40",
"output": "black"
},
{
"input": "128 -504",
"output": "black"
},
{
"input": "0 72",
"output": "black"
},
{
"input": "168 -26",
"output": "black"
},
{
"input": "72 -154",
"output": "black"
},
{
"input": "117 -44",
"output": "black"
},
{
"input": "-72 -646",
"output": "black"
},
{
"input": "253 -204",
"output": "black"
},
{
"input": "-40 198",
"output": "black"
},
{
"input": "-216 -90",
"output": "black"
},
{
"input": "15 -8",
"output": "black"
},
{
"input": "-180 -432",
"output": "black"
},
{
"input": "280 342",
"output": "black"
},
{
"input": "132 224",
"output": "black"
},
{
"input": "-192 -256",
"output": "black"
},
{
"input": "351 -280",
"output": "black"
}
] | 1,660,464,849
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
x,y = map(int,input().split())
if x>0 and y>0:
#first
a = max(x,y)
if a%2:
print('white')
else:
print('black')
elif x>0 and y<0:
#fourth
a = max(x, abs(y))
if a%2:
print('black')
else:
print('white')
elif x<0 and y>0:
a = max(abs(x), y)
if a%2:
print('black')
else:
print('white')
#second
else:
#third
a = max(abs(x),abs(y))
if a%2:
print('white')
else:
print('black')
|
Title: Find Color
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture:
The picture shows only the central part of the clock. This coloring naturally extends to infinity.
The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball.
All the points located on the border of one of the areas have to be considered painted black.
Input Specification:
The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000.
Output Specification:
Find the required color.
All the points between which and the origin of coordinates the distance is integral-value are painted black.
Demo Input:
['-2 1\n', '2 1\n', '4 3\n']
Demo Output:
['white\n', 'black\n', 'black\n']
Note:
none
|
```python
x,y = map(int,input().split())
if x>0 and y>0:
#first
a = max(x,y)
if a%2:
print('white')
else:
print('black')
elif x>0 and y<0:
#fourth
a = max(x, abs(y))
if a%2:
print('black')
else:
print('white')
elif x<0 and y>0:
a = max(abs(x), y)
if a%2:
print('black')
else:
print('white')
#second
else:
#third
a = max(abs(x),abs(y))
if a%2:
print('white')
else:
print('black')
```
| 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"
}
] | 1,562,369,859
| 279
|
PyPy 3
|
OK
|
TESTS
| 16
| 140
| 0
|
n = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
count = 0
while(n[1] > 0):
n[1] -= (86400 - a[count])
count+=1
print(count)
|
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
n = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
count = 0
while(n[1] > 0):
n[1] -= (86400 - a[count])
count+=1
print(count)
```
| 3
|
|
339
|
A
|
Helpful Maths
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings",
"strings"
] | null | null |
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
|
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
|
Print the new sum that Xenia can count.
|
[
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] |
[
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "1+2+1+2+2+2+2+1+3+3",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3",
"output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3",
"output": "3"
},
{
"input": "1+1",
"output": "1+1"
},
{
"input": "1+2",
"output": "1+2"
},
{
"input": "1+3",
"output": "1+3"
},
{
"input": "2+1",
"output": "1+2"
},
{
"input": "2+2",
"output": "2+2"
},
{
"input": "2+3",
"output": "2+3"
},
{
"input": "3+1",
"output": "1+3"
},
{
"input": "3+2",
"output": "2+3"
},
{
"input": "3+3",
"output": "3+3"
}
] | 1,694,965,763
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 21
| 122
| 0
|
# Helpful maths
s = input().split("+")
s.sort()
result = ""
for i in range(len(s)):
if (i == (len(s) - 1)):
result += s[i]
else :
result += f"{s[i]}+"
print(result)
|
Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none
|
```python
# Helpful maths
s = input().split("+")
s.sort()
result = ""
for i in range(len(s)):
if (i == (len(s) - 1)):
result += s[i]
else :
result += f"{s[i]}+"
print(result)
```
| 3
|
|
513
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"math"
] | null | null |
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
|
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
|
Output "First" if the first player wins and "Second" otherwise.
|
[
"2 2 1 2\n",
"2 1 1 1\n"
] |
[
"Second\n",
"First\n"
] |
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely.
| 3
|
[
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"output": "Second"
},
{
"input": "1 1 1 1",
"output": "Second"
},
{
"input": "50 50 50 50",
"output": "Second"
},
{
"input": "50 49 1 2",
"output": "First"
},
{
"input": "50 48 3 1",
"output": "First"
},
{
"input": "48 50 12 11",
"output": "Second"
},
{
"input": "49 50 11 12",
"output": "Second"
},
{
"input": "49 49 4 1",
"output": "Second"
},
{
"input": "49 49 3 3",
"output": "Second"
},
{
"input": "1 50 1 50",
"output": "Second"
},
{
"input": "1 50 50 50",
"output": "Second"
},
{
"input": "50 1 1 1",
"output": "First"
},
{
"input": "50 1 1 50",
"output": "First"
},
{
"input": "32 31 10 9",
"output": "First"
},
{
"input": "32 4 17 3",
"output": "First"
}
] | 1,589,054,856
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 140
| 20,172,800
|
n1, n2, k1, k2 = map(int, input().split())
c = 0
c2 = 0
while k1 != 0 and n1 != 0: #n1 = 5, n2 = 7, k1 = 3, k2 = 5
n1 -= 1
c += 1
k1 -= 1
while k2 != 0 and n2 != 0:
n2 -= 1
c2 += 1
k2 -= 1
if c < c2:
print('Second')
else:
print('First')
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely.
|
```python
n1, n2, k1, k2 = map(int, input().split())
c = 0
c2 = 0
while k1 != 0 and n1 != 0: #n1 = 5, n2 = 7, k1 = 3, k2 = 5
n1 -= 1
c += 1
k1 -= 1
while k2 != 0 and n2 != 0:
n2 -= 1
c2 += 1
k2 -= 1
if c < c2:
print('Second')
else:
print('First')
```
| 0
|
|
711
|
A
|
Bus to Udayland
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
|
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
|
[
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] |
[
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] |
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX
| 500
|
[
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
},
{
"input": "1\nXO|OX",
"output": "NO"
},
{
"input": "1\nOO|OO",
"output": "YES\n++|OO"
},
{
"input": "4\nXO|XX\nXX|XO\nOX|XX\nXO|XO",
"output": "NO"
},
{
"input": "9\nOX|XO\nOX|XO\nXO|OX\nOX|OX\nXO|OX\nXX|OO\nOX|OX\nOX|XO\nOX|OX",
"output": "YES\nOX|XO\nOX|XO\nXO|OX\nOX|OX\nXO|OX\nXX|++\nOX|OX\nOX|XO\nOX|OX"
},
{
"input": "61\nOX|XX\nOX|XX\nOX|XX\nXO|XO\nXX|XO\nXX|XX\nXX|XX\nOX|XX\nXO|XO\nOX|XO\nXO|OX\nXX|XX\nXX|XX\nOX|OX\nXX|OX\nOX|XO\nOX|XO\nXO|OX\nXO|XX\nOX|XX\nOX|XX\nXO|OX\nXO|XX\nXO|XX\nOX|XX\nXX|XX\nXX|XO\nXO|XX\nXX|XX\nXO|OX\nXX|XO\nXO|XX\nXO|XO\nXO|OX\nXX|OX\nXO|OX\nOX|XX\nXX|OX\nXX|XX\nOX|XO\nOX|XX\nXO|OX\nOX|XX\nOX|XX\nXO|XO\nXO|XX\nOX|XX\nXO|XO\nOX|XX\nXX|XX\nOX|XO\nXO|XO\nXO|XO\nOX|OX\nXX|OX\nXX|OX\nOX|XO\nOX|XX\nOX|OX\nXO|XX\nOX|XX",
"output": "NO"
},
{
"input": "1\nOO|XX",
"output": "YES\n++|XX"
},
{
"input": "10\nOO|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|XX",
"output": "YES\n++|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|XX"
},
{
"input": "10\nXX|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|OO",
"output": "YES\nXX|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|++"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nOO|OX\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\n++|OX\nOX|XO"
},
{
"input": "6\nOO|XX\nXO|XX\nOX|OO\nXX|OX\nOO|XX\nOX|XX",
"output": "YES\n++|XX\nXO|XX\nOX|OO\nXX|OX\nOO|XX\nOX|XX"
}
] | 1,629,810,000
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 77
| 20,172,800
|
n=int(input())
a=[]
kt=True
for i in range(n):
a.append('')
s=input()
if(kt==True):
if(s[0]=='O' and s[1]=='O'):
a[i]+="++"+s[2]+s[3]+s[4]
kt=False
elif(s[3]=='O' and s[4]=='O'):
a[i]+=s[0]+s[1]+s[2]+"++"
kt=False
else: a.append(s)
else:
a.append(s)
if(kt==False):
print("YES")
for i in a: print(i)
else: print("NO")
|
Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX
|
```python
n=int(input())
a=[]
kt=True
for i in range(n):
a.append('')
s=input()
if(kt==True):
if(s[0]=='O' and s[1]=='O'):
a[i]+="++"+s[2]+s[3]+s[4]
kt=False
elif(s[3]=='O' and s[4]=='O'):
a[i]+=s[0]+s[1]+s[2]+"++"
kt=False
else: a.append(s)
else:
a.append(s)
if(kt==False):
print("YES")
for i in a: print(i)
else: print("NO")
```
| 0
|
|
275
|
A
|
Lights Out
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
|
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
|
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
|
[
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] |
[
"001\n010\n100\n",
"010\n011\n100\n"
] |
none
| 500
|
[
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{
"input": "24 54 37\n60 63 6\n1 84 26",
"output": "110\n101\n011"
},
{
"input": "23 10 40\n15 6 40\n92 80 77",
"output": "101\n100\n000"
},
{
"input": "62 74 80\n95 74 93\n2 47 95",
"output": "010\n001\n110"
},
{
"input": "80 83 48\n26 0 66\n47 76 37",
"output": "000\n000\n010"
},
{
"input": "32 15 65\n7 54 36\n5 51 3",
"output": "111\n101\n001"
},
{
"input": "22 97 12\n71 8 24\n100 21 64",
"output": "100\n001\n100"
},
{
"input": "46 37 13\n87 0 50\n90 8 55",
"output": "111\n011\n000"
},
{
"input": "57 43 58\n20 82 83\n66 16 52",
"output": "111\n010\n110"
},
{
"input": "45 56 93\n47 51 59\n18 51 63",
"output": "101\n011\n100"
},
{
"input": "47 66 67\n14 1 37\n27 81 69",
"output": "001\n001\n110"
},
{
"input": "26 69 69\n85 18 23\n14 22 74",
"output": "110\n001\n010"
},
{
"input": "10 70 65\n94 27 25\n74 66 30",
"output": "111\n010\n100"
},
{
"input": "97 1 74\n15 99 1\n88 68 86",
"output": "001\n011\n000"
},
{
"input": "36 48 42\n45 41 66\n26 64 1",
"output": "001\n111\n010"
},
{
"input": "52 81 97\n29 77 71\n66 11 2",
"output": "100\n100\n111"
},
{
"input": "18 66 33\n19 49 49\n48 46 26",
"output": "011\n100\n000"
},
{
"input": "68 79 52\n51 39 100\n29 14 26",
"output": "110\n000\n111"
},
{
"input": "91 69 77\n91 26 64\n91 88 57",
"output": "001\n011\n110"
},
{
"input": "16 69 64\n48 21 80\n81 51 51",
"output": "010\n101\n111"
},
{
"input": "96 14 2\n100 18 12\n65 34 89",
"output": "111\n010\n010"
},
{
"input": "93 95 90\n8 59 42\n53 13 19",
"output": "100\n001\n111"
},
{
"input": "71 84 18\n100 19 67\n9 76 15",
"output": "010\n010\n001"
},
{
"input": "38 93 85\n21 88 64\n4 96 25",
"output": "111\n011\n000"
},
{
"input": "75 20 20\n60 5 78\n77 4 69",
"output": "011\n001\n000"
},
{
"input": "65 70 96\n19 6 83\n33 37 82",
"output": "100\n000\n011"
},
{
"input": "11 13 60\n17 13 46\n42 21 39",
"output": "000\n011\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 0",
"output": "111\n111\n111"
},
{
"input": "0 0 0\n0 1 0\n0 0 0",
"output": "101\n000\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 1",
"output": "111\n110\n100"
}
] | 1,631,183,770
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 108
| 20,172,800
|
on=[[1,1,1],[1,1,1],[1,1,1]]
l=[]
for i in range(3):
a,b,c=map(int,input().split())
t=[a,b,c]
l.append(t)
for i in range(3):
for j in range(3):
if l[i][j]%2:
on[i][j]=(on[i][j]+1)%2 #if the element is 1, it is toggled to (1+1)%2=0. else if it is 0 it is toggled to (0+1)%2=1
if j!=2:
on[i][j+1]=(on[i][j+1]+1)%2
if j!=0:
on[i][j-1]=(on[i][j-1]+1)%2
if i!=0:
on[i-1][j]=(on[i-1][j]+1)%2
if i!=2:
on[i+1][j]=(on[i+1][j]+1)%2
for i in on:
for j in i:
print(j, end='')
print()
|
Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none
|
```python
on=[[1,1,1],[1,1,1],[1,1,1]]
l=[]
for i in range(3):
a,b,c=map(int,input().split())
t=[a,b,c]
l.append(t)
for i in range(3):
for j in range(3):
if l[i][j]%2:
on[i][j]=(on[i][j]+1)%2 #if the element is 1, it is toggled to (1+1)%2=0. else if it is 0 it is toggled to (0+1)%2=1
if j!=2:
on[i][j+1]=(on[i][j+1]+1)%2
if j!=0:
on[i][j-1]=(on[i][j-1]+1)%2
if i!=0:
on[i-1][j]=(on[i-1][j]+1)%2
if i!=2:
on[i+1][j]=(on[i+1][j]+1)%2
for i in on:
for j in i:
print(j, end='')
print()
```
| 3
|
|
165
|
A
|
Supercentral Point
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
|
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
|
Print the only number — the number of supercentral points of the given set.
|
[
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
| 500
|
[
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,666,786,288
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 26
| 310
| 3,379,200
|
n = int(input())
plst = []
for _ in range(n):
p = tuple(map(int,input().split()))
plst.append(p)
count =0
for i in plst:
f1=f2=f3=f4 = False
for j in plst:
if (i[0]==j[0]) & (i[1]>j[1]):
f1=True
if (i[0]==j[0]) & (i[1]<j[1]):
f2=True
if (i[0]>j[0]) & (i[1]==j[1]):
f3=True
if (i[0]<j[0]) & (i[1]==j[1]):
f4=True
if f1&f2&f3&f4:
count +=1
break
print(count)
|
Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
|
```python
n = int(input())
plst = []
for _ in range(n):
p = tuple(map(int,input().split()))
plst.append(p)
count =0
for i in plst:
f1=f2=f3=f4 = False
for j in plst:
if (i[0]==j[0]) & (i[1]>j[1]):
f1=True
if (i[0]==j[0]) & (i[1]<j[1]):
f2=True
if (i[0]>j[0]) & (i[1]==j[1]):
f3=True
if (i[0]<j[0]) & (i[1]==j[1]):
f4=True
if f1&f2&f3&f4:
count +=1
break
print(count)
```
| 3
|
|
143
|
A
|
Help Vasilisa the Wise 2
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
|
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
|
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
|
[
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] |
[
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] |
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
| 500
|
[
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
},
{
"input": "12 17\n10 19\n13 16",
"output": "-1"
},
{
"input": "11 11\n17 5\n12 10",
"output": "9 2\n8 3"
},
{
"input": "12 11\n11 12\n16 7",
"output": "-1"
},
{
"input": "5 9\n7 7\n8 6",
"output": "3 2\n4 5"
},
{
"input": "10 7\n4 13\n11 6",
"output": "-1"
},
{
"input": "18 10\n16 12\n12 16",
"output": "-1"
},
{
"input": "13 6\n10 9\n6 13",
"output": "-1"
},
{
"input": "14 16\n16 14\n18 12",
"output": "-1"
},
{
"input": "16 10\n16 10\n12 14",
"output": "-1"
},
{
"input": "11 9\n12 8\n11 9",
"output": "-1"
},
{
"input": "5 14\n10 9\n10 9",
"output": "-1"
},
{
"input": "2 4\n1 5\n3 3",
"output": "-1"
},
{
"input": "17 16\n14 19\n18 15",
"output": "-1"
},
{
"input": "12 12\n14 10\n16 8",
"output": "9 3\n5 7"
},
{
"input": "15 11\n16 10\n9 17",
"output": "7 8\n9 2"
},
{
"input": "8 10\n9 9\n13 5",
"output": "6 2\n3 7"
},
{
"input": "13 7\n10 10\n5 15",
"output": "4 9\n6 1"
},
{
"input": "14 11\n9 16\n16 9",
"output": "-1"
},
{
"input": "12 8\n14 6\n8 12",
"output": "-1"
},
{
"input": "10 6\n6 10\n4 12",
"output": "-1"
},
{
"input": "10 8\n10 8\n4 14",
"output": "-1"
},
{
"input": "14 13\n9 18\n14 13",
"output": "-1"
},
{
"input": "9 14\n8 15\n8 15",
"output": "-1"
},
{
"input": "3 8\n2 9\n6 5",
"output": "-1"
},
{
"input": "14 17\n18 13\n15 16",
"output": "-1"
},
{
"input": "16 14\n15 15\n17 13",
"output": "9 7\n6 8"
},
{
"input": "14 11\n16 9\n13 12",
"output": "9 5\n7 4"
},
{
"input": "13 10\n11 12\n7 16",
"output": "4 9\n7 3"
},
{
"input": "14 8\n11 11\n13 9",
"output": "8 6\n3 5"
},
{
"input": "12 11\n13 10\n10 13",
"output": "-1"
},
{
"input": "6 5\n2 9\n5 6",
"output": "-1"
},
{
"input": "7 8\n8 7\n12 3",
"output": "-1"
},
{
"input": "7 11\n7 11\n6 12",
"output": "-1"
},
{
"input": "8 5\n11 2\n8 5",
"output": "-1"
},
{
"input": "10 16\n14 12\n14 12",
"output": "-1"
},
{
"input": "7 9\n4 12\n5 11",
"output": "-1"
},
{
"input": "11 13\n19 5\n12 12",
"output": "-1"
},
{
"input": "8 12\n5 15\n11 9",
"output": "2 6\n3 9"
},
{
"input": "16 5\n13 8\n10 11",
"output": "9 7\n4 1"
},
{
"input": "3 14\n8 9\n10 7",
"output": "2 1\n6 8"
},
{
"input": "12 14\n11 15\n9 17",
"output": "3 9\n8 6"
},
{
"input": "13 7\n9 11\n14 6",
"output": "8 5\n1 6"
},
{
"input": "13 8\n15 6\n11 10",
"output": "9 4\n6 2"
},
{
"input": "9 6\n5 10\n3 12",
"output": "1 8\n4 2"
},
{
"input": "12 7\n5 14\n8 11",
"output": "3 9\n2 5"
},
{
"input": "9 12\n3 17\n10 10",
"output": "-1"
},
{
"input": "3 8\n4 6\n5 5",
"output": "-1"
}
] | 1,607,253,484
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 52
| 278
| 0
|
r1,r2=map(int,input().split())
c1,c2=map(int,input().split())
d1,d2=map(int,input().split())
a=(r1+c1-d2)//2
b=r1-a
c=c1-a
d=c2-b
if a+d==d1 and b+c==d2 and a+b==r1 and c+d==r2 and a+c==c1 and b+d==c2 and len(set([a,b,c,d]))==4 and max(a,b,c,d)<10 and min(a,b,c,d)>0:
print(a,b)
print(c,d)
else:
print(-1)
|
Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
|
```python
r1,r2=map(int,input().split())
c1,c2=map(int,input().split())
d1,d2=map(int,input().split())
a=(r1+c1-d2)//2
b=r1-a
c=c1-a
d=c2-b
if a+d==d1 and b+c==d2 and a+b==r1 and c+d==r2 and a+c==c1 and b+d==c2 and len(set([a,b,c,d]))==4 and max(a,b,c,d)<10 and min(a,b,c,d)>0:
print(a,b)
print(c,d)
else:
print(-1)
```
| 3
|
|
988
|
A
|
Diverse Team
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
|
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
|
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$.
|
[
"5 3\n15 13 15 15 12\n",
"5 4\n15 13 15 15 12\n",
"4 4\n20 10 40 30\n"
] |
[
"YES\n1 2 5 \n",
"NO\n",
"YES\n1 2 3 4 \n"
] |
All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter.
| 0
|
[
{
"input": "5 3\n15 13 15 15 12",
"output": "YES\n1 2 5 "
},
{
"input": "5 4\n15 13 15 15 12",
"output": "NO"
},
{
"input": "4 4\n20 10 40 30",
"output": "YES\n1 2 3 4 "
},
{
"input": "1 1\n1",
"output": "YES\n1 "
},
{
"input": "100 53\n16 17 1 2 27 5 9 9 53 24 17 33 35 24 20 48 56 73 12 14 39 55 58 13 59 73 29 26 40 33 22 29 34 22 55 38 63 66 36 13 60 42 10 15 21 9 11 5 23 37 79 47 26 3 79 53 44 8 71 75 42 11 34 39 79 33 10 26 23 23 17 14 54 41 60 31 83 5 45 4 14 35 6 60 28 48 23 18 60 36 21 28 7 34 9 25 52 43 54 19",
"output": "YES\n1 2 3 4 5 6 7 9 10 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 31 33 36 37 38 39 41 42 43 44 45 47 49 50 51 52 54 57 58 59 60 73 74 76 77 79 80 83 "
},
{
"input": "2 2\n100 100",
"output": "NO"
},
{
"input": "2 2\n100 99",
"output": "YES\n1 2 "
},
{
"input": "100 100\n63 100 75 32 53 24 73 98 76 15 70 48 8 81 88 58 95 78 27 92 14 16 72 43 46 39 66 38 64 42 59 9 22 51 4 6 10 94 28 99 68 80 35 50 45 20 47 7 30 26 49 91 77 19 96 57 65 1 11 13 31 12 82 87 93 34 62 3 21 79 56 41 89 18 44 23 74 86 2 33 69 36 61 67 25 83 5 84 90 37 40 29 97 60 52 55 54 71 17 85",
"output": "YES\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100 41\n54 16 42 3 45 6 9 72 100 13 24 57 35 5 89 13 97 27 43 9 73 89 48 16 48 55 18 15 55 28 30 6 18 41 100 61 9 42 35 54 57 25 73 15 42 54 49 5 72 48 30 55 4 43 94 5 60 92 93 23 89 75 53 92 74 93 89 28 69 6 3 49 15 28 49 57 54 55 30 57 69 18 89 6 25 23 93 74 30 13 87 53 6 42 4 54 60 30 4 35",
"output": "NO"
},
{
"input": "100 2\n70 64 70 32 70 64 32 70 64 32 32 64 70 64 64 32 64 64 64 70 70 64 64 64 64 70 32 64 70 64 32 70 70 70 64 70 64 70 64 32 70 32 70 64 64 64 32 70 64 70 70 32 70 32 32 32 70 32 70 32 64 64 70 32 32 64 70 64 32 32 64 64 32 32 70 70 32 70 32 64 32 70 64 64 32 64 32 64 70 32 70 32 70 64 64 64 70 70 64 70",
"output": "YES\n1 2 "
}
] | 1,618,165,650
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 10
| 62
| 307,200
|
def total(lista,k):
a = {}
indices = []
indice = 0
largo =0
for i in lista:
if a.get(i)==None:
a[i]=0
indices.append(indice)
largo+=1
if k == largo:
print("YES")
for j in indices:
print(j+1,end=" ")
return None
indice+=1
print("NO")
def prueba():
n,k = [int(i) for i in input().split(" ")]
lista = [int(i) for i in input().split(" ")]
total(lista,k)
prueba()
|
Title: Diverse Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
Output Specification:
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$.
Demo Input:
['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n']
Demo Output:
['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n']
Note:
All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter.
|
```python
def total(lista,k):
a = {}
indices = []
indice = 0
largo =0
for i in lista:
if a.get(i)==None:
a[i]=0
indices.append(indice)
largo+=1
if k == largo:
print("YES")
for j in indices:
print(j+1,end=" ")
return None
indice+=1
print("NO")
def prueba():
n,k = [int(i) for i in input().split(" ")]
lista = [int(i) for i in input().split(" ")]
total(lista,k)
prueba()
```
| 3
|
|
894
|
A
|
QAQ
|
PROGRAMMING
| 800
|
[
"brute force",
"dp"
] | null | null |
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
|
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
|
Print a single integer — the number of subsequences "QAQ" in the string.
|
[
"QAQAQYSYIOIWIN\n",
"QAQQQZZYNOIWIN\n"
] |
[
"4\n",
"3\n"
] |
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
| 500
|
[
{
"input": "QAQAQYSYIOIWIN",
"output": "4"
},
{
"input": "QAQQQZZYNOIWIN",
"output": "3"
},
{
"input": "QA",
"output": "0"
},
{
"input": "IAQVAQZLQBQVQFTQQQADAQJA",
"output": "24"
},
{
"input": "QQAAQASGAYAAAAKAKAQIQEAQAIAAIAQQQQQ",
"output": "378"
},
{
"input": "AMVFNFJIAVNQJWIVONQOAOOQSNQSONOASONAONQINAONAOIQONANOIQOANOQINAONOQINAONOXJCOIAQOAOQAQAQAQAQWWWAQQAQ",
"output": "1077"
},
{
"input": "AAQQAXBQQBQQXBNQRJAQKQNAQNQVDQASAGGANQQQQTJFFQQQTQQA",
"output": "568"
},
{
"input": "KAZXAVLPJQBQVQQQQQAPAQQGQTQVZQAAAOYA",
"output": "70"
},
{
"input": "W",
"output": "0"
},
{
"input": "DBA",
"output": "0"
},
{
"input": "RQAWNACASAAKAGAAAAQ",
"output": "10"
},
{
"input": "QJAWZAAOAAGIAAAAAOQATASQAEAAAAQFQQHPA",
"output": "111"
},
{
"input": "QQKWQAQAAAAAAAAGAAVAQUEQQUMQMAQQQNQLAMAAAUAEAAEMAAA",
"output": "411"
},
{
"input": "QQUMQAYAUAAGWAAAQSDAVAAQAAAASKQJJQQQQMAWAYYAAAAAAEAJAXWQQ",
"output": "625"
},
{
"input": "QORZOYAQ",
"output": "1"
},
{
"input": "QCQAQAGAWAQQQAQAVQAQQQQAQAQQQAQAAATQAAVAAAQQQQAAAUUQAQQNQQWQQWAQAAQQKQYAQAAQQQAAQRAQQQWBQQQQAPBAQGQA",
"output": "13174"
},
{
"input": "QQAQQAKQFAQLQAAWAMQAZQAJQAAQQOACQQAAAYANAQAQQAQAAQQAOBQQJQAQAQAQQQAAAAABQQQAVNZAQQQQAMQQAFAAEAQAQHQT",
"output": "10420"
},
{
"input": "AQEGQHQQKQAQQPQKAQQQAAAAQQQAQEQAAQAAQAQFSLAAQQAQOQQAVQAAAPQQAWAQAQAFQAXAQQQQTRLOQAQQJQNQXQQQQSQVDQQQ",
"output": "12488"
},
{
"input": "QNQKQQQLASQBAVQQQQAAQQOQRJQQAQQQEQZUOANAADAAQQJAQAQARAAAQQQEQBHTQAAQAAAAQQMKQQQIAOJJQQAQAAADADQUQQQA",
"output": "9114"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "35937"
},
{
"input": "AMQQAAQAAQAAAAAAQQQBOAAANAAKQJCYQAE",
"output": "254"
},
{
"input": "AYQBAEQGAQEOAKGIXLQJAIAKQAAAQPUAJAKAATFWQQAOQQQUFQYAQQMQHOKAAJXGFCARAQSATHAUQQAATQJJQDQRAANQQAE",
"output": "2174"
},
{
"input": "AAQXAAQAYQAAAAGAQHVQYAGIVACADFAAQAAAAQZAAQMAKZAADQAQDAAQDAAAMQQOXYAQQQAKQBAAQQKAXQBJZDDLAAHQQ",
"output": "2962"
},
{
"input": "AYQQYAVAMNIAUAAKBBQVACWKTQSAQZAAQAAASZJAWBCAALAARHACQAKQQAQAARPAQAAQAQAAZQUSHQAMFVFZQQQQSAQQXAA",
"output": "2482"
},
{
"input": "LQMAQQARQAQBJQQQAGAAZQQXALQQAARQAQQQQAAQQAQQQAQQCAQQAQQAYQQQRAAZATQALYQQAAHHAAQHAAAAAAAAQQMAAQNAKQ",
"output": "7768"
},
{
"input": "MAQQWAQOYQMAAAQAQPQZAOAAQAUAQNAAQAAAITQSAQAKAQKAQQWSQAAQQAGUCDQMQWKQUXKWQQAAQQAAQQZQDQQQAABXQUUXQOA",
"output": "5422"
},
{
"input": "QTAAQDAQXAQQJQQQGAAAQQQQSBQZKAQQAQQQQEAQNUQBZCQLYQZQEQQAAQHQVAORKQVAQYQNASZQAARZAAGAAAAOQDCQ",
"output": "3024"
},
{
"input": "QQWAQQGQQUZQQQLZAAQYQXQVAQFQUAQZUQZZQUKBHSHTQYLQAOQXAQQGAQQTQOAQARQADAJRAAQPQAQQUQAUAMAUVQAAAQQAWQ",
"output": "4527"
},
{
"input": "QQAAQQAQVAQZQQQQAOEAQZPQIBQZACQQAFQQLAAQDATZQANHKYQQAQTAAFQRQAIQAJPWQAQTEIRXAEQQAYWAAAUKQQAQAQQQSQQH",
"output": "6416"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAAAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQ",
"output": "13136"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQQAA",
"output": "14231"
},
{
"input": "QQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "15296"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQA",
"output": "20825"
},
{
"input": "AQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQ",
"output": "20825"
},
{
"input": "Q",
"output": "0"
},
{
"input": "A",
"output": "0"
},
{
"input": "FFF",
"output": "0"
},
{
"input": "AAAAAA",
"output": "0"
}
] | 1,645,077,178
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 46
| 0
|
#check stevegregg's solution for more insight on dp solution
zoop=input()
qcount=[0]*len(zoop)
qacount=[0]*len(zoop)
qaqcount=[0]*len(zoop)
if zoop[0]=='Q':
qcount[0]=1
for i in range(1,len(zoop)):
zp=zoop[i]
if zp=='Q':
qcount[i]=qcount[i-1]+1
qacount[i]=qacount[i-1]
qaqcount[i]=qaqcount[i-1]+qacount[i-1]
elif zp=='A':
qcount[i]=qcount[i-1]
qacount[i]=qacount[i-1]+qcount[i-1]
qaqcount[i]=qaqcount[i-1]
else:
qcount[i]=qcount[i-1]
qacount[i]=qacount[i-1]
qaqcount[i]=qaqcount[i-1]
print(qaqcount[-1])
|
Title: QAQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input Specification:
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
Output Specification:
Print a single integer — the number of subsequences "QAQ" in the string.
Demo Input:
['QAQAQYSYIOIWIN\n', 'QAQQQZZYNOIWIN\n']
Demo Output:
['4\n', '3\n']
Note:
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
|
```python
#check stevegregg's solution for more insight on dp solution
zoop=input()
qcount=[0]*len(zoop)
qacount=[0]*len(zoop)
qaqcount=[0]*len(zoop)
if zoop[0]=='Q':
qcount[0]=1
for i in range(1,len(zoop)):
zp=zoop[i]
if zp=='Q':
qcount[i]=qcount[i-1]+1
qacount[i]=qacount[i-1]
qaqcount[i]=qaqcount[i-1]+qacount[i-1]
elif zp=='A':
qcount[i]=qcount[i-1]
qacount[i]=qacount[i-1]+qcount[i-1]
qaqcount[i]=qaqcount[i-1]
else:
qcount[i]=qcount[i-1]
qacount[i]=qacount[i-1]
qaqcount[i]=qaqcount[i-1]
print(qaqcount[-1])
```
| 3
|
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,680,426,567
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
nums = list(input().split())
nums.sort()
a , b = int(nums[0]) , int(nums[1])
while a != 0 :
b = a
a = b % a
print(b)
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
nums = list(input().split())
nums.sort()
a , b = int(nums[0]) , int(nums[1])
while a != 0 :
b = a
a = b % a
print(b)
```
| 0
|
|
368
|
B
|
Sereja and Suffixes
|
PROGRAMMING
| 1,100
|
[
"data structures",
"dp"
] | null | null |
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
|
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
|
[
"10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n"
] |
[
"6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n"
] |
none
| 1,000
|
[
{
"input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1"
},
{
"input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2",
"output": "3\n4\n5"
},
{
"input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4",
"output": "3\n5\n2\n4\n3\n3\n2\n5\n1\n3"
},
{
"input": "10 2\n2 6 5 7 2 2 3 2 4 8\n1\n2",
"output": "7\n7"
},
{
"input": "7 1\n68346 10956 76708 23018 84063 34833 80407\n1",
"output": "7"
},
{
"input": "2 2\n8 4\n1\n1",
"output": "2\n2"
},
{
"input": "1 5\n5\n1\n1\n1\n1\n1",
"output": "1\n1\n1\n1\n1"
},
{
"input": "4 7\n3 1 4 2\n4\n1\n2\n3\n2\n4\n4",
"output": "1\n4\n3\n2\n3\n1\n1"
},
{
"input": "4 3\n9 1 7 1\n1\n4\n2",
"output": "3\n1\n2"
},
{
"input": "8 3\n9280 6676 2720 6172 8329 10413 3975 1394\n5\n6\n7",
"output": "4\n3\n2"
},
{
"input": "1 1\n1\n1",
"output": "1"
}
] | 1,668,419,356
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 15
| 0
|
n,m=map(int,input().split())
l=[int(i) for i in input().split()]
for i in range(m):
new=[]
for j in l[i:]:
if j not in new:
new.append(j)
print(len(new))
|
Title: Sereja and Suffixes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
Output Specification:
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
Demo Input:
['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n']
Demo Output:
['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n']
Note:
none
|
```python
n,m=map(int,input().split())
l=[int(i) for i in input().split()]
for i in range(m):
new=[]
for j in l[i:]:
if j not in new:
new.append(j)
print(len(new))
```
| 0
|
|
980
|
B
|
Marlin
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms"
] | null | null |
The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$.
The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
|
The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.
|
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.
|
[
"7 2\n",
"5 3\n"
] |
[
"YES\n.......\n.#.....\n.#.....\n.......\n",
"YES\n.....\n.###.\n.....\n.....\n"
] |
none
| 1,000
|
[
{
"input": "7 2",
"output": "YES\n.......\n.#.....\n.#.....\n......."
},
{
"input": "5 3",
"output": "YES\n.....\n.###.\n.....\n....."
},
{
"input": "3 2",
"output": "YES\n...\n.#.\n.#.\n..."
},
{
"input": "3 0",
"output": "YES\n...\n...\n...\n..."
},
{
"input": "49 1",
"output": "YES\n.................................................\n........................#........................\n.................................................\n................................................."
},
{
"input": "9 4",
"output": "YES\n.........\n.##......\n.##......\n........."
},
{
"input": "9 5",
"output": "YES\n.........\n.#.#.....\n.###.....\n........."
},
{
"input": "99 193",
"output": "YES\n...................................................................................................\n.###############################################################################################.#.\n.#################################################################################################.\n..................................................................................................."
},
{
"input": "99 14",
"output": "YES\n...................................................................................................\n.#######...........................................................................................\n.#######...........................................................................................\n..................................................................................................."
},
{
"input": "57 15",
"output": "YES\n.........................................................\n.######.#................................................\n.########................................................\n........................................................."
},
{
"input": "99 3",
"output": "YES\n...................................................................................................\n................................................###................................................\n...................................................................................................\n..................................................................................................."
},
{
"input": "3 1",
"output": "YES\n...\n.#.\n...\n..."
},
{
"input": "9 9",
"output": "YES\n.........\n.###.#...\n.#####...\n........."
},
{
"input": "67 9",
"output": "YES\n...................................................................\n.###.#.............................................................\n.#####.............................................................\n..................................................................."
},
{
"input": "99 99",
"output": "YES\n...................................................................................................\n.################################################.#................................................\n.##################################################................................................\n..................................................................................................."
},
{
"input": "31 32",
"output": "YES\n...............................\n.################..............\n.################..............\n..............................."
},
{
"input": "5 1",
"output": "YES\n.....\n..#..\n.....\n....."
},
{
"input": "5 2",
"output": "YES\n.....\n.#...\n.#...\n....."
},
{
"input": "5 4",
"output": "YES\n.....\n.##..\n.##..\n....."
},
{
"input": "5 6",
"output": "YES\n.....\n.###.\n.###.\n....."
},
{
"input": "5 5",
"output": "YES\n.....\n.#.#.\n.###.\n....."
},
{
"input": "7 9",
"output": "YES\n.......\n.###.#.\n.#####.\n......."
},
{
"input": "7 10",
"output": "YES\n.......\n.#####.\n.#####.\n......."
},
{
"input": "19 12",
"output": "YES\n...................\n.######............\n.######............\n..................."
},
{
"input": "19 3",
"output": "YES\n...................\n........###........\n...................\n..................."
},
{
"input": "37 14",
"output": "YES\n.....................................\n.#######.............................\n.#######.............................\n....................................."
},
{
"input": "37 15",
"output": "YES\n.....................................\n.######.#............................\n.########............................\n....................................."
},
{
"input": "37 37",
"output": "YES\n.....................................\n.#################.#.................\n.###################.................\n....................................."
},
{
"input": "37 36",
"output": "YES\n.....................................\n.##################..................\n.##################..................\n....................................."
},
{
"input": "37 35",
"output": "YES\n.....................................\n.################.#..................\n.##################..................\n....................................."
},
{
"input": "37 34",
"output": "YES\n.....................................\n.#################...................\n.#################...................\n....................................."
},
{
"input": "37 38",
"output": "YES\n.....................................\n.###################.................\n.###################.................\n....................................."
},
{
"input": "37 39",
"output": "YES\n.....................................\n.##################.#................\n.####################................\n....................................."
},
{
"input": "37 40",
"output": "YES\n.....................................\n.####################................\n.####################................\n....................................."
},
{
"input": "5 0",
"output": "YES\n.....\n.....\n.....\n....."
},
{
"input": "67 1",
"output": "YES\n...................................................................\n.................................#.................................\n...................................................................\n..................................................................."
},
{
"input": "37 19",
"output": "YES\n.....................................\n.########.#..........................\n.##########..........................\n....................................."
},
{
"input": "77 7",
"output": "YES\n.............................................................................\n.##.#........................................................................\n.####........................................................................\n............................................................................."
},
{
"input": "33 47",
"output": "YES\n.................................\n.######################.#........\n.########################........\n................................."
},
{
"input": "33 48",
"output": "YES\n.................................\n.########################........\n.########################........\n................................."
},
{
"input": "23 40",
"output": "YES\n.......................\n.####################..\n.####################..\n......................."
},
{
"input": "23 39",
"output": "YES\n.......................\n.##################.#..\n.####################..\n......................."
},
{
"input": "49 3",
"output": "YES\n.................................................\n.......................###.......................\n.................................................\n................................................."
},
{
"input": "99 1",
"output": "YES\n...................................................................................................\n.................................................#.................................................\n...................................................................................................\n..................................................................................................."
},
{
"input": "77 0",
"output": "YES\n.............................................................................\n.............................................................................\n.............................................................................\n............................................................................."
},
{
"input": "99 0",
"output": "YES\n...................................................................................................\n...................................................................................................\n...................................................................................................\n..................................................................................................."
},
{
"input": "99 5",
"output": "YES\n...................................................................................................\n.#.#...............................................................................................\n.###...............................................................................................\n..................................................................................................."
},
{
"input": "99 4",
"output": "YES\n...................................................................................................\n.##................................................................................................\n.##................................................................................................\n..................................................................................................."
},
{
"input": "99 20",
"output": "YES\n...................................................................................................\n.##########........................................................................................\n.##########........................................................................................\n..................................................................................................."
},
{
"input": "99 194",
"output": "YES\n...................................................................................................\n.#################################################################################################.\n.#################################################################################################.\n..................................................................................................."
},
{
"input": "99 192",
"output": "YES\n...................................................................................................\n.################################################################################################..\n.################################################################################################..\n..................................................................................................."
},
{
"input": "99 190",
"output": "YES\n...................................................................................................\n.###############################################################################################...\n.###############################################################################################...\n..................................................................................................."
},
{
"input": "99 189",
"output": "YES\n...................................................................................................\n.#############################################################################################.#...\n.###############################################################################################...\n..................................................................................................."
},
{
"input": "99 177",
"output": "YES\n...................................................................................................\n.#######################################################################################.#.........\n.#########################################################################################.........\n..................................................................................................."
},
{
"input": "99 154",
"output": "YES\n...................................................................................................\n.#############################################################################.....................\n.#############################################################################.....................\n..................................................................................................."
},
{
"input": "99 127",
"output": "YES\n...................................................................................................\n.##############################################################.#..................................\n.################################################################..................................\n..................................................................................................."
},
{
"input": "99 55",
"output": "YES\n...................................................................................................\n.##########################.#......................................................................\n.############################......................................................................\n..................................................................................................."
},
{
"input": "99 40",
"output": "YES\n...................................................................................................\n.####################..............................................................................\n.####################..............................................................................\n..................................................................................................."
},
{
"input": "97 190",
"output": "YES\n.................................................................................................\n.###############################################################################################.\n.###############################################################################################.\n................................................................................................."
},
{
"input": "97 100",
"output": "YES\n.................................................................................................\n.##################################################..............................................\n.##################################################..............................................\n................................................................................................."
},
{
"input": "97 111",
"output": "YES\n.................................................................................................\n.######################################################.#........................................\n.########################################################........................................\n................................................................................................."
},
{
"input": "97 64",
"output": "YES\n.................................................................................................\n.################################................................................................\n.################################................................................................\n................................................................................................."
},
{
"input": "97 77",
"output": "YES\n.................................................................................................\n.#####################################.#.........................................................\n.#######################################.........................................................\n................................................................................................."
},
{
"input": "91 77",
"output": "YES\n...........................................................................................\n.#####################################.#...................................................\n.#######################################...................................................\n..........................................................................................."
},
{
"input": "91 128",
"output": "YES\n...........................................................................................\n.################################################################..........................\n.################################################################..........................\n..........................................................................................."
},
{
"input": "91 113",
"output": "YES\n...........................................................................................\n.#######################################################.#.................................\n.#########################################################.................................\n..........................................................................................."
},
{
"input": "55 55",
"output": "YES\n.......................................................\n.##########################.#..........................\n.############################..........................\n......................................................."
},
{
"input": "43 34",
"output": "YES\n...........................................\n.#################.........................\n.#################.........................\n..........................................."
},
{
"input": "13 21",
"output": "YES\n.............\n.#########.#.\n.###########.\n............."
},
{
"input": "27 50",
"output": "YES\n...........................\n.#########################.\n.#########################.\n..........................."
},
{
"input": "27 49",
"output": "YES\n...........................\n.#######################.#.\n.#########################.\n..........................."
},
{
"input": "27 48",
"output": "YES\n...........................\n.########################..\n.########################..\n..........................."
},
{
"input": "27 40",
"output": "YES\n...........................\n.####################......\n.####################......\n..........................."
},
{
"input": "87 80",
"output": "YES\n.......................................................................................\n.########################################..............................................\n.########################################..............................................\n......................................................................................."
},
{
"input": "69 17",
"output": "YES\n.....................................................................\n.#######.#...........................................................\n.#########...........................................................\n....................................................................."
},
{
"input": "39 73",
"output": "YES\n.......................................\n.###################################.#.\n.#####################################.\n......................................."
},
{
"input": "39 72",
"output": "YES\n.......................................\n.####################################..\n.####################################..\n......................................."
},
{
"input": "33 57",
"output": "YES\n.................................\n.###########################.#...\n.#############################...\n................................."
},
{
"input": "33 54",
"output": "YES\n.................................\n.###########################.....\n.###########################.....\n................................."
},
{
"input": "99 2",
"output": "YES\n...................................................................................................\n.#.................................................................................................\n.#.................................................................................................\n..................................................................................................."
},
{
"input": "99 7",
"output": "YES\n...................................................................................................\n.##.#..............................................................................................\n.####..............................................................................................\n..................................................................................................."
},
{
"input": "99 37",
"output": "YES\n...................................................................................................\n.#################.#...............................................................................\n.###################...............................................................................\n..................................................................................................."
},
{
"input": "57 67",
"output": "YES\n.........................................................\n.################################.#......................\n.##################################......................\n........................................................."
},
{
"input": "57 34",
"output": "YES\n.........................................................\n.#################.......................................\n.#################.......................................\n........................................................."
},
{
"input": "73 17",
"output": "YES\n.........................................................................\n.#######.#...............................................................\n.#########...............................................................\n........................................................................."
},
{
"input": "73 43",
"output": "YES\n.........................................................................\n.####################.#..................................................\n.######################..................................................\n........................................................................."
},
{
"input": "7 3",
"output": "YES\n.......\n..###..\n.......\n......."
},
{
"input": "7 7",
"output": "YES\n.......\n.##.#..\n.####..\n......."
}
] | 1,525,794,208
| 2,308
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 62
| 7,065,600
|
n, k = map(int, input().strip().split())
if k <= n-2:
print('YES')
print('.'*n)
print('.'+'#'*k+'.'*(n-k-1))
print('.'*n)
print('.'*n)
elif k%2 == 0:
print('YES')
print('.'*n)
for _ in range(2):
print('.'+'#'*(k//2)+'.'*(n-k//2-1))
print('.'*n)
else:
print('NO')
|
Title: Marlin
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$.
The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
Input Specification:
The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.
Output Specification:
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.
Demo Input:
['7 2\n', '5 3\n']
Demo Output:
['YES\n.......\n.#.....\n.#.....\n.......\n', 'YES\n.....\n.###.\n.....\n.....\n']
Note:
none
|
```python
n, k = map(int, input().strip().split())
if k <= n-2:
print('YES')
print('.'*n)
print('.'+'#'*k+'.'*(n-k-1))
print('.'*n)
print('.'*n)
elif k%2 == 0:
print('YES')
print('.'*n)
for _ in range(2):
print('.'+'#'*(k//2)+'.'*(n-k//2-1))
print('.'*n)
else:
print('NO')
```
| 0
|
|
110
|
A
|
Nearly Lucky Number
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Nearly Lucky Number
|
2
|
256
|
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
|
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
|
[
"40047\n",
"7747774\n",
"1000000000000000000\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO".
| 500
|
[
{
"input": "40047",
"output": "NO"
},
{
"input": "7747774",
"output": "YES"
},
{
"input": "1000000000000000000",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "4",
"output": "NO"
},
{
"input": "474404774",
"output": "NO"
},
{
"input": "4744000695826",
"output": "YES"
},
{
"input": "10000000004744744",
"output": "YES"
},
{
"input": "446486416781684178",
"output": "YES"
},
{
"input": "999999999",
"output": "NO"
},
{
"input": "7777",
"output": "YES"
},
{
"input": "87414417444",
"output": "NO"
},
{
"input": "111222333444555667",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "4700",
"output": "NO"
},
{
"input": "3794555488744477",
"output": "NO"
},
{
"input": "444444444444444444",
"output": "NO"
},
{
"input": "474447447774444774",
"output": "NO"
},
{
"input": "777777777777777",
"output": "NO"
},
{
"input": "34777745021000000",
"output": "NO"
},
{
"input": "963",
"output": "NO"
},
{
"input": "855474448854788540",
"output": "NO"
},
{
"input": "999999999999994744",
"output": "YES"
},
{
"input": "400000000474",
"output": "YES"
},
{
"input": "123456789123456789",
"output": "YES"
},
{
"input": "740577777584945874",
"output": "NO"
},
{
"input": "7777777",
"output": "YES"
},
{
"input": "4444000111222333",
"output": "YES"
},
{
"input": "9847745885202111",
"output": "YES"
},
{
"input": "123456000000",
"output": "NO"
},
{
"input": "4744447444444",
"output": "NO"
},
{
"input": "7477",
"output": "YES"
},
{
"input": "4747477",
"output": "YES"
},
{
"input": "777777777444444444",
"output": "NO"
}
] | 1,698,134,320
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
x = input()
if x.count('4') + x.count('7') in (4, 7):
print("YES")
else:
print("NO")
|
Title: Nearly Lucky Number
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
Demo Input:
['40047\n', '7747774\n', '1000000000000000000\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO".
|
```python
x = input()
if x.count('4') + x.count('7') in (4, 7):
print("YES")
else:
print("NO")
```
| 3.977
|
300
|
C
|
Beautiful Numbers
|
PROGRAMMING
| 1,800
|
[
"brute force",
"combinatorics"
] | null | null |
Vitaly is a very weird man. He's got two favorite digits *a* and *b*. Vitaly calls a positive integer good, if the decimal representation of this integer only contains digits *a* and *b*. Vitaly calls a good number excellent, if the sum of its digits is a good number.
For example, let's say that Vitaly's favourite digits are 1 and 3, then number 12 isn't good and numbers 13 or 311 are. Also, number 111 is excellent and number 11 isn't.
Now Vitaly is wondering, how many excellent numbers of length exactly *n* are there. As this number can be rather large, he asks you to count the remainder after dividing it by 1000000007 (109<=+<=7).
A number's length is the number of digits in its decimal representation without leading zeroes.
|
The first line contains three integers: *a*, *b*, *n* (1<=≤<=*a*<=<<=*b*<=≤<=9,<=1<=≤<=*n*<=≤<=106).
|
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
|
[
"1 3 3\n",
"2 3 10\n"
] |
[
"1\n",
"165\n"
] |
none
| 2,000
|
[
{
"input": "1 3 3",
"output": "1"
},
{
"input": "2 3 10",
"output": "165"
},
{
"input": "6 8 14215",
"output": "651581472"
},
{
"input": "4 9 104671",
"output": "329390901"
},
{
"input": "6 7 78755",
"output": "0"
},
{
"input": "1 8 265",
"output": "461320265"
},
{
"input": "3 9 37413",
"output": "461358757"
},
{
"input": "1 7 49055",
"output": "461364774"
},
{
"input": "3 4 11028",
"output": "461668105"
},
{
"input": "2 6 32377",
"output": "887598327"
},
{
"input": "3 5 80791",
"output": "999993599"
},
{
"input": "4 8 11857",
"output": "999991923"
},
{
"input": "1 3 10785",
"output": "999952603"
},
{
"input": "4 6 11808",
"output": "999925731"
},
{
"input": "1 2 11857",
"output": "999991923"
},
{
"input": "2 4 88193",
"output": "999976846"
},
{
"input": "1 4 37226",
"output": "999970594"
},
{
"input": "2 5 53049",
"output": "259705254"
},
{
"input": "3 6 1000000",
"output": "786609214"
},
{
"input": "7 9 999999",
"output": "53911803"
},
{
"input": "8 9 999999",
"output": "447886447"
},
{
"input": "3 8 1000000",
"output": "0"
},
{
"input": "2 8 999999",
"output": "0"
},
{
"input": "1 6 997695",
"output": "0"
},
{
"input": "1 5 997694",
"output": "0"
},
{
"input": "5 9 997693",
"output": "0"
},
{
"input": "5 8 997690",
"output": "21735480"
},
{
"input": "7 8 2",
"output": "0"
},
{
"input": "6 9 1",
"output": "2"
},
{
"input": "8 9 111111",
"output": "900401372"
},
{
"input": "8 9 1000000",
"output": "573697309"
},
{
"input": "1 2 1000000",
"output": "786609214"
}
] | 1,692,609,243
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 2,150,400
|
MOD = 1000000007
a, b, n = map(int, input().split())
fac, inv = [1] * (n + 1), [1] * (n + 1)
for i in range(1, n + 1): fac[i] = i * fac[i - 1] % MOD
inv[n] = pow(fac[n], -1, MOD)
for i in range(n, 0, -1): inv[i - 1] = i * inv[i] % MOD
def comb(n, r):
return fac[n] * inv[r] * inv[n - r] % MOD
def good(x):
return all([c == str(a) or c == str(b) for c in str(x)])
ans = 0
for i in range(n + 1):
if good(i * a + (n - i) * b): ans += comb(n, i)
print(ans % MOD)
|
Title: Beautiful Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly is a very weird man. He's got two favorite digits *a* and *b*. Vitaly calls a positive integer good, if the decimal representation of this integer only contains digits *a* and *b*. Vitaly calls a good number excellent, if the sum of its digits is a good number.
For example, let's say that Vitaly's favourite digits are 1 and 3, then number 12 isn't good and numbers 13 or 311 are. Also, number 111 is excellent and number 11 isn't.
Now Vitaly is wondering, how many excellent numbers of length exactly *n* are there. As this number can be rather large, he asks you to count the remainder after dividing it by 1000000007 (109<=+<=7).
A number's length is the number of digits in its decimal representation without leading zeroes.
Input Specification:
The first line contains three integers: *a*, *b*, *n* (1<=≤<=*a*<=<<=*b*<=≤<=9,<=1<=≤<=*n*<=≤<=106).
Output Specification:
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
Demo Input:
['1 3 3\n', '2 3 10\n']
Demo Output:
['1\n', '165\n']
Note:
none
|
```python
MOD = 1000000007
a, b, n = map(int, input().split())
fac, inv = [1] * (n + 1), [1] * (n + 1)
for i in range(1, n + 1): fac[i] = i * fac[i - 1] % MOD
inv[n] = pow(fac[n], -1, MOD)
for i in range(n, 0, -1): inv[i - 1] = i * inv[i] % MOD
def comb(n, r):
return fac[n] * inv[r] * inv[n - r] % MOD
def good(x):
return all([c == str(a) or c == str(b) for c in str(x)])
ans = 0
for i in range(n + 1):
if good(i * a + (n - i) * b): ans += comb(n, i)
print(ans % MOD)
```
| -1
|
|
450
|
B
|
Jzzhu and Sequences
|
PROGRAMMING
| 1,300
|
[
"implementation",
"math"
] | null | null |
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
|
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
|
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
|
[
"2 3\n3\n",
"0 -1\n2\n"
] |
[
"1\n",
"1000000006\n"
] |
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
| 1,000
|
[
{
"input": "2 3\n3",
"output": "1"
},
{
"input": "0 -1\n2",
"output": "1000000006"
},
{
"input": "-9 -11\n12345",
"output": "1000000005"
},
{
"input": "0 0\n1000000000",
"output": "0"
},
{
"input": "-1000000000 1000000000\n2000000000",
"output": "1000000000"
},
{
"input": "-12345678 12345678\n1912345678",
"output": "12345678"
},
{
"input": "728374857 678374857\n1928374839",
"output": "950000007"
},
{
"input": "278374837 992837483\n1000000000",
"output": "721625170"
},
{
"input": "-693849384 502938493\n982838498",
"output": "502938493"
},
{
"input": "-783928374 983738273\n992837483",
"output": "16261734"
},
{
"input": "-872837483 -682738473\n999999999",
"output": "190099010"
},
{
"input": "-892837483 -998273847\n999283948",
"output": "892837483"
},
{
"input": "-283938494 738473848\n1999999999",
"output": "716061513"
},
{
"input": "-278374857 819283838\n1",
"output": "721625150"
},
{
"input": "-1000000000 123456789\n1",
"output": "7"
},
{
"input": "-529529529 -524524524\n2",
"output": "475475483"
},
{
"input": "1 2\n2000000000",
"output": "2"
},
{
"input": "-1 -2\n2000000000",
"output": "1000000005"
},
{
"input": "1 2\n1999999999",
"output": "1"
},
{
"input": "1 2\n1999999998",
"output": "1000000006"
},
{
"input": "1 2\n1999999997",
"output": "1000000005"
},
{
"input": "1 2\n1999999996",
"output": "1000000006"
},
{
"input": "69975122 366233206\n1189460676",
"output": "703741923"
},
{
"input": "812229413 904420051\n806905621",
"output": "812229413"
},
{
"input": "872099024 962697902\n1505821695",
"output": "90598878"
},
{
"input": "887387283 909670917\n754835014",
"output": "112612724"
},
{
"input": "37759824 131342932\n854621399",
"output": "868657075"
},
{
"input": "-246822123 800496170\n626323615",
"output": "753177884"
},
{
"input": "-861439463 974126967\n349411083",
"output": "835566423"
},
{
"input": "-69811049 258093841\n1412447",
"output": "741906166"
},
{
"input": "844509330 -887335829\n123329059",
"output": "844509330"
},
{
"input": "83712471 -876177148\n1213284777",
"output": "40110388"
},
{
"input": "598730524 -718984219\n1282749880",
"output": "401269483"
},
{
"input": "-474244697 -745885656\n1517883612",
"output": "271640959"
},
{
"input": "-502583588 -894906953\n1154189557",
"output": "497416419"
},
{
"input": "-636523651 -873305815\n154879215",
"output": "763217843"
},
{
"input": "721765550 594845720\n78862386",
"output": "126919830"
},
{
"input": "364141461 158854993\n1337196589",
"output": "364141461"
},
{
"input": "878985260 677031952\n394707801",
"output": "798046699"
},
{
"input": "439527072 -24854079\n1129147002",
"output": "464381151"
},
{
"input": "840435009 -612103127\n565968986",
"output": "387896880"
},
{
"input": "875035447 -826471373\n561914518",
"output": "124964560"
},
{
"input": "-342526698 305357084\n70776744",
"output": "352116225"
},
{
"input": "-903244186 899202229\n1527859274",
"output": "899202229"
},
{
"input": "-839482546 815166320\n1127472130",
"output": "839482546"
},
{
"input": "-976992569 -958313041\n1686580818",
"output": "981320479"
},
{
"input": "-497338894 -51069176\n737081851",
"output": "502661113"
},
{
"input": "-697962643 -143148799\n1287886520",
"output": "856851208"
},
{
"input": "-982572938 -482658433\n1259858332",
"output": "982572938"
},
{
"input": "123123 78817\n2000000000",
"output": "78817"
},
{
"input": "1000000000 -1000000000\n3",
"output": "14"
},
{
"input": "-1000000000 1000000000\n6",
"output": "14"
},
{
"input": "2 3\n6",
"output": "1000000006"
},
{
"input": "0 -1\n6",
"output": "1"
},
{
"input": "500000000 -1000000000\n600000003",
"output": "500000014"
},
{
"input": "-1000000000 1000000000\n3",
"output": "999999993"
},
{
"input": "1 3\n6",
"output": "1000000005"
},
{
"input": "1 2\n12",
"output": "1000000006"
},
{
"input": "7 -1000000000\n3",
"output": "0"
},
{
"input": "-999999997 999999997\n6",
"output": "20"
},
{
"input": "3 4\n6",
"output": "1000000006"
},
{
"input": "-1 2\n6",
"output": "1000000004"
},
{
"input": "2 3\n12",
"output": "1000000006"
},
{
"input": "4 18\n6",
"output": "999999993"
},
{
"input": "1 2\n6",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n6",
"output": "999999993"
},
{
"input": "999999999 -999999999\n3",
"output": "16"
},
{
"input": "-1 0\n1",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n9",
"output": "14"
},
{
"input": "999999999 -1000000000\n12",
"output": "999999992"
},
{
"input": "1000000000 -7\n3",
"output": "0"
},
{
"input": "-5 5\n6",
"output": "999999997"
},
{
"input": "5 9\n6",
"output": "1000000003"
},
{
"input": "-15 -10\n1",
"output": "999999992"
}
] | 1,645,955,729
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 15
| 0
|
x, y = map(int, input().split())
arr = [(x-y)%1000000007, x, y, (y-x)%1000000007, -x, -y]
print(arr[int(input())%6])
|
Title: Jzzhu and Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
Output Specification:
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\n3\n', '0 -1\n2\n']
Demo Output:
['1\n', '1000000006\n']
Note:
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
|
```python
x, y = map(int, input().split())
arr = [(x-y)%1000000007, x, y, (y-x)%1000000007, -x, -y]
print(arr[int(input())%6])
```
| 0
|
|
27
|
A
|
Next Test
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] |
A. Next Test
|
2
|
256
|
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
|
Output the required default value for the next test index.
|
[
"3\n1 7 2\n"
] |
[
"3\n"
] |
none
| 500
|
[
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "3\n3 4 1",
"output": "2"
},
{
"input": "4\n6 4 3 5",
"output": "1"
},
{
"input": "5\n3 2 1 7 4",
"output": "5"
},
{
"input": "6\n4 1 2 5 3 7",
"output": "6"
},
{
"input": "7\n3 2 1 6 5 7 4",
"output": "8"
},
{
"input": "8\n2 8 3 7 6 9 1 5",
"output": "4"
},
{
"input": "9\n10 5 9 3 8 7 1 2 4",
"output": "6"
},
{
"input": "10\n7 2 3 8 9 6 5 4 1 10",
"output": "11"
},
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "3\n2 4 1",
"output": "3"
},
{
"input": "4\n4 2 3 1",
"output": "5"
},
{
"input": "5\n3 1 4 2 5",
"output": "6"
},
{
"input": "6\n1 3 6 7 2 4",
"output": "5"
},
{
"input": "7\n1 5 4 7 2 3 6",
"output": "8"
},
{
"input": "8\n12 1 6 5 2 8 3 4",
"output": "7"
},
{
"input": "9\n3 2 7 5 6 4 1 9 10",
"output": "8"
},
{
"input": "10\n1 7 13 6 5 10 3 8 2 4",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "1\n3000",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3000 1",
"output": "2"
}
] | 1,528,123,483
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 22
| 186
| 7,065,600
|
def main():
n = input()
arr = list(map(int, input().split()))
s = set()
for item in arr:
s.add(item)
idx = 1
while idx < 3001:
if idx not in s:
print(idx)
exit(0)
idx += 1
if __name__ == "__main__":
main()
|
Title: Next Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
Output Specification:
Output the required default value for the next test index.
Demo Input:
['3\n1 7 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
def main():
n = input()
arr = list(map(int, input().split()))
s = set()
for item in arr:
s.add(item)
idx = 1
while idx < 3001:
if idx not in s:
print(idx)
exit(0)
idx += 1
if __name__ == "__main__":
main()
```
| 0
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.
Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons.
The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*.
|
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.
|
[
"4\n1 9\n3 1\n6 1\n7 4\n",
"7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n"
] |
[
"1\n",
"3\n"
] |
For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2.
For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.
| 0
|
[
{
"input": "4\n1 9\n3 1\n6 1\n7 4",
"output": "1"
},
{
"input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1",
"output": "3"
},
{
"input": "1\n0 1",
"output": "0"
},
{
"input": "1\n0 1000000",
"output": "0"
},
{
"input": "1\n1000000 1000000",
"output": "0"
},
{
"input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 6\n7 7",
"output": "4"
},
{
"input": "5\n1 1\n3 1\n5 1\n7 10\n8 10",
"output": "2"
},
{
"input": "11\n110 90\n100 70\n90 10\n80 10\n70 1\n60 1\n50 10\n40 1\n30 1\n10 1\n20 1",
"output": "4"
}
] | 1,450,892,015
| 3,515
|
Python 3
|
OK
|
TESTS
| 41
| 748
| 10,444,800
|
import math
import itertools
from bisect import bisect
n = int(input())
lamps = []
for i in range(n):
a, b = [int(x) for x in input().split()]
lamps.append((a, b))
lamps.sort()
saved_lamps = []
for lamp in lamps:
i = bisect(lamps, (lamp[0]-lamp[1], 0))
if i == len(lamps) or i == 0:
saved_lamps.append(1)
else:
saved_lamps.append(saved_lamps[i-1] + 1)
print(n - max(saved_lamps))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.
Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons.
The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*.
Output Specification:
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.
Demo Input:
['4\n1 9\n3 1\n6 1\n7 4\n', '7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n']
Demo Output:
['1\n', '3\n']
Note:
For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2.
For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.
|
```python
import math
import itertools
from bisect import bisect
n = int(input())
lamps = []
for i in range(n):
a, b = [int(x) for x in input().split()]
lamps.append((a, b))
lamps.sort()
saved_lamps = []
for lamp in lamps:
i = bisect(lamps, (lamp[0]-lamp[1], 0))
if i == len(lamps) or i == 0:
saved_lamps.append(1)
else:
saved_lamps.append(saved_lamps[i-1] + 1)
print(n - max(saved_lamps))
```
| 3
|
|
722
|
A
|
Broken Clock
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation"
] | null | null |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.
You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.
For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.
|
The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.
The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.
|
The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.
|
[
"24\n17:30\n",
"12\n17:30\n",
"24\n99:99\n"
] |
[
"17:30\n",
"07:30\n",
"09:09\n"
] |
none
| 500
|
[
{
"input": "24\n17:30",
"output": "17:30"
},
{
"input": "12\n17:30",
"output": "07:30"
},
{
"input": "24\n99:99",
"output": "09:09"
},
{
"input": "12\n05:54",
"output": "05:54"
},
{
"input": "12\n00:05",
"output": "01:05"
},
{
"input": "24\n23:80",
"output": "23:00"
},
{
"input": "24\n73:16",
"output": "03:16"
},
{
"input": "12\n03:77",
"output": "03:07"
},
{
"input": "12\n47:83",
"output": "07:03"
},
{
"input": "24\n23:88",
"output": "23:08"
},
{
"input": "24\n51:67",
"output": "01:07"
},
{
"input": "12\n10:33",
"output": "10:33"
},
{
"input": "12\n00:01",
"output": "01:01"
},
{
"input": "12\n07:74",
"output": "07:04"
},
{
"input": "12\n00:60",
"output": "01:00"
},
{
"input": "24\n08:32",
"output": "08:32"
},
{
"input": "24\n42:59",
"output": "02:59"
},
{
"input": "24\n19:87",
"output": "19:07"
},
{
"input": "24\n26:98",
"output": "06:08"
},
{
"input": "12\n12:91",
"output": "12:01"
},
{
"input": "12\n11:30",
"output": "11:30"
},
{
"input": "12\n90:32",
"output": "10:32"
},
{
"input": "12\n03:69",
"output": "03:09"
},
{
"input": "12\n33:83",
"output": "03:03"
},
{
"input": "24\n10:45",
"output": "10:45"
},
{
"input": "24\n65:12",
"output": "05:12"
},
{
"input": "24\n22:64",
"output": "22:04"
},
{
"input": "24\n48:91",
"output": "08:01"
},
{
"input": "12\n02:51",
"output": "02:51"
},
{
"input": "12\n40:11",
"output": "10:11"
},
{
"input": "12\n02:86",
"output": "02:06"
},
{
"input": "12\n99:96",
"output": "09:06"
},
{
"input": "24\n19:24",
"output": "19:24"
},
{
"input": "24\n55:49",
"output": "05:49"
},
{
"input": "24\n01:97",
"output": "01:07"
},
{
"input": "24\n39:68",
"output": "09:08"
},
{
"input": "24\n24:00",
"output": "04:00"
},
{
"input": "12\n91:00",
"output": "01:00"
},
{
"input": "24\n00:30",
"output": "00:30"
},
{
"input": "12\n13:20",
"output": "03:20"
},
{
"input": "12\n13:00",
"output": "03:00"
},
{
"input": "12\n42:35",
"output": "02:35"
},
{
"input": "12\n20:00",
"output": "10:00"
},
{
"input": "12\n21:00",
"output": "01:00"
},
{
"input": "24\n10:10",
"output": "10:10"
},
{
"input": "24\n30:40",
"output": "00:40"
},
{
"input": "24\n12:00",
"output": "12:00"
},
{
"input": "12\n10:60",
"output": "10:00"
},
{
"input": "24\n30:00",
"output": "00:00"
},
{
"input": "24\n34:00",
"output": "04:00"
},
{
"input": "12\n22:00",
"output": "02:00"
},
{
"input": "12\n20:20",
"output": "10:20"
}
] | 1,475,332,495
| 1,794
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 21
| 62
| 0
|
def main():
t = int(input())
s = input()
h0 = int(s[0])
h1 = int(s[1])
m0 = int(s[3])
m1 = int(s[4])
if m0 >= 6:
m0 = 0
if t == 24:
if 10*h0 + h1 > 23:
h0 = 0
if t == 12:
if 10*h0 + h1 == 0:
h0 = 1
elif 10*h0 + h1 > 12:
if h1 == 0:
h0 = 0
h1 = 1
else:
h0 = 0
print(h0,h1,':',m0,m1,sep='')
main()
|
Title: Broken Clock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.
You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.
For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.
Input Specification:
The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.
The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.
Output Specification:
The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.
Demo Input:
['24\n17:30\n', '12\n17:30\n', '24\n99:99\n']
Demo Output:
['17:30\n', '07:30\n', '09:09\n']
Note:
none
|
```python
def main():
t = int(input())
s = input()
h0 = int(s[0])
h1 = int(s[1])
m0 = int(s[3])
m1 = int(s[4])
if m0 >= 6:
m0 = 0
if t == 24:
if 10*h0 + h1 > 23:
h0 = 0
if t == 12:
if 10*h0 + h1 == 0:
h0 = 1
elif 10*h0 + h1 > 12:
if h1 == 0:
h0 = 0
h1 = 1
else:
h0 = 0
print(h0,h1,':',m0,m1,sep='')
main()
```
| 0
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,693,422,608
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
import math
a = input().split(" ")
j = 1
j *= -1 * math.floor(-1 * int(a[0])/int(a[2]))
j *= -1 * math.floor(-1 * int(a[1])/int(a[2]))
print(j)
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
import math
a = input().split(" ")
j = 1
j *= -1 * math.floor(-1 * int(a[0])/int(a[2]))
j *= -1 * math.floor(-1 * int(a[1])/int(a[2]))
print(j)
```
| 3.977
|
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"
}
] | 1,600,441,123
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 108
| 0
|
d1, d2, d3 = map(int, input().split())
if d1 >= d2:
print(d2 + min(d1+d2, d3) + min(d1, d3+d2))
else:
print(d1 + min(d1+d2, d3) + min(d2, d3+d1))
|
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
d1, d2, d3 = map(int, input().split())
if d1 >= d2:
print(d2 + min(d1+d2, d3) + min(d1, d3+d2))
else:
print(d1 + min(d1+d2, d3) + min(d2, d3+d1))
```
| 3
|
|
767
|
A
|
Snacktower
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
|
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
|
[
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] |
[
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] |
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
| 500
|
[
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,620,884,799
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 280
| 35,328,000
|
n = int(input())
arr = list(map(int, input().split()))
snack = [0]*(n+1)
cr = n
m = n+1
op=[]
for i in range(n):
snack[arr[i]] = 1
if arr[i] == cr:
while snack[cr] == 1:
op.append(cr)
snack[cr]=2
cr-=1
print(*op)
op.clear()
|
Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
|
```python
n = int(input())
arr = list(map(int, input().split()))
snack = [0]*(n+1)
cr = n
m = n+1
op=[]
for i in range(n):
snack[arr[i]] = 1
if arr[i] == cr:
while snack[cr] == 1:
op.append(cr)
snack[cr]=2
cr-=1
print(*op)
op.clear()
```
| 3
|
|
637
|
B
|
Chat Order
|
PROGRAMMING
| 1,200
|
[
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null |
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
|
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
|
[
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] |
[
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] |
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
| 1,000
|
[
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"output": "ypg"
},
{
"input": "3\nexhll\nexhll\narruapexj",
"output": "arruapexj\nexhll"
},
{
"input": "3\nfv\nle\nle",
"output": "le\nfv"
},
{
"input": "8\nm\nm\nm\nm\nm\nm\nm\nm",
"output": "m"
},
{
"input": "10\nr\nr\ni\nw\nk\nr\nb\nu\nu\nr",
"output": "r\nu\nb\nk\nw\ni"
},
{
"input": "7\ne\nfau\ncmk\nnzs\nby\nwx\ntjmok",
"output": "tjmok\nwx\nby\nnzs\ncmk\nfau\ne"
},
{
"input": "6\nklrj\nwe\nklrj\nwe\nwe\nwe",
"output": "we\nklrj"
},
{
"input": "8\nzncybqmh\naeebef\nzncybqmh\nn\naeebef\nzncybqmh\nzncybqmh\nzncybqmh",
"output": "zncybqmh\naeebef\nn"
},
{
"input": "30\nkqqcbs\nvap\nkymomn\nj\nkqqcbs\nfuzlzoum\nkymomn\ndbh\nfuzlzoum\nkymomn\nvap\nvlgzs\ndbh\nvlgzs\nbvy\ndbh\nkymomn\nkymomn\neoqql\nkymomn\nkymomn\nkqqcbs\nvlgzs\nkqqcbs\nkqqcbs\nfuzlzoum\nvlgzs\nrylgdoo\nvlgzs\nrylgdoo",
"output": "rylgdoo\nvlgzs\nfuzlzoum\nkqqcbs\nkymomn\neoqql\ndbh\nbvy\nvap\nj"
},
{
"input": "40\nji\nv\nv\nns\nji\nn\nji\nv\nfvy\nvje\nns\nvje\nv\nhas\nv\nusm\nhas\nfvy\nvje\nkdb\nn\nv\nji\nji\nn\nhas\nv\nji\nkdb\nr\nvje\nns\nv\nusm\nn\nvje\nhas\nns\nhas\nn",
"output": "n\nhas\nns\nvje\nusm\nv\nr\nkdb\nji\nfvy"
},
{
"input": "50\njcg\nvle\njopb\nepdb\nnkef\nfv\nxj\nufe\nfuy\noqta\ngbc\nyuz\nec\nyji\nkuux\ncwm\ntq\nnno\nhp\nzry\nxxpp\ntjvo\ngyz\nkwo\nvwqz\nyaqc\njnj\nwoav\nqcv\ndcu\ngc\nhovn\nop\nevy\ndc\ntrpu\nyb\nuzfa\npca\noq\nnhxy\nsiqu\nde\nhphy\nc\nwovu\nf\nbvv\ndsik\nlwyg",
"output": "lwyg\ndsik\nbvv\nf\nwovu\nc\nhphy\nde\nsiqu\nnhxy\noq\npca\nuzfa\nyb\ntrpu\ndc\nevy\nop\nhovn\ngc\ndcu\nqcv\nwoav\njnj\nyaqc\nvwqz\nkwo\ngyz\ntjvo\nxxpp\nzry\nhp\nnno\ntq\ncwm\nkuux\nyji\nec\nyuz\ngbc\noqta\nfuy\nufe\nxj\nfv\nnkef\nepdb\njopb\nvle\njcg"
},
{
"input": "100\nvhh\nvhh\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh",
"output": "vhh\nfa"
},
{
"input": "2\naa\nbb",
"output": "bb\naa"
},
{
"input": "2\naa\na",
"output": "a\naa"
},
{
"input": "3\naa\naa\naa",
"output": "aa"
},
{
"input": "5\naa\na\naa\na\naa",
"output": "aa\na"
},
{
"input": "7\naaaa\naaaa\naaa\na\naa\naaaaaaa\naaa",
"output": "aaa\naaaaaaa\naa\na\naaaa"
},
{
"input": "5\na\naa\naaa\naaaa\na",
"output": "a\naaaa\naaa\naa"
},
{
"input": "12\naaaaa\naaaaaa\naaaa\naaaaaa\naa\naaaa\naaaa\naaaaaa\na\naaa\naaaaaaaa\naa",
"output": "aa\naaaaaaaa\naaa\na\naaaaaa\naaaa\naaaaa"
},
{
"input": "3\na\naa\naaa",
"output": "aaa\naa\na"
},
{
"input": "9\nzzz\nzzzzz\nzzz\nzzzz\nzz\nzzzz\nzzzzz\nzzzz\nzzzzzzz",
"output": "zzzzzzz\nzzzz\nzzzzz\nzz\nzzz"
}
] | 1,543,081,505
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 95
| 857
| 27,033,600
|
import math
n = int(input())
names = {}
for i in range(0,n):
x = input()
names[x] = i
names = sorted(names.items(), key = lambda x: x[1])
ans = []
for key in names:
ans.append(key[0])
for i in range(len(ans)-1,-1,-1):
print(ans[i])
|
Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
|
```python
import math
n = int(input())
names = {}
for i in range(0,n):
x = input()
names[x] = i
names = sorted(names.items(), key = lambda x: x[1])
ans = []
for key in names:
ans.append(key[0])
for i in range(len(ans)-1,-1,-1):
print(ans[i])
```
| 3
|
|
510
|
A
|
Fox And Snake
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
|
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
|
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
|
[
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] |
[
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] |
none
| 500
|
[
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########"
},
{
"input": "3 5",
"output": "#####\n....#\n#####"
},
{
"input": "3 6",
"output": "######\n.....#\n######"
},
{
"input": "7 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "7 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####"
},
{
"input": "49 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 27",
"output": "###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n....................."
},
{
"input": "11 15",
"output": "###############\n..............#\n###############\n#..............\n###############\n..............#\n###############\n#..............\n###############\n..............#\n###############"
},
{
"input": "11 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "19 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "23 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "49 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "33 43",
"output": "###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#.................."
},
{
"input": "33 44",
"output": "############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#......."
},
{
"input": "45 45",
"output": "#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#########################################..."
},
{
"input": "45 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "45 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###"
},
{
"input": "45 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####"
},
{
"input": "5 4",
"output": "####\n...#\n####\n#...\n####"
},
{
"input": "5 5",
"output": "#####\n....#\n#####\n#....\n#####"
},
{
"input": "5 6",
"output": "######\n.....#\n######\n#.....\n######"
},
{
"input": "7 5",
"output": "#####\n....#\n#####\n#....\n#####\n....#\n#####"
},
{
"input": "7 6",
"output": "######\n.....#\n######\n#.....\n######\n.....#\n######"
},
{
"input": "7 7",
"output": "#######\n......#\n#######\n#......\n#######\n......#\n#######"
}
] | 1,686,631,789
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 307,200
|
import sys
import math
import datetime
def main():
#n,k = map(int, input().split())
#a = [int(x) for x in sys.stdin.readline().split()]
#t = int(input())
m, n = map(int, input().split())
flag = False
i=0
j=0
while i<m:
if i%2 == 0:
while(j<n):
print('#', end = ' ')
j+=1
j=0
print()
else:
if flag:
print('#', end = ' ')
while(j<n-1):
print('.', end = ' ')
j+=1
j=0
else:
j+=1
while(j<n):
print('.', end = ' ')
j+=1
print('#', end = ' ')
j=0
flag = not(flag)
print()
i += 1
main()
|
Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none
|
```python
import sys
import math
import datetime
def main():
#n,k = map(int, input().split())
#a = [int(x) for x in sys.stdin.readline().split()]
#t = int(input())
m, n = map(int, input().split())
flag = False
i=0
j=0
while i<m:
if i%2 == 0:
while(j<n):
print('#', end = ' ')
j+=1
j=0
print()
else:
if flag:
print('#', end = ' ')
while(j<n-1):
print('.', end = ' ')
j+=1
j=0
else:
j+=1
while(j<n):
print('.', end = ' ')
j+=1
print('#', end = ' ')
j=0
flag = not(flag)
print()
i += 1
main()
```
| 0
|
|
260
|
A
|
Adding Digits
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
|
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
|
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
|
[
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] |
[
"524848\n",
"121\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,674,214,010
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 5
| 93
| 2,150,400
|
a, b, n = [int(x) for x in input().split()]
if a%b==0:
print(a+('0'*n))
else:
x=0
while x<=9:
if(a*10+x)%b==0:
a=a*10+x
print(str(a)+('0'*(n-1)))
break;
x+=1
if x==10:
print(-1)
|
Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none
|
```python
a, b, n = [int(x) for x in input().split()]
if a%b==0:
print(a+('0'*n))
else:
x=0
while x<=9:
if(a*10+x)%b==0:
a=a*10+x
print(str(a)+('0'*(n-1)))
break;
x+=1
if x==10:
print(-1)
```
| -1
|
|
716
|
A
|
Crazy Computer
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
|
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
|
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
|
[
"6 5\n1 3 8 14 19 20\n",
"6 1\n1 3 5 7 9 10\n"
] |
[
"3",
"2"
] |
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.
| 500
|
[
{
"input": "6 5\n1 3 8 14 19 20",
"output": "3"
},
{
"input": "6 1\n1 3 5 7 9 10",
"output": "2"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 7 12 13 14",
"output": "4"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "2"
},
{
"input": "3 5\n1 10 20",
"output": "1"
},
{
"input": "3 10\n1 2 3",
"output": "3"
},
{
"input": "2 1\n1 100",
"output": "1"
},
{
"input": "3 1\n1 2 10",
"output": "1"
},
{
"input": "2 1\n1 2",
"output": "2"
}
] | 1,551,725,512
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 171
| 8,499,200
|
n, c = (int(i) for i in input().split())
l = [int(i) for i in input().split()]
t = 1
for i in range(1, n):
if l[i] - l[i - 1] > c:
t = 1
else:
t += 1
print(t)
|
Title: Crazy Computer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Output Specification:
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
Demo Input:
['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n']
Demo Output:
['3', '2']
Note:
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.
|
```python
n, c = (int(i) for i in input().split())
l = [int(i) for i in input().split()]
t = 1
for i in range(1, n):
if l[i] - l[i - 1] > c:
t = 1
else:
t += 1
print(t)
```
| 3
|
|
126
|
B
|
Password
|
PROGRAMMING
| 1,700
|
[
"binary search",
"dp",
"hashing",
"string suffix structures",
"strings"
] | null | null |
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
|
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
|
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
|
[
"fixprefixsuffix\n",
"abcdabc\n"
] |
[
"fix",
"Just a legend"
] |
none
| 1,000
|
[
{
"input": "fixprefixsuffix",
"output": "fix"
},
{
"input": "abcdabc",
"output": "Just a legend"
},
{
"input": "qwertyqwertyqwerty",
"output": "qwerty"
},
{
"input": "papapapap",
"output": "papap"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaa"
},
{
"input": "ghbdtn",
"output": "Just a legend"
},
{
"input": "a",
"output": "Just a legend"
},
{
"input": "aa",
"output": "Just a legend"
},
{
"input": "ab",
"output": "Just a legend"
},
{
"input": "aaa",
"output": "a"
},
{
"input": "aba",
"output": "Just a legend"
},
{
"input": "aab",
"output": "Just a legend"
},
{
"input": "abb",
"output": "Just a legend"
},
{
"input": "abc",
"output": "Just a legend"
},
{
"input": "aaabaabaaaaab",
"output": "Just a legend"
},
{
"input": "aabaaabaaaaab",
"output": "aab"
},
{
"input": "aaabaaaabab",
"output": "Just a legend"
},
{
"input": "abcabcabcabcabc",
"output": "abcabcabc"
},
{
"input": "aaaaabaaaa",
"output": "aaaa"
},
{
"input": "aaaabaaaaaaa",
"output": "aaaa"
},
{
"input": "ghghghgxghghghg",
"output": "ghghg"
},
{
"input": "kincenvizh",
"output": "Just a legend"
},
{
"input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw",
"output": "Just a legend"
},
{
"input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx",
"output": "Just a legend"
},
{
"input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb",
"output": "nfbdzgdlbjhrlvfryyjb"
},
{
"input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp",
"output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp"
},
{
"input": "abcazabca",
"output": "a"
}
] | 1,671,234,250
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
#papapapap
string = input(); lst = []; flag = False
for i in range(int(len(string) / 2) + 1):
string_ = string[0: i]
if string[0: i] == string[len(string) - i:len(string)]:
new_string = string.replace(string[0], "", 1)
new_string = new_string.replace(string[-1], "", 1)
if string_ in new_string:
lst.append(string_)
flag = True
if flag:print(lst[-1])
else:print("Just a legend")
# for i in range(len(string), 0, -1):
# dict[(string[0: i])] = 0 #добавляем в словарь ключи(возможные повторяющиеся подстроки и количество их повторения в будущем)
# #и счетчик их повторений
# for item in dict: #считаем количество повторений
# new_string = item
# #print(item, "xxxxxxxxxxxx")
# for i in range(len(string)): #
# if string[i:i + len(new_string)] == item: dict[item] += 1
# for item in dict:
# if dict[item] >= 3: answer[item] = dict[item]
# for item in answer:
# print(item)
# flag = True
# break
# if flag == False: print("Just a legend")
|
Title: Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
Input Specification:
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Output Specification:
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
Demo Input:
['fixprefixsuffix\n', 'abcdabc\n']
Demo Output:
['fix', 'Just a legend']
Note:
none
|
```python
#papapapap
string = input(); lst = []; flag = False
for i in range(int(len(string) / 2) + 1):
string_ = string[0: i]
if string[0: i] == string[len(string) - i:len(string)]:
new_string = string.replace(string[0], "", 1)
new_string = new_string.replace(string[-1], "", 1)
if string_ in new_string:
lst.append(string_)
flag = True
if flag:print(lst[-1])
else:print("Just a legend")
# for i in range(len(string), 0, -1):
# dict[(string[0: i])] = 0 #добавляем в словарь ключи(возможные повторяющиеся подстроки и количество их повторения в будущем)
# #и счетчик их повторений
# for item in dict: #считаем количество повторений
# new_string = item
# #print(item, "xxxxxxxxxxxx")
# for i in range(len(string)): #
# if string[i:i + len(new_string)] == item: dict[item] += 1
# for item in dict:
# if dict[item] >= 3: answer[item] = dict[item]
# for item in answer:
# print(item)
# flag = True
# break
# if flag == False: print("Just a legend")
```
| 0
|
|
353
|
A
|
Domino
|
PROGRAMMING
| 1,200
|
[
"implementation",
"math"
] | null | null |
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
|
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
|
[
"2\n4 2\n6 4\n",
"1\n2 3\n",
"3\n1 4\n2 3\n4 4\n"
] |
[
"0\n",
"-1\n",
"1\n"
] |
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8.
| 500
|
[
{
"input": "2\n4 2\n6 4",
"output": "0"
},
{
"input": "1\n2 3",
"output": "-1"
},
{
"input": "3\n1 4\n2 3\n4 4",
"output": "1"
},
{
"input": "5\n5 4\n5 4\n1 5\n5 5\n3 3",
"output": "1"
},
{
"input": "20\n1 3\n5 2\n5 2\n2 6\n2 4\n1 1\n1 3\n1 4\n2 6\n4 2\n5 6\n2 2\n6 2\n4 3\n2 1\n6 2\n6 5\n4 5\n2 4\n1 4",
"output": "-1"
},
{
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},
{
"input": "90\n1 4\n3 5\n4 2\n2 5\n4 3\n2 6\n2 6\n3 2\n4 4\n6 1\n4 3\n2 3\n5 3\n6 6\n2 2\n6 3\n4 1\n4 4\n5 6\n6 4\n4 2\n5 6\n4 6\n4 4\n6 4\n4 1\n5 3\n3 2\n4 4\n5 2\n5 4\n6 4\n1 2\n3 3\n3 4\n6 4\n1 6\n4 2\n3 2\n1 1\n2 2\n5 1\n6 6\n4 1\n5 2\n3 6\n2 1\n2 2\n4 6\n6 5\n4 4\n5 5\n5 6\n1 6\n1 4\n5 6\n3 6\n6 3\n5 6\n6 5\n5 1\n6 1\n6 6\n6 3\n1 5\n4 5\n3 1\n6 6\n3 4\n6 2\n1 4\n2 2\n3 2\n5 6\n2 4\n1 4\n6 3\n4 6\n1 4\n5 2\n1 2\n6 5\n1 5\n1 4\n4 2\n2 5\n3 2\n5 1\n5 4\n5 3",
"output": "-1"
},
{
"input": "95\n4 3\n3 2\n5 5\n5 3\n1 6\n4 4\n5 5\n6 5\n3 5\n1 5\n4 2\n5 1\n1 2\n2 3\n6 4\n2 3\n6 3\n6 5\n5 6\n1 4\n2 6\n2 6\n2 5\n2 1\n3 1\n3 5\n2 2\n6 1\n2 4\n4 6\n6 6\n6 4\n3 2\n5 1\n4 3\n6 5\n2 3\n4 1\n2 5\n6 5\n6 5\n6 5\n5 1\n5 4\n4 6\n3 2\n2 5\n2 6\n4 6\n6 3\n6 4\n5 6\n4 6\n2 4\n3 4\n1 4\n2 4\n2 3\n5 6\n6 4\n3 1\n5 1\n3 6\n3 5\n2 6\n6 3\n4 3\n3 1\n6 1\n2 2\n6 3\n2 2\n2 2\n6 4\n6 1\n2 1\n5 6\n5 4\n5 2\n3 4\n3 6\n2 1\n1 6\n5 5\n2 6\n2 3\n3 6\n1 3\n1 5\n5 1\n1 2\n2 2\n5 3\n6 4\n4 5",
"output": "0"
},
{
"input": "95\n4 5\n5 6\n3 2\n5 1\n4 3\n4 1\n6 1\n5 2\n2 4\n5 3\n2 3\n6 4\n4 1\n1 6\n2 6\n2 3\n4 6\n2 4\n3 4\n4 2\n5 5\n1 1\n1 5\n4 3\n4 5\n6 2\n6 1\n6 3\n5 5\n4 1\n5 1\n2 3\n5 1\n3 6\n6 6\n4 5\n4 4\n4 3\n1 6\n6 6\n4 6\n6 4\n1 2\n6 2\n4 6\n6 6\n5 5\n6 1\n5 2\n4 5\n6 6\n6 5\n4 4\n1 5\n4 6\n4 1\n3 6\n5 1\n3 1\n4 6\n4 5\n1 3\n5 4\n4 5\n2 2\n6 1\n5 2\n6 5\n2 2\n1 1\n6 3\n6 1\n2 6\n3 3\n2 1\n4 6\n2 4\n5 5\n5 2\n3 2\n1 2\n6 6\n6 2\n5 1\n2 6\n5 2\n2 2\n5 5\n3 5\n3 3\n2 6\n5 3\n4 3\n1 6\n5 4",
"output": "-1"
},
{
"input": "100\n1 1\n3 5\n2 1\n1 2\n3 4\n5 6\n5 6\n6 1\n5 5\n2 4\n5 5\n5 6\n6 2\n6 6\n2 6\n1 4\n2 2\n3 2\n1 3\n5 5\n6 3\n5 6\n1 1\n1 2\n1 2\n2 1\n2 3\n1 6\n4 3\n1 1\n2 5\n2 4\n4 4\n1 5\n3 3\n6 1\n3 5\n1 1\n3 6\n3 1\n4 2\n4 3\n3 6\n6 6\n1 6\n6 2\n2 5\n5 4\n6 3\n1 4\n2 6\n6 2\n3 4\n6 1\n6 5\n4 6\n6 5\n4 4\n3 1\n6 3\n5 1\n2 4\n5 1\n1 2\n2 4\n2 1\n6 6\n5 3\n4 6\n6 3\n5 5\n3 3\n1 1\n6 5\n4 3\n2 6\n1 5\n3 5\n2 4\n4 5\n1 6\n2 3\n6 3\n5 5\n2 6\n2 6\n3 4\n3 2\n6 1\n3 4\n6 4\n3 3\n2 3\n5 1\n3 1\n6 2\n2 3\n6 4\n1 4\n1 2",
"output": "-1"
},
{
"input": "100\n1 1\n5 5\n1 2\n5 3\n5 5\n2 2\n1 5\n3 4\n3 2\n1 3\n5 6\n4 5\n2 1\n5 5\n2 2\n1 6\n6 1\n5 1\n4 1\n4 6\n3 5\n6 1\n2 3\n5 6\n3 6\n2 3\n5 6\n1 6\n3 2\n2 2\n3 3\n6 5\n5 5\n1 4\n5 6\n6 4\n1 4\n1 2\n2 6\n3 2\n6 4\n5 3\n3 3\n6 4\n4 6\n2 2\n5 6\n5 1\n1 2\n3 4\n4 5\n1 1\n3 4\n5 2\n4 5\n3 3\n1 1\n3 4\n1 6\n2 4\n1 3\n3 2\n6 5\n1 6\n3 6\n2 3\n2 6\n5 1\n5 5\n5 6\n4 1\n6 2\n3 6\n5 3\n2 2\n2 4\n6 6\n3 6\n4 6\n2 5\n5 3\n1 2\n3 4\n3 4\n6 2\n2 4\n2 2\n4 6\n3 5\n4 2\n5 6\n4 2\n2 3\n6 2\n5 6\n2 1\n3 3\n6 6\n4 3\n4 2",
"output": "1"
},
{
"input": "1\n2 2",
"output": "0"
},
{
"input": "3\n2 4\n6 6\n3 3",
"output": "-1"
},
{
"input": "2\n3 6\n4 1",
"output": "1"
},
{
"input": "3\n1 1\n1 1\n3 3",
"output": "-1"
},
{
"input": "3\n2 3\n1 1\n2 3",
"output": "1"
},
{
"input": "3\n2 2\n2 1\n1 2",
"output": "1"
},
{
"input": "3\n1 1\n1 1\n1 1",
"output": "-1"
}
] | 1,626,321,143
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 59
| 216
| 20,172,800
|
c=0;n=int(input());o1,o2=0,0;l,w=[],[]
for i in range(n):
x,y=map(int,input().split())
l.append(x);w.append(y)
if x%2:o1+=1
if y%2:o2+=1
if o1%2 and o2%2:
for i in range(n):
if l[i]%2==0 and w[i]%2 :print(1);break
elif l[i]%2 and w[i]%2==0:print(1);break
else:print(-1)
elif o1%2==0 and o2%2==0:print(0)
else:print(-1)
|
Title: Domino
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
Output Specification:
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
Demo Input:
['2\n4 2\n6 4\n', '1\n2 3\n', '3\n1 4\n2 3\n4 4\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8.
|
```python
c=0;n=int(input());o1,o2=0,0;l,w=[],[]
for i in range(n):
x,y=map(int,input().split())
l.append(x);w.append(y)
if x%2:o1+=1
if y%2:o2+=1
if o1%2 and o2%2:
for i in range(n):
if l[i]%2==0 and w[i]%2 :print(1);break
elif l[i]%2 and w[i]%2==0:print(1);break
else:print(-1)
elif o1%2==0 and o2%2==0:print(0)
else:print(-1)
```
| 3
|
|
985
|
D
|
Sand Fortress
|
PROGRAMMING
| 2,100
|
[
"binary search",
"constructive algorithms",
"math"
] | null | null |
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought *n* packs of sand with you. Let height *h**i* of the sand pillar on some spot *i* be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with *H* sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
- *h*1<=≤<=*H*: no sand from the leftmost spot should go over the fence; - For any |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen; - : you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
|
The only line contains two integer numbers *n* and *H* (1<=≤<=*n*,<=*H*<=≤<=1018) — the number of sand packs you have and the height of the fence, respectively.
|
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
|
[
"5 2\n",
"6 8\n"
] |
[
"3\n",
"3\n"
] |
Here are the heights of some valid castles:
- *n* = 5, *H* = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...] - *n* = 6, *H* = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
- *n* = 5, *H* = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...] - *n* = 6, *H* = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
| 0
|
[
{
"input": "5 2",
"output": "3"
},
{
"input": "6 8",
"output": "3"
},
{
"input": "20 4",
"output": "7"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1414213562"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 1000000000000000000",
"output": "1"
},
{
"input": "1000000000000000000 1",
"output": "1999999999"
},
{
"input": "1036191544337895 45523434",
"output": "45523434"
},
{
"input": "1036191544337896 45523434",
"output": "45523435"
},
{
"input": "1036191544337895 45523433",
"output": "45523435"
},
{
"input": "1036191544337895 1",
"output": "64379858"
},
{
"input": "1036191544337895 1000000000000000000",
"output": "45523434"
},
{
"input": "30 3",
"output": "9"
},
{
"input": "30 4",
"output": "8"
},
{
"input": "6 100",
"output": "3"
},
{
"input": "7 100",
"output": "4"
},
{
"input": "1000000000000000000 99999999999",
"output": "1414213562"
},
{
"input": "999999997351043581 1000000000000000000",
"output": "1414213561"
},
{
"input": "911343366122896086 1416605974",
"output": "1350069158"
},
{
"input": "828974163639871882 2010864527",
"output": "1287613423"
},
{
"input": "696616491401388220 958775125",
"output": "1191798158"
},
{
"input": "999999999000000000 1",
"output": "1999999998"
},
{
"input": "961245465290770608 1687994843",
"output": "1386539192"
},
{
"input": "692106376966414549 974053139",
"output": "1186035874"
},
{
"input": "12 1",
"output": "6"
},
{
"input": "806680349368385877 1068656310",
"output": "1278847474"
}
] | 1,626,850,210
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 6
| 93
| 20,172,800
|
import math
n, h = map(int, input().split())
x = int((math.sqrt(1+n*8)-1)/2)
if (x <= h):
if (x * (x + 1) / 2 == n):
print(x)
else:
print(x+1)
else:
y = int(math.sqrt((2 * n + h * h - h) / 2.0))
sum = (h + x) * (x - h + 1) / 2+x*(x-1)/2
if (sum == n):
print(y+y-h)
else:
print(y+y-h+1)
|
Title: Sand Fortress
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought *n* packs of sand with you. Let height *h**i* of the sand pillar on some spot *i* be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with *H* sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
- *h*1<=≤<=*H*: no sand from the leftmost spot should go over the fence; - For any |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen; - : you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input Specification:
The only line contains two integer numbers *n* and *H* (1<=≤<=*n*,<=*H*<=≤<=1018) — the number of sand packs you have and the height of the fence, respectively.
Output Specification:
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Demo Input:
['5 2\n', '6 8\n']
Demo Output:
['3\n', '3\n']
Note:
Here are the heights of some valid castles:
- *n* = 5, *H* = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...] - *n* = 6, *H* = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
- *n* = 5, *H* = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...] - *n* = 6, *H* = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
|
```python
import math
n, h = map(int, input().split())
x = int((math.sqrt(1+n*8)-1)/2)
if (x <= h):
if (x * (x + 1) / 2 == n):
print(x)
else:
print(x+1)
else:
y = int(math.sqrt((2 * n + h * h - h) / 2.0))
sum = (h + x) * (x - h + 1) / 2+x*(x-1)/2
if (sum == n):
print(y+y-h)
else:
print(y+y-h+1)
```
| 0
|
|
534
|
A
|
Exam
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
An exam for *n* students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (*i* and *i*<=+<=1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.
Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.
|
A single line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of students at an exam.
|
In the first line print integer *k* — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.
In the second line print *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is the number of the student on the *i*-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |*a**i*<=-<=*a**i*<=+<=1|<=≠<=1 for all *i* from 1 to *k*<=-<=1.
If there are several possible answers, output any of them.
|
[
"6",
"3\n"
] |
[
"6\n1 5 3 6 2 4",
"2\n1 3"
] |
none
| 500
|
[
{
"input": "6",
"output": "6\n5 3 1 6 4 2 "
},
{
"input": "3",
"output": "2\n1 3"
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "4",
"output": "4\n3 1 4 2 "
},
{
"input": "5",
"output": "5\n5 3 1 4 2 "
},
{
"input": "7",
"output": "7\n7 5 3 1 6 4 2 "
},
{
"input": "8",
"output": "8\n7 5 3 1 8 6 4 2 "
},
{
"input": "9",
"output": "9\n9 7 5 3 1 8 6 4 2 "
},
{
"input": "10",
"output": "10\n9 7 5 3 1 10 8 6 4 2 "
},
{
"input": "13",
"output": "13\n13 11 9 7 5 3 1 12 10 8 6 4 2 "
},
{
"input": "16",
"output": "16\n15 13 11 9 7 5 3 1 16 14 12 10 8 6 4 2 "
},
{
"input": "25",
"output": "25\n25 23 21 19 17 15 13 11 9 7 5 3 1 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "29",
"output": "29\n29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "120",
"output": "120\n119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 120 118 116 114 112 110 108 106 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "128",
"output": "128\n127 125 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 128 126 124 122 120 118 116 114 112 110 108 106 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "216",
"output": "216\n215 213 211 209 207 205 203 201 199 197 195 193 191 189 187 185 183 181 179 177 175 173 171 169 167 165 163 161 159 157 155 153 151 149 147 145 143 141 139 137 135 133 131 129 127 125 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 216 214 212 210 208 206 204 202 200 198 196 194 192 190 188 186 184 182 180 178 176 174 172 170 168 166 164 162 160 158 156 154 1..."
},
{
"input": "729",
"output": "729\n729 727 725 723 721 719 717 715 713 711 709 707 705 703 701 699 697 695 693 691 689 687 685 683 681 679 677 675 673 671 669 667 665 663 661 659 657 655 653 651 649 647 645 643 641 639 637 635 633 631 629 627 625 623 621 619 617 615 613 611 609 607 605 603 601 599 597 595 593 591 589 587 585 583 581 579 577 575 573 571 569 567 565 563 561 559 557 555 553 551 549 547 545 543 541 539 537 535 533 531 529 527 525 523 521 519 517 515 513 511 509 507 505 503 501 499 497 495 493 491 489 487 485 483 481 479 47..."
},
{
"input": "1111",
"output": "1111\n1111 1109 1107 1105 1103 1101 1099 1097 1095 1093 1091 1089 1087 1085 1083 1081 1079 1077 1075 1073 1071 1069 1067 1065 1063 1061 1059 1057 1055 1053 1051 1049 1047 1045 1043 1041 1039 1037 1035 1033 1031 1029 1027 1025 1023 1021 1019 1017 1015 1013 1011 1009 1007 1005 1003 1001 999 997 995 993 991 989 987 985 983 981 979 977 975 973 971 969 967 965 963 961 959 957 955 953 951 949 947 945 943 941 939 937 935 933 931 929 927 925 923 921 919 917 915 913 911 909 907 905 903 901 899 897 895 893 891 889 8..."
},
{
"input": "1597",
"output": "1597\n1597 1595 1593 1591 1589 1587 1585 1583 1581 1579 1577 1575 1573 1571 1569 1567 1565 1563 1561 1559 1557 1555 1553 1551 1549 1547 1545 1543 1541 1539 1537 1535 1533 1531 1529 1527 1525 1523 1521 1519 1517 1515 1513 1511 1509 1507 1505 1503 1501 1499 1497 1495 1493 1491 1489 1487 1485 1483 1481 1479 1477 1475 1473 1471 1469 1467 1465 1463 1461 1459 1457 1455 1453 1451 1449 1447 1445 1443 1441 1439 1437 1435 1433 1431 1429 1427 1425 1423 1421 1419 1417 1415 1413 1411 1409 1407 1405 1403 1401 1399 1397 ..."
},
{
"input": "1777",
"output": "1777\n1777 1775 1773 1771 1769 1767 1765 1763 1761 1759 1757 1755 1753 1751 1749 1747 1745 1743 1741 1739 1737 1735 1733 1731 1729 1727 1725 1723 1721 1719 1717 1715 1713 1711 1709 1707 1705 1703 1701 1699 1697 1695 1693 1691 1689 1687 1685 1683 1681 1679 1677 1675 1673 1671 1669 1667 1665 1663 1661 1659 1657 1655 1653 1651 1649 1647 1645 1643 1641 1639 1637 1635 1633 1631 1629 1627 1625 1623 1621 1619 1617 1615 1613 1611 1609 1607 1605 1603 1601 1599 1597 1595 1593 1591 1589 1587 1585 1583 1581 1579 1577 ..."
},
{
"input": "2048",
"output": "2048\n2047 2045 2043 2041 2039 2037 2035 2033 2031 2029 2027 2025 2023 2021 2019 2017 2015 2013 2011 2009 2007 2005 2003 2001 1999 1997 1995 1993 1991 1989 1987 1985 1983 1981 1979 1977 1975 1973 1971 1969 1967 1965 1963 1961 1959 1957 1955 1953 1951 1949 1947 1945 1943 1941 1939 1937 1935 1933 1931 1929 1927 1925 1923 1921 1919 1917 1915 1913 1911 1909 1907 1905 1903 1901 1899 1897 1895 1893 1891 1889 1887 1885 1883 1881 1879 1877 1875 1873 1871 1869 1867 1865 1863 1861 1859 1857 1855 1853 1851 1849 1847 ..."
},
{
"input": "2999",
"output": "2999\n2999 2997 2995 2993 2991 2989 2987 2985 2983 2981 2979 2977 2975 2973 2971 2969 2967 2965 2963 2961 2959 2957 2955 2953 2951 2949 2947 2945 2943 2941 2939 2937 2935 2933 2931 2929 2927 2925 2923 2921 2919 2917 2915 2913 2911 2909 2907 2905 2903 2901 2899 2897 2895 2893 2891 2889 2887 2885 2883 2881 2879 2877 2875 2873 2871 2869 2867 2865 2863 2861 2859 2857 2855 2853 2851 2849 2847 2845 2843 2841 2839 2837 2835 2833 2831 2829 2827 2825 2823 2821 2819 2817 2815 2813 2811 2809 2807 2805 2803 2801 2799 ..."
},
{
"input": "3001",
"output": "3001\n3001 2999 2997 2995 2993 2991 2989 2987 2985 2983 2981 2979 2977 2975 2973 2971 2969 2967 2965 2963 2961 2959 2957 2955 2953 2951 2949 2947 2945 2943 2941 2939 2937 2935 2933 2931 2929 2927 2925 2923 2921 2919 2917 2915 2913 2911 2909 2907 2905 2903 2901 2899 2897 2895 2893 2891 2889 2887 2885 2883 2881 2879 2877 2875 2873 2871 2869 2867 2865 2863 2861 2859 2857 2855 2853 2851 2849 2847 2845 2843 2841 2839 2837 2835 2833 2831 2829 2827 2825 2823 2821 2819 2817 2815 2813 2811 2809 2807 2805 2803 2801 ..."
},
{
"input": "4181",
"output": "4181\n4181 4179 4177 4175 4173 4171 4169 4167 4165 4163 4161 4159 4157 4155 4153 4151 4149 4147 4145 4143 4141 4139 4137 4135 4133 4131 4129 4127 4125 4123 4121 4119 4117 4115 4113 4111 4109 4107 4105 4103 4101 4099 4097 4095 4093 4091 4089 4087 4085 4083 4081 4079 4077 4075 4073 4071 4069 4067 4065 4063 4061 4059 4057 4055 4053 4051 4049 4047 4045 4043 4041 4039 4037 4035 4033 4031 4029 4027 4025 4023 4021 4019 4017 4015 4013 4011 4009 4007 4005 4003 4001 3999 3997 3995 3993 3991 3989 3987 3985 3983 3981 ..."
},
{
"input": "4990",
"output": "4990\n4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 4789 ..."
},
{
"input": "4991",
"output": "4991\n4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 ..."
},
{
"input": "4992",
"output": "4992\n4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 ..."
},
{
"input": "4993",
"output": "4993\n4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 ..."
},
{
"input": "4994",
"output": "4994\n4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 ..."
},
{
"input": "4995",
"output": "4995\n4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 ..."
},
{
"input": "4996",
"output": "4996\n4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 ..."
},
{
"input": "4997",
"output": "4997\n4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 ..."
},
{
"input": "4998",
"output": "4998\n4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 ..."
},
{
"input": "4999",
"output": "4999\n4999 4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 ..."
},
{
"input": "5000",
"output": "5000\n4999 4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 ..."
}
] | 1,497,273,271
| 871
|
Python 3
|
OK
|
TESTS
| 36
| 62
| 409,600
|
n = int(input())
b = []
for i in range(2, n + 1, 2):
b.append(i)
for i in range(1, n + 1, 2):
b.append(i)
for i in range(n - 1):
if abs(b[i] - b[i + 1]) == 1:
b.pop(i)
break
print("%d\n" % (len(b)) + " ".join([str(x) for x in b]))
|
Title: Exam
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An exam for *n* students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (*i* and *i*<=+<=1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.
Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.
Input Specification:
A single line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of students at an exam.
Output Specification:
In the first line print integer *k* — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.
In the second line print *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is the number of the student on the *i*-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |*a**i*<=-<=*a**i*<=+<=1|<=≠<=1 for all *i* from 1 to *k*<=-<=1.
If there are several possible answers, output any of them.
Demo Input:
['6', '3\n']
Demo Output:
['6\n1 5 3 6 2 4', '2\n1 3']
Note:
none
|
```python
n = int(input())
b = []
for i in range(2, n + 1, 2):
b.append(i)
for i in range(1, n + 1, 2):
b.append(i)
for i in range(n - 1):
if abs(b[i] - b[i + 1]) == 1:
b.pop(i)
break
print("%d\n" % (len(b)) + " ".join([str(x) for x in b]))
```
| 3
|
|
365
|
A
|
Good Number
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
|
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the number of *k*-good numbers in *a*.
|
[
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] |
[
"10\n",
"1\n"
] |
none
| 500
|
[
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,670,257,950
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,150,400
|
def Alldidgits(s):
for i in '0123456789':
if s[i] not in s:
return False
return True
def notOverKCorner(x,k):
if '0' in str(x):
return True
return False
def NotOverK(x, k):
for i in str(x):
if int(i) <= k:
return True
return False
def main():
#inputs
lst = input().split(' ')
n = lst[0]
k = lst[1]
X= []
counter = 0
for i in range(0, n):
X.append(int(input()))
#processing
for i in X:
if notOverKCorner(i,k):
counter += 1
if Alldidgits(i) and NotOverK(i, k):
counter += 1
print(counter)
main()
|
Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none
|
```python
def Alldidgits(s):
for i in '0123456789':
if s[i] not in s:
return False
return True
def notOverKCorner(x,k):
if '0' in str(x):
return True
return False
def NotOverK(x, k):
for i in str(x):
if int(i) <= k:
return True
return False
def main():
#inputs
lst = input().split(' ')
n = lst[0]
k = lst[1]
X= []
counter = 0
for i in range(0, n):
X.append(int(input()))
#processing
for i in X:
if notOverKCorner(i,k):
counter += 1
if Alldidgits(i) and NotOverK(i, k):
counter += 1
print(counter)
main()
```
| -1
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,677,773,660
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
t=int(input('how many lines?'))
print(t)
s=0
while s<=t:
w = str(input('your line:'))
c= 0
if len(w) <=10:
print(w)
else:
for i in w:
c+=1
c-=2
print(w[0]+str(c)+w[-1])
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
t=int(input('how many lines?'))
print(t)
s=0
while s<=t:
w = str(input('your line:'))
c= 0
if len(w) <=10:
print(w)
else:
for i in w:
c+=1
c-=2
print(w[0]+str(c)+w[-1])
```
| -1
|
58
|
C
|
Trees
|
PROGRAMMING
| 1,800
|
[
"brute force"
] |
C. Trees
|
2
|
256
|
On Bertown's main street *n* trees are growing, the tree number *i* has the height of *a**i* meters (1<=≤<=*i*<=≤<=*n*). By the arrival of the President of Berland these trees were decided to be changed so that their heights formed a beautiful sequence. This means that the heights of trees on ends (the 1st one and the *n*-th one) should be equal to each other, the heights of the 2-nd and the (*n*<=-<=1)-th tree must also be equal to each other, at that the height of the 2-nd tree should be larger than the height of the first tree by 1, and so on. In other words, the heights of the trees, standing at equal distance from the edge (of one end of the sequence) must be equal to each other, and with the increasing of the distance from the edge by 1 the tree height must also increase by 1. For example, the sequences "2 3 4 5 5 4 3 2" and "1 2 3 2 1" are beautiful, and '1 3 3 1" and "1 2 3 1" are not.
Changing the height of a tree is a very expensive operation, using advanced technologies invented by Berland scientists. In one operation you can choose any tree and change its height to any number, either increase or decrease. Note that even after the change the height should remain a positive integer, i. e, it can't be less than or equal to zero. Identify the smallest number of changes of the trees' height needed for the sequence of their heights to become beautiful.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) which is the number of trees. The second line contains integers *a**i* (1<=≤<=*a**i*<=≤<=105) which are the heights of the trees.
|
Print a single number which is the minimal number of trees whose heights will have to be changed for the sequence to become beautiful.
|
[
"3\n2 2 2\n",
"4\n1 2 2 1\n"
] |
[
"1\n",
"0\n"
] |
none
| 1,500
|
[
{
"input": "3\n2 2 2",
"output": "1"
},
{
"input": "4\n1 2 2 1",
"output": "0"
},
{
"input": "3\n61452 50974 73849",
"output": "2"
},
{
"input": "4\n86002 1199 86003 86002",
"output": "1"
},
{
"input": "5\n92605 92606 41969 98774 92605",
"output": "2"
},
{
"input": "10\n1 1 2 3 4 4 3 2 1 10",
"output": "9"
},
{
"input": "10\n73905 73906 73907 85732 73909 73909 73908 73907 73906 73905",
"output": "1"
},
{
"input": "10\n60718 99414 65042 65043 65044 38495 95782 65042 65041 65040",
"output": "4"
},
{
"input": "7\n1 2 3 7 6 5 4",
"output": "3"
},
{
"input": "5\n4 5 6 5 1",
"output": "1"
},
{
"input": "1\n100000",
"output": "0"
},
{
"input": "5\n1 1 1 1 1",
"output": "3"
},
{
"input": "7\n1 2 3 7 6 5 4",
"output": "3"
},
{
"input": "1\n2727",
"output": "0"
},
{
"input": "5\n1 2 3 2 4",
"output": "1"
},
{
"input": "9\n100 12 13 14 15 14 13 12 11",
"output": "1"
},
{
"input": "5\n2 4 5 4 2",
"output": "2"
},
{
"input": "5\n3 3 4 3 2",
"output": "1"
}
] | 1,602,005,793
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 374
| 7,680,000
|
n=int(input())
a= list(map(int, input().split()))
x=[n]*100001
for i in range(n):
x[a[i]-min(n-i-1,i)]-=1
print (min(x[1:]))
|
Title: Trees
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
On Bertown's main street *n* trees are growing, the tree number *i* has the height of *a**i* meters (1<=≤<=*i*<=≤<=*n*). By the arrival of the President of Berland these trees were decided to be changed so that their heights formed a beautiful sequence. This means that the heights of trees on ends (the 1st one and the *n*-th one) should be equal to each other, the heights of the 2-nd and the (*n*<=-<=1)-th tree must also be equal to each other, at that the height of the 2-nd tree should be larger than the height of the first tree by 1, and so on. In other words, the heights of the trees, standing at equal distance from the edge (of one end of the sequence) must be equal to each other, and with the increasing of the distance from the edge by 1 the tree height must also increase by 1. For example, the sequences "2 3 4 5 5 4 3 2" and "1 2 3 2 1" are beautiful, and '1 3 3 1" and "1 2 3 1" are not.
Changing the height of a tree is a very expensive operation, using advanced technologies invented by Berland scientists. In one operation you can choose any tree and change its height to any number, either increase or decrease. Note that even after the change the height should remain a positive integer, i. e, it can't be less than or equal to zero. Identify the smallest number of changes of the trees' height needed for the sequence of their heights to become beautiful.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) which is the number of trees. The second line contains integers *a**i* (1<=≤<=*a**i*<=≤<=105) which are the heights of the trees.
Output Specification:
Print a single number which is the minimal number of trees whose heights will have to be changed for the sequence to become beautiful.
Demo Input:
['3\n2 2 2\n', '4\n1 2 2 1\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
n=int(input())
a= list(map(int, input().split()))
x=[n]*100001
for i in range(n):
x[a[i]-min(n-i-1,i)]-=1
print (min(x[1:]))
```
| 3.892195
|
560
|
A
|
Currency System in Geraldion
|
PROGRAMMING
| 1,000
|
[
"implementation",
"sortings"
] | null | null |
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
|
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
|
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
|
[
"5\n1 2 3 4 5\n"
] |
[
"-1\n"
] |
none
| 500
|
[
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"output": "1"
},
{
"input": "50\n110876 835020 859879 999908 712969 788264 287153 921820 330355 499311 209594 484829 296329 940051 174081 931503 1 780512 390075 97866 124255 950067 697612 244256 782385 789882 37608 82153 399889 598867 416717 377988 535636 511221 792568 683271 131077 290194 496712 330720 587436 563481 645817 942562 654093 980561 382937 48293 582608 116156",
"output": "-1"
},
{
"input": "50\n474421 421097 217233 156339 27075 733996 281778 863492 184707 956857 288561 70997 393786 337382 663642 131184 637 273801 799870 295017 392338 842567 161819 297705 102013 930684 375703 838048 154915 138503 629056 256591 893619 19263 787927 684541 320265 841090 421423 490879 394582 493952 619247 633202 612928 50907 276653 407819 489945 153173",
"output": "1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "1\n1000000",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
}
] | 1,599,266,363
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 2,048,000
|
import sys
input = sys.stdin.readline
b= input()
p1= input()
p2= input()
result=[[(lambda :True if p1[i]+p2[j]<=b[0] else False)() for i in range(2)] for j in range(2) for k in range(2)].count([True,True])>1
print(result)
|
Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
b= input()
p1= input()
p2= input()
result=[[(lambda :True if p1[i]+p2[j]<=b[0] else False)() for i in range(2)] for j in range(2) for k in range(2)].count([True,True])>1
print(result)
```
| -1
|
|
962
|
C
|
Make a Square
|
PROGRAMMING
| 1,400
|
[
"brute force",
"implementation",
"math"
] | null | null |
You are given a positive integer $n$, written without leading zeroes (for example, the number 04 is incorrect).
In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.
Determine the minimum number of operations that you need to consistently apply to the given integer $n$ to make from it the square of some positive integer or report that it is impossible.
An integer $x$ is the square of some positive integer if and only if $x=y^2$ for some positive integer $y$.
|
The first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^{9}$). The number is given without leading zeroes.
|
If it is impossible to make the square of some positive integer from $n$, print -1. In the other case, print the minimal number of operations required to do it.
|
[
"8314\n",
"625\n",
"333\n"
] |
[
"2\n",
"0\n",
"-1\n"
] |
In the first example we should delete from $8314$ the digits $3$ and $4$. After that $8314$ become equals to $81$, which is the square of the integer $9$.
In the second example the given $625$ is the square of the integer $25$, so you should not delete anything.
In the third example it is impossible to make the square from $333$, so the answer is -1.
| 0
|
[
{
"input": "8314",
"output": "2"
},
{
"input": "625",
"output": "0"
},
{
"input": "333",
"output": "-1"
},
{
"input": "1881388645",
"output": "6"
},
{
"input": "1059472069",
"output": "3"
},
{
"input": "1354124829",
"output": "4"
},
{
"input": "149723943",
"output": "4"
},
{
"input": "101",
"output": "2"
},
{
"input": "1999967841",
"output": "0"
},
{
"input": "2000000000",
"output": "-1"
},
{
"input": "1999431225",
"output": "0"
},
{
"input": "30",
"output": "-1"
},
{
"input": "1000",
"output": "1"
},
{
"input": "3081",
"output": "2"
},
{
"input": "10",
"output": "1"
},
{
"input": "2003064",
"output": "3"
},
{
"input": "701",
"output": "2"
},
{
"input": "1234567891",
"output": "4"
},
{
"input": "10625",
"output": "2"
},
{
"input": "13579",
"output": "4"
},
{
"input": "1999999999",
"output": "9"
},
{
"input": "150000",
"output": "1"
},
{
"input": "8010902",
"output": "3"
},
{
"input": "20100",
"output": "2"
},
{
"input": "40404",
"output": "2"
},
{
"input": "70000729",
"output": "5"
},
{
"input": "1899933124",
"output": "5"
},
{
"input": "1999999081",
"output": "8"
},
{
"input": "326700",
"output": "2"
},
{
"input": "1",
"output": "0"
},
{
"input": "1000000990",
"output": "3"
},
{
"input": "10000",
"output": "0"
},
{
"input": "100001",
"output": "1"
},
{
"input": "1410065408",
"output": "7"
},
{
"input": "1409865409",
"output": "5"
},
{
"input": "1000050001",
"output": "3"
},
{
"input": "1044435556",
"output": "2"
},
{
"input": "520993450",
"output": "6"
},
{
"input": "131073",
"output": "5"
},
{
"input": "500040004",
"output": "6"
},
{
"input": "237555493",
"output": "7"
},
{
"input": "1120671621",
"output": "5"
},
{
"input": "298755045",
"output": "5"
},
{
"input": "1476838469",
"output": "5"
},
{
"input": "654921893",
"output": "4"
},
{
"input": "1538038021",
"output": "4"
},
{
"input": "716121445",
"output": "6"
},
{
"input": "1894204869",
"output": "5"
},
{
"input": "1800098866",
"output": "7"
},
{
"input": "890665277",
"output": "8"
},
{
"input": "1686264392",
"output": "6"
},
{
"input": "1336639314",
"output": "6"
},
{
"input": "132238429",
"output": "5"
},
{
"input": "927837544",
"output": "4"
},
{
"input": "18403955",
"output": "4"
},
{
"input": "1668778878",
"output": "8"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "-1"
},
{
"input": "4",
"output": "0"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "-1"
},
{
"input": "7",
"output": "-1"
},
{
"input": "8",
"output": "-1"
},
{
"input": "9",
"output": "0"
},
{
"input": "11",
"output": "1"
},
{
"input": "12",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "14",
"output": "1"
},
{
"input": "15",
"output": "1"
},
{
"input": "16",
"output": "0"
}
] | 1,523,564,663
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 70
| 109
| 7,065,600
|
import itertools
import math
n=input()
res=[seq for i in range(len(n),0,-1) for seq in itertools.combinations(n,i) if math.sqrt(int(''.join(seq))).is_integer()]
if res==[]:
print(-1)
else:
flag=0
for i in range(len(res)):
if res[i][0]!='0':
flag=1
print(len(n)-len(res[i]))
break
if(flag==0):
print(-1)
|
Title: Make a Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a positive integer $n$, written without leading zeroes (for example, the number 04 is incorrect).
In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.
Determine the minimum number of operations that you need to consistently apply to the given integer $n$ to make from it the square of some positive integer or report that it is impossible.
An integer $x$ is the square of some positive integer if and only if $x=y^2$ for some positive integer $y$.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^{9}$). The number is given without leading zeroes.
Output Specification:
If it is impossible to make the square of some positive integer from $n$, print -1. In the other case, print the minimal number of operations required to do it.
Demo Input:
['8314\n', '625\n', '333\n']
Demo Output:
['2\n', '0\n', '-1\n']
Note:
In the first example we should delete from $8314$ the digits $3$ and $4$. After that $8314$ become equals to $81$, which is the square of the integer $9$.
In the second example the given $625$ is the square of the integer $25$, so you should not delete anything.
In the third example it is impossible to make the square from $333$, so the answer is -1.
|
```python
import itertools
import math
n=input()
res=[seq for i in range(len(n),0,-1) for seq in itertools.combinations(n,i) if math.sqrt(int(''.join(seq))).is_integer()]
if res==[]:
print(-1)
else:
flag=0
for i in range(len(res)):
if res[i][0]!='0':
flag=1
print(len(n)-len(res[i]))
break
if(flag==0):
print(-1)
```
| 3
|
|
558
|
A
|
Lala Land and Apple Trees
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Amr lives in Lala Land. Lala Land is a very beautiful country that is located on a coordinate line. Lala Land is famous with its apple trees growing everywhere.
Lala Land has exactly *n* apple trees. Tree number *i* is located in a position *x**i* and has *a**i* apples growing on it. Amr wants to collect apples from the apple trees. Amr currently stands in *x*<==<=0 position. At the beginning, he can choose whether to go right or left. He'll continue in his direction until he meets an apple tree he didn't visit before. He'll take all of its apples and then reverse his direction, continue walking in this direction until he meets another apple tree he didn't visit before and so on. In the other words, Amr reverses his direction when visiting each new apple tree. Amr will stop collecting apples when there are no more trees he didn't visit in the direction he is facing.
What is the maximum number of apples he can collect?
|
The first line contains one number *n* (1<=≤<=*n*<=≤<=100), the number of apple trees in Lala Land.
The following *n* lines contains two integers each *x**i*, *a**i* (<=-<=105<=≤<=*x**i*<=≤<=105, *x**i*<=≠<=0, 1<=≤<=*a**i*<=≤<=105), representing the position of the *i*-th tree and number of apples on it.
It's guaranteed that there is at most one apple tree at each coordinate. It's guaranteed that no tree grows in point 0.
|
Output the maximum number of apples Amr can collect.
|
[
"2\n-1 5\n1 5\n",
"3\n-2 2\n1 4\n-1 3\n",
"3\n1 9\n3 5\n7 10\n"
] |
[
"10",
"9",
"9"
] |
In the first sample test it doesn't matter if Amr chose at first to go left or right. In both cases he'll get all the apples.
In the second sample test the optimal solution is to go left to *x* = - 1, collect apples from there, then the direction will be reversed, Amr has to go to *x* = 1, collect apples from there, then the direction will be reversed and Amr goes to the final tree *x* = - 2.
In the third sample test the optimal solution is to go right to *x* = 1, collect apples from there, then the direction will be reversed and Amr will not be able to collect anymore apples because there are no apple trees to his left.
| 500
|
[
{
"input": "2\n-1 5\n1 5",
"output": "10"
},
{
"input": "3\n-2 2\n1 4\n-1 3",
"output": "9"
},
{
"input": "3\n1 9\n3 5\n7 10",
"output": "9"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "4\n10000 100000\n-1000 100000\n-2 100000\n-1 100000",
"output": "300000"
},
{
"input": "1\n-1 1",
"output": "1"
},
{
"input": "27\n-30721 24576\n-6620 92252\n88986 24715\n-94356 10509\n-6543 29234\n-68554 69530\n39176 96911\n67266 99669\n95905 51002\n-94093 92134\n65382 23947\n-6525 79426\n-448 67531\n-70083 26921\n-86333 50029\n48924 8036\n-27228 5349\n6022 10691\n-13840 56735\n50398 58794\n-63258 45557\n-27792 77057\n98295 1203\n-51294 18757\n35037 61941\n-30112 13076\n82334 20463",
"output": "1036452"
},
{
"input": "18\n-18697 44186\n56333 51938\n-75688 49735\n77762 14039\n-43996 81060\n69700 49107\n74532 45568\n-94476 203\n-92347 90745\n58921 44650\n57563 63561\n44630 8486\n35750 5999\n3249 34202\n75358 68110\n-33245 60458\n-88148 2342\n87856 85532",
"output": "632240"
},
{
"input": "28\n49728 91049\n-42863 4175\n-89214 22191\n77977 16965\n-42960 87627\n-84329 97494\n89270 75906\n-13695 28908\n-72279 13607\n-97327 87062\n-58682 32094\n39108 99936\n29304 93784\n-63886 48237\n-77359 57648\n-87013 79017\n-41086 35033\n-60613 83555\n-48955 56816\n-20568 26802\n52113 25160\n-88885 45294\n22601 42971\n62693 65662\n-15985 5357\n86671 8522\n-59921 11271\n-79304 25044",
"output": "891593"
},
{
"input": "25\n5704 67795\n6766 31836\n-41715 89987\n76854 9848\n11648 90020\n-79763 10107\n96971 92636\n-64205 71937\n87997 38273\n-9782 57187\n22186 6905\n-41130 40258\n-28403 66579\n19578 43375\n35735 52929\n-52417 89388\n-89430 1939\n9401 43491\n-11228 10112\n-86859 16024\n-51486 33467\n-80578 65080\n-52820 98445\n-89165 7657\n-97106 79422",
"output": "1109655"
},
{
"input": "16\n-41732 47681\n44295 28942\n-75194 99827\n69982 18020\n-75378 22026\n80032 22908\n-34879 41113\n36257 48574\n-35882 84333\n29646 71151\n-86214 80886\n72724 39364\n-42529 60880\n29150 29921\n-8471 80781\n79387 70834",
"output": "847241"
},
{
"input": "3\n-94146 4473\n28707 99079\n-4153 8857",
"output": "112409"
},
{
"input": "3\n-3 3\n-2 2\n-1 1",
"output": "1"
},
{
"input": "2\n100000 3\n-100000 9",
"output": "12"
},
{
"input": "2\n-100000 100000\n100000 99999",
"output": "199999"
}
] | 1,660,075,199
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 46
| 62
| 0
|
n = int(input())
p_apples = 0
n_apples = 0
p_trees = []
n_trees = []
for i in range(n):
x,a = map(int,input().split())
if x > 0:
p_trees.append((x,a))
else :
n_trees.append((x,a))
n_trees.sort(reverse=1,key= lambda x : x[0])
p_trees.sort(key= lambda x : x[0])
# try left
i= 0
while True :
if i >= len(n_trees):
break
n_apples += n_trees[i][1]
if i >= len(p_trees):
break
n_apples += p_trees[i][1]
i +=1
# try right
i= 0
while True :
if i >= len(p_trees):
break
p_apples += p_trees[i][1]
if i >= len(n_trees):
break
p_apples += n_trees[i][1]
i +=1
print(max(n_apples,p_apples))
|
Title: Lala Land and Apple Trees
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr lives in Lala Land. Lala Land is a very beautiful country that is located on a coordinate line. Lala Land is famous with its apple trees growing everywhere.
Lala Land has exactly *n* apple trees. Tree number *i* is located in a position *x**i* and has *a**i* apples growing on it. Amr wants to collect apples from the apple trees. Amr currently stands in *x*<==<=0 position. At the beginning, he can choose whether to go right or left. He'll continue in his direction until he meets an apple tree he didn't visit before. He'll take all of its apples and then reverse his direction, continue walking in this direction until he meets another apple tree he didn't visit before and so on. In the other words, Amr reverses his direction when visiting each new apple tree. Amr will stop collecting apples when there are no more trees he didn't visit in the direction he is facing.
What is the maximum number of apples he can collect?
Input Specification:
The first line contains one number *n* (1<=≤<=*n*<=≤<=100), the number of apple trees in Lala Land.
The following *n* lines contains two integers each *x**i*, *a**i* (<=-<=105<=≤<=*x**i*<=≤<=105, *x**i*<=≠<=0, 1<=≤<=*a**i*<=≤<=105), representing the position of the *i*-th tree and number of apples on it.
It's guaranteed that there is at most one apple tree at each coordinate. It's guaranteed that no tree grows in point 0.
Output Specification:
Output the maximum number of apples Amr can collect.
Demo Input:
['2\n-1 5\n1 5\n', '3\n-2 2\n1 4\n-1 3\n', '3\n1 9\n3 5\n7 10\n']
Demo Output:
['10', '9', '9']
Note:
In the first sample test it doesn't matter if Amr chose at first to go left or right. In both cases he'll get all the apples.
In the second sample test the optimal solution is to go left to *x* = - 1, collect apples from there, then the direction will be reversed, Amr has to go to *x* = 1, collect apples from there, then the direction will be reversed and Amr goes to the final tree *x* = - 2.
In the third sample test the optimal solution is to go right to *x* = 1, collect apples from there, then the direction will be reversed and Amr will not be able to collect anymore apples because there are no apple trees to his left.
|
```python
n = int(input())
p_apples = 0
n_apples = 0
p_trees = []
n_trees = []
for i in range(n):
x,a = map(int,input().split())
if x > 0:
p_trees.append((x,a))
else :
n_trees.append((x,a))
n_trees.sort(reverse=1,key= lambda x : x[0])
p_trees.sort(key= lambda x : x[0])
# try left
i= 0
while True :
if i >= len(n_trees):
break
n_apples += n_trees[i][1]
if i >= len(p_trees):
break
n_apples += p_trees[i][1]
i +=1
# try right
i= 0
while True :
if i >= len(p_trees):
break
p_apples += p_trees[i][1]
if i >= len(n_trees):
break
p_apples += n_trees[i][1]
i +=1
print(max(n_apples,p_apples))
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,687,864,915
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
a = list(map(int,input().split()))
print((a[0]*a[1])//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
a = list(map(int,input().split()))
print((a[0]*a[1])//2)
```
| 3.977
|
625
|
B
|
War of the Corporations
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
|
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
|
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
|
[
"intellect\ntell\n",
"google\napple\n",
"sirisiri\nsir\n"
] |
[
"1",
"0",
"2"
] |
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
| 750
|
[
{
"input": "intellect\ntell",
"output": "1"
},
{
"input": "google\napple",
"output": "0"
},
{
"input": "sirisiri\nsir",
"output": "2"
},
{
"input": "sirisiri\nsiri",
"output": "2"
},
{
"input": "aaaaaaa\naaaa",
"output": "1"
},
{
"input": "bbbbbb\nbb",
"output": "3"
},
{
"input": "abc\nabcabc",
"output": "0"
},
{
"input": "kek\nkekekek",
"output": "0"
},
{
"input": "aaaaa\naaa",
"output": "1"
},
{
"input": "abcdabcv\nabcd",
"output": "1"
},
{
"input": "abcabcabczabcabcabcz\ncab",
"output": "4"
},
{
"input": "aatopotopotopotaa\ntopot",
"output": "2"
},
{
"input": "abcabcabcabcabcabcabcabcabcabc\nabcabcabcabcabcabcabcabcabcabc",
"output": "1"
},
{
"input": "sosossosos\nsos",
"output": "2"
},
{
"input": "sosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosos\nsos",
"output": "20"
},
{
"input": "tatatx\ntatx",
"output": "1"
},
{
"input": "sxxsxxsxxd\nsxxsxxd",
"output": "1"
}
] | 1,681,498,063
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 109
| 0
|
gogol = input()
pineapple = input()
ans = 0
sl = len(gogol)
spl = len(pineapple)
l = spl
i=0
while i<sl:
if gogol[i:i+l]==pineapple[0:spl+1]:
ans+=1
i+=l-1
i+=1
print(ans)
|
Title: War of the Corporations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input Specification:
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output Specification:
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Demo Input:
['intellect\ntell\n', 'google\napple\n', 'sirisiri\nsir\n']
Demo Output:
['1', '0', '2']
Note:
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
|
```python
gogol = input()
pineapple = input()
ans = 0
sl = len(gogol)
spl = len(pineapple)
l = spl
i=0
while i<sl:
if gogol[i:i+l]==pineapple[0:spl+1]:
ans+=1
i+=l-1
i+=1
print(ans)
```
| 3
|
|
68
|
B
|
Energy exchange
|
PROGRAMMING
| 1,600
|
[
"binary search"
] |
B. Energy exchange
|
2
|
256
|
It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the *i*-th accumulator has *a**i* units of energy. Energy can be transferred from one accumulator to the other. Every time *x* units of energy are transferred (*x* is not necessarily an integer) *k* percent of it is lost. That is, if *x* units were transferred from one accumulator to the other, amount of energy in the first one decreased by *x* units and in other increased by units.
Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.
|
First line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10000,<=0<=≤<=*k*<=≤<=99) — number of accumulators and the percent of energy that is lost during transfers.
Next line contains *n* integers *a*1,<=*a*2,<=... ,<=*a**n* — amounts of energy in the first, second, .., *n*-th accumulator respectively (0<=≤<=*a**i*<=≤<=1000,<=1<=≤<=*i*<=≤<=*n*).
|
Output maximum possible amount of energy that can remain in each of accumulators after the transfers of energy.
The absolute or relative error in the answer should not exceed 10<=-<=6.
|
[
"3 50\n4 2 1\n",
"2 90\n1 11\n"
] |
[
"2.000000000\n",
"1.909090909\n"
] |
none
| 1,000
|
[
{
"input": "3 50\n4 2 1",
"output": "2.000000000"
},
{
"input": "2 90\n1 11",
"output": "1.909090909"
},
{
"input": "5 26\n42 65 23 43 64",
"output": "45.415178571"
},
{
"input": "5 45\n964 515 454 623 594",
"output": "594.109756098"
},
{
"input": "1 20\n784",
"output": "784.000000000"
},
{
"input": "10 20\n812 896 36 596 709 641 679 778 738 302",
"output": "597.255813953"
},
{
"input": "10 83\n689 759 779 927 15 231 976 943 604 917",
"output": "406.839285714"
},
{
"input": "11 1\n235 280 196 670 495 379 391 280 847 875 506",
"output": "467.586301370"
},
{
"input": "12 71\n472 111 924 103 975 527 807 618 400 523 607 424",
"output": "413.249554367"
},
{
"input": "13 89\n19 944 341 846 764 676 222 957 953 481 708 920 950",
"output": "361.924390244"
},
{
"input": "14 6\n256 465 759 589 242 824 638 985 506 128 809 105 301 827",
"output": "523.427098675"
},
{
"input": "100 95\n154 444 715 98 35 347 799 313 40 821 118 786 31 587 888 84 88 751 98 86 321 720 201 247 302 518 663 904 482 385 139 646 581 995 847 775 173 252 508 722 380 922 634 911 102 384 346 212 705 380 220 221 492 421 244 591 758 631 370 866 536 872 294 152 337 810 761 235 789 839 365 366 623 897 905 249 685 838 380 873 702 379 865 68 215 168 425 264 652 228 167 498 733 41 502 21 565 956 430 171",
"output": "179.075000000"
},
{
"input": "101 71\n113 551 568 26 650 547 89 668 64 651 110 515 482 401 170 971 623 672 135 106 985 751 286 255 82 588 122 568 751 867 335 488 324 122 829 256 675 471 255 723 630 802 667 665 206 774 573 499 361 202 620 522 72 220 739 868 101 135 254 519 896 227 224 968 263 826 466 377 360 24 124 874 877 513 130 79 630 786 265 150 232 783 449 914 815 557 646 367 733 576 840 683 417 709 569 432 515 702 811 877 286",
"output": "343.047284817"
},
{
"input": "102 99\n73 348 420 956 955 436 69 714 87 480 102 555 933 215 452 167 157 593 863 816 337 471 371 574 862 967 581 543 330 348 221 640 378 250 500 428 866 379 1 723 880 992 9 419 0 163 800 96 16 25 19 513 653 19 924 144 135 950 449 481 255 582 844 473 189 841 862 520 242 210 573 381 130 820 357 911 884 735 460 428 764 187 344 760 413 636 868 780 123 614 822 869 792 66 636 843 465 449 191 891 819 30",
"output": "68.702920443"
},
{
"input": "103 26\n33 455 273 884 569 636 360 69 802 310 405 594 693 339 43 53 692 514 590 835 1000 191 456 582 641 35 731 207 600 830 416 483 431 377 481 910 367 597 58 413 128 873 42 173 104 553 26 383 673 849 728 503 924 819 108 422 169 454 333 134 926 247 464 289 115 547 567 663 123 396 21 890 385 436 584 432 829 683 345 706 294 901 238 606 12 24 89 882 203 962 804 745 166 425 393 252 415 195 571 596 41 486 445",
"output": "419.922659430"
},
{
"input": "104 54\n683 252 125 813 874 835 651 424 826 139 397 323 143 153 326 941 536 435 317 854 353 222 851 591 420 415 190 872 178 311 612 635 174 505 153 81 559 815 805 414 378 62 75 929 208 942 254 670 329 671 127 494 504 618 292 699 203 959 218 788 285 602 83 104 41 562 272 806 4 582 780 87 639 743 811 263 83 632 230 984 826 304 133 142 612 413 310 985 594 309 787 930 541 92 461 663 675 942 952 610 574 633 758 999",
"output": "399.430903462"
}
] | 1,589,987,745
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 374
| 716,800
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
sumEn = sum(a)
left = 0
right = 1000
eps = 1e-7
while right - left > eps:
mid = (left + right)/2
sumLoss = 0
for i in a:
if i > mid:
sumLoss += i - mid
sumLoss = sumLoss * (k/100)
if mid*n < sumEn - sumLoss:
left = mid
else:
right = mid
print('%.7f' % left)
|
Title: Energy exchange
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the *i*-th accumulator has *a**i* units of energy. Energy can be transferred from one accumulator to the other. Every time *x* units of energy are transferred (*x* is not necessarily an integer) *k* percent of it is lost. That is, if *x* units were transferred from one accumulator to the other, amount of energy in the first one decreased by *x* units and in other increased by units.
Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.
Input Specification:
First line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10000,<=0<=≤<=*k*<=≤<=99) — number of accumulators and the percent of energy that is lost during transfers.
Next line contains *n* integers *a*1,<=*a*2,<=... ,<=*a**n* — amounts of energy in the first, second, .., *n*-th accumulator respectively (0<=≤<=*a**i*<=≤<=1000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Output maximum possible amount of energy that can remain in each of accumulators after the transfers of energy.
The absolute or relative error in the answer should not exceed 10<=-<=6.
Demo Input:
['3 50\n4 2 1\n', '2 90\n1 11\n']
Demo Output:
['2.000000000\n', '1.909090909\n']
Note:
none
|
```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
sumEn = sum(a)
left = 0
right = 1000
eps = 1e-7
while right - left > eps:
mid = (left + right)/2
sumLoss = 0
for i in a:
if i > mid:
sumLoss += i - mid
sumLoss = sumLoss * (k/100)
if mid*n < sumEn - sumLoss:
left = mid
else:
right = mid
print('%.7f' % left)
```
| 3.905165
|
508
|
A
|
Pasha and Pixels
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
|
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
|
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
|
[
"2 2 4\n1 1\n1 2\n2 1\n2 2\n",
"2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n",
"5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n"
] |
[
"4\n",
"5\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2 2 4\n1 1\n1 2\n2 1\n2 2",
"output": "4"
},
{
"input": "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1",
"output": "5"
},
{
"input": "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2",
"output": "0"
},
{
"input": "3 3 11\n2 1\n3 1\n1 1\n1 3\n1 2\n2 3\n3 3\n3 2\n2 2\n1 3\n3 3",
"output": "9"
},
{
"input": "2 2 5\n1 1\n2 1\n2 1\n1 2\n2 2",
"output": "5"
},
{
"input": "518 518 10\n37 97\n47 278\n17 467\n158 66\n483 351\n83 123\n285 219\n513 187\n380 75\n304 352",
"output": "0"
},
{
"input": "1 1 5\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "0"
},
{
"input": "1 5 5\n1 1\n1 2\n1 3\n1 4\n1 5",
"output": "0"
},
{
"input": "5 1 5\n1 1\n2 1\n3 1\n4 1\n5 1",
"output": "0"
},
{
"input": "1 1 1\n1 1",
"output": "0"
},
{
"input": "10 10 4\n5 9\n6 9\n6 10\n5 10",
"output": "4"
},
{
"input": "1000 1000 4\n999 999\n999 1000\n1000 999\n1000 1000",
"output": "4"
},
{
"input": "2 3 5\n2 3\n1 3\n1 2\n1 1\n2 2",
"output": "5"
},
{
"input": "1000 1000 4\n1000 1000\n999 999\n1000 999\n999 1000",
"output": "4"
}
] | 1,698,052,688
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 546
| 15,974,400
|
# 读取输入
n, m, k = map(int, input().split())
black_cells = set()
# 用于检查是否形成 2x2 黑色方块的函数
def check_for_black_square(x, y):
return ((x, y) in black_cells and
(x+1, y) in black_cells and
(x, y+1) in black_cells and
(x+1, y+1) in black_cells)
# 用于检查当前操作是否导致形成黑色方块的函数
def check_move(i, j):
for x in range(i-1, i+2):
for y in range(j-1, j+2):
if check_for_black_square(x, y):
return True
return False
for move in range(1, k + 1):
i, j = map(int, input().split())
black_cells.add((i, j)) # 标记当前像素为黑色
if check_move(i, j):
print(move) # 如果当前操作导致形成黑色方块，输出操作次数
break
else:
print(0) # 没有形成黑色方块，输出0
|
Title: Pasha and Pixels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
Input Specification:
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
Output Specification:
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
Demo Input:
['2 2 4\n1 1\n1 2\n2 1\n2 2\n', '2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n', '5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n']
Demo Output:
['4\n', '5\n', '0\n']
Note:
none
|
```python
# 读取输入
n, m, k = map(int, input().split())
black_cells = set()
# 用于检查是否形成 2x2 黑色方块的函数
def check_for_black_square(x, y):
return ((x, y) in black_cells and
(x+1, y) in black_cells and
(x, y+1) in black_cells and
(x+1, y+1) in black_cells)
# 用于检查当前操作是否导致形成黑色方块的函数
def check_move(i, j):
for x in range(i-1, i+2):
for y in range(j-1, j+2):
if check_for_black_square(x, y):
return True
return False
for move in range(1, k + 1):
i, j = map(int, input().split())
black_cells.add((i, j)) # 标记当前像素为黑色
if check_move(i, j):
print(move) # 如果当前操作导致形成黑色方块，输出操作次数
break
else:
print(0) # 没有形成黑色方块，输出0
```
| 3
|
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,612,006,963
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 307,200
|
import math
def solve(arr, x):
print(arr)
count = 0
for i in arr:
if x+i < 0:
count += 1
else:
x += i
return x,count
def main():
n,x =list(map(int,input().split(' ')))
#n = int(input())
arr = []
for j in range(n):
# i = list(map(int,input().split(' ')))
# i = input().split(' ')
i = int(''.join(input().split(' ')))
arr.append(i)
print(*solve(arr, x))
main()
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
import math
def solve(arr, x):
print(arr)
count = 0
for i in arr:
if x+i < 0:
count += 1
else:
x += i
return x,count
def main():
n,x =list(map(int,input().split(' ')))
#n = int(input())
arr = []
for j in range(n):
# i = list(map(int,input().split(' ')))
# i = input().split(' ')
i = int(''.join(input().split(' ')))
arr.append(i)
print(*solve(arr, x))
main()
```
| 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"
}
] | 1,651,396,900
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 56
| 93
| 1,433,600
|
n=int(input())
arra=[]
flag=False
for i in range(n):
arra.append(input())
for i in range(n):
for j in range(n-1):
if len(arra[j])>len(arra[j+1]):
a=arra[j]
arra[j]=arra[j+1]
arra[j+1]=a
for i in range(n-1):
if arra[i+1].find(arra[i])==-1:
print("NO")
flag=True
break
if flag==False:
print("YES")
for i in range(n):
print(arra[i])
|
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
n=int(input())
arra=[]
flag=False
for i in range(n):
arra.append(input())
for i in range(n):
for j in range(n-1):
if len(arra[j])>len(arra[j+1]):
a=arra[j]
arra[j]=arra[j+1]
arra[j+1]=a
for i in range(n-1):
if arra[i+1].find(arra[i])==-1:
print("NO")
flag=True
break
if flag==False:
print("YES")
for i in range(n):
print(arra[i])
```
| 3
|
|
365
|
A
|
Good Number
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
|
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the number of *k*-good numbers in *a*.
|
[
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] |
[
"10\n",
"1\n"
] |
none
| 500
|
[
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,647,598,087
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 77
| 0
|
import sys
input = lambda: sys.stdin.readline().rstrip()
def main():
n,k = [int(inp) for inp in input().split()]
ans=0
for _ in range (n):
s = [int(inp) for inp in list(input())]
minim = min(s)
if(minim<k):
ans +=1
print(ans)
if __name__ == '__main__':
main()
|
Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none
|
```python
import sys
input = lambda: sys.stdin.readline().rstrip()
def main():
n,k = [int(inp) for inp in input().split()]
ans=0
for _ in range (n):
s = [int(inp) for inp in list(input())]
minim = min(s)
if(minim<k):
ans +=1
print(ans)
if __name__ == '__main__':
main()
```
| 0
|
|
475
|
A
|
Bayan Bus
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
The final round of Bayan Programming Contest will be held in Tehran, and the participants will be carried around with a yellow bus. The bus has 34 passenger seats: 4 seats in the last row and 3 seats in remaining rows.
The event coordinator has a list of *k* participants who should be picked up at the airport. When a participant gets on the bus, he will sit in the last row with an empty seat. If there is more than one empty seat in that row, he will take the leftmost one.
In order to keep track of the people who are on the bus, the event coordinator needs a figure showing which seats are going to be taken by *k* participants. Your task is to draw the figure representing occupied seats.
|
The only line of input contains integer *k*, (0<=≤<=*k*<=≤<=34), denoting the number of participants.
|
Print the figure of a bus with *k* passengers as described in sample tests. Character '#' denotes an empty seat, while 'O' denotes a taken seat. 'D' is the bus driver and other characters in the output are for the purpose of beautifying the figure. Strictly follow the sample test cases output format. Print exactly six lines. Do not output extra space or other characters.
|
[
"9\n",
"20\n"
] |
[
"+------------------------+\n|O.O.O.#.#.#.#.#.#.#.#.|D|)\n|O.O.O.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+\n",
"+------------------------+\n|O.O.O.O.O.O.O.#.#.#.#.|D|)\n|O.O.O.O.O.O.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.#.#.#.#.#.|.|)\n+------------------------+\n"
] |
none
| 500
|
[
{
"input": "9",
"output": "+------------------------+\n|O.O.O.#.#.#.#.#.#.#.#.|D|)\n|O.O.O.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "20",
"output": "+------------------------+\n|O.O.O.O.O.O.O.#.#.#.#.|D|)\n|O.O.O.O.O.O.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "30",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.O.#.|D|)\n|O.O.O.O.O.O.O.O.O.O.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.#.#.|.|)\n+------------------------+"
},
{
"input": "5",
"output": "+------------------------+\n|O.O.#.#.#.#.#.#.#.#.#.|D|)\n|O.#.#.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "0",
"output": "+------------------------+\n|#.#.#.#.#.#.#.#.#.#.#.|D|)\n|#.#.#.#.#.#.#.#.#.#.#.|.|\n|#.......................|\n|#.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "1",
"output": "+------------------------+\n|O.#.#.#.#.#.#.#.#.#.#.|D|)\n|#.#.#.#.#.#.#.#.#.#.#.|.|\n|#.......................|\n|#.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "2",
"output": "+------------------------+\n|O.#.#.#.#.#.#.#.#.#.#.|D|)\n|O.#.#.#.#.#.#.#.#.#.#.|.|\n|#.......................|\n|#.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "3",
"output": "+------------------------+\n|O.#.#.#.#.#.#.#.#.#.#.|D|)\n|O.#.#.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|#.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "4",
"output": "+------------------------+\n|O.#.#.#.#.#.#.#.#.#.#.|D|)\n|O.#.#.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "6",
"output": "+------------------------+\n|O.O.#.#.#.#.#.#.#.#.#.|D|)\n|O.O.#.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.#.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "7",
"output": "+------------------------+\n|O.O.#.#.#.#.#.#.#.#.#.|D|)\n|O.O.#.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "8",
"output": "+------------------------+\n|O.O.O.#.#.#.#.#.#.#.#.|D|)\n|O.O.#.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "10",
"output": "+------------------------+\n|O.O.O.#.#.#.#.#.#.#.#.|D|)\n|O.O.O.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "11",
"output": "+------------------------+\n|O.O.O.O.#.#.#.#.#.#.#.|D|)\n|O.O.O.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "12",
"output": "+------------------------+\n|O.O.O.O.#.#.#.#.#.#.#.|D|)\n|O.O.O.O.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.#.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "13",
"output": "+------------------------+\n|O.O.O.O.#.#.#.#.#.#.#.|D|)\n|O.O.O.O.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "14",
"output": "+------------------------+\n|O.O.O.O.O.#.#.#.#.#.#.|D|)\n|O.O.O.O.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "15",
"output": "+------------------------+\n|O.O.O.O.O.#.#.#.#.#.#.|D|)\n|O.O.O.O.O.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.#.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "16",
"output": "+------------------------+\n|O.O.O.O.O.#.#.#.#.#.#.|D|)\n|O.O.O.O.O.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "17",
"output": "+------------------------+\n|O.O.O.O.O.O.#.#.#.#.#.|D|)\n|O.O.O.O.O.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "18",
"output": "+------------------------+\n|O.O.O.O.O.O.#.#.#.#.#.|D|)\n|O.O.O.O.O.O.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.#.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "19",
"output": "+------------------------+\n|O.O.O.O.O.O.#.#.#.#.#.|D|)\n|O.O.O.O.O.O.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "21",
"output": "+------------------------+\n|O.O.O.O.O.O.O.#.#.#.#.|D|)\n|O.O.O.O.O.O.O.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.#.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "22",
"output": "+------------------------+\n|O.O.O.O.O.O.O.#.#.#.#.|D|)\n|O.O.O.O.O.O.O.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "23",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.#.#.#.|D|)\n|O.O.O.O.O.O.O.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "24",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.#.#.#.|D|)\n|O.O.O.O.O.O.O.O.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.#.#.#.#.|.|)\n+------------------------+"
},
{
"input": "25",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.#.#.#.|D|)\n|O.O.O.O.O.O.O.O.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.#.#.#.|.|)\n+------------------------+"
},
{
"input": "26",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.#.#.|D|)\n|O.O.O.O.O.O.O.O.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.#.#.#.|.|)\n+------------------------+"
},
{
"input": "27",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.#.#.|D|)\n|O.O.O.O.O.O.O.O.O.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.#.#.#.|.|)\n+------------------------+"
},
{
"input": "28",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.#.#.|D|)\n|O.O.O.O.O.O.O.O.O.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.#.#.|.|)\n+------------------------+"
},
{
"input": "29",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.O.#.|D|)\n|O.O.O.O.O.O.O.O.O.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.#.#.|.|)\n+------------------------+"
},
{
"input": "31",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.O.#.|D|)\n|O.O.O.O.O.O.O.O.O.O.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.O.#.|.|)\n+------------------------+"
},
{
"input": "32",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.O.O.|D|)\n|O.O.O.O.O.O.O.O.O.O.#.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.O.#.|.|)\n+------------------------+"
},
{
"input": "33",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.O.O.|D|)\n|O.O.O.O.O.O.O.O.O.O.O.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.O.#.|.|)\n+------------------------+"
},
{
"input": "34",
"output": "+------------------------+\n|O.O.O.O.O.O.O.O.O.O.O.|D|)\n|O.O.O.O.O.O.O.O.O.O.O.|.|\n|O.......................|\n|O.O.O.O.O.O.O.O.O.O.O.|.|)\n+------------------------+"
}
] | 1,412,520,328
| 6,328
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 61
| 0
|
#from itertools import chain
#from pprint import pprint
class Row:
@classmethod
def createDottedRow(cls):
r = cls(islast=False)
for x in range(3):
r.append('.')
return r
def __init__(self, islast):
self._isLast = islast
self.isFull = False
self.rowContent = []
def append(self, char='O'):
l = len(self.rowContent)
assert len(self.rowContent) < 4
if l == 2 and self._isLast is False:
self.rowContent.append('.')
#else:
self.rowContent.append(char)
if len(self.rowContent) == 4:
self.isFull = True
def __str__(self):
assert len(self.rowContent) <= 4
#result = [passngr for passngr in self.rowContent]
#for passenger in rowContent:
tmp = ''.join(self.rowContent)
result = tmp.ljust(4, '#')
if self._isLast is False and result[2] == '#':
return result[:2] + '.' + result[-1]
return result
class Matrix:
def __init__(self):
r = Row.createDottedRow()
#print(r)
self.matrixContent = []
self.currentRow = None
self.firstTime = True
self._len = 0
def append(self, char='O'):
self._len += 1
if self.firstTime is True:
r = Row(islast=True)
r.append(char)
self.matrixContent.append(r)
self.firstTime = False
return
r = self.matrixContent[-1]
if r.isFull is True:
tmp = Row(islast=False)
self.matrixContent.append(Row.createDottedRow())
self.matrixContent.append(tmp)
tmp.append(char)
return
r.append(char)
def fill(self):
amount = 34 - len(self)
for x in range(amount):
self.append('#')
def __len__(self):
return self._len
bus = Matrix()
amount = int(input())
for x in range(amount):
bus.append()
#for row in m.matrixContent:
# print(row)
bus.fill()
mat = list(str(row) for row in bus.matrixContent)
l = list(
zip(*mat))
def result():
i = 0
for row in l:
yield '|'
for point in row:
yield point
if i == 0 :yield '.|D|)\n'
elif i == 3: yield '.|.|)\n'
else: yield '.|.|\n'
i+=1
print('+------------------------+\n' + ''.join(result()) +
'+------------------------+'
)
|
Title: Bayan Bus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The final round of Bayan Programming Contest will be held in Tehran, and the participants will be carried around with a yellow bus. The bus has 34 passenger seats: 4 seats in the last row and 3 seats in remaining rows.
The event coordinator has a list of *k* participants who should be picked up at the airport. When a participant gets on the bus, he will sit in the last row with an empty seat. If there is more than one empty seat in that row, he will take the leftmost one.
In order to keep track of the people who are on the bus, the event coordinator needs a figure showing which seats are going to be taken by *k* participants. Your task is to draw the figure representing occupied seats.
Input Specification:
The only line of input contains integer *k*, (0<=≤<=*k*<=≤<=34), denoting the number of participants.
Output Specification:
Print the figure of a bus with *k* passengers as described in sample tests. Character '#' denotes an empty seat, while 'O' denotes a taken seat. 'D' is the bus driver and other characters in the output are for the purpose of beautifying the figure. Strictly follow the sample test cases output format. Print exactly six lines. Do not output extra space or other characters.
Demo Input:
['9\n', '20\n']
Demo Output:
['+------------------------+\n|O.O.O.#.#.#.#.#.#.#.#.|D|)\n|O.O.O.#.#.#.#.#.#.#.#.|.|\n|O.......................|\n|O.O.#.#.#.#.#.#.#.#.#.|.|)\n+------------------------+\n', '+------------------------+\n|O.O.O.O.O.O.O.#.#.#.#.|D|)\n|O.O.O.O.O.O.#.#.#.#.#.|.|\n|O.......................|\n|O.O.O.O.O.O.#.#.#.#.#.|.|)\n+------------------------+\n']
Note:
none
|
```python
#from itertools import chain
#from pprint import pprint
class Row:
@classmethod
def createDottedRow(cls):
r = cls(islast=False)
for x in range(3):
r.append('.')
return r
def __init__(self, islast):
self._isLast = islast
self.isFull = False
self.rowContent = []
def append(self, char='O'):
l = len(self.rowContent)
assert len(self.rowContent) < 4
if l == 2 and self._isLast is False:
self.rowContent.append('.')
#else:
self.rowContent.append(char)
if len(self.rowContent) == 4:
self.isFull = True
def __str__(self):
assert len(self.rowContent) <= 4
#result = [passngr for passngr in self.rowContent]
#for passenger in rowContent:
tmp = ''.join(self.rowContent)
result = tmp.ljust(4, '#')
if self._isLast is False and result[2] == '#':
return result[:2] + '.' + result[-1]
return result
class Matrix:
def __init__(self):
r = Row.createDottedRow()
#print(r)
self.matrixContent = []
self.currentRow = None
self.firstTime = True
self._len = 0
def append(self, char='O'):
self._len += 1
if self.firstTime is True:
r = Row(islast=True)
r.append(char)
self.matrixContent.append(r)
self.firstTime = False
return
r = self.matrixContent[-1]
if r.isFull is True:
tmp = Row(islast=False)
self.matrixContent.append(Row.createDottedRow())
self.matrixContent.append(tmp)
tmp.append(char)
return
r.append(char)
def fill(self):
amount = 34 - len(self)
for x in range(amount):
self.append('#')
def __len__(self):
return self._len
bus = Matrix()
amount = int(input())
for x in range(amount):
bus.append()
#for row in m.matrixContent:
# print(row)
bus.fill()
mat = list(str(row) for row in bus.matrixContent)
l = list(
zip(*mat))
def result():
i = 0
for row in l:
yield '|'
for point in row:
yield point
if i == 0 :yield '.|D|)\n'
elif i == 3: yield '.|.|)\n'
else: yield '.|.|\n'
i+=1
print('+------------------------+\n' + ''.join(result()) +
'+------------------------+'
)
```
| 0
|
|
983
|
A
|
Finite or not?
|
PROGRAMMING
| 1,700
|
[
"implementation",
"math"
] | null | null |
You are given several queries. Each query consists of three integers $p$, $q$ and $b$. You need to answer whether the result of $p/q$ in notation with base $b$ is a finite fraction.
A fraction in notation with base $b$ is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
|
The first line contains a single integer $n$ ($1 \le n \le 10^5$) — the number of queries.
Next $n$ lines contain queries, one per line. Each line contains three integers $p$, $q$, and $b$ ($0 \le p \le 10^{18}$, $1 \le q \le 10^{18}$, $2 \le b \le 10^{18}$). All numbers are given in notation with base $10$.
|
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
|
[
"2\n6 12 10\n4 3 10\n",
"4\n1 1 2\n9 36 2\n4 12 3\n3 5 4\n"
] |
[
"Finite\nInfinite\n",
"Finite\nFinite\nFinite\nInfinite\n"
] |
$\frac{6}{12} = \frac{1}{2} = 0,5_{10}$
$\frac{4}{3} = 1,(3)_{10}$
$\frac{9}{36} = \frac{1}{4} = 0,01_2$
$\frac{4}{12} = \frac{1}{3} = 0,1_3$
| 500
|
[
{
"input": "2\n6 12 10\n4 3 10",
"output": "Finite\nInfinite"
},
{
"input": "4\n1 1 2\n9 36 2\n4 12 3\n3 5 4",
"output": "Finite\nFinite\nFinite\nInfinite"
},
{
"input": "10\n10 5 3\n1 7 10\n7 5 7\n4 4 9\n6 5 2\n6 7 5\n9 9 7\n7 5 5\n6 6 4\n10 8 2",
"output": "Finite\nInfinite\nInfinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nFinite"
},
{
"input": "10\n1 3 10\n6 2 6\n2 3 9\n7 8 4\n5 6 10\n1 2 7\n0 3 6\n9 3 4\n4 4 9\n10 9 10",
"output": "Infinite\nFinite\nFinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nInfinite"
},
{
"input": "10\n10 8 5\n0 6 9\n0 7 6\n5 7 3\n7 6 8\n0 4 8\n2 6 3\n10 2 9\n6 7 9\n9 1 4",
"output": "Infinite\nFinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nInfinite\nFinite"
},
{
"input": "10\n5 8 2\n0 5 8\n5 9 7\n0 7 2\n6 7 2\n10 3 7\n8 1 10\n9 1 8\n0 7 10\n9 1 4",
"output": "Finite\nFinite\nInfinite\nFinite\nInfinite\nInfinite\nFinite\nFinite\nFinite\nFinite"
},
{
"input": "1\n1 864691128455135232 2",
"output": "Infinite"
},
{
"input": "11\n1 1000000000000000000 10000000\n2 999 9\n2 999 333111\n0 9 7\n17 128 2\n13 311992186885373952 18\n1971402979058461 750473176484995605 75\n14 19 23\n3 21914624432020321 23\n3 21914624432020321 46\n3 21914624432020321 47",
"output": "Finite\nInfinite\nFinite\nFinite\nFinite\nFinite\nFinite\nInfinite\nFinite\nFinite\nInfinite"
},
{
"input": "1\n1 100000000000000000 10000000000000000",
"output": "Finite"
},
{
"input": "1\n1 4294967297 4294967296",
"output": "Infinite"
},
{
"input": "1\n1 5244319080000 30030",
"output": "Finite"
}
] | 1,584,450,508
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 139
| 0
|
from math import gcd
def solve():
p, q, b = map(int, input().split())
g = gcd(p, q)
p //= g
q //= g
if gcd(q, b) in (q, b):
print("Finite")
else:
print("Infinte")
n = int(input())
for i in range(n):
solve()
|
Title: Finite or not?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given several queries. Each query consists of three integers $p$, $q$ and $b$. You need to answer whether the result of $p/q$ in notation with base $b$ is a finite fraction.
A fraction in notation with base $b$ is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 10^5$) — the number of queries.
Next $n$ lines contain queries, one per line. Each line contains three integers $p$, $q$, and $b$ ($0 \le p \le 10^{18}$, $1 \le q \le 10^{18}$, $2 \le b \le 10^{18}$). All numbers are given in notation with base $10$.
Output Specification:
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Demo Input:
['2\n6 12 10\n4 3 10\n', '4\n1 1 2\n9 36 2\n4 12 3\n3 5 4\n']
Demo Output:
['Finite\nInfinite\n', 'Finite\nFinite\nFinite\nInfinite\n']
Note:
$\frac{6}{12} = \frac{1}{2} = 0,5_{10}$
$\frac{4}{3} = 1,(3)_{10}$
$\frac{9}{36} = \frac{1}{4} = 0,01_2$
$\frac{4}{12} = \frac{1}{3} = 0,1_3$
|
```python
from math import gcd
def solve():
p, q, b = map(int, input().split())
g = gcd(p, q)
p //= g
q //= g
if gcd(q, b) in (q, b):
print("Finite")
else:
print("Infinte")
n = int(input())
for i in range(n):
solve()
```
| 0
|
|
625
|
B
|
War of the Corporations
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
|
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
|
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
|
[
"intellect\ntell\n",
"google\napple\n",
"sirisiri\nsir\n"
] |
[
"1",
"0",
"2"
] |
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
| 750
|
[
{
"input": "intellect\ntell",
"output": "1"
},
{
"input": "google\napple",
"output": "0"
},
{
"input": "sirisiri\nsir",
"output": "2"
},
{
"input": "sirisiri\nsiri",
"output": "2"
},
{
"input": "aaaaaaa\naaaa",
"output": "1"
},
{
"input": "bbbbbb\nbb",
"output": "3"
},
{
"input": "abc\nabcabc",
"output": "0"
},
{
"input": "kek\nkekekek",
"output": "0"
},
{
"input": "aaaaa\naaa",
"output": "1"
},
{
"input": "abcdabcv\nabcd",
"output": "1"
},
{
"input": "abcabcabczabcabcabcz\ncab",
"output": "4"
},
{
"input": "aatopotopotopotaa\ntopot",
"output": "2"
},
{
"input": "abcabcabcabcabcabcabcabcabcabc\nabcabcabcabcabcabcabcabcabcabc",
"output": "1"
},
{
"input": "sosossosos\nsos",
"output": "2"
},
{
"input": "sosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosos\nsos",
"output": "20"
},
{
"input": "tatatx\ntatx",
"output": "1"
},
{
"input": "sxxsxxsxxd\nsxxsxxd",
"output": "1"
}
] | 1,544,615,398
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 109
| 0
|
s = input()
s1 = input()
i = 0
j = 0
ans = 0
while i < len(s):
if s[i] == s1[j]:
j += 1
i += 1
if j == len(s1):
j = 0
ans += 1
print(ans)
|
Title: War of the Corporations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input Specification:
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output Specification:
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Demo Input:
['intellect\ntell\n', 'google\napple\n', 'sirisiri\nsir\n']
Demo Output:
['1', '0', '2']
Note:
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
|
```python
s = input()
s1 = input()
i = 0
j = 0
ans = 0
while i < len(s):
if s[i] == s1[j]:
j += 1
i += 1
if j == len(s1):
j = 0
ans += 1
print(ans)
```
| 0
|
|
14
|
B
|
Young Photographer
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Young Photographer
|
2
|
64
|
Among other things, Bob is keen on photography. Especially he likes to take pictures of sportsmen. That was the reason why he placed himself in position *x*0 of a long straight racetrack and got ready to take pictures. But the problem was that not all the runners passed him. The total amount of sportsmen, training at that racetrack, equals *n*. And each of them regularly runs distances within a particular segment of the racetrack, which is the same for each sportsman. For example, the first sportsman runs from position *a*1 to position *b*1, the second — from *a*2 to *b*2
What is the minimum distance that Bob should move to have a chance to take pictures of each sportsman? Bob can take a picture of a sportsman, if he stands within the segment that this sportsman covers on the racetrack.
|
The first line of the input file contains integers *n* and *x*0 (1<=≤<=*n*<=≤<=100; 0<=≤<=*x*0<=≤<=1000). The following *n* lines contain pairs of integers *a**i*,<=*b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000; *a**i*<=≠<=*b**i*).
|
Output the required minimum distance in the same units as the positions on the racetrack. If there is no such a position, output -1.
|
[
"3 3\n0 7\n14 2\n4 6\n"
] |
[
"1\n"
] |
none
| 0
|
[
{
"input": "3 3\n0 7\n14 2\n4 6",
"output": "1"
},
{
"input": "1 1\n0 10",
"output": "0"
},
{
"input": "2 2\n1 2\n3 2",
"output": "0"
},
{
"input": "3 2\n1 2\n2 3\n3 4",
"output": "-1"
},
{
"input": "2 4\n10 4\n1 5",
"output": "0"
},
{
"input": "1 10\n1 9",
"output": "1"
},
{
"input": "1 10\n123 12",
"output": "2"
},
{
"input": "1 17\n10 17",
"output": "0"
},
{
"input": "1 22\n22 33",
"output": "0"
},
{
"input": "1 3\n1 2",
"output": "1"
},
{
"input": "2 5\n0 3\n2 1",
"output": "3"
},
{
"input": "3 3\n7 3\n6 4\n3 7",
"output": "1"
},
{
"input": "4 9\n8 6\n11 5\n5 11\n8 3",
"output": "1"
},
{
"input": "2 4\n1 4\n4 0",
"output": "0"
},
{
"input": "3 7\n5 8\n7 5\n4 7",
"output": "0"
},
{
"input": "4 7\n8 2\n5 7\n8 2\n5 8",
"output": "0"
},
{
"input": "2 3\n4 1\n4 1",
"output": "0"
},
{
"input": "3 8\n7 2\n3 7\n5 2",
"output": "3"
},
{
"input": "4 0\n9 1\n8 1\n8 4\n4 5",
"output": "4"
},
{
"input": "4 7\n2 5\n3 6\n3 5\n7 4",
"output": "2"
},
{
"input": "10 16\n4 18\n6 19\n22 1\n23 0\n1 22\n9 22\n4 19\n0 14\n6 14\n0 16",
"output": "2"
},
{
"input": "20 1\n35 8\n40 6\n49 5\n48 18\n46 16\n45 16\n44 10\n16 44\n8 46\n2 45\n38 3\n42 1\n13 35\n35 18\n12 33\n32 11\n31 3\n50 20\n47 6\n38 2",
"output": "19"
},
{
"input": "30 43\n17 72\n75 26\n23 69\n83 30\n15 82\n4 67\n83 27\n33 62\n26 83\n70 26\n69 25\n16 67\n77 26\n66 33\n7 88\n70 9\n10 79\n76 9\n30 77\n77 28\n21 68\n81 14\n13 72\n88 15\n60 29\n87 28\n16 58\n6 58\n71 9\n83 18",
"output": "0"
},
{
"input": "40 69\n29 109\n28 87\n52 106\n101 34\n32 92\n91 60\n90 47\n62 102\n33 72\n27 87\n45 78\n103 37\n94 33\n56 98\n38 79\n31 83\n105 53\n47 89\n50 83\n93 62\n96 49\n47 75\n89 47\n89 61\n93 54\n46 100\n110 41\n103 28\n101 57\n100 62\n71 37\n65 80\n86 28\n73 42\n96 44\n33 111\n98 39\n87 55\n108 65\n31 101",
"output": "0"
},
{
"input": "50 77\n95 55\n113 33\n101 17\n109 56\n117 7\n77 12\n14 84\n57 101\n96 28\n108 22\n105 12\n17 114\n51 115\n18 112\n104 25\n50 115\n14 111\n55 113\n124 20\n101 37\n18 121\n41 90\n77 41\n117 16\n8 83\n92 45\n48 86\n16 84\n13 98\n40 107\n14 94\n23 111\n36 121\n50 100\n35 90\n103 37\n96 51\n109 15\n13 117\n117 42\n112 45\n88 36\n51 121\n127 49\n112 15\n9 95\n122 46\n126 40\n57 93\n56 88",
"output": "0"
},
{
"input": "5 12\n2 7\n7 5\n3 10\n11 3\n2 11",
"output": "5"
},
{
"input": "15 15\n12 37\n40 4\n38 8\n5 36\n11 31\n21 33\n9 37\n4 38\n8 33\n5 39\n7 39\n38 16\n16 41\n38 9\n5 32",
"output": "6"
},
{
"input": "25 40\n66 26\n56 19\n64 38\n64 23\n25 49\n51 26\n67 20\n65 35\n33 66\n28 63\n27 57\n40 56\n59 26\n35 56\n39 67\n30 63\n69 22\n21 63\n67 22\n20 66\n26 65\n64 26\n44 57\n57 41\n35 50",
"output": "4"
},
{
"input": "50 77\n24 119\n43 119\n102 22\n117 30\n127 54\n93 19\n120 9\n118 27\n98 16\n17 105\n22 127\n109 52\n115 40\n11 121\n12 120\n113 30\n13 108\n33 124\n31 116\n112 39\n37 108\n127 28\n127 39\n120 29\n19 114\n103 18\n106 16\n24 121\n93 10\n36 112\n104 40\n39 100\n36 97\n83 9\n14 114\n126 12\n85 47\n25 84\n105 29\n35 113\n102 19\n8 110\n111 28\n94 12\n11 115\n40 124\n39 85\n47 93\n94 31\n17 121",
"output": "0"
},
{
"input": "1 21\n973 373",
"output": "352"
},
{
"input": "2 212\n831 551\n810 753",
"output": "541"
},
{
"input": "3 404\n690 728\n820 260\n186 402",
"output": "-1"
},
{
"input": "4 906\n548 906\n830 457\n228 638\n464 167",
"output": "-1"
},
{
"input": "5 97\n97 393\n840 965\n269 183\n596 49\n975 62",
"output": "-1"
},
{
"input": "3 183\n416 335\n773 648\n434 198",
"output": "-1"
},
{
"input": "3 868\n251 927\n862 464\n157 756",
"output": "112"
},
{
"input": "3 242\n397 208\n951 279\n570 622",
"output": "-1"
},
{
"input": "3 618\n543 800\n38 94\n293 179",
"output": "-1"
},
{
"input": "3 993\n378 81\n127 911\n16 737",
"output": "615"
},
{
"input": "5 12\n11 1\n9 6\n1 11\n3 8\n874 842",
"output": "-1"
},
{
"input": "15 16\n11 40\n5 32\n5 31\n36 10\n34 9\n43 6\n28 6\n34 8\n43 15\n9 28\n14 34\n34 6\n7 31\n31 14\n68 478",
"output": "-1"
},
{
"input": "25 57\n47 31\n64 21\n43 56\n47 19\n70 27\n28 61\n41 61\n39 45\n46 21\n55 35\n70 22\n22 69\n30 67\n55 42\n37 58\n50 28\n57 42\n35 48\n68 40\n38 50\n62 20\n31 52\n38 70\n64 35\n666 393",
"output": "-1"
},
{
"input": "50 118\n83 55\n101 33\n89 17\n97 56\n105 7\n65 12\n14 72\n57 89\n84 28\n96 22\n93 12\n17 102\n51 103\n18 100\n92 25\n50 103\n14 99\n55 101\n112 20\n89 37\n18 109\n41 78\n65 41\n105 16\n8 71\n80 45\n48 74\n16 72\n13 86\n40 95\n14 82\n23 99\n36 109\n50 88\n35 78\n91 37\n84 51\n97 15\n13 105\n105 42\n100 45\n76 36\n51 109\n115 49\n100 15\n9 83\n110 46\n114 40\n57 81\n528 348",
"output": "-1"
},
{
"input": "1 21\n0 1000",
"output": "0"
}
] | 1,639,479,003
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 186
| 0
|
n, m = map(int, input().split())
arr = []
for i in range(n):
a, b = map(int, input().split())
arr.append([min(a, b), max(a, b)])
arr.sort(key=lambda x: x[0])
# print(arr)
m1 = 0
m2 = 0
for i in arr:
if i[0] <= m <= i[1]:
continue
else:
m1 = max(m1, m - i[1])
m2 = max(m2, i[0]-m)
print(m1 + m2)
|
Title: Young Photographer
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Among other things, Bob is keen on photography. Especially he likes to take pictures of sportsmen. That was the reason why he placed himself in position *x*0 of a long straight racetrack and got ready to take pictures. But the problem was that not all the runners passed him. The total amount of sportsmen, training at that racetrack, equals *n*. And each of them regularly runs distances within a particular segment of the racetrack, which is the same for each sportsman. For example, the first sportsman runs from position *a*1 to position *b*1, the second — from *a*2 to *b*2
What is the minimum distance that Bob should move to have a chance to take pictures of each sportsman? Bob can take a picture of a sportsman, if he stands within the segment that this sportsman covers on the racetrack.
Input Specification:
The first line of the input file contains integers *n* and *x*0 (1<=≤<=*n*<=≤<=100; 0<=≤<=*x*0<=≤<=1000). The following *n* lines contain pairs of integers *a**i*,<=*b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000; *a**i*<=≠<=*b**i*).
Output Specification:
Output the required minimum distance in the same units as the positions on the racetrack. If there is no such a position, output -1.
Demo Input:
['3 3\n0 7\n14 2\n4 6\n']
Demo Output:
['1\n']
Note:
none
|
```python
n, m = map(int, input().split())
arr = []
for i in range(n):
a, b = map(int, input().split())
arr.append([min(a, b), max(a, b)])
arr.sort(key=lambda x: x[0])
# print(arr)
m1 = 0
m2 = 0
for i in arr:
if i[0] <= m <= i[1]:
continue
else:
m1 = max(m1, m - i[1])
m2 = max(m2, i[0]-m)
print(m1 + m2)
```
| 0
|
758
|
A
|
Holiday Of Equality
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
|
In the only line print the integer *S* — the minimum number of burles which are had to spend.
|
[
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] |
[
"10",
"1",
"4",
"0"
] |
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
| 500
|
[
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281",
"output": "5464380"
},
{
"input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353",
"output": "9535765"
},
{
"input": "3\n1 3 2",
"output": "3"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n228503",
"output": "0"
},
{
"input": "2\n32576 550340",
"output": "517764"
},
{
"input": "3\n910648 542843 537125",
"output": "741328"
},
{
"input": "4\n751720 572344 569387 893618",
"output": "787403"
},
{
"input": "6\n433864 631347 597596 794426 713555 231193",
"output": "1364575"
},
{
"input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107",
"output": "4647430"
},
{
"input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025",
"output": "13488674"
},
{
"input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763",
"output": "21993384"
},
{
"input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351",
"output": "26984185"
},
{
"input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642",
"output": "32816391"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1000000",
"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": "0"
},
{
"input": "100\n1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99000000"
},
{
"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 1000000",
"output": "99000000"
},
{
"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 1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99000000"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "10"
},
{
"input": "5\n2 10 0 0 0",
"output": "38"
}
] | 1,652,473,478
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 41
| 77
| 0
|
n = int(input())
cit = list(map(int, input().split()))
ans = 0
maxN = max(cit)
for i in cit:
ans += maxN - i
print(ans)
|
Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
|
```python
n = int(input())
cit = list(map(int, input().split()))
ans = 0
maxN = max(cit)
for i in cit:
ans += maxN - i
print(ans)
```
| 3
|
|
171
|
E
|
MYSTERIOUS LANGUAGE
|
PROGRAMMING
| 2,000
|
[
"*special"
] | null | null |
You are given a mysterious language (codenamed "Secret") available in "Custom Test" tab. Find out what this language is and write a program which outputs its name. Note that the program must be written in this language.
|
This program has only one test, and it's empty (it doesn't give your program anything to read).
|
Output the name of the mysterious language.
|
[] |
[] |
none
| 0
|
[
{
"input": "1",
"output": "INTERCAL"
}
] | 1,688,977,040
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print('PyPy 3.9.10 (7.3.9, 64-bit)')
|
Title: MYSTERIOUS LANGUAGE
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a mysterious language (codenamed "Secret") available in "Custom Test" tab. Find out what this language is and write a program which outputs its name. Note that the program must be written in this language.
Input Specification:
This program has only one test, and it's empty (it doesn't give your program anything to read).
Output Specification:
Output the name of the mysterious language.
Note:
none
|
```python
print('PyPy 3.9.10 (7.3.9, 64-bit)')
```
| 0
|
|
887
|
C
|
Solution for Cube
|
PROGRAMMING
| 1,500
|
[
"brute force",
"implementation"
] | null | null |
During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2.
It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction.
To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above.
Cube is called solved if for each face of cube all squares on it has the same color.
https://en.wikipedia.org/wiki/Rubik's_Cube
|
In first line given a sequence of 24 integers *a**i* (1<=≤<=*a**i*<=≤<=6), where *a**i* denotes color of *i*-th square. There are exactly 4 occurrences of all colors in this sequence.
|
Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise.
|
[
"2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4\n",
"5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3\n"
] |
[
"NO",
"YES"
] |
In first test case cube looks like this:
In second test case cube looks like this:
It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16.
| 1,500
|
[
{
"input": "2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4",
"output": "NO"
},
{
"input": "5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3",
"output": "YES"
},
{
"input": "2 6 3 3 5 5 2 6 1 1 6 4 4 4 2 4 6 5 3 1 2 5 3 1",
"output": "NO"
},
{
"input": "3 4 2 3 5 5 6 6 4 5 4 6 5 1 1 1 6 2 1 3 3 2 4 2",
"output": "NO"
},
{
"input": "5 5 2 5 3 3 2 6 6 4 2 4 6 1 4 3 1 6 2 1 3 4 5 1",
"output": "NO"
},
{
"input": "6 6 1 2 6 1 1 3 5 4 3 4 3 5 5 2 4 4 6 2 1 5 3 2",
"output": "NO"
},
{
"input": "2 2 1 1 5 5 5 5 3 3 4 4 1 4 1 4 2 3 2 3 6 6 6 6",
"output": "YES"
},
{
"input": "1 1 1 1 5 5 3 3 4 4 4 4 3 3 2 2 6 6 5 5 2 2 6 6",
"output": "YES"
},
{
"input": "1 1 1 1 3 3 3 3 5 5 5 5 2 2 2 2 4 4 4 4 6 6 6 6",
"output": "NO"
},
{
"input": "5 4 5 4 4 6 4 6 6 3 6 3 1 1 1 1 2 2 2 2 5 3 5 3",
"output": "YES"
},
{
"input": "3 3 5 5 2 2 2 2 6 6 4 4 6 3 6 3 4 5 4 5 1 1 1 1",
"output": "YES"
},
{
"input": "6 6 6 6 2 2 5 5 1 1 1 1 4 4 2 2 5 5 3 3 3 3 4 4",
"output": "YES"
},
{
"input": "4 6 4 6 6 1 6 1 1 3 1 3 2 2 2 2 5 5 5 5 4 3 4 3",
"output": "YES"
},
{
"input": "6 6 2 2 3 3 3 3 4 4 5 5 4 6 4 6 5 2 5 2 1 1 1 1",
"output": "YES"
},
{
"input": "3 3 3 3 4 4 5 5 1 1 1 1 2 2 4 4 5 5 6 6 6 6 2 2",
"output": "YES"
},
{
"input": "2 5 2 5 4 2 4 2 1 4 1 4 6 6 6 6 3 3 3 3 1 5 1 5",
"output": "YES"
},
{
"input": "4 4 3 3 5 5 5 5 1 1 6 6 3 6 3 6 4 1 4 1 2 2 2 2",
"output": "YES"
},
{
"input": "5 5 5 5 6 6 2 2 3 3 3 3 2 2 1 1 4 4 6 6 1 1 4 4",
"output": "YES"
},
{
"input": "1 4 3 4 2 6 5 2 1 5 1 6 3 4 3 6 5 5 1 3 2 6 4 2",
"output": "NO"
},
{
"input": "4 4 2 5 3 2 4 2 5 3 6 4 6 5 1 3 1 5 6 3 1 1 6 2",
"output": "NO"
},
{
"input": "4 5 3 4 5 5 6 3 2 5 1 6 2 1 6 3 1 4 2 3 2 6 1 4",
"output": "NO"
},
{
"input": "3 3 2 3 6 4 4 4 1 2 1 3 2 5 6 6 1 2 6 5 4 5 1 5",
"output": "NO"
},
{
"input": "5 6 1 1 4 5 6 5 4 6 2 1 4 2 6 5 3 2 3 2 3 1 3 4",
"output": "NO"
},
{
"input": "4 4 4 5 2 3 4 1 3 3 1 5 6 5 6 6 1 3 6 2 5 2 1 2",
"output": "NO"
},
{
"input": "3 2 5 6 1 4 3 4 6 5 4 3 2 3 2 2 1 4 1 1 6 5 6 5",
"output": "NO"
},
{
"input": "5 4 6 2 5 6 4 1 6 3 3 1 3 2 4 1 1 6 2 3 5 2 4 5",
"output": "NO"
},
{
"input": "6 6 3 1 5 6 5 3 2 5 3 1 2 4 1 6 4 5 2 2 4 1 3 4",
"output": "NO"
},
{
"input": "6 5 4 1 6 5 2 3 3 5 3 6 4 2 6 5 4 2 1 1 4 1 3 2",
"output": "NO"
},
{
"input": "1 3 5 6 4 4 4 3 5 2 2 2 3 1 5 6 3 4 6 5 1 2 1 6",
"output": "NO"
},
{
"input": "3 6 5 4 4 6 1 4 3 2 5 2 1 2 6 2 5 4 1 3 1 6 5 3",
"output": "NO"
},
{
"input": "5 2 6 1 5 3 5 3 1 1 3 6 6 2 4 2 5 4 4 2 1 3 4 6",
"output": "NO"
},
{
"input": "2 5 6 2 3 6 5 6 2 3 1 3 6 4 5 4 1 1 1 5 3 4 4 2",
"output": "NO"
},
{
"input": "4 5 4 4 3 3 1 2 3 1 1 5 2 2 5 6 6 4 3 2 6 5 1 6",
"output": "NO"
},
{
"input": "5 2 5 2 3 5 3 5 4 3 4 3 6 6 6 6 1 1 1 1 4 2 4 2",
"output": "YES"
},
{
"input": "2 4 2 4 4 5 4 5 5 1 5 1 3 3 3 3 6 6 6 6 2 1 2 1",
"output": "YES"
},
{
"input": "3 5 3 5 5 1 5 1 1 4 1 4 6 6 6 6 2 2 2 2 3 4 3 4",
"output": "YES"
},
{
"input": "2 1 2 1 4 2 4 2 6 4 6 4 5 5 5 5 3 3 3 3 6 1 6 1",
"output": "YES"
},
{
"input": "4 4 2 2 1 1 1 1 5 5 6 6 2 6 2 6 4 5 4 5 3 3 3 3",
"output": "YES"
},
{
"input": "1 1 2 2 4 4 4 4 5 5 6 6 5 1 5 1 6 2 6 2 3 3 3 3",
"output": "YES"
},
{
"input": "2 2 6 6 4 4 4 4 1 1 5 5 1 2 1 2 5 6 5 6 3 3 3 3",
"output": "YES"
},
{
"input": "2 2 3 3 6 6 6 6 4 4 1 1 3 1 3 1 2 4 2 4 5 5 5 5",
"output": "YES"
},
{
"input": "6 6 6 6 4 4 3 3 5 5 5 5 3 3 1 1 2 2 4 4 1 1 2 2",
"output": "YES"
},
{
"input": "2 2 2 2 4 4 5 5 3 3 3 3 6 6 4 4 5 5 1 1 1 1 6 6",
"output": "YES"
},
{
"input": "1 1 1 1 5 5 6 6 3 3 3 3 4 4 5 5 6 6 2 2 2 2 4 4",
"output": "YES"
},
{
"input": "4 4 4 4 2 2 3 3 1 1 1 1 3 3 6 6 5 5 2 2 6 6 5 5",
"output": "YES"
},
{
"input": "1 1 1 1 2 2 3 3 6 6 6 6 5 5 4 4 3 3 2 2 4 4 5 5",
"output": "NO"
},
{
"input": "1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6",
"output": "NO"
},
{
"input": "5 5 5 5 1 1 2 2 6 6 6 6 4 4 3 3 3 3 4 4 2 2 1 1",
"output": "NO"
}
] | 1,513,171,520
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 5,632,000
|
l=list(map(int,input().split()))
U=l[0:4]
F=l[4:8]
D=l[8:12]
L=l[12:16]
R=l[16:20]
B=l[20:24]
if (L.count(L[0])==4 and R.count(R[0])==4) :
if D[1]==D[3]==F[0]==F[2] and F[1]==F[3]==U[0]==U[2] and U[1]==U[3]==B[1]==B[3] and B[0]==B[2]==D[0]==D[2]:
print("YES")
elif D[1]==D[3]==B[1]==B[3] and B[0]==B[2]==U[0]==U[2] and U[1]==U[3]==F[0]==F[2] and F[1]==F[3]==D[0]==D[2]:
print("YES")
else:
print("NO")
elif (U.count(U[0])==4 and D.count(D[0])==4):
if L[1]==L[2]==F[3]==F[4] and F[1]==F[2]==R[3]==R[4] and R[1]==R[2]==B[3]==B[4] and B[1]==B[2]==L[3]==L[4]:
print("YES")
elif L[1]==L[2]==B[3]==B[4] and B[1]==B[2]==R[3]==R[4] and R[1]==R[2]==F[3]==F[4] and F[1]==F[2]==L[3]==L[4]:
print("YES")
else:
print("NO")
elif (F.count(L[0])==4 and B.count(R[0])==4):
if L[1]==L[3]==U[3]==U[4] and U[1]==U[2]==R[1]==R[3] and R[2]==R[4]==D[1]==D[2] and D[3]==B[4]==L[2]==L[4]:
print("YES")
elif L[1]==L[3]==D[1]==D[2] and D[3]==D[4]==R[1]==R[3] and R[2]==R[4]==U[4]==U[3] and U[1]==U[2]==L[2]==L[4]:
print("YES")
else:
print("NO")
else:
print("NO")
|
Title: Solution for Cube
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2.
It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction.
To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above.
Cube is called solved if for each face of cube all squares on it has the same color.
https://en.wikipedia.org/wiki/Rubik's_Cube
Input Specification:
In first line given a sequence of 24 integers *a**i* (1<=≤<=*a**i*<=≤<=6), where *a**i* denotes color of *i*-th square. There are exactly 4 occurrences of all colors in this sequence.
Output Specification:
Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise.
Demo Input:
['2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4\n', '5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3\n']
Demo Output:
['NO', 'YES']
Note:
In first test case cube looks like this:
In second test case cube looks like this:
It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16.
|
```python
l=list(map(int,input().split()))
U=l[0:4]
F=l[4:8]
D=l[8:12]
L=l[12:16]
R=l[16:20]
B=l[20:24]
if (L.count(L[0])==4 and R.count(R[0])==4) :
if D[1]==D[3]==F[0]==F[2] and F[1]==F[3]==U[0]==U[2] and U[1]==U[3]==B[1]==B[3] and B[0]==B[2]==D[0]==D[2]:
print("YES")
elif D[1]==D[3]==B[1]==B[3] and B[0]==B[2]==U[0]==U[2] and U[1]==U[3]==F[0]==F[2] and F[1]==F[3]==D[0]==D[2]:
print("YES")
else:
print("NO")
elif (U.count(U[0])==4 and D.count(D[0])==4):
if L[1]==L[2]==F[3]==F[4] and F[1]==F[2]==R[3]==R[4] and R[1]==R[2]==B[3]==B[4] and B[1]==B[2]==L[3]==L[4]:
print("YES")
elif L[1]==L[2]==B[3]==B[4] and B[1]==B[2]==R[3]==R[4] and R[1]==R[2]==F[3]==F[4] and F[1]==F[2]==L[3]==L[4]:
print("YES")
else:
print("NO")
elif (F.count(L[0])==4 and B.count(R[0])==4):
if L[1]==L[3]==U[3]==U[4] and U[1]==U[2]==R[1]==R[3] and R[2]==R[4]==D[1]==D[2] and D[3]==B[4]==L[2]==L[4]:
print("YES")
elif L[1]==L[3]==D[1]==D[2] and D[3]==D[4]==R[1]==R[3] and R[2]==R[4]==U[4]==U[3] and U[1]==U[2]==L[2]==L[4]:
print("YES")
else:
print("NO")
else:
print("NO")
```
| 0
|
|
590
|
B
|
Chip 'n Dale Rescue Rangers
|
PROGRAMMING
| 2,100
|
[
"binary search",
"geometry",
"math"
] | null | null |
A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress. In a few moments they were ready, and the dirigible of the rescue chipmunks hit the road.
We assume that the action takes place on a Cartesian plane. The headquarters of the rescuers is located at point (*x*1,<=*y*1), and the distress signal came from the point (*x*2,<=*y*2).
Due to Gadget's engineering talent, the rescuers' dirigible can instantly change its current velocity and direction of movement at any moment and as many times as needed. The only limitation is: the speed of the aircraft relative to the air can not exceed meters per second.
Of course, Gadget is a true rescuer and wants to reach the destination as soon as possible. The matter is complicated by the fact that the wind is blowing in the air and it affects the movement of the dirigible. According to the weather forecast, the wind will be defined by the vector (*v**x*,<=*v**y*) for the nearest *t* seconds, and then will change to (*w**x*,<=*w**y*). These vectors give both the direction and velocity of the wind. Formally, if a dirigible is located at the point (*x*,<=*y*), while its own velocity relative to the air is equal to zero and the wind (*u**x*,<=*u**y*) is blowing, then after seconds the new position of the dirigible will be .
Gadget is busy piloting the aircraft, so she asked Chip to calculate how long will it take them to reach the destination if they fly optimally. He coped with the task easily, but Dale is convinced that Chip has given the random value, aiming only not to lose the face in front of Gadget. Dale has asked you to find the right answer.
It is guaranteed that the speed of the wind at any moment of time is strictly less than the maximum possible speed of the airship relative to the air.
|
The first line of the input contains four integers *x*1, *y*1, *x*2, *y*2 (|*x*1|,<=<=|*y*1|,<=<=|*x*2|,<=<=|*y*2|<=≤<=10<=000) — the coordinates of the rescuers' headquarters and the point, where signal of the distress came from, respectively.
The second line contains two integers and *t* (0<=<<=*v*,<=*t*<=≤<=1000), which are denoting the maximum speed of the chipmunk dirigible relative to the air and the moment of time when the wind changes according to the weather forecast, respectively.
Next follow one per line two pairs of integer (*v**x*,<=*v**y*) and (*w**x*,<=*w**y*), describing the wind for the first *t* seconds and the wind that will blow at all the remaining time, respectively. It is guaranteed that and .
|
Print a single real value — the minimum time the rescuers need to get to point (*x*2,<=*y*2). You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"0 0 5 5\n3 2\n-1 -1\n-1 0\n",
"0 0 0 1000\n100 1000\n-50 0\n50 0\n"
] |
[
"3.729935587093555327\n",
"11.547005383792516398\n"
] |
none
| 1,000
|
[
{
"input": "0 0 5 5\n3 2\n-1 -1\n-1 0",
"output": "3.729935587093555327"
},
{
"input": "0 0 0 1000\n100 1000\n-50 0\n50 0",
"output": "11.547005383792516398"
},
{
"input": "0 0 0 1000\n100 5\n0 -50\n0 50",
"output": "10"
},
{
"input": "0 1000 0 0\n50 10\n-49 0\n49 0",
"output": "20"
},
{
"input": "0 1000 0 0\n50 10\n0 -48\n0 -49",
"output": "10.202020202020200657"
},
{
"input": "0 0 0 -5000\n100 20\n-50 0\n50 0",
"output": "50.262613427796381416"
},
{
"input": "0 0 0 -350\n55 5\n0 -50\n0 50",
"output": "3.3333333333333330373"
},
{
"input": "0 -1000 0 0\n11 10\n-10 0\n10 0",
"output": "146.8240957550254393"
},
{
"input": "0 -1000 0 0\n22 10\n0 -12\n0 -10",
"output": "85"
},
{
"input": "0 7834 -1 902\n432 43\n22 22\n-22 -22",
"output": "16.930588983107490719"
},
{
"input": "0 -10000 -10000 0\n1 777\n0 0\n0 0",
"output": "14142.13562373095192"
},
{
"input": "0 0 0 750\n25 30\n0 -1\n0 24",
"output": "30.612244897959186574"
},
{
"input": "-10000 10000 10000 10000\n2 1000\n0 -1\n-1 0",
"output": "19013.151067740152939"
},
{
"input": "-1 -1 1 1\n1 1\n0 0\n0 0",
"output": "2.8284271247461898469"
},
{
"input": "1 1 0 0\n2 1\n0 1\n0 1",
"output": "1.2152504370215302387"
},
{
"input": "-1 -1 0 0\n2 1\n-1 0\n0 -1",
"output": "1.1547005383792514621"
},
{
"input": "-1 -1 1 1\n2 1\n-1 0\n0 -1",
"output": "2.1892547876100074689"
},
{
"input": "-1 -1 2 2\n5 1\n-2 -1\n-1 -2",
"output": "1.4770329614269006591"
},
{
"input": "-5393 -8779 7669 9721\n613 13\n-313 -37\n-23 -257",
"output": "57.962085855983815463"
},
{
"input": "10000 10000 -10000 -10000\n1 999\n0 0\n0 0",
"output": "28284.27124746190384"
},
{
"input": "10000 -10000 -10000 10000\n1000 999\n0 -999\n999 0",
"output": "1018.7770495642339483"
},
{
"input": "10000 -10000 -10000 10000\n2 999\n1 0\n0 0",
"output": "14499.637935134793224"
},
{
"input": "10000 10000 -10000 -10000\n2 999\n-1 0\n0 0",
"output": "13793.458603628027049"
},
{
"input": "-10000 -10000 10000 10000\n1000 1000\n700 700\n0 999",
"output": "14.213562373095047775"
},
{
"input": "0 0 0 0\n1000 1\n0 0\n0 0",
"output": "3.9443045261050590271e-019"
},
{
"input": "10000 10000 10000 10000\n1 1000\n0 0\n0 0",
"output": "3.9443045261050590271e-019"
},
{
"input": "-999 -999 -999 -999\n1000 1000\n999 0\n0 999",
"output": "7.4941785995996121514e-018"
},
{
"input": "0 0 0 1\n1000 1\n0 999\n0 999",
"output": "0.00050025012506253112125"
},
{
"input": "-753 8916 -754 8915\n1000 1000\n-999 -44\n999 44",
"output": "0.0009587450100212672327"
},
{
"input": "-753 8916 -754 8915\n1000 1000\n999 44\n999 44",
"output": "33.112069856121181033"
},
{
"input": "-753 8916 -754 8915\n1000 33\n999 44\n-44 -999",
"output": "33.000003244932003099"
},
{
"input": "-753 8916 -754 8915\n1000 33\n999 44\n998 44",
"output": "33.003425712046578155"
},
{
"input": "-10000 10000 10000 -10000\n1000 1000\n-891 454\n-891 454",
"output": "17933348.203209973872"
},
{
"input": "-10000 10000 10000 -10000\n1000 1\n-890 455\n-891 454",
"output": "17933056.870118297637"
},
{
"input": "-10000 10000 10000 -10000\n1000 10\n-890 455\n-891 454",
"output": "17930434.872296713293"
},
{
"input": "-10000 10000 10000 -10000\n1000 100\n-890 455\n-891 454",
"output": "17904214.89444501698"
},
{
"input": "-9810 1940 9810 -1940\n1000 1000\n-981 194\n-981 194",
"output": "13333303.333326257765"
},
{
"input": "10000 10000 -6470 -10000\n969 1000\n616 748\n616 748",
"output": "50211053.368792012334"
},
{
"input": "-1000 -1000 1000 1000\n577 1\n-408 -408\n-408 -408",
"output": "3264002.4509786106646"
},
{
"input": "-10000 -10000 10000 10000\n577 1\n-408 -408\n-408 -408",
"output": "32640024.509786851704"
},
{
"input": "0 0 1940 9810\n1000 5\n-194 -981\n-194 -981",
"output": "6666651.6666631288826"
},
{
"input": "-10000 -10000 10000 10000\n1000 47\n-194 -981\n-194 -981",
"output": "15666683.687925204635"
},
{
"input": "-10000 -10000 10000 10000\n1000 5\n-194 -981\n-194 -981",
"output": "15666683.687925204635"
},
{
"input": "10000 10000 9120 -9360\n969 1000\n44 968\n44 968",
"output": "37558410"
},
{
"input": "10000 10000 -10000 -10000\n969 873\n44 968\n44 968",
"output": "40480019.762838959694"
},
{
"input": "0 0 1940 9810\n1000 1000\n-194 -981\n-194 -981",
"output": "6666651.6666631288826"
},
{
"input": "0 0 0 0\n5 10\n-1 -1\n-1 -1",
"output": "3.9443045261050590271e-019"
},
{
"input": "-10000 -10000 10000 10000\n1000 1000\n-454 -891\n-454 -891",
"output": "17933348.203209973872"
},
{
"input": "-9680 -440 9680 440\n969 1000\n-968 -44\n-968 -44",
"output": "37558410"
},
{
"input": "0 0 4540 8910\n1000 1000\n-454 -891\n-454 -891",
"output": "6666651.6666630636901"
},
{
"input": "0 0 0 0\n10 10\n0 0\n0 0",
"output": "3.9443045261050590271e-019"
}
] | 1,645,118,626
| 5,026
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
def process(x1, y1, x2, y2, v_max, t, vx, vy, wx, wy):
if x1==x2 and y1==y2:
return 0
x2-=x1
y2-=y1
d_total = (x2**2+y2**2)**0.5
d_x = x2/d_total
d_y = y2/d_total
"""
(cos(A)*v_max+vx) = C*d_x
(sin(A)*v_max+vy) = C*d_y
(C*d_x-vx)**2+(C*d_y-vy)**2 = v_max**2?
C**2+(-2*d_x*vx-2*d_y*v_y)*C+(v_x**2+v_y**2) = v_max**2
"""
b1 = -2*(d_x*vx+d_y*vy)
c1 = vx**2+vy**2-v_max**2
C = (-1*b1+(b1**2-4*c1)**0.5)/2
if C*t >= d_total:
return d_total/C
b2 = -2*(d_x*wx+d_y*wy)
c2 = (wx**2+wy**2-v_max**2)
C2 = (-1*b2+(b2**2-4*c2)**0.5)/2
"""
I go
C*t for the first t
then
I need to go C2
for the remaining d_total-C*t
"""
return t+(d_total-C*t)/C2
x1, y1, x2, y2 = [int(x) for x in input().split()]
v_max, t = [int(x) for x in input().split()]
vx, vy = [int(x) for x in input().split()]
wx, wy = [int(x) for x in input().split()]
print(process(x1, y1, x2, y2, v_max, t, vx, vy, wx, wy))
|
Title: Chip 'n Dale Rescue Rangers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress. In a few moments they were ready, and the dirigible of the rescue chipmunks hit the road.
We assume that the action takes place on a Cartesian plane. The headquarters of the rescuers is located at point (*x*1,<=*y*1), and the distress signal came from the point (*x*2,<=*y*2).
Due to Gadget's engineering talent, the rescuers' dirigible can instantly change its current velocity and direction of movement at any moment and as many times as needed. The only limitation is: the speed of the aircraft relative to the air can not exceed meters per second.
Of course, Gadget is a true rescuer and wants to reach the destination as soon as possible. The matter is complicated by the fact that the wind is blowing in the air and it affects the movement of the dirigible. According to the weather forecast, the wind will be defined by the vector (*v**x*,<=*v**y*) for the nearest *t* seconds, and then will change to (*w**x*,<=*w**y*). These vectors give both the direction and velocity of the wind. Formally, if a dirigible is located at the point (*x*,<=*y*), while its own velocity relative to the air is equal to zero and the wind (*u**x*,<=*u**y*) is blowing, then after seconds the new position of the dirigible will be .
Gadget is busy piloting the aircraft, so she asked Chip to calculate how long will it take them to reach the destination if they fly optimally. He coped with the task easily, but Dale is convinced that Chip has given the random value, aiming only not to lose the face in front of Gadget. Dale has asked you to find the right answer.
It is guaranteed that the speed of the wind at any moment of time is strictly less than the maximum possible speed of the airship relative to the air.
Input Specification:
The first line of the input contains four integers *x*1, *y*1, *x*2, *y*2 (|*x*1|,<=<=|*y*1|,<=<=|*x*2|,<=<=|*y*2|<=≤<=10<=000) — the coordinates of the rescuers' headquarters and the point, where signal of the distress came from, respectively.
The second line contains two integers and *t* (0<=<<=*v*,<=*t*<=≤<=1000), which are denoting the maximum speed of the chipmunk dirigible relative to the air and the moment of time when the wind changes according to the weather forecast, respectively.
Next follow one per line two pairs of integer (*v**x*,<=*v**y*) and (*w**x*,<=*w**y*), describing the wind for the first *t* seconds and the wind that will blow at all the remaining time, respectively. It is guaranteed that and .
Output Specification:
Print a single real value — the minimum time the rescuers need to get to point (*x*2,<=*y*2). You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['0 0 5 5\n3 2\n-1 -1\n-1 0\n', '0 0 0 1000\n100 1000\n-50 0\n50 0\n']
Demo Output:
['3.729935587093555327\n', '11.547005383792516398\n']
Note:
none
|
```python
def process(x1, y1, x2, y2, v_max, t, vx, vy, wx, wy):
if x1==x2 and y1==y2:
return 0
x2-=x1
y2-=y1
d_total = (x2**2+y2**2)**0.5
d_x = x2/d_total
d_y = y2/d_total
"""
(cos(A)*v_max+vx) = C*d_x
(sin(A)*v_max+vy) = C*d_y
(C*d_x-vx)**2+(C*d_y-vy)**2 = v_max**2?
C**2+(-2*d_x*vx-2*d_y*v_y)*C+(v_x**2+v_y**2) = v_max**2
"""
b1 = -2*(d_x*vx+d_y*vy)
c1 = vx**2+vy**2-v_max**2
C = (-1*b1+(b1**2-4*c1)**0.5)/2
if C*t >= d_total:
return d_total/C
b2 = -2*(d_x*wx+d_y*wy)
c2 = (wx**2+wy**2-v_max**2)
C2 = (-1*b2+(b2**2-4*c2)**0.5)/2
"""
I go
C*t for the first t
then
I need to go C2
for the remaining d_total-C*t
"""
return t+(d_total-C*t)/C2
x1, y1, x2, y2 = [int(x) for x in input().split()]
v_max, t = [int(x) for x in input().split()]
vx, vy = [int(x) for x in input().split()]
wx, wy = [int(x) for x in input().split()]
print(process(x1, y1, x2, y2, v_max, t, vx, vy, wx, wy))
```
| 0
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,692,791,197
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 71
| 186
| 1,843,200
|
counter = 0
max = 0
n = int(input())
for i in range(n):
a, b = map(int, input().split())
counter += (b-a)
if counter > max:
max = counter
print(max)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
counter = 0
max = 0
n = int(input())
for i in range(n):
a, b = map(int, input().split())
counter += (b-a)
if counter > max:
max = counter
print(max)
```
| 3
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,652,869,794
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 1,000
| 4,505,600
|
i=list(map(int,input().split(' ')))
n,m,a=i[0],i[1],i[2]
cr,cc,s=0,0,0
if s==1:
print(n*m)
else:
while cr<m:
while cc<n:
s+=1
cc+=a
cc=0
cr+=a
print(s)
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
i=list(map(int,input().split(' ')))
n,m,a=i[0],i[1],i[2]
cr,cc,s=0,0,0
if s==1:
print(n*m)
else:
while cr<m:
while cc<n:
s+=1
cc+=a
cc=0
cr+=a
print(s)
```
| 0
|
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