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float64 -1
3.99
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
287
|
A
|
IQ Test
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.
|
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
|
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
|
[
"####\n.#..\n####\n....\n",
"####\n....\n####\n....\n"
] |
[
"YES\n",
"NO\n"
] |
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.
| 500
|
[
{
"input": "###.\n...#\n###.\n...#",
"output": "NO"
},
{
"input": ".##.\n#..#\n.##.\n#..#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.#.",
"output": "NO"
},
{
"input": "##..\n..##\n##..\n..##",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n.#.#\n.#.#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n#.#.\n#.#.",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n#.#.\n.#.#",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n#.#.\n#.#.",
"output": "NO"
},
{
"input": ".#.#\n.#.#\n.#.#\n.#.#",
"output": "NO"
},
{
"input": "#..#\n.##.\n####\n####",
"output": "YES"
},
{
"input": "#.#.\n.###\n#.#.\n.###",
"output": "YES"
},
{
"input": "#..#\n.##.\n.##.\n#..#",
"output": "YES"
},
{
"input": ".##.\n.#..\n##.#\n#..#",
"output": "YES"
},
{
"input": ".##.\n##..\n#..#\n..##",
"output": "YES"
},
{
"input": "##..\n##..\n..##\n..##",
"output": "YES"
},
{
"input": ".#..\n###.\n.#.#\n..#.",
"output": "YES"
},
{
"input": "###.\n###.\n...#\n...#",
"output": "YES"
},
{
"input": "#.##\n##.#\n#.##\n##.#",
"output": "YES"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.##",
"output": "YES"
},
{
"input": "##..\n..##\n##..\n...#",
"output": "YES"
},
{
"input": ".#..\n..##\n##..\n..##",
"output": "YES"
},
{
"input": "##..\n...#\n##..\n...#",
"output": "YES"
},
{
"input": ".#..\n..#.\n.#..\n..#.",
"output": "YES"
},
{
"input": "....\n....\n....\n.#.#",
"output": "YES"
},
{
"input": "....\n....\n....\n...#",
"output": "YES"
}
] | 1,642,139,444
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 108
| 0
|
a=[]
for i in range(4):
b=input()
b=b.replace('#','1')
b=b.replace('.','0')
a.append(b)
f=0
for i in range(3):
for j in range(3):
s=int(a[i][j])+int(a[i][j+1])+int(a[i+1][j])+int(a[i+1][j+1])
if(s!=2):
f=1
break
if(f==1):
print("YES")
else:
print('NO')
|
Title: IQ Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.
Input Specification:
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
Output Specification:
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
Demo Input:
['####\n.#..\n####\n....\n', '####\n....\n####\n....\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.
|
```python
a=[]
for i in range(4):
b=input()
b=b.replace('#','1')
b=b.replace('.','0')
a.append(b)
f=0
for i in range(3):
for j in range(3):
s=int(a[i][j])+int(a[i][j+1])+int(a[i+1][j])+int(a[i+1][j+1])
if(s!=2):
f=1
break
if(f==1):
print("YES")
else:
print('NO')
```
| 3
|
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,434,128,193
| 693
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 5
| 405
| 102,400
|
n = int(input())
a = input().split()
pos = []
for i in range(n):
pos.append(1)
for x in range(n):
if a[i] < a[x]:
pos[i] += 1
out = ''
for i in pos:
out += str(i)+' '
print(out)
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
n = int(input())
a = input().split()
pos = []
for i in range(n):
pos.append(1)
for x in range(n):
if a[i] < a[x]:
pos[i] += 1
out = ''
for i in pos:
out += str(i)+' '
print(out)
```
| 0
|
|
892
|
B
|
Wrath
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw.
|
Print one integer — the total number of alive people after the bell rings.
|
[
"4\n0 1 0 10\n",
"2\n0 0\n",
"10\n1 1 3 0 0 0 2 1 0 3\n"
] |
[
"1\n",
"2\n",
"3\n"
] |
In first sample the last person kills everyone in front of him.
| 1,000
|
[
{
"input": "4\n0 1 0 10",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "10\n1 1 3 0 0 0 2 1 0 3",
"output": "3"
},
{
"input": "10\n0 0 2 0 0 3 3 2 2 0",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "5\n0 0 0 1 0",
"output": "4"
},
{
"input": "6\n3 1 1 0 3 3",
"output": "1"
},
{
"input": "8\n0 0 0 1 0 0 1 2",
"output": "5"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "2\n1 3",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "1"
},
{
"input": "11\n1 0 0 1 1 3 2 0 0 2 3",
"output": "4"
},
{
"input": "1\n1",
"output": "1"
}
] | 1,524,391,567
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 1,325
| 83,456,000
|
n = int(input())
l = [0] + list(map(int, input().strip().split()))
totalDead = 0
currentLow = n + 1
currentHigh = n + 1
for i in range(n, 0, -1):
low = max(1, i - l[i])
high = i - 1
# valid
if (low <= high):
if (low <= currentLow and high >= currentLow):
currentLow = low
elif (high < currentLow):
totalDead += currentHigh - currentLow + 1
currentLow = low
currentHigh = high
totalDead += currentHigh - currentLow + 1
print(n - totalDead + 1)
|
Title: Wrath
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw.
Output Specification:
Print one integer — the total number of alive people after the bell rings.
Demo Input:
['4\n0 1 0 10\n', '2\n0 0\n', '10\n1 1 3 0 0 0 2 1 0 3\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In first sample the last person kills everyone in front of him.
|
```python
n = int(input())
l = [0] + list(map(int, input().strip().split()))
totalDead = 0
currentLow = n + 1
currentHigh = n + 1
for i in range(n, 0, -1):
low = max(1, i - l[i])
high = i - 1
# valid
if (low <= high):
if (low <= currentLow and high >= currentLow):
currentLow = low
elif (high < currentLow):
totalDead += currentHigh - currentLow + 1
currentLow = low
currentHigh = high
totalDead += currentHigh - currentLow + 1
print(n - totalDead + 1)
```
| 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"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,637,670,215
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
m=int(input(),2)
n=int(input(),2)
print(int(bin(m^n)[2::]))
|
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
m=int(input(),2)
n=int(input(),2)
print(int(bin(m^n)[2::]))
```
| 0
|
26
|
A
|
Almost Prime
|
PROGRAMMING
| 900
|
[
"number theory"
] |
A. Almost Prime
|
2
|
256
|
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
|
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
|
Output the amount of almost prime numbers between 1 and *n*, inclusive.
|
[
"10\n",
"21\n"
] |
[
"2\n",
"8\n"
] |
none
| 500
|
[
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,603,912,392
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 186
| 307,200
|
def sieve(_numb):
first_dividers = []
for i in range(_numb + 1):
first_dividers.append(0)
first_dividers[0] = 1
first_dividers[1] = 1
for j in range(2, _numb + 1):
if first_dividers[j] != 0:
continue
if first_dividers[j] == 0:
first_dividers[j] = j
for x in range(j * 2, _numb + 1, j):
if first_dividers[x] == 0:
first_dividers[x] = j
return first_dividers
n = int(input())
counter = 0
dividers = sieve(n)
for i in range(6, n + 1):
alm_first = True
first_divider = dividers[i]
sec_divider = 0
numb = i
while True:
numb = numb // dividers[numb]
divider = dividers[numb]
if numb <= 1:
break
if divider != first_divider:
sec_divider = divider
if divider != sec_divider and sec_divider != 0:
alm_first = False
break
if sec_divider == 0 or alm_first is False:
continue
else:
counter += 1
print(counter)
|
Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none
|
```python
def sieve(_numb):
first_dividers = []
for i in range(_numb + 1):
first_dividers.append(0)
first_dividers[0] = 1
first_dividers[1] = 1
for j in range(2, _numb + 1):
if first_dividers[j] != 0:
continue
if first_dividers[j] == 0:
first_dividers[j] = j
for x in range(j * 2, _numb + 1, j):
if first_dividers[x] == 0:
first_dividers[x] = j
return first_dividers
n = int(input())
counter = 0
dividers = sieve(n)
for i in range(6, n + 1):
alm_first = True
first_divider = dividers[i]
sec_divider = 0
numb = i
while True:
numb = numb // dividers[numb]
divider = dividers[numb]
if numb <= 1:
break
if divider != first_divider:
sec_divider = divider
if divider != sec_divider and sec_divider != 0:
alm_first = False
break
if sec_divider == 0 or alm_first is False:
continue
else:
counter += 1
print(counter)
```
| 0
|
946
|
D
|
Timetable
|
PROGRAMMING
| 1,800
|
[
"dp"
] | null | null |
Ivan is a student at Berland State University (BSU). There are *n* days in Berland week, and each of these days Ivan might have some classes at the university.
There are *m* working hours during each Berland day, and each lesson at the university lasts exactly one hour. If at some day Ivan's first lesson is during *i*-th hour, and last lesson is during *j*-th hour, then he spends *j*<=-<=*i*<=+<=1 hours in the university during this day. If there are no lessons during some day, then Ivan stays at home and therefore spends 0 hours in the university.
Ivan doesn't like to spend a lot of time in the university, so he has decided to skip some lessons. He cannot skip more than *k* lessons during the week. After deciding which lessons he should skip and which he should attend, every day Ivan will enter the university right before the start of the first lesson he does not skip, and leave it after the end of the last lesson he decides to attend. If Ivan skips all lessons during some day, he doesn't go to the university that day at all.
Given *n*, *m*, *k* and Ivan's timetable, can you determine the minimum number of hours he has to spend in the university during one week, if he cannot skip more than *k* lessons?
|
The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=≤<=500) — the number of days in the Berland week, the number of working hours during each day, and the number of lessons Ivan can skip, respectively.
Then *n* lines follow, *i*-th line containing a binary string of *m* characters. If *j*-th character in *i*-th line is 1, then Ivan has a lesson on *i*-th day during *j*-th hour (if it is 0, there is no such lesson).
|
Print the minimum number of hours Ivan has to spend in the university during the week if he skips not more than *k* lessons.
|
[
"2 5 1\n01001\n10110\n",
"2 5 0\n01001\n10110\n"
] |
[
"5\n",
"8\n"
] |
In the first example Ivan can skip any of two lessons during the first day, so he spends 1 hour during the first day and 4 hours during the second day.
In the second example Ivan can't skip any lessons, so he spends 4 hours every day.
| 0
|
[
{
"input": "2 5 1\n01001\n10110",
"output": "5"
},
{
"input": "2 5 0\n01001\n10110",
"output": "8"
},
{
"input": "3 4 0\n0000\n0000\n0000",
"output": "0"
},
{
"input": "3 4 12\n1111\n1111\n1111",
"output": "0"
},
{
"input": "3 4 6\n1111\n1111\n1111",
"output": "6"
},
{
"input": "10 10 0\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001",
"output": "100"
},
{
"input": "10 10 5\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001",
"output": "55"
},
{
"input": "10 10 10\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001",
"output": "10"
},
{
"input": "10 10 20\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001",
"output": "0"
},
{
"input": "10 10 19\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001",
"output": "1"
},
{
"input": "4 16 11\n1111011101111111\n0111110111111111\n0011101111011000\n0000010000001000",
"output": "30"
},
{
"input": "1 1 5\n1",
"output": "0"
},
{
"input": "4 113 370\n01110100100010110100110000000011001111110000100111111001011001110111110111001011010110000110110010101010110001000\n01101001111110001010001100101100111100111001001001001101110101100110110110001110100010111011101011101110011110100\n01100110001001111010000010101100111100011111010000101010011011111111000111111001001010110110011011111110110010111\n11100111000100010000100111010101110110100101100100001111000001001010001000101110011100101011101100011010111010000",
"output": "0"
},
{
"input": "3 3 4\n000\n000\n000",
"output": "0"
},
{
"input": "1 5 1\n10001",
"output": "1"
},
{
"input": "1 1 1\n0",
"output": "0"
},
{
"input": "10 10 100\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001\n1000000001",
"output": "0"
},
{
"input": "5 1 6\n1\n1\n1\n0\n1",
"output": "0"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "8 4 8\n0001\n0010\n0100\n0000\n1001\n1101\n0010\n0001",
"output": "2"
},
{
"input": "1 1 2\n1",
"output": "0"
},
{
"input": "1 1 0\n1",
"output": "1"
},
{
"input": "1 1 2\n0",
"output": "0"
}
] | 1,553,977,676
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 11
| 2,000
| 1,740,800
|
inp = input()
n,m,k = [int(x) for x in inp.split(' ')]
schedules = []
for i in range(n):
schedules.append([int(x) for x in input()])
# m = 16
# k = 4
def all_shortcuts(day_schedule):
best_solutions = [99999999 for _ in range(k+1)]
for i in range(m):
for j in reversed(range(i, m+1)):
used = sum(day_schedule[:i]) + sum(day_schedule[j:])
if used <= k:
left = j-i
if left < best_solutions[used]:
best_solutions[used] = left
return best_solutions
# a = [int(x) for x in '0011011110111000']
#
# print(all_shortcuts(a))
#
# schedules = [
# [int(x) for x in '0011011110111000'],
# [int(x) for x in '0011011110111000']
# ]
best_solutions = [0 for _ in range(k+1)]
for day_schedule in schedules:
current_schedule = all_shortcuts(day_schedule)
worst_case = [9999999 for _ in range(k+1)]
for i in range(0, k + 1):
for j in range(k + 1 - i):
new_turn = current_schedule[i] + best_solutions[j]
if new_turn < worst_case[i + j]:
worst_case[i + j] = new_turn
# print(i+j, , worst_case[i + j])
best_solutions = worst_case
print(best_solutions[-1])
|
Title: Timetable
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan is a student at Berland State University (BSU). There are *n* days in Berland week, and each of these days Ivan might have some classes at the university.
There are *m* working hours during each Berland day, and each lesson at the university lasts exactly one hour. If at some day Ivan's first lesson is during *i*-th hour, and last lesson is during *j*-th hour, then he spends *j*<=-<=*i*<=+<=1 hours in the university during this day. If there are no lessons during some day, then Ivan stays at home and therefore spends 0 hours in the university.
Ivan doesn't like to spend a lot of time in the university, so he has decided to skip some lessons. He cannot skip more than *k* lessons during the week. After deciding which lessons he should skip and which he should attend, every day Ivan will enter the university right before the start of the first lesson he does not skip, and leave it after the end of the last lesson he decides to attend. If Ivan skips all lessons during some day, he doesn't go to the university that day at all.
Given *n*, *m*, *k* and Ivan's timetable, can you determine the minimum number of hours he has to spend in the university during one week, if he cannot skip more than *k* lessons?
Input Specification:
The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=≤<=500) — the number of days in the Berland week, the number of working hours during each day, and the number of lessons Ivan can skip, respectively.
Then *n* lines follow, *i*-th line containing a binary string of *m* characters. If *j*-th character in *i*-th line is 1, then Ivan has a lesson on *i*-th day during *j*-th hour (if it is 0, there is no such lesson).
Output Specification:
Print the minimum number of hours Ivan has to spend in the university during the week if he skips not more than *k* lessons.
Demo Input:
['2 5 1\n01001\n10110\n', '2 5 0\n01001\n10110\n']
Demo Output:
['5\n', '8\n']
Note:
In the first example Ivan can skip any of two lessons during the first day, so he spends 1 hour during the first day and 4 hours during the second day.
In the second example Ivan can't skip any lessons, so he spends 4 hours every day.
|
```python
inp = input()
n,m,k = [int(x) for x in inp.split(' ')]
schedules = []
for i in range(n):
schedules.append([int(x) for x in input()])
# m = 16
# k = 4
def all_shortcuts(day_schedule):
best_solutions = [99999999 for _ in range(k+1)]
for i in range(m):
for j in reversed(range(i, m+1)):
used = sum(day_schedule[:i]) + sum(day_schedule[j:])
if used <= k:
left = j-i
if left < best_solutions[used]:
best_solutions[used] = left
return best_solutions
# a = [int(x) for x in '0011011110111000']
#
# print(all_shortcuts(a))
#
# schedules = [
# [int(x) for x in '0011011110111000'],
# [int(x) for x in '0011011110111000']
# ]
best_solutions = [0 for _ in range(k+1)]
for day_schedule in schedules:
current_schedule = all_shortcuts(day_schedule)
worst_case = [9999999 for _ in range(k+1)]
for i in range(0, k + 1):
for j in range(k + 1 - i):
new_turn = current_schedule[i] + best_solutions[j]
if new_turn < worst_case[i + j]:
worst_case[i + j] = new_turn
# print(i+j, , worst_case[i + j])
best_solutions = worst_case
print(best_solutions[-1])
```
| 0
|
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,600,774,411
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 202
| 2,150,400
|
mas=[" "]*5;
mas[0]="Sheldon";
mas[1]="Leonard";
mas[2]="Penny";
mas[3]="Rajesh";
mas[4]="Howard";
n=int(input());
n-=1
while n >= 5:
n = n - 5
n = n / 2
print(mas[n]);
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
mas=[" "]*5;
mas[0]="Sheldon";
mas[1]="Leonard";
mas[2]="Penny";
mas[3]="Rajesh";
mas[4]="Howard";
n=int(input());
n-=1
while n >= 5:
n = n - 5
n = n / 2
print(mas[n]);
```
| -1
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,572,026,650
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 218
| 0
|
str1=input()
str2=input()
eq1=int(str1)+int(str2)
eq1=str(eq1)
wz1=str1.replace("0","")
wz2=str2.replace("0","")
eq2=eq1.replace("0","")
b=int(wz1)+int(wz2)
a=int(eq2)
if a==b:
print("YES")
else:
print("NO")
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
str1=input()
str2=input()
eq1=int(str1)+int(str2)
eq1=str(eq1)
wz1=str1.replace("0","")
wz2=str2.replace("0","")
eq2=eq1.replace("0","")
b=int(wz1)+int(wz2)
a=int(eq2)
if a==b:
print("YES")
else:
print("NO")
```
| 3.9455
|
19
|
B
|
Checkout Assistant
|
PROGRAMMING
| 1,900
|
[
"dp"
] |
B. Checkout Assistant
|
1
|
256
|
Bob came to a cash & carry store, put *n* items into his trolley, and went to the checkout counter to pay. Each item is described by its price *c**i* and time *t**i* in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.
|
The first input line contains number *n* (1<=≤<=*n*<=≤<=2000). In each of the following *n* lines each item is described by a pair of numbers *t**i*, *c**i* (0<=≤<=*t**i*<=≤<=2000,<=1<=≤<=*c**i*<=≤<=109). If *t**i* is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item *i*.
|
Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.
|
[
"4\n2 10\n0 20\n1 5\n1 3\n",
"3\n0 1\n0 10\n0 100\n"
] |
[
"8\n",
"111\n"
] |
none
| 0
|
[
{
"input": "4\n2 10\n0 20\n1 5\n1 3",
"output": "8"
},
{
"input": "3\n0 1\n0 10\n0 100",
"output": "111"
},
{
"input": "2\n0 635254032\n0 75159864",
"output": "710413896"
},
{
"input": "2\n0 861438648\n1 469893784",
"output": "469893784"
},
{
"input": "2\n2 87623264\n0 864627704",
"output": "87623264"
},
{
"input": "5\n0 51690939\n0 62436494\n0 39009053\n0 70426298\n0 631569476",
"output": "855132260"
},
{
"input": "5\n0 968804136\n0 736567537\n2 343136264\n0 259899572\n1 425744418",
"output": "768880682"
},
{
"input": "5\n4 33400980\n2 410698581\n4 794747123\n0 301889198\n3 219919361",
"output": "33400980"
},
{
"input": "5\n5 491957578\n4 747764103\n3 446810728\n4 896625835\n4 190748656",
"output": "190748656"
},
{
"input": "10\n3 512981548\n0 214269975\n2 603901234\n3 772872647\n0 224281389\n4 561877930\n1 809519308\n4 883486551\n1 114469023\n2 184038037",
"output": "860384990"
}
] | 1,542,471,174
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 93
| 0
|
qtd = int(input())
buys = []
for i in range(qtd):
buys.append(tuple([int(x) for x in input().split()]))
buys = sorted(buys, key=lambda tup: tup[1]) #sortList
buysLeft = len(buys)
lixocara = []
cashLixo = 0
cash = 0
for i in range(len(buys)):
if buysLeft <= 0:
break
if buys[i][0] == 0:
lixocara.append(i)
continue
else:
cash += buys[i][1]
buysLeft -= buys[i][0]
buysLeft -= 1
if buysLeft > 0:
for i in lixocara:
cash += buys[i][1]
buysLeft -= 1
if buysLeft == 0:
break
print(cash)
|
Title: Checkout Assistant
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob came to a cash & carry store, put *n* items into his trolley, and went to the checkout counter to pay. Each item is described by its price *c**i* and time *t**i* in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.
Input Specification:
The first input line contains number *n* (1<=≤<=*n*<=≤<=2000). In each of the following *n* lines each item is described by a pair of numbers *t**i*, *c**i* (0<=≤<=*t**i*<=≤<=2000,<=1<=≤<=*c**i*<=≤<=109). If *t**i* is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item *i*.
Output Specification:
Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.
Demo Input:
['4\n2 10\n0 20\n1 5\n1 3\n', '3\n0 1\n0 10\n0 100\n']
Demo Output:
['8\n', '111\n']
Note:
none
|
```python
qtd = int(input())
buys = []
for i in range(qtd):
buys.append(tuple([int(x) for x in input().split()]))
buys = sorted(buys, key=lambda tup: tup[1]) #sortList
buysLeft = len(buys)
lixocara = []
cashLixo = 0
cash = 0
for i in range(len(buys)):
if buysLeft <= 0:
break
if buys[i][0] == 0:
lixocara.append(i)
continue
else:
cash += buys[i][1]
buysLeft -= buys[i][0]
buysLeft -= 1
if buysLeft > 0:
for i in lixocara:
cash += buys[i][1]
buysLeft -= 1
if buysLeft == 0:
break
print(cash)
```
| 0
|
44
|
A
|
Indian Summer
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. Indian Summer
|
2
|
256
|
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
|
Output the single number — the number of Alyona's leaves.
|
[
"5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n",
"3\noak yellow\noak yellow\noak yellow\n"
] |
[
"4\n",
"1\n"
] |
none
| 0
|
[
{
"input": "5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green",
"output": "4"
},
{
"input": "3\noak yellow\noak yellow\noak yellow",
"output": "1"
},
{
"input": "5\nxbnbkzn hp\nkaqkl vrgzbvqstu\nj aqidx\nhos gyul\nwefxmh tygpluae",
"output": "5"
},
{
"input": "1\nqvwli hz",
"output": "1"
},
{
"input": "4\nsrhk x\nsrhk x\nqfoe vnrjuab\nqfoe vnrjuab",
"output": "2"
},
{
"input": "4\nsddqllmmpk syded\nfprsq fnenjnaz\nn hdej\nsddqllmmpk syded",
"output": "3"
},
{
"input": "17\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw",
"output": "1"
},
{
"input": "18\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nb d\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp",
"output": "2"
},
{
"input": "13\nsvpzwtwn rykzfdce\nqweiyeck jkreouy\nhk nnli\ntwxrnbbdt vtuv\nxokqjtylly sz\nesdt dbfidjslq\ng ybqgomvw\nxcpfjmf zcqvz\nifajadhj invzueip\nvdemdnxifb hckwebmi\nsdpnhipam wvowzavh\nuqdlfskhgo vunbpghae\ne dtigwnb",
"output": "13"
},
{
"input": "20\nm vkfh\nvptikamead mvx\nitu mf\nklruxckw aqega\nekqkd enjllwol\ncc uybfdh\nimrfdngvo u\ne uh\ntwt jsslcfuogk\nbljwqsag tuqbdn\nqcv q\nasx gzhvwwmajj\nqcv q\nekqkd enjllwol\nasx gzhvwwmajj\nks vv\nkzyfi cn\ncc uybfdh\nitu mf\ncjbjhtbyvk vatwfmux",
"output": "15"
},
{
"input": "2\nab ab\na bab",
"output": "2"
}
] | 1,645,889,599
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 29
| 124
| 0
|
s = set()
for _ in range(int(input())):
s.add(input())
print(len(s))
|
Title: Indian Summer
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
Output Specification:
Output the single number — the number of Alyona's leaves.
Demo Input:
['5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n', '3\noak yellow\noak yellow\noak yellow\n']
Demo Output:
['4\n', '1\n']
Note:
none
|
```python
s = set()
for _ in range(int(input())):
s.add(input())
print(len(s))
```
| 3.969
|
701
|
C
|
They Are Everywhere
|
PROGRAMMING
| 1,500
|
[
"binary search",
"strings",
"two pointers"
] | null | null |
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
|
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
|
[
"3\nAaA\n",
"7\nbcAAcbc\n",
"6\naaBCCe\n"
] |
[
"2\n",
"3\n",
"5\n"
] |
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
| 1,000
|
[
{
"input": "3\nAaA",
"output": "2"
},
{
"input": "7\nbcAAcbc",
"output": "3"
},
{
"input": "6\naaBCCe",
"output": "5"
},
{
"input": "1\nA",
"output": "1"
},
{
"input": "1\ng",
"output": "1"
},
{
"input": "52\nabcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "52"
},
{
"input": "2\nAA",
"output": "1"
},
{
"input": "4\nqqqE",
"output": "2"
},
{
"input": "10\nrrrrroooro",
"output": "2"
},
{
"input": "15\nOCOCCCCiCOCCCOi",
"output": "3"
},
{
"input": "20\nVEVnVVnWnVEVVnEVBEWn",
"output": "5"
},
{
"input": "25\ncpcyPPjPPcPPPPcppPcPpppcP",
"output": "6"
},
{
"input": "30\nsssssAsesssssssssssssessssssss",
"output": "3"
},
{
"input": "35\ngdXdddgddddddddggggXdbgdggdgddddddb",
"output": "4"
},
{
"input": "40\nIgsggIiIggzgigIIiiIIIiIgIggIzgIiiiggggIi",
"output": "9"
},
{
"input": "45\neteeeeeteaattaeetaetteeettoetettteyeteeeotaae",
"output": "9"
},
{
"input": "50\nlUlUllUlUllllUllllUllllUlUlllUlllUlllllUUlllUUlkUl",
"output": "3"
},
{
"input": "55\nAAAAASAAAASAASAAAAAAAAAAAAASAAAAAAAAAAAAAAAASAAAAAAAAAA",
"output": "2"
},
{
"input": "60\nRRRrSRRRRRRRRRSSRRRSRRRRRRRRrRSRRRRRRRRRRRRRRSRRRRRSSRSRrRRR",
"output": "3"
},
{
"input": "65\nhhMhMhhhhhhhhhhhMhhMMMhhhhBhhhhMhhhhMhhhhhMhhhBhhhhhhhhhhBhhhhhhh",
"output": "5"
},
{
"input": "70\nwAwwwAwwwwwwwwwwwwwwAwAAwwAwwwwwwwwAwAAAwAAwwwwwwwwwAwwwwwwwwwwwwAAwww",
"output": "2"
},
{
"input": "75\niiiXXiiyiiiXyXiiyXiiXiiiiiiXXyiiiiXXiiXiiXifiXiXXiifiiiiiiXfXiyiXXiXiiiiXiX",
"output": "4"
},
{
"input": "80\nSrSrrrrrrrrrrrrrrSSSrrrrrrSrrrrSrrrrrrrrrrSSrrrrrrrrrrrSrrrSrrrrSrrrrSrrrrSSrSSr",
"output": "2"
},
{
"input": "85\nwkMMMwMMkMMMMMMMkkkkMMMMzkkMMwMMkkwMkMwkMMkMMwwMzMMMkkMwwMMMMMMkMMkMzMMMkMMkwMkMMMkMM",
"output": "6"
},
{
"input": "90\nZllZZZyZlZlllZlylllZlllZZllllllllZZllllllllllyylZZyvZvZlllZZlZllZlZlllZyllZyZlllZlllllllZl",
"output": "5"
},
{
"input": "95\nEmuBuEBmmEBBElBlElmmBEmmmEmmEuBEEmummmEmBBBBEWBBBmEEBmmummBBmmlluBBmElmEBEmBmBBmBmuLmEBBmlEBmBu",
"output": "39"
},
{
"input": "100\nfAfhfNNNhfffAfNffNANhffNffffffNfffffhNfffNfffNfffNfNNNffNfffNfhNNNffffNfNfffAffNffNffNNfhfffNhNffNNN",
"output": "4"
},
{
"input": "7\nabcbbab",
"output": "3"
},
{
"input": "65\nkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkjjjjkkkkkkkkkkkkkkkkkkkllllllljk",
"output": "3"
},
{
"input": "9\nbcdddbddc",
"output": "3"
},
{
"input": "71\nlqcrdopdpfagcilpcmheqhjbmnciqmeoabflrenbkrsaabaanpehdmpqcdqnljcrsplnlco",
"output": "36"
},
{
"input": "8\nabbbbbbc",
"output": "8"
}
] | 1,654,475,888
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 80
| 187
| 614,400
|
"""
They Are Everywhere
Sergei B., the young coach of Pokemons, has found the big house which consists of
n flats ordered in a row from left to right. It is possible to enter each flat
from the street. It is possible to go out from each flat. Also, each flat is
connected with the flat to the left and the flat to the right. Flat number 1
is only connected with the flat number 2 and the flat number n is only connected
with the flat number n - 1.
There is exactly one Pokemon of some type in each of these flats. Sergei B.
asked residents of the house to let him enter their flats in order to catch Pokemons.
After consulting the residents of the house decided to let Sergei B. enter one flat
from the street, visit several flats and then go out from some flat. But they won't
let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in
order to collect Pokemons of all types that appear in this house. Your task is to
help him and determine this minimum number of flats he has to visit.
"""
#import sys
#sys.stdin=open('theyare_everywhere_test.txt','r')
n = map(int, input())
s = input()
L = len(set(list(s)))
i = 0
j = 0
res = len(s)
cur_dict = {}
for i in range(len(s)):
while j<len(s) and len(cur_dict)<L:
if s[j] not in cur_dict:
cur_dict[s[j]] = 1
else:
cur_dict[s[j]] += 1
j += 1
if len(cur_dict)==L:
res = min(res,j-i)
cur_dict[s[i]] -= 1
if cur_dict[s[i]]==0:
del cur_dict[s[i]]
print(res)
|
Title: They Are Everywhere
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
Output Specification:
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Demo Input:
['3\nAaA\n', '7\nbcAAcbc\n', '6\naaBCCe\n']
Demo Output:
['2\n', '3\n', '5\n']
Note:
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
|
```python
"""
They Are Everywhere
Sergei B., the young coach of Pokemons, has found the big house which consists of
n flats ordered in a row from left to right. It is possible to enter each flat
from the street. It is possible to go out from each flat. Also, each flat is
connected with the flat to the left and the flat to the right. Flat number 1
is only connected with the flat number 2 and the flat number n is only connected
with the flat number n - 1.
There is exactly one Pokemon of some type in each of these flats. Sergei B.
asked residents of the house to let him enter their flats in order to catch Pokemons.
After consulting the residents of the house decided to let Sergei B. enter one flat
from the street, visit several flats and then go out from some flat. But they won't
let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in
order to collect Pokemons of all types that appear in this house. Your task is to
help him and determine this minimum number of flats he has to visit.
"""
#import sys
#sys.stdin=open('theyare_everywhere_test.txt','r')
n = map(int, input())
s = input()
L = len(set(list(s)))
i = 0
j = 0
res = len(s)
cur_dict = {}
for i in range(len(s)):
while j<len(s) and len(cur_dict)<L:
if s[j] not in cur_dict:
cur_dict[s[j]] = 1
else:
cur_dict[s[j]] += 1
j += 1
if len(cur_dict)==L:
res = min(res,j-i)
cur_dict[s[i]] -= 1
if cur_dict[s[i]]==0:
del cur_dict[s[i]]
print(res)
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,644,907,962
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
cumle=input()
bharf=0
for n in cumle:
if n.isupper():
bharf=bharf+1
if bharf>len(cumle)-bharf:
print(cumle.upper())
else:
print(cumle.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
cumle=input()
bharf=0
for n in cumle:
if n.isupper():
bharf=bharf+1
if bharf>len(cumle)-bharf:
print(cumle.upper())
else:
print(cumle.lower())
```
| 3.977
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,676,822,410
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 124
| 0
|
s1=input()
s2=input()
flag=True
for i in range(len(s1)):
if s1[i]!=s2[len(s2)-1-i]:
flag=False
break
if flag:
print('YES')
else:
print('NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s1=input()
s2=input()
flag=True
for i in range(len(s1)):
if s1[i]!=s2[len(s2)-1-i]:
flag=False
break
if flag:
print('YES')
else:
print('NO')
```
| 3.969
|
919
|
D
|
Substring
|
PROGRAMMING
| 1,700
|
[
"dfs and similar",
"dp",
"graphs"
] | null | null |
You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest.
|
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges.
The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node.
Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected.
|
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
|
[
"5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n",
"6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n",
"10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n"
] |
[
"3\n",
"-1\n",
"4\n"
] |
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
| 1,500
|
[
{
"input": "5 4\nabaca\n1 2\n1 3\n3 4\n4 5",
"output": "3"
},
{
"input": "6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4",
"output": "-1"
},
{
"input": "10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7",
"output": "4"
},
{
"input": "1 1\nf\n1 1",
"output": "-1"
},
{
"input": "10 50\nebibwbjihv\n1 10\n1 2\n5 4\n1 8\n9 7\n5 6\n1 8\n8 7\n2 6\n5 4\n1 9\n3 2\n8 3\n5 6\n5 9\n2 4\n2 7\n3 9\n1 2\n1 7\n1 10\n3 7\n1 8\n3 10\n8 6\n1 7\n10 6\n1 6\n5 8\n1 5\n2 10\n3 9\n5 8\n8 3\n3 7\n5 2\n1 10\n1 4\n5 3\n3 2\n1 2\n5 8\n10 4\n2 10\n8 2\n1 9\n1 8\n1 2\n3 4\n1 8",
"output": "2"
},
{
"input": "13 37\ndwpzcppjmhkmz\n2 6\n3 6\n6 7\n6 7\n6 7\n6 7\n6 8\n6 8\n6 8\n6 8\n4 6\n4 6\n5 6\n4 6\n4 6\n6 9\n6 9\n6 10\n6 10\n6 10\n6 10\n4 6\n1 6\n1 6\n10 11\n6 11\n1 6\n6 12\n6 12\n6 12\n6 13\n6 13\n6 13\n6 13\n3 6\n2 6\n2 6",
"output": "3"
},
{
"input": "5 8\ntetqw\n2 1\n4 4\n5 5\n5 2\n4 5\n1 5\n1 5\n1 1",
"output": "-1"
},
{
"input": "5 8\nreeet\n4 3\n2 5\n4 2\n2 4\n4 2\n5 2\n3 3\n3 4",
"output": "-1"
}
] | 1,626,099,308
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 1,575
| 29,798,400
|
def detect_cycle(graph,s,visited):
visited[s]=True
for i in graph[s]:
if visited[i]==False:
if detect_cycle(graph,i,visited):
return True
else:
return False
elif(visited[i]==True):
return True
return False
def dfs_path(graph,s,visited,path,maxx,string,char_ar):
visited[s]=True
char_ar[ord(string[s-1])-ord('a')]+=1
maxx.append(max(char_ar))
for i in graph[s]:
if visited[i]==False:
dfs_path(graph,i,visited,path,maxx,string,char_ar)
char_ar[ord(string[s-1])-ord('a')]-=1
n,m=map(int,input().split())
string=input()
graph=[[] for i in range(n+1)]
for i in range(m):
a,b=map(int,input().split())
graph[a].append(b)
s=1
visited=[False]*len(graph)
if(detect_cycle(graph,s,visited)):
print(-1)
#continue
else:
visited=[False]*len(graph)
ans=0
for i in range(1,n+1):
if visited[i]==False:
maxx=[]
char_ar=[0]*26
dfs_path(graph,s,visited,[],maxx,string,char_ar)
ans=max(ans,max(maxx))
print(ans)
|
Title: Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest.
Input Specification:
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges.
The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node.
Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected.
Output Specification:
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
Demo Input:
['5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n', '6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n', '10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n']
Demo Output:
['3\n', '-1\n', '4\n']
Note:
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
|
```python
def detect_cycle(graph,s,visited):
visited[s]=True
for i in graph[s]:
if visited[i]==False:
if detect_cycle(graph,i,visited):
return True
else:
return False
elif(visited[i]==True):
return True
return False
def dfs_path(graph,s,visited,path,maxx,string,char_ar):
visited[s]=True
char_ar[ord(string[s-1])-ord('a')]+=1
maxx.append(max(char_ar))
for i in graph[s]:
if visited[i]==False:
dfs_path(graph,i,visited,path,maxx,string,char_ar)
char_ar[ord(string[s-1])-ord('a')]-=1
n,m=map(int,input().split())
string=input()
graph=[[] for i in range(n+1)]
for i in range(m):
a,b=map(int,input().split())
graph[a].append(b)
s=1
visited=[False]*len(graph)
if(detect_cycle(graph,s,visited)):
print(-1)
#continue
else:
visited=[False]*len(graph)
ans=0
for i in range(1,n+1):
if visited[i]==False:
maxx=[]
char_ar=[0]*26
dfs_path(graph,s,visited,[],maxx,string,char_ar)
ans=max(ans,max(maxx))
print(ans)
```
| 0
|
|
202
|
A
|
LLPS
|
PROGRAMMING
| 800
|
[
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null |
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
|
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
|
Print the lexicographically largest palindromic subsequence of string *s*.
|
[
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] |
[
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] |
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
| 500
|
[
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,601,852,450
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 248
| 0
|
# LLPS
s = input()
ch_max = None
for ch in s:
if ch_max is None:
ch_max = ch
num = 1
elif ch > ch_max:
ch_max = ch
num = 1
elif ch == ch_max:
num += 1
print(ch_max * num)
|
Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
```python
# LLPS
s = input()
ch_max = None
for ch in s:
if ch_max is None:
ch_max = ch
num = 1
elif ch > ch_max:
ch_max = ch
num = 1
elif ch == ch_max:
num += 1
print(ch_max * num)
```
| 3
|
|
883
|
M
|
Quadcopter Competition
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Polycarp takes part in a quadcopter competition. According to the rules a flying robot should:
- start the race from some point of a field, - go around the flag, - close cycle returning back to the starting point.
Polycarp knows the coordinates of the starting point (*x*1,<=*y*1) and the coordinates of the point where the flag is situated (*x*2,<=*y*2). Polycarp’s quadcopter can fly only parallel to the sides of the field each tick changing exactly one coordinate by 1. It means that in one tick the quadcopter can fly from the point (*x*,<=*y*) to any of four points: (*x*<=-<=1,<=*y*), (*x*<=+<=1,<=*y*), (*x*,<=*y*<=-<=1) or (*x*,<=*y*<=+<=1).
Thus the quadcopter path is a closed cycle starting and finishing in (*x*1,<=*y*1) and containing the point (*x*2,<=*y*2) strictly inside.
What is the minimal length of the quadcopter path?
|
The first line contains two integer numbers *x*1 and *y*1 (<=-<=100<=≤<=*x*1,<=*y*1<=≤<=100) — coordinates of the quadcopter starting (and finishing) point.
The second line contains two integer numbers *x*2 and *y*2 (<=-<=100<=≤<=*x*2,<=*y*2<=≤<=100) — coordinates of the flag.
It is guaranteed that the quadcopter starting point and the flag do not coincide.
|
Print the length of minimal path of the quadcopter to surround the flag and return back.
|
[
"1 5\n5 2\n",
"0 1\n0 0\n"
] |
[
"18\n",
"8\n"
] |
none
| 0
|
[
{
"input": "1 5\n5 2",
"output": "18"
},
{
"input": "0 1\n0 0",
"output": "8"
},
{
"input": "-100 -100\n100 100",
"output": "804"
},
{
"input": "-100 -100\n-100 100",
"output": "406"
},
{
"input": "-100 -100\n100 -100",
"output": "406"
},
{
"input": "100 -100\n-100 -100",
"output": "406"
},
{
"input": "100 -100\n-100 100",
"output": "804"
},
{
"input": "100 -100\n100 100",
"output": "406"
},
{
"input": "-100 100\n-100 -100",
"output": "406"
},
{
"input": "-100 100\n100 -100",
"output": "804"
},
{
"input": "-100 100\n100 100",
"output": "406"
},
{
"input": "100 100\n-100 -100",
"output": "804"
},
{
"input": "100 100\n-100 100",
"output": "406"
},
{
"input": "100 100\n100 -100",
"output": "406"
},
{
"input": "45 -43\n45 -44",
"output": "8"
},
{
"input": "76 76\n75 75",
"output": "8"
},
{
"input": "-34 -56\n-35 -56",
"output": "8"
},
{
"input": "56 -7\n55 -6",
"output": "8"
},
{
"input": "43 -11\n43 -10",
"output": "8"
},
{
"input": "1 -3\n2 -2",
"output": "8"
},
{
"input": "55 71\n56 71",
"output": "8"
},
{
"input": "54 -87\n55 -88",
"output": "8"
},
{
"input": "22 98\n100 33",
"output": "290"
},
{
"input": "37 84\n-83 5",
"output": "402"
},
{
"input": "52 74\n-73 -39",
"output": "480"
},
{
"input": "66 51\n51 -71",
"output": "278"
},
{
"input": "-31 44\n73 86",
"output": "296"
},
{
"input": "-20 34\n-9 55",
"output": "68"
},
{
"input": "-5 19\n-91 -86",
"output": "386"
},
{
"input": "-82 5\n28 -17",
"output": "268"
},
{
"input": "-90 -100\n55 48",
"output": "590"
},
{
"input": "-75 -14\n-32 8",
"output": "134"
},
{
"input": "-53 -28\n-13 -28",
"output": "86"
},
{
"input": "-42 -46\n10 -64",
"output": "144"
},
{
"input": "55 -42\n25 2",
"output": "152"
},
{
"input": "70 -64\n-54 70",
"output": "520"
},
{
"input": "93 -78\n-32 -75",
"output": "260"
},
{
"input": "8 -93\n79 -6",
"output": "320"
},
{
"input": "50 43\n54 10",
"output": "78"
},
{
"input": "65 32\n-37 71",
"output": "286"
},
{
"input": "80 18\n-15 -58",
"output": "346"
},
{
"input": "94 92\n4 -1",
"output": "370"
},
{
"input": "-10 96\n27 64",
"output": "142"
},
{
"input": "-96 78\n-56 32",
"output": "176"
},
{
"input": "-81 64\n-37 -8",
"output": "236"
},
{
"input": "-58 49\n74 -40",
"output": "446"
},
{
"input": "-62 -55\n1 18",
"output": "276"
},
{
"input": "-51 -69\n-78 86",
"output": "368"
},
{
"input": "-29 -80\n-56 -47",
"output": "124"
},
{
"input": "-14 -94\n55 -90",
"output": "150"
},
{
"input": "83 -2\n82 83",
"output": "176"
},
{
"input": "98 -16\n-96 40",
"output": "504"
},
{
"input": "17 -34\n-86 -93",
"output": "328"
},
{
"input": "32 -48\n33 -37",
"output": "28"
},
{
"input": "74 87\n3 92",
"output": "156"
},
{
"input": "89 73\n-80 49",
"output": "390"
},
{
"input": "4 58\n-61 -80",
"output": "410"
},
{
"input": "15 48\n50 -20",
"output": "210"
},
{
"input": "-82 45\n81 46",
"output": "332"
},
{
"input": "-68 26\n-2 6",
"output": "176"
},
{
"input": "-53 4\n-92 -31",
"output": "152"
},
{
"input": "-30 94\n31 -58",
"output": "430"
},
{
"input": "-38 -11\n58 99",
"output": "416"
},
{
"input": "-27 -25\n-28 68",
"output": "192"
},
{
"input": "-5 -39\n-10 -77",
"output": "90"
},
{
"input": "-90 -54\n9 -9",
"output": "292"
},
{
"input": "7 -57\n28 61",
"output": "282"
},
{
"input": "18 -67\n-51 21",
"output": "318"
},
{
"input": "41 -82\n-33 -15",
"output": "286"
},
{
"input": "56 -8\n91 -55",
"output": "168"
},
{
"input": "-23 -13\n-24 -12",
"output": "8"
},
{
"input": "1 32\n1 33",
"output": "8"
},
{
"input": "25 76\n24 76",
"output": "8"
},
{
"input": "-29 -78\n-28 -79",
"output": "8"
},
{
"input": "-77 19\n-76 19",
"output": "8"
},
{
"input": "-53 63\n-53 62",
"output": "8"
},
{
"input": "86 12\n86 11",
"output": "8"
},
{
"input": "14 56\n13 56",
"output": "8"
},
{
"input": "63 41\n62 40",
"output": "8"
},
{
"input": "0 -4\n1 -4",
"output": "8"
},
{
"input": "24 41\n24 42",
"output": "8"
},
{
"input": "48 85\n49 86",
"output": "8"
},
{
"input": "0 0\n0 1",
"output": "8"
},
{
"input": "0 0\n1 0",
"output": "8"
},
{
"input": "0 0\n1 1",
"output": "8"
},
{
"input": "0 1\n0 0",
"output": "8"
},
{
"input": "0 1\n1 0",
"output": "8"
},
{
"input": "0 1\n1 1",
"output": "8"
},
{
"input": "1 0\n0 0",
"output": "8"
},
{
"input": "1 0\n0 1",
"output": "8"
},
{
"input": "1 0\n1 1",
"output": "8"
},
{
"input": "1 1\n0 0",
"output": "8"
},
{
"input": "1 1\n0 1",
"output": "8"
},
{
"input": "1 1\n1 0",
"output": "8"
},
{
"input": "100 100\n99 -100",
"output": "406"
},
{
"input": "100 100\n-100 99",
"output": "406"
},
{
"input": "-100 -100\n-99 100",
"output": "406"
},
{
"input": "-100 -100\n100 -99",
"output": "406"
},
{
"input": "0 0\n1 2",
"output": "10"
},
{
"input": "0 0\n2 1",
"output": "10"
}
] | 1,585,334,940
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 100
| 109
| 0
|
x1,y1=map(int,input().split())
x2,y2=map(int,input().split())
print(2*(max(1,abs((x2)-x1))+max(1,abs((y2-y1)))+2))
|
Title: Quadcopter Competition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp takes part in a quadcopter competition. According to the rules a flying robot should:
- start the race from some point of a field, - go around the flag, - close cycle returning back to the starting point.
Polycarp knows the coordinates of the starting point (*x*1,<=*y*1) and the coordinates of the point where the flag is situated (*x*2,<=*y*2). Polycarp’s quadcopter can fly only parallel to the sides of the field each tick changing exactly one coordinate by 1. It means that in one tick the quadcopter can fly from the point (*x*,<=*y*) to any of four points: (*x*<=-<=1,<=*y*), (*x*<=+<=1,<=*y*), (*x*,<=*y*<=-<=1) or (*x*,<=*y*<=+<=1).
Thus the quadcopter path is a closed cycle starting and finishing in (*x*1,<=*y*1) and containing the point (*x*2,<=*y*2) strictly inside.
What is the minimal length of the quadcopter path?
Input Specification:
The first line contains two integer numbers *x*1 and *y*1 (<=-<=100<=≤<=*x*1,<=*y*1<=≤<=100) — coordinates of the quadcopter starting (and finishing) point.
The second line contains two integer numbers *x*2 and *y*2 (<=-<=100<=≤<=*x*2,<=*y*2<=≤<=100) — coordinates of the flag.
It is guaranteed that the quadcopter starting point and the flag do not coincide.
Output Specification:
Print the length of minimal path of the quadcopter to surround the flag and return back.
Demo Input:
['1 5\n5 2\n', '0 1\n0 0\n']
Demo Output:
['18\n', '8\n']
Note:
none
|
```python
x1,y1=map(int,input().split())
x2,y2=map(int,input().split())
print(2*(max(1,abs((x2)-x1))+max(1,abs((y2-y1)))+2))
```
| 3
|
|
791
|
A
|
Bear and Big Brother
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
|
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
|
[
"4 7\n",
"4 9\n",
"1 1\n"
] |
[
"2\n",
"3\n",
"1\n"
] |
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
| 500
|
[
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "4"
},
{
"input": "1 6",
"output": "5"
},
{
"input": "1 7",
"output": "5"
},
{
"input": "1 8",
"output": "6"
},
{
"input": "1 9",
"output": "6"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "2 4",
"output": "2"
},
{
"input": "2 5",
"output": "3"
},
{
"input": "2 6",
"output": "3"
},
{
"input": "2 7",
"output": "4"
},
{
"input": "2 8",
"output": "4"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "2 10",
"output": "4"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "3 4",
"output": "1"
},
{
"input": "3 5",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "3 7",
"output": "3"
},
{
"input": "3 8",
"output": "3"
},
{
"input": "3 9",
"output": "3"
},
{
"input": "3 10",
"output": "3"
},
{
"input": "4 4",
"output": "1"
},
{
"input": "4 5",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "4 7",
"output": "2"
},
{
"input": "4 8",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "4 10",
"output": "3"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "5 6",
"output": "1"
},
{
"input": "5 7",
"output": "1"
},
{
"input": "5 8",
"output": "2"
},
{
"input": "5 9",
"output": "2"
},
{
"input": "5 10",
"output": "2"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "6 7",
"output": "1"
},
{
"input": "6 8",
"output": "1"
},
{
"input": "6 9",
"output": "2"
},
{
"input": "6 10",
"output": "2"
},
{
"input": "7 7",
"output": "1"
},
{
"input": "7 8",
"output": "1"
},
{
"input": "7 9",
"output": "1"
},
{
"input": "7 10",
"output": "1"
},
{
"input": "8 8",
"output": "1"
},
{
"input": "8 9",
"output": "1"
},
{
"input": "8 10",
"output": "1"
},
{
"input": "9 9",
"output": "1"
},
{
"input": "9 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "1 2",
"output": "2"
}
] | 1,698,289,297
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 62
| 46
| 0
|
def bear():
user = input()
sep_weights = user.split(" ")
a = int(sep_weights[0])
b = int(sep_weights[1])
t = 0
while a <= b:
a *= 3
b *= 2
t += 1
print(t)
bear()
|
Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
|
```python
def bear():
user = input()
sep_weights = user.split(" ")
a = int(sep_weights[0])
b = int(sep_weights[1])
t = 0
while a <= b:
a *= 3
b *= 2
t += 1
print(t)
bear()
```
| 3
|
|
49
|
A
|
Sleuth
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Sleuth
|
2
|
256
|
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
|
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
|
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
|
[
"Is it a melon?\n",
"Is it an apple?\n",
"Is it a banana ?\n",
"Is it an apple and a banana simultaneouSLY?\n"
] |
[
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "Is it a melon?",
"output": "NO"
},
{
"input": "Is it an apple?",
"output": "YES"
},
{
"input": " Is it a banana ?",
"output": "YES"
},
{
"input": "Is it an apple and a banana simultaneouSLY?",
"output": "YES"
},
{
"input": "oHtSbDwzHb?",
"output": "NO"
},
{
"input": "sZecYdUvZHrXx?",
"output": "NO"
},
{
"input": "uMtXK?",
"output": "NO"
},
{
"input": "U?",
"output": "YES"
},
{
"input": "aqFDkCUKeHMyvZFcAyWlMUSQTFomtaWjoKLVyxLCw vcufPBFbaljOuHWiDCROYTcmbgzbaqHXKPOYEbuEtRqqoxBbOETCsQzhw?",
"output": "NO"
},
{
"input": "dJcNqQiFXzcbsj fItCpBLyXOnrSBPebwyFHlxUJHqCUzzCmcAvMiKL NunwOXnKeIxUZmBVwiCUfPkjRAkTPbkYCmwRRnDSLaz?",
"output": "NO"
},
{
"input": "gxzXbdcAQMuFKuuiPohtMgeypr wpDIoDSyOYTdvylcg SoEBZjnMHHYZGEqKgCgBeTbyTwyGuPZxkxsnSczotBdYyfcQsOVDVC?",
"output": "NO"
},
{
"input": "FQXBisXaJFMiHFQlXjixBDMaQuIbyqSBKGsBfTmBKCjszlGVZxEOqYYqRTUkGpSDDAoOXyXcQbHcPaegeOUBNeSD JiKOdECPOF?",
"output": "NO"
},
{
"input": "YhCuZnrWUBEed?",
"output": "NO"
},
{
"input": "hh?",
"output": "NO"
},
{
"input": "whU?",
"output": "YES"
},
{
"input": "fgwg?",
"output": "NO"
},
{
"input": "GlEmEPKrYcOnBNJUIFjszWUyVdvWw DGDjoCMtRJUburkPToCyDrOtMr?",
"output": "NO"
},
{
"input": "n?",
"output": "NO"
},
{
"input": "BueDOlxgzeNlxrzRrMbKiQdmGujEKmGxclvaPpTuHmTqBp?",
"output": "NO"
},
{
"input": "iehvZNQXDGCuVmJPOEysLyUryTdfaIxIuTzTadDbqRQGoCLXkxnyfWSGoLXebNnQQNTqAQJebbyYvHOfpUnXeWdjx?",
"output": "NO"
},
{
"input": " J ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " o ?",
"output": "YES"
},
{
"input": " T ?",
"output": "NO"
},
{
"input": " q ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " c ?",
"output": "NO"
},
{
"input": " B ?",
"output": "NO"
},
{
"input": "LuhxDHVwMPTtUIUMIQTuQETgXCOQPsfdFlyHvpfOVedjUTpGLAZGOHloIjJJtOLAlHPivzA?",
"output": "YES"
},
{
"input": "wmztmzFfwbGyOmNHENUFMTsFEMWYA?",
"output": "YES"
},
{
"input": "wGsfZCSwN PEUhNUrLfABrxA?",
"output": "YES"
},
{
"input": "mCDHENXjYbgMdBimAdPnewaHfpGWowjWrVAdvWczjw iDcUbyzMsmsnwbviiKiAyGVA?",
"output": "YES"
},
{
"input": "ARIWnwqFqxsQXsXXzHqvFjxOCttAGPUzDtWzsenPYdNXuFOIUGYZsLLK IaoxiyjBBRThoelwdPTkuCQfcBLUEJpCPIrVZlvUWA?",
"output": "YES"
},
{
"input": " PslvVpgpN BXkMFBEVXsyZFIQbBEFxGkYTeXKrOdcmhbiTUatYRUoYAayrchqbksswIlfIjerZPqptvCGnMUhyrQSvwltRhFzA?",
"output": "YES"
},
{
"input": "HpBkttwSjBXDmyleGiRWNUMPaAIE uzTrp KJDzaUiCdsMYOoWKHoUhWUoecCPmACymMUUbGav UMRpCytPETwNFAObZJA?",
"output": "YES"
}
] | 1,527,864,477
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 156
| 7,065,600
|
# python 3
"""
"""
def sleuth(question_str) -> str:
question_lc = question_str.lower()
alphabet = "abcdefghijklmnopqrstuvwxyz"
vowel = "aeiouy"
for i in range(len(question_lc) - 1, -1, -1):
if question_lc[i] in alphabet:
if question_lc[i] in vowel:
return "YES"
else:
return "NO"
if __name__ == "__main__":
"""
Inside of this is the test.
Outside is the API
"""
question = input()
print(sleuth(question))
|
Title: Sleuth
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input Specification:
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output Specification:
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Demo Input:
['Is it a melon?\n', 'Is it an apple?\n', 'Is it a banana ?\n', 'Is it an apple and a banana simultaneouSLY?\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
none
|
```python
# python 3
"""
"""
def sleuth(question_str) -> str:
question_lc = question_str.lower()
alphabet = "abcdefghijklmnopqrstuvwxyz"
vowel = "aeiouy"
for i in range(len(question_lc) - 1, -1, -1):
if question_lc[i] in alphabet:
if question_lc[i] in vowel:
return "YES"
else:
return "NO"
if __name__ == "__main__":
"""
Inside of this is the test.
Outside is the API
"""
question = input()
print(sleuth(question))
```
| 3.947839
|
50
|
B
|
Choosing Symbol Pairs
|
PROGRAMMING
| 1,500
|
[
"strings"
] |
B. Choosing Symbol Pairs
|
2
|
256
|
There is a given string *S* consisting of *N* symbols. Your task is to find the number of ordered pairs of integers *i* and *j* such that
1. 1<=≤<=*i*,<=*j*<=≤<=*N*
2. *S*[*i*]<==<=*S*[*j*], that is the *i*-th symbol of string *S* is equal to the *j*-th.
|
The single input line contains *S*, consisting of lowercase Latin letters and digits. It is guaranteed that string *S* in not empty and its length does not exceed 105.
|
Print a single number which represents the number of pairs *i* and *j* with the needed property. Pairs (*x*,<=*y*) and (*y*,<=*x*) should be considered different, i.e. the ordered pairs count.
|
[
"great10\n",
"aaaaaaaaaa\n"
] |
[
"7\n",
"100\n"
] |
none
| 1,000
|
[
{
"input": "great10",
"output": "7"
},
{
"input": "aaaaaaaaaa",
"output": "100"
},
{
"input": "great10",
"output": "7"
},
{
"input": "aaaaaaaaaa",
"output": "100"
},
{
"input": "aabb",
"output": "8"
},
{
"input": "w",
"output": "1"
},
{
"input": "129a",
"output": "4"
},
{
"input": "233444",
"output": "14"
},
{
"input": "abacaba",
"output": "21"
},
{
"input": "abcdefghijklmnopqrstuvwxyz0987654321abcdefghijklmnopqrstuvwxyz0987654321abcdefghijklmnopqrstuvwxyz0987654321",
"output": "324"
},
{
"input": "zazaeeeeeeeq34443333444tttttt",
"output": "155"
},
{
"input": "00000000000000000000000",
"output": "529"
},
{
"input": "999000888775646453342311",
"output": "62"
}
] | 1,636,119,100
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 122
| 4,505,600
|
s= input()
s1 = list(set(s))
d={}
for i in s1:
d[i]=s.count(i)
ans=0
for i in s1:
ans += d[i]**2
print(ans)
|
Title: Choosing Symbol Pairs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given string *S* consisting of *N* symbols. Your task is to find the number of ordered pairs of integers *i* and *j* such that
1. 1<=≤<=*i*,<=*j*<=≤<=*N*
2. *S*[*i*]<==<=*S*[*j*], that is the *i*-th symbol of string *S* is equal to the *j*-th.
Input Specification:
The single input line contains *S*, consisting of lowercase Latin letters and digits. It is guaranteed that string *S* in not empty and its length does not exceed 105.
Output Specification:
Print a single number which represents the number of pairs *i* and *j* with the needed property. Pairs (*x*,<=*y*) and (*y*,<=*x*) should be considered different, i.e. the ordered pairs count.
Demo Input:
['great10\n', 'aaaaaaaaaa\n']
Demo Output:
['7\n', '100\n']
Note:
none
|
```python
s= input()
s1 = list(set(s))
d={}
for i in s1:
d[i]=s.count(i)
ans=0
for i in s1:
ans += d[i]**2
print(ans)
```
| 3.961108
|
520
|
A
|
Pangram
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
|
Output "YES", if the string is a pangram and "NO" otherwise.
|
[
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "YES"
},
{
"input": "48\nthereisasyetinsufficientdataforameaningfulanswer",
"output": "NO"
},
{
"input": "30\nToBeOrNotToBeThatIsTheQuestion",
"output": "NO"
},
{
"input": "30\njackdawslovemybigsphinxofquarz",
"output": "NO"
},
{
"input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY",
"output": "YES"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx",
"output": "YES"
},
{
"input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ",
"output": "YES"
},
{
"input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ",
"output": "YES"
},
{
"input": "25\nnxYTzLFwzNolAumjgcAboyxAj",
"output": "NO"
},
{
"input": "26\npRWdodGdxUESvcScPGbUoooZsC",
"output": "NO"
},
{
"input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj",
"output": "NO"
},
{
"input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa",
"output": "YES"
},
{
"input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK",
"output": "NO"
},
{
"input": "26\nvCUFRKElZOnjmXGylWQaHDiPst",
"output": "NO"
},
{
"input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY",
"output": "NO"
},
{
"input": "26\npGiFluRteQwkaVoPszJyNBChxM",
"output": "NO"
},
{
"input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY",
"output": "NO"
},
{
"input": "26\nLndjgvAEuICHKxPwqYztosrmBN",
"output": "NO"
},
{
"input": "26\nMdaXJrCipnOZLykfqHWEStevbU",
"output": "NO"
},
{
"input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba",
"output": "NO"
},
{
"input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo",
"output": "NO"
},
{
"input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa",
"output": "NO"
},
{
"input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO",
"output": "NO"
},
{
"input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe",
"output": "NO"
},
{
"input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG",
"output": "NO"
},
{
"input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF",
"output": "NO"
},
{
"input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu",
"output": "NO"
},
{
"input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf",
"output": "NO"
},
{
"input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE",
"output": "NO"
},
{
"input": "26\nDqspXZJTuONYieKgaHLMBwfVSC",
"output": "NO"
},
{
"input": "26\necOyUkqNljFHRVXtIpWabGMLDz",
"output": "NO"
},
{
"input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI",
"output": "NO"
},
{
"input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA",
"output": "NO"
},
{
"input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz",
"output": "NO"
},
{
"input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ",
"output": "NO"
},
{
"input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF",
"output": "NO"
},
{
"input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB",
"output": "NO"
},
{
"input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg",
"output": "NO"
},
{
"input": "26\nmnbvcxxlkjhgfdsapoiuytrewq",
"output": "NO"
},
{
"input": "26\naAbcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl",
"output": "NO"
},
{
"input": "25\nabcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "26\nabcdefhijklmnopqrstVxyzABC",
"output": "NO"
},
{
"input": "25\nqwertyuiopasdfghjklxcvbnm",
"output": "NO"
},
{
"input": "34\nTheQuickBrownFoxJumpsOverTheLayDog",
"output": "NO"
},
{
"input": "26\nabcdefghigklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyA",
"output": "NO"
},
{
"input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog",
"output": "NO"
},
{
"input": "25\nbcdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa",
"output": "NO"
},
{
"input": "26\nabcdefghiklmnopqrstvxyzABC",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxzZ",
"output": "NO"
},
{
"input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "NO"
}
] | 1,696,402,837
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 46
| 0
|
# your code goes here
n=int(input())
v=input()
s='abcdefghijklmnopqrstuvwxyz'
v=v.lower()
count=0
for i in s:
if i in v:
count+=1
if count==26:
print("YES")
else:
print("NO")
|
Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
# your code goes here
n=int(input())
v=input()
s='abcdefghijklmnopqrstuvwxyz'
v=v.lower()
count=0
for i in s:
if i in v:
count+=1
if count==26:
print("YES")
else:
print("NO")
```
| 3
|
|
160
|
A
|
Twins
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] | null | null |
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
|
In the single line print the single number — the minimum needed number of coins.
|
[
"2\n3 3\n",
"3\n2 1 2\n"
] |
[
"2\n",
"2\n"
] |
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
| 500
|
[
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "4"
},
{
"input": "7\n10 10 5 5 5 5 1",
"output": "3"
},
{
"input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4",
"output": "8"
},
{
"input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40",
"output": "8"
},
{
"input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90",
"output": "6"
},
{
"input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28",
"output": "14"
},
{
"input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42",
"output": "22"
},
{
"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": "51"
},
{
"input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2",
"output": "37"
},
{
"input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "36"
},
{
"input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3",
"output": "33"
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "30"
},
{
"input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2",
"output": "29"
},
{
"input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "30"
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "29"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "30"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "4\n4 1 2 3",
"output": "2"
},
{
"input": "5\n2 2 2 4 4",
"output": "2"
},
{
"input": "4\n100 99 99 100",
"output": "2"
}
] | 1,697,091,482
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
#罗誉城 化学与分子工程学院 2300011776
s=int(input())
n0=input().split()
n=sorted([int(i) for i in n0])
n.reverse()
for i in range(len(n)):
if n[:i+1]>n[i+1:]:
print(i+1)
break
|
Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
|
```python
#罗誉城 化学与分子工程学院 2300011776
s=int(input())
n0=input().split()
n=sorted([int(i) for i in n0])
n.reverse()
for i in range(len(n)):
if n[:i+1]>n[i+1:]:
print(i+1)
break
```
| 0
|
|
37
|
A
|
Towers
|
PROGRAMMING
| 1,000
|
[
"sortings"
] |
A. Towers
|
2
|
256
|
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
|
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
|
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
|
[
"3\n1 2 3\n",
"4\n6 5 6 7\n"
] |
[
"1 3\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3\n1 2 3",
"output": "1 3"
},
{
"input": "4\n6 5 6 7",
"output": "2 3"
},
{
"input": "4\n3 2 1 1",
"output": "2 3"
},
{
"input": "4\n1 2 3 3",
"output": "2 3"
},
{
"input": "3\n20 22 36",
"output": "1 3"
},
{
"input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162",
"output": "2 23"
},
{
"input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193",
"output": "2 43"
},
{
"input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111",
"output": "2 57"
},
{
"input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642",
"output": "2 80"
},
{
"input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188",
"output": "2 92"
},
{
"input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233",
"output": "3 101"
},
{
"input": "5\n5 5 5 5 5",
"output": "5 1"
},
{
"input": "3\n1000 1000 1000",
"output": "3 1"
},
{
"input": "1\n1000",
"output": "1 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "5\n1 1000 1000 1000 1000",
"output": "4 2"
},
{
"input": "5\n1000 1000 1000 8 7",
"output": "3 3"
}
] | 1,398,195,735
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 154
| 0
|
n = int(input())
s = list(map(int, input().split()))
d = [0 for i in range(1001)]
maxh = 0
for i in range(n):
d[s[i]] += 1
maxh = max(maxh, d[s[i]])
print(maxh, 1001 - d.count(0))
|
Title: Towers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
Input Specification:
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
Output Specification:
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
Demo Input:
['3\n1 2 3\n', '4\n6 5 6 7\n']
Demo Output:
['1 3\n', '2 3\n']
Note:
none
|
```python
n = int(input())
s = list(map(int, input().split()))
d = [0 for i in range(1001)]
maxh = 0
for i in range(n):
d[s[i]] += 1
maxh = max(maxh, d[s[i]])
print(maxh, 1001 - d.count(0))
```
| 3.9615
|
11
|
A
|
Increasing Sequence
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"implementation",
"math"
] |
A. Increasing Sequence
|
1
|
64
|
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
|
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
|
Output the minimal number of moves needed to make the sequence increasing.
|
[
"4 2\n1 3 3 2\n"
] |
[
"3\n"
] |
none
| 0
|
[
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
{
"input": "2 7\n1 1",
"output": "1"
},
{
"input": "3 3\n18 1 9",
"output": "10"
},
{
"input": "3 3\n15 17 9",
"output": "3"
},
{
"input": "3 3\n10 9 12",
"output": "2"
},
{
"input": "10 3\n2 1 17 10 5 16 8 4 15 17",
"output": "31"
},
{
"input": "10 3\n6 11 4 12 22 15 23 26 24 26",
"output": "13"
},
{
"input": "10 3\n10 24 13 15 18 14 15 26 33 35",
"output": "29"
},
{
"input": "100 3\n529 178 280 403 326 531 671 427 188 866 669 646 421 804 494 609 53 1012 211 243 887 833 900 543 226 42 859 718 454 372 971 692 846 770 511 395 499 479 641 756 115 269 206 45 1039 727 400 779 859 614 146 214 196 919 702 959 380 830 535 878 859 784 316 305 782 924 536 243 236 978 564 150 291 877 808 983 537 839 490 120 168 838 267 650 900 170 211 504 326 771 895 984 994 483 776 100 471 1078 317 580",
"output": "15717"
},
{
"input": "100 3\n329 226 331 909 962 112 837 1005 194 818 506 416 125 648 367 459 400 582 989 547 329 438 234 121 272 226 821 376 834 427 718 164 834 113 654 177 737 212 169 696 744 180 89 944 233 147 667 990 809 1072 1085 1093 814 265 1067 312 833 572 303 901 1032 504 185 817 389 158 613 723 239 269 911 352 769 404 225 822 897 606 947 323 913 804 923 1084 552 901 486 249 209 898 847 610 728 1122 986 669 1116 1076 367 327",
"output": "16133"
}
] | 1,635,324,637
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 7
| 154
| 17,715,200
|
n, d = map(int, input().split())
b = list(map(int, input().split()[:n]))
b_addlowest = min(b) + d
b.insert(0, b_addlowest)
b.remove(min(b))
k = len(b)
moves = 0
for i in range(k):
already_sorted = True
for j in range(k - i - 1):
if b[j] > b[j + 1]:
b[j], b[j + 1] = b[j + 1], b[j]
moves += 1
already_sorted = False
if already_sorted:
break
print(moves)
|
Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
n, d = map(int, input().split())
b = list(map(int, input().split()[:n]))
b_addlowest = min(b) + d
b.insert(0, b_addlowest)
b.remove(min(b))
k = len(b)
moves = 0
for i in range(k):
already_sorted = True
for j in range(k - i - 1):
if b[j] > b[j + 1]:
b[j], b[j + 1] = b[j + 1], b[j]
moves += 1
already_sorted = False
if already_sorted:
break
print(moves)
```
| 0
|
675
|
A
|
Infinite Sequence
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
|
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
|
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
|
[
"1 7 3\n",
"10 10 0\n",
"1 -4 5\n",
"0 60 50\n"
] |
[
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] |
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer.
| 500
|
[
{
"input": "1 7 3",
"output": "YES"
},
{
"input": "10 10 0",
"output": "YES"
},
{
"input": "1 -4 5",
"output": "NO"
},
{
"input": "0 60 50",
"output": "NO"
},
{
"input": "1 -4 -5",
"output": "YES"
},
{
"input": "0 1 0",
"output": "NO"
},
{
"input": "10 10 42",
"output": "YES"
},
{
"input": "-1000000000 1000000000 -1",
"output": "NO"
},
{
"input": "10 16 4",
"output": "NO"
},
{
"input": "-1000000000 1000000000 5",
"output": "YES"
},
{
"input": "1000000000 -1000000000 5",
"output": "NO"
},
{
"input": "1000000000 -1000000000 0",
"output": "NO"
},
{
"input": "1000000000 1000000000 0",
"output": "YES"
},
{
"input": "115078364 -899474523 -1",
"output": "YES"
},
{
"input": "-245436499 416383245 992",
"output": "YES"
},
{
"input": "-719636354 536952440 2",
"output": "YES"
},
{
"input": "-198350539 963391024 68337739",
"output": "YES"
},
{
"input": "-652811055 875986516 1091",
"output": "YES"
},
{
"input": "119057893 -516914539 -39748277",
"output": "YES"
},
{
"input": "989140430 731276607 -36837689",
"output": "YES"
},
{
"input": "677168390 494583489 -985071853",
"output": "NO"
},
{
"input": "58090193 777423708 395693923",
"output": "NO"
},
{
"input": "479823846 -403424770 -653472589",
"output": "NO"
},
{
"input": "-52536829 -132023273 -736287999",
"output": "NO"
},
{
"input": "-198893776 740026818 -547885271",
"output": "NO"
},
{
"input": "-2 -2 -2",
"output": "YES"
},
{
"input": "-2 -2 -1",
"output": "YES"
},
{
"input": "-2 -2 0",
"output": "YES"
},
{
"input": "-2 -2 1",
"output": "YES"
},
{
"input": "-2 -2 2",
"output": "YES"
},
{
"input": "-2 -1 -2",
"output": "NO"
},
{
"input": "-2 -1 -1",
"output": "NO"
},
{
"input": "-2 -1 0",
"output": "NO"
},
{
"input": "-2 -1 1",
"output": "YES"
},
{
"input": "-2 -1 2",
"output": "NO"
},
{
"input": "-2 0 -2",
"output": "NO"
},
{
"input": "-2 0 -1",
"output": "NO"
},
{
"input": "-2 0 0",
"output": "NO"
},
{
"input": "-2 0 1",
"output": "YES"
},
{
"input": "-2 0 2",
"output": "YES"
},
{
"input": "-2 1 -2",
"output": "NO"
},
{
"input": "-2 1 -1",
"output": "NO"
},
{
"input": "-2 1 0",
"output": "NO"
},
{
"input": "-2 1 1",
"output": "YES"
},
{
"input": "-2 1 2",
"output": "NO"
},
{
"input": "-2 2 -2",
"output": "NO"
},
{
"input": "-2 2 -1",
"output": "NO"
},
{
"input": "-2 2 0",
"output": "NO"
},
{
"input": "-2 2 1",
"output": "YES"
},
{
"input": "-2 2 2",
"output": "YES"
},
{
"input": "-1 -2 -2",
"output": "NO"
},
{
"input": "-1 -2 -1",
"output": "YES"
},
{
"input": "-1 -2 0",
"output": "NO"
},
{
"input": "-1 -2 1",
"output": "NO"
},
{
"input": "-1 -2 2",
"output": "NO"
},
{
"input": "-1 -1 -2",
"output": "YES"
},
{
"input": "-1 -1 -1",
"output": "YES"
},
{
"input": "-1 -1 0",
"output": "YES"
},
{
"input": "-1 -1 1",
"output": "YES"
},
{
"input": "-1 -1 2",
"output": "YES"
},
{
"input": "-1 0 -2",
"output": "NO"
},
{
"input": "-1 0 -1",
"output": "NO"
},
{
"input": "-1 0 0",
"output": "NO"
},
{
"input": "-1 0 1",
"output": "YES"
},
{
"input": "-1 0 2",
"output": "NO"
},
{
"input": "-1 1 -2",
"output": "NO"
},
{
"input": "-1 1 -1",
"output": "NO"
},
{
"input": "-1 1 0",
"output": "NO"
},
{
"input": "-1 1 1",
"output": "YES"
},
{
"input": "-1 1 2",
"output": "YES"
},
{
"input": "-1 2 -2",
"output": "NO"
},
{
"input": "-1 2 -1",
"output": "NO"
},
{
"input": "-1 2 0",
"output": "NO"
},
{
"input": "-1 2 1",
"output": "YES"
},
{
"input": "-1 2 2",
"output": "NO"
},
{
"input": "0 -2 -2",
"output": "YES"
},
{
"input": "0 -2 -1",
"output": "YES"
},
{
"input": "0 -2 0",
"output": "NO"
},
{
"input": "0 -2 1",
"output": "NO"
},
{
"input": "0 -2 2",
"output": "NO"
},
{
"input": "0 -1 -2",
"output": "NO"
},
{
"input": "0 -1 -1",
"output": "YES"
},
{
"input": "0 -1 0",
"output": "NO"
},
{
"input": "0 -1 1",
"output": "NO"
},
{
"input": "0 -1 2",
"output": "NO"
},
{
"input": "0 0 -2",
"output": "YES"
},
{
"input": "0 0 -1",
"output": "YES"
},
{
"input": "0 0 0",
"output": "YES"
},
{
"input": "0 0 1",
"output": "YES"
},
{
"input": "0 0 2",
"output": "YES"
},
{
"input": "0 1 -2",
"output": "NO"
},
{
"input": "0 1 -1",
"output": "NO"
},
{
"input": "0 1 0",
"output": "NO"
},
{
"input": "0 1 1",
"output": "YES"
},
{
"input": "0 1 2",
"output": "NO"
},
{
"input": "0 2 -2",
"output": "NO"
},
{
"input": "0 2 -1",
"output": "NO"
},
{
"input": "0 2 0",
"output": "NO"
},
{
"input": "0 2 1",
"output": "YES"
},
{
"input": "0 2 2",
"output": "YES"
},
{
"input": "1 -2 -2",
"output": "NO"
},
{
"input": "1 -2 -1",
"output": "YES"
},
{
"input": "1 -2 0",
"output": "NO"
},
{
"input": "1 -2 1",
"output": "NO"
},
{
"input": "1 -2 2",
"output": "NO"
},
{
"input": "1 -1 -2",
"output": "YES"
},
{
"input": "1 -1 -1",
"output": "YES"
},
{
"input": "1 -1 0",
"output": "NO"
},
{
"input": "1 -1 1",
"output": "NO"
},
{
"input": "1 -1 2",
"output": "NO"
},
{
"input": "1 0 -2",
"output": "NO"
},
{
"input": "1 0 -1",
"output": "YES"
},
{
"input": "1 0 0",
"output": "NO"
},
{
"input": "1 0 1",
"output": "NO"
},
{
"input": "1 0 2",
"output": "NO"
},
{
"input": "1 1 -2",
"output": "YES"
},
{
"input": "1 1 -1",
"output": "YES"
},
{
"input": "1 1 0",
"output": "YES"
},
{
"input": "1 1 1",
"output": "YES"
},
{
"input": "1 1 2",
"output": "YES"
},
{
"input": "1 2 -2",
"output": "NO"
},
{
"input": "1 2 -1",
"output": "NO"
},
{
"input": "1 2 0",
"output": "NO"
},
{
"input": "1 2 1",
"output": "YES"
},
{
"input": "1 2 2",
"output": "NO"
},
{
"input": "2 -2 -2",
"output": "YES"
},
{
"input": "2 -2 -1",
"output": "YES"
},
{
"input": "2 -2 0",
"output": "NO"
},
{
"input": "2 -2 1",
"output": "NO"
},
{
"input": "2 -2 2",
"output": "NO"
},
{
"input": "2 -1 -2",
"output": "NO"
},
{
"input": "2 -1 -1",
"output": "YES"
},
{
"input": "2 -1 0",
"output": "NO"
},
{
"input": "2 -1 1",
"output": "NO"
},
{
"input": "2 -1 2",
"output": "NO"
},
{
"input": "2 0 -2",
"output": "YES"
},
{
"input": "2 0 -1",
"output": "YES"
},
{
"input": "2 0 0",
"output": "NO"
},
{
"input": "2 0 1",
"output": "NO"
},
{
"input": "2 0 2",
"output": "NO"
},
{
"input": "2 1 -2",
"output": "NO"
},
{
"input": "2 1 -1",
"output": "YES"
},
{
"input": "2 1 0",
"output": "NO"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "2 1 2",
"output": "NO"
},
{
"input": "2 2 -2",
"output": "YES"
},
{
"input": "2 2 -1",
"output": "YES"
},
{
"input": "2 2 0",
"output": "YES"
},
{
"input": "2 2 1",
"output": "YES"
},
{
"input": "2 2 2",
"output": "YES"
},
{
"input": "-1000000000 1000000000 1",
"output": "YES"
},
{
"input": "-1000000000 1000000000 2",
"output": "YES"
},
{
"input": "1000000000 -1000000000 -1",
"output": "YES"
},
{
"input": "5 2 3",
"output": "NO"
},
{
"input": "2 1 -1",
"output": "YES"
},
{
"input": "3 2 1",
"output": "NO"
},
{
"input": "0 -5 -3",
"output": "NO"
},
{
"input": "2 5 5",
"output": "NO"
},
{
"input": "0 10 1",
"output": "YES"
},
{
"input": "15 5 -5",
"output": "YES"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "20 10 0",
"output": "NO"
},
{
"input": "20 15 5",
"output": "NO"
},
{
"input": "1 6 1",
"output": "YES"
},
{
"input": "1000000000 0 -1000000000",
"output": "YES"
},
{
"input": "1 1 -5",
"output": "YES"
},
{
"input": "4 6 1",
"output": "YES"
},
{
"input": "-5 -10 -5",
"output": "YES"
},
{
"input": "2 0 0",
"output": "NO"
},
{
"input": "10 9 -1",
"output": "YES"
},
{
"input": "-2 -1 -1",
"output": "NO"
},
{
"input": "1 13 3",
"output": "YES"
},
{
"input": "2 3 0",
"output": "NO"
},
{
"input": "1 1 -1",
"output": "YES"
},
{
"input": "5 -10 -5",
"output": "YES"
},
{
"input": "5 3 1",
"output": "NO"
},
{
"input": "1 1000000000 1",
"output": "YES"
},
{
"input": "-1000000000 1000000000 1000000000",
"output": "YES"
}
] | 1,590,307,281
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 178
| 155
| 0
|
a,b,c = map(int, input().rstrip().split(" "))
if b < a and c > 0:
print("NO")
elif c==0:
if b!=a:
print("NO")
else:
print("YES")
elif b > a and c < 0:
print("NO")
else:
if (b - a)%c:
print("NO")
else:
print("YES")
|
Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
Input Specification:
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
Output Specification:
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
Demo Input:
['1 7 3\n', '10 10 0\n', '1 -4 5\n', '0 60 50\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer.
|
```python
a,b,c = map(int, input().rstrip().split(" "))
if b < a and c > 0:
print("NO")
elif c==0:
if b!=a:
print("NO")
else:
print("YES")
elif b > a and c < 0:
print("NO")
else:
if (b - a)%c:
print("NO")
else:
print("YES")
```
| 3
|
|
999
|
A
|
Mishka and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
|
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
|
Print one integer — the maximum number of problems Mishka can solve.
|
[
"8 4\n4 2 3 1 5 1 6 4\n",
"5 2\n3 1 2 1 3\n",
"5 100\n12 34 55 43 21\n"
] |
[
"5\n",
"0\n",
"5\n"
] |
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
| 0
|
[
{
"input": "8 4\n4 2 3 1 5 1 6 4",
"output": "5"
},
{
"input": "5 2\n3 1 2 1 3",
"output": "0"
},
{
"input": "5 100\n12 34 55 43 21",
"output": "5"
},
{
"input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62",
"output": "100"
},
{
"input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98",
"output": "98"
},
{
"input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2",
"output": "39"
},
{
"input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16",
"output": "99"
},
{
"input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1",
"output": "1"
},
{
"input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "3 2\n1 4 1",
"output": "2"
},
{
"input": "1 2\n100",
"output": "0"
},
{
"input": "7 4\n4 2 3 4 4 2 3",
"output": "7"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "1 2\n15",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "5 3\n3 4 3 2 1",
"output": "4"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "1 5\n1",
"output": "1"
},
{
"input": "6 6\n7 1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n6 5 5 5 5",
"output": "4"
},
{
"input": "1 4\n2",
"output": "1"
},
{
"input": "9 4\n1 2 1 2 4 2 1 2 1",
"output": "9"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 10\n5",
"output": "1"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4",
"output": "71"
},
{
"input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90",
"output": "63"
},
{
"input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8",
"output": "61"
},
{
"input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22",
"output": "72"
},
{
"input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2",
"output": "52"
},
{
"input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90",
"output": "60"
},
{
"input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10",
"output": "56"
},
{
"input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90",
"output": "61"
},
{
"input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24",
"output": "99"
},
{
"input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50",
"output": "99"
},
{
"input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48",
"output": "99"
},
{
"input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57",
"output": "99"
},
{
"input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93",
"output": "99"
},
{
"input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33",
"output": "3"
},
{
"input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51",
"output": "3"
},
{
"input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49",
"output": "6"
},
{
"input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8",
"output": "66"
},
{
"input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95",
"output": "0"
},
{
"input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66",
"output": "0"
},
{
"input": "2 8\n8 8",
"output": "2"
},
{
"input": "1 6\n3",
"output": "1"
},
{
"input": "1 5\n5",
"output": "1"
}
] | 1,613,748,425
| 725
|
Python 3
|
OK
|
TESTS
| 48
| 77
| 409,600
|
from sys import stdin, stdout
#from math import pow
#import numpy as np
from collections import Counter
# int(stdin.readline()) stdout.write(str())
# stdin.readline()
# map(int, stdin.readline().split())
# list(map(int, stdin.readline().split()))
a, b = map(int, stdin.readline().split())
l = list(map(int, stdin.readline().split()))
count = 0
for k in range(len(l)):
if l[k]<=b:
count+=1
else:
break
l = l[::-1]
for j in range(len(l)):
if l[j]<=b:
count+=1
else:
break
stdout.write(str(min(a,count)))
|
Title: Mishka and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Output Specification:
Print one integer — the maximum number of problems Mishka can solve.
Demo Input:
['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n']
Demo Output:
['5\n', '0\n', '5\n']
Note:
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
|
```python
from sys import stdin, stdout
#from math import pow
#import numpy as np
from collections import Counter
# int(stdin.readline()) stdout.write(str())
# stdin.readline()
# map(int, stdin.readline().split())
# list(map(int, stdin.readline().split()))
a, b = map(int, stdin.readline().split())
l = list(map(int, stdin.readline().split()))
count = 0
for k in range(len(l)):
if l[k]<=b:
count+=1
else:
break
l = l[::-1]
for j in range(len(l)):
if l[j]<=b:
count+=1
else:
break
stdout.write(str(min(a,count)))
```
| 3
|
|
579
|
A
|
Raising Bacteria
|
PROGRAMMING
| 1,000
|
[
"bitmasks"
] | null | null |
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
|
The only line containing one integer *x* (1<=≤<=*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"
}
] | 1,668,407,185
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 46
| 0
|
n = int(input())
s = bin(n)[2:]
c = 0
for i in s:
if i == '1':
c += 1
print(c)
|
Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
|
```python
n = int(input())
s = bin(n)[2:]
c = 0
for i in s:
if i == '1':
c += 1
print(c)
```
| 3
|
|
801
|
A
|
Vicious Keyboard
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.
|
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
|
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
|
[
"VK\n",
"VV\n",
"V\n",
"VKKKKKKKKKVVVVVVVVVK\n",
"KVKV\n"
] |
[
"1\n",
"1\n",
"0\n",
"3\n",
"1\n"
] |
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
| 500
|
[
{
"input": "VK",
"output": "1"
},
{
"input": "VV",
"output": "1"
},
{
"input": "V",
"output": "0"
},
{
"input": "VKKKKKKKKKVVVVVVVVVK",
"output": "3"
},
{
"input": "KVKV",
"output": "1"
},
{
"input": "VKKVVVKVKVK",
"output": "5"
},
{
"input": "VKVVKVKVVKVKKKKVVVVVVVVKVKVVVVVVKKVKKVKVVKVKKVVVVKV",
"output": "14"
},
{
"input": "VVKKVKKVVKKVKKVKVVKKVKKVVKKVKVVKKVKKVKVVKKVVKKVKVVKKVKVVKKVVKVVKKVKKVKKVKKVKKVKVVKKVKKVKKVKKVKKVVKVK",
"output": "32"
},
{
"input": "KVVKKVKVKVKVKVKKVKVKVVKVKVVKVVKVKKVKVKVKVKVKVKVKVKVKVKVKVKVKVKVVKVKVVKKVKVKK",
"output": "32"
},
{
"input": "KVVVVVKKVKVVKVVVKVVVKKKVKKKVVKVKKKVKKKKVKVVVVVKKKVVVVKKVVVVKKKVKVVVVVVVKKVKVKKKVVKVVVKVVKK",
"output": "21"
},
{
"input": "VVVVVKKVKVKVKVVKVVKKVVKVKKKKKKKVKKKVVVVVVKKVVVKVKVVKVKKVVKVVVKKKKKVVVVVKVVVVKVVVKKVKKVKKKVKKVKKVVKKV",
"output": "25"
},
{
"input": "KKVVKVVKVVKKVVKKVKVVKKV",
"output": "7"
},
{
"input": "KKVVKKVVVKKVKKVKKVVVKVVVKKVKKVVVKKVVVKVVVKVVVKKVVVKKVVVKVVVKKVVVKVVKKVVVKKVVVKKVVKVVVKKVVKKVKKVVVKKV",
"output": "24"
},
{
"input": "KVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVKVKVVKVKVKKVKVKVKVKVVKVKVKVKVKVKVKVKVKKVKVKVV",
"output": "35"
},
{
"input": "VKVVVKKKVKVVKVKVKVKVKVV",
"output": "9"
},
{
"input": "KKKKVKKVKVKVKKKVVVVKK",
"output": "6"
},
{
"input": "KVKVKKVVVVVVKKKVKKKKVVVVKVKKVKVVK",
"output": "9"
},
{
"input": "KKVKKVKKKVKKKVKKKVKVVVKKVVVVKKKVKKVVKVKKVKVKVKVVVKKKVKKKKKVVKVVKVVVKKVVKVVKKKKKVK",
"output": "22"
},
{
"input": "VVVKVKVKVVVVVKVVVKKVVVKVVVVVKKVVKVVVKVVVKVKKKVVKVVVVVKVVVVKKVVKVKKVVKKKVKVVKVKKKKVVKVVVKKKVKVKKKKKK",
"output": "25"
},
{
"input": "VKVVKVVKKKVVKVKKKVVKKKVVKVVKVVKKVKKKVKVKKKVVKVKKKVVKVVKKKVVKKKVKKKVVKKVVKKKVKVKKKVKKKVKKKVKVKKKVVKVK",
"output": "29"
},
{
"input": "KKVKVVVKKVV",
"output": "3"
},
{
"input": "VKVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVK",
"output": "16"
},
{
"input": "VVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVV",
"output": "13"
},
{
"input": "VVKKVKVKKKVVVKVVVKVKKVKKKVVVKVVKVKKVKKVKVKVVKKVVKKVKVVKKKVVKKVVVKVKVVVKVKVVKVKKVKKV",
"output": "26"
},
{
"input": "VVKVKKVVKKVVKKVVKKVVKKVKKVVKVKKVVKKVVKKVVKKVVKKVVKVVKKVVKVVKKVVKVVKKVVKKVKKVVKVVKKVVKVVKKVV",
"output": "26"
},
{
"input": "K",
"output": "0"
},
{
"input": "VKVK",
"output": "2"
},
{
"input": "VKVV",
"output": "2"
},
{
"input": "KV",
"output": "0"
},
{
"input": "KK",
"output": "1"
},
{
"input": "KKVK",
"output": "2"
},
{
"input": "KKKK",
"output": "1"
},
{
"input": "KKV",
"output": "1"
},
{
"input": "KKVKVK",
"output": "3"
},
{
"input": "VKKVK",
"output": "2"
},
{
"input": "VKKK",
"output": "2"
},
{
"input": "KKK",
"output": "1"
},
{
"input": "KVV",
"output": "1"
},
{
"input": "KKVKV",
"output": "2"
},
{
"input": "VVK",
"output": "1"
},
{
"input": "VVVKVKVKVKVKVKVK",
"output": "8"
},
{
"input": "KVVVK",
"output": "2"
},
{
"input": "VVVKK",
"output": "2"
},
{
"input": "KKVV",
"output": "1"
},
{
"input": "KKKKKKK",
"output": "1"
},
{
"input": "VKKKVK",
"output": "3"
},
{
"input": "KKVVV",
"output": "1"
},
{
"input": "VVVVVV",
"output": "1"
},
{
"input": "KKKV",
"output": "1"
},
{
"input": "VVKVV",
"output": "2"
},
{
"input": "VKVKKK",
"output": "3"
},
{
"input": "VKKV",
"output": "1"
},
{
"input": "VKKVV",
"output": "2"
},
{
"input": "VVKKVV",
"output": "2"
},
{
"input": "KKVVKKV",
"output": "2"
},
{
"input": "KKKKK",
"output": "1"
},
{
"input": "VKVVKKVKKVVKVKKVKKKVKKVKVKK",
"output": "10"
},
{
"input": "VKVKVV",
"output": "3"
},
{
"input": "VKVVKVV",
"output": "3"
},
{
"input": "VVV",
"output": "1"
},
{
"input": "VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV",
"output": "1"
},
{
"input": "VVKKKKKKVKK",
"output": "3"
},
{
"input": "KVKVKVV",
"output": "3"
}
] | 1,492,357,660
| 760
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 1
| 46
| 4,608,000
|
s=list(' '+input())
m=0
for i in range(len(s)-1):
ss=[]
for i0 in range(len(s)):
ss.append(s[i0])
if ss[i]=='V':
ss[i]='K'
elif ss[i]=='K':
ss[i]='V'
mm=0
for i0 in range(len(s)-1):
if ss[i0]=='V'and ss[i0+1]=='K':
mm=mm+1
if mm>m:
m=mm
print(m)
|
Title: Vicious Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.
Input Specification:
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
Output Specification:
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
Demo Input:
['VK\n', 'VV\n', 'V\n', 'VKKKKKKKKKVVVVVVVVVK\n', 'KVKV\n']
Demo Output:
['1\n', '1\n', '0\n', '3\n', '1\n']
Note:
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
|
```python
s=list(' '+input())
m=0
for i in range(len(s)-1):
ss=[]
for i0 in range(len(s)):
ss.append(s[i0])
if ss[i]=='V':
ss[i]='K'
elif ss[i]=='K':
ss[i]='V'
mm=0
for i0 in range(len(s)-1):
if ss[i0]=='V'and ss[i0+1]=='K':
mm=mm+1
if mm>m:
m=mm
print(m)
```
| 0
|
|
592
|
C
|
The Big Race
|
PROGRAMMING
| 1,800
|
[
"math"
] | null | null |
Vector Willman and Array Bolt are the two most famous athletes of Byteforces. They are going to compete in a race with a distance of *L* meters today.
Willman and Bolt have exactly the same speed, so when they compete the result is always a tie. That is a problem for the organizers because they want a winner.
While watching previous races the organizers have noticed that Willman can perform only steps of length equal to *w* meters, and Bolt can perform only steps of length equal to *b* meters. Organizers decided to slightly change the rules of the race. Now, at the end of the racetrack there will be an abyss, and the winner will be declared the athlete, who manages to run farther from the starting point of the the racetrack (which is not the subject to change by any of the athletes).
Note that none of the athletes can run infinitely far, as they both will at some moment of time face the point, such that only one step further will cause them to fall in the abyss. In other words, the athlete will not fall into the abyss if the total length of all his steps will be less or equal to the chosen distance *L*.
Since the organizers are very fair, the are going to set the length of the racetrack as an integer chosen randomly and uniformly in range from 1 to *t* (both are included). What is the probability that Willman and Bolt tie again today?
|
The first line of the input contains three integers *t*, *w* and *b* (1<=≤<=*t*,<=*w*,<=*b*<=≤<=5·1018) — the maximum possible length of the racetrack, the length of Willman's steps and the length of Bolt's steps respectively.
|
Print the answer to the problem as an irreducible fraction . Follow the format of the samples output.
The fraction (*p* and *q* are integers, and both *p*<=≥<=0 and *q*<=><=0 holds) is called irreducible, if there is no such integer *d*<=><=1, that both *p* and *q* are divisible by *d*.
|
[
"10 3 2\n",
"7 1 2\n"
] |
[
"3/10\n",
"3/7\n"
] |
In the first sample Willman and Bolt will tie in case 1, 6 or 7 are chosen as the length of the racetrack.
| 1,500
|
[
{
"input": "10 3 2",
"output": "3/10"
},
{
"input": "7 1 2",
"output": "3/7"
},
{
"input": "1 1 1",
"output": "1/1"
},
{
"input": "5814 31 7",
"output": "94/2907"
},
{
"input": "94268 813 766",
"output": "765/94268"
},
{
"input": "262610 5583 4717",
"output": "2358/131305"
},
{
"input": "3898439 96326 71937",
"output": "71936/3898439"
},
{
"input": "257593781689876390 32561717 4411677",
"output": "7914548537/257593781689876390"
},
{
"input": "111319886766128339 7862842484895022 3003994959686829",
"output": "3003994959686828/111319886766128339"
},
{
"input": "413850294331656955 570110918058849723 409853735661743839",
"output": "409853735661743838/413850294331656955"
},
{
"input": "3000000000000000000 2999999999999999873 2999999999999999977",
"output": "23437499999999999/23437500000000000"
},
{
"input": "9 6 1",
"output": "1/9"
},
{
"input": "32 9 2",
"output": "3/32"
},
{
"input": "976 5 6",
"output": "41/244"
},
{
"input": "5814 31 7",
"output": "94/2907"
},
{
"input": "94268 714 345",
"output": "689/94268"
},
{
"input": "262610 5583 4717",
"output": "2358/131305"
},
{
"input": "3898439 96326 71937",
"output": "71936/3898439"
},
{
"input": "54682301 778668 253103",
"output": "253102/54682301"
},
{
"input": "329245015 1173508 8918834",
"output": "1173507/329245015"
},
{
"input": "321076647734423976 7 7",
"output": "1/1"
},
{
"input": "455227494055672047 92 28",
"output": "19792499741550983/455227494055672047"
},
{
"input": "595779167455745259 6954 8697",
"output": "205511958419723/595779167455745259"
},
{
"input": "1000000000000000000 1000000000 2000000000",
"output": "1/2"
},
{
"input": "462643382718281828 462643382718281507 462643382718281701",
"output": "33045955908448679/33045955908448702"
},
{
"input": "4000000000000000000 9999999999999997 99999999999999999",
"output": "2499999999999999/1000000000000000000"
},
{
"input": "4003000100004000000 9999999099999999 99999999999999999",
"output": "4999999549999999/2001500050002000000"
},
{
"input": "4903000100004000000 58997960959949999 99933992929999999",
"output": "29498980479974999/2451500050002000000"
},
{
"input": "257593781689876390 32561717 4411677",
"output": "7914548537/257593781689876390"
},
{
"input": "111319886766128339 7862842484895022 3003994959686829",
"output": "3003994959686828/111319886766128339"
},
{
"input": "413850294331656955 570110918058849723 409853735661743839",
"output": "409853735661743838/413850294331656955"
},
{
"input": "232 17 83",
"output": "2/29"
},
{
"input": "5496272 63 200",
"output": "13765/2748136"
},
{
"input": "180 174 53",
"output": "13/45"
},
{
"input": "1954 190 537",
"output": "189/1954"
},
{
"input": "146752429 510 514",
"output": "571199/146752429"
},
{
"input": "579312860 55 70",
"output": "10344881/144828215"
},
{
"input": "1 9 9",
"output": "1/1"
},
{
"input": "95 19 19",
"output": "1/1"
},
{
"input": "404 63 441",
"output": "31/202"
},
{
"input": "5566 4798 4798",
"output": "1/1"
},
{
"input": "118289676 570846883 570846883",
"output": "1/1"
},
{
"input": "763 358 358",
"output": "1/1"
},
{
"input": "85356138 7223 482120804",
"output": "3611/42678069"
},
{
"input": "674664088 435395270 5",
"output": "9/674664088"
},
{
"input": "762200126044291557 370330636048898430 6",
"output": "17/762200126044291557"
},
{
"input": "917148533938841535 47 344459175789842163",
"output": "28/183429706787768307"
},
{
"input": "360212127113008697 877228952036215545 5259",
"output": "5258/360212127113008697"
},
{
"input": "683705963104411677 89876390 116741460012229240",
"output": "539258339/683705963104411677"
},
{
"input": "573003994959686829 275856334120822851 1319886766128339",
"output": "3959660298385016/573003994959686829"
},
{
"input": "409853735661743839 413850294331656955 413850294331656955",
"output": "1/1"
},
{
"input": "19 1 19",
"output": "1/19"
},
{
"input": "576 18 32",
"output": "1/16"
},
{
"input": "9540 10 954",
"output": "1/477"
},
{
"input": "101997840 6 16999640",
"output": "1/8499820"
},
{
"input": "955944 1278 748",
"output": "1/639"
},
{
"input": "482120804 66748 7223",
"output": "1/66748"
},
{
"input": "370330636048898430 61721772674816405 6",
"output": "1/61721772674816405"
},
{
"input": "344459175789842163 7328918633826429 47",
"output": "1/7328918633826429"
},
{
"input": "877228952036215545 166805277055755 5259",
"output": "1/55601759018585"
},
{
"input": "116741460012229240 1298911316 89876390",
"output": "1/649455658"
},
{
"input": "275856334120822851 209 1319886766128339",
"output": "1/1319886766128339"
},
{
"input": "413850294331656955 1 413850294331656955",
"output": "1/413850294331656955"
},
{
"input": "54682301 778668 253103",
"output": "253102/54682301"
},
{
"input": "329245015 3931027 6443236",
"output": "357366/29931365"
},
{
"input": "321076647734423976 7 8",
"output": "1672274206950125/13378193655600999"
},
{
"input": "455227494055672047 71 60",
"output": "6411654845854559/455227494055672047"
},
{
"input": "595779167455745259 9741 9331",
"output": "61162012885196/595779167455745259"
},
{
"input": "6470 80 160",
"output": "327/647"
},
{
"input": "686325 828 1656",
"output": "114511/228775"
},
{
"input": "4535304 2129 4258",
"output": "755973/1511768"
},
{
"input": "40525189 6365 12730",
"output": "20265394/40525189"
},
{
"input": "675297075 25986 51972",
"output": "112553659/225099025"
},
{
"input": "5681598412 75376 226128",
"output": "1893897375/5681598412"
},
{
"input": "384118571739435733 619773000 1859319000",
"output": "128039524053435733/384118571739435733"
},
{
"input": "391554751752251913 625743359 1877230077",
"output": "130518250652782079/391554751752251913"
},
{
"input": "390728504279201198 625082797 1250165594",
"output": "195364252413988195/390728504279201198"
},
{
"input": "389902265396085075 624421544 1248843088",
"output": "64983710976697837/129967421798695025"
},
{
"input": "734812071040507372 857211800 2571635400",
"output": "61234339274051543/183703017760126843"
},
{
"input": "1 1 2",
"output": "0/1"
},
{
"input": "3 1 4",
"output": "0/1"
},
{
"input": "8 2 3",
"output": "3/8"
},
{
"input": "64 32 16",
"output": "1/2"
},
{
"input": "1 1 1000000000",
"output": "0/1"
},
{
"input": "1000000000 1 1",
"output": "1/1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1/1"
},
{
"input": "1000000000 2 4",
"output": "1/2"
},
{
"input": "1000000000 123 456",
"output": "6579023/1000000000"
},
{
"input": "1000000000 123123 654",
"output": "24851/1000000000"
},
{
"input": "123456 123 456",
"output": "215/30864"
},
{
"input": "123456 1234567 123",
"output": "61/61728"
},
{
"input": "314159265 271 8281",
"output": "37939/314159265"
},
{
"input": "11071994 4231 1324",
"output": "2647/11071994"
},
{
"input": "961748927 961748941 982451653",
"output": "1/1"
},
{
"input": "15485221 1259 90863",
"output": "1258/15485221"
},
{
"input": "5000000000000000000 4999999999999999837 4999999999999999963",
"output": "1249999999999999959/1250000000000000000"
},
{
"input": "4000000000000000000 3999999999999999691 3999999999999999887",
"output": "399999999999999969/400000000000000000"
},
{
"input": "999999999999999999 999999999999999709 999999999999999737",
"output": "333333333333333236/333333333333333333"
},
{
"input": "799999999999999999 799999999999999969 799999999999999991",
"output": "799999999999999968/799999999999999999"
},
{
"input": "812312312312312222 812312312312311897 812312312312312029",
"output": "406156156156155948/406156156156156111"
},
{
"input": "500000000000000000 499999999999999927 499999999999999931",
"output": "249999999999999963/250000000000000000"
},
{
"input": "555555555555555555 555555555555555083 555555555555555229",
"output": "50505050505050462/50505050505050505"
},
{
"input": "199419941994199419 199419941994199369 199419941994199391",
"output": "66473313998066456/66473313998066473"
},
{
"input": "145685485411238588 145685485411238483 145685485411238573",
"output": "72842742705619241/72842742705619294"
},
{
"input": "314159265358979323 314159265358979167 314159265358979213",
"output": "314159265358979166/314159265358979323"
},
{
"input": "10 1000000000000000000 1000000000000000001",
"output": "1/1"
},
{
"input": "5 100000000000000000 99999999999999999",
"output": "1/1"
},
{
"input": "5 1000000000000 1000000000001",
"output": "1/1"
},
{
"input": "5 1000000000000000000 1000000000000000001",
"output": "1/1"
},
{
"input": "2 1000000000000000000 1000000000000000001",
"output": "1/1"
},
{
"input": "2 10 11",
"output": "1/1"
},
{
"input": "10 123456789123456789 723456789123456781",
"output": "1/1"
},
{
"input": "12345678910 123456789101112131 123456789101112132",
"output": "1/1"
},
{
"input": "5 499999999999999999 499999999999999998",
"output": "1/1"
}
] | 1,631,185,912
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 23
| 77
| 6,963,200
|
def gcd(a,b):
if b==0: return a
return gcd(b,a%b)
def lcm(a,b):
return max(a,b)//gcd(a,b)*min(a,b)
t,w,b=[int(i) for i in input().split()]
l=lcm(b,w)
d=min(b,w)-1
if b==w:
print('1/1')
else:
L = t//l*(d+1) + min(t-t//l*l, d+1) - 1
g = gcd(L,t)
print(L//g,'/',t//g,sep='')
|
Title: The Big Race
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vector Willman and Array Bolt are the two most famous athletes of Byteforces. They are going to compete in a race with a distance of *L* meters today.
Willman and Bolt have exactly the same speed, so when they compete the result is always a tie. That is a problem for the organizers because they want a winner.
While watching previous races the organizers have noticed that Willman can perform only steps of length equal to *w* meters, and Bolt can perform only steps of length equal to *b* meters. Organizers decided to slightly change the rules of the race. Now, at the end of the racetrack there will be an abyss, and the winner will be declared the athlete, who manages to run farther from the starting point of the the racetrack (which is not the subject to change by any of the athletes).
Note that none of the athletes can run infinitely far, as they both will at some moment of time face the point, such that only one step further will cause them to fall in the abyss. In other words, the athlete will not fall into the abyss if the total length of all his steps will be less or equal to the chosen distance *L*.
Since the organizers are very fair, the are going to set the length of the racetrack as an integer chosen randomly and uniformly in range from 1 to *t* (both are included). What is the probability that Willman and Bolt tie again today?
Input Specification:
The first line of the input contains three integers *t*, *w* and *b* (1<=≤<=*t*,<=*w*,<=*b*<=≤<=5·1018) — the maximum possible length of the racetrack, the length of Willman's steps and the length of Bolt's steps respectively.
Output Specification:
Print the answer to the problem as an irreducible fraction . Follow the format of the samples output.
The fraction (*p* and *q* are integers, and both *p*<=≥<=0 and *q*<=><=0 holds) is called irreducible, if there is no such integer *d*<=><=1, that both *p* and *q* are divisible by *d*.
Demo Input:
['10 3 2\n', '7 1 2\n']
Demo Output:
['3/10\n', '3/7\n']
Note:
In the first sample Willman and Bolt will tie in case 1, 6 or 7 are chosen as the length of the racetrack.
|
```python
def gcd(a,b):
if b==0: return a
return gcd(b,a%b)
def lcm(a,b):
return max(a,b)//gcd(a,b)*min(a,b)
t,w,b=[int(i) for i in input().split()]
l=lcm(b,w)
d=min(b,w)-1
if b==w:
print('1/1')
else:
L = t//l*(d+1) + min(t-t//l*l, d+1) - 1
g = gcd(L,t)
print(L//g,'/',t//g,sep='')
```
| 0
|
|
1,008
|
A
|
Romaji
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
|
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
|
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
|
[
"sumimasen\n",
"ninja\n",
"codeforces\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
| 500
|
[
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"input": "n",
"output": "YES"
},
{
"input": "necnei",
"output": "NO"
},
{
"input": "nternn",
"output": "NO"
},
{
"input": "aucunuohja",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "b",
"output": "NO"
},
{
"input": "nn",
"output": "YES"
},
{
"input": "nnnzaaa",
"output": "YES"
},
{
"input": "zn",
"output": "NO"
},
{
"input": "ab",
"output": "NO"
},
{
"input": "aaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaab",
"output": "NO"
},
{
"input": "aaaaaaaaan",
"output": "YES"
},
{
"input": "baaaaaaaaa",
"output": "YES"
},
{
"input": "naaaaaaaaa",
"output": "YES"
},
{
"input": "nbaaaaaaaa",
"output": "YES"
},
{
"input": "bbaaaaaaaa",
"output": "NO"
},
{
"input": "bnaaaaaaaa",
"output": "NO"
},
{
"input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu",
"output": "YES"
},
{
"input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino",
"output": "NO"
},
{
"input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul",
"output": "NO"
},
{
"input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu",
"output": "YES"
},
{
"input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo",
"output": "NO"
},
{
"input": "y",
"output": "NO"
},
{
"input": "by",
"output": "NO"
},
{
"input": "yy",
"output": "NO"
},
{
"input": "nbn",
"output": "NO"
},
{
"input": "nz",
"output": "NO"
},
{
"input": "king",
"output": "NO"
},
{
"input": "g",
"output": "NO"
},
{
"input": "az",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "yn",
"output": "NO"
},
{
"input": "aeo",
"output": "YES"
},
{
"input": "nb",
"output": "NO"
},
{
"input": "npn",
"output": "NO"
},
{
"input": "kini",
"output": "YES"
},
{
"input": "pya",
"output": "NO"
},
{
"input": "m",
"output": "NO"
},
{
"input": "p",
"output": "NO"
},
{
"input": "aaaaaaaak",
"output": "NO"
},
{
"input": "aab",
"output": "NO"
},
{
"input": "d",
"output": "NO"
},
{
"input": "at",
"output": "NO"
},
{
"input": "aaaaaak",
"output": "NO"
},
{
"input": "aaz",
"output": "NO"
},
{
"input": "aaab",
"output": "NO"
},
{
"input": "s",
"output": "NO"
},
{
"input": "nzzen",
"output": "NO"
},
{
"input": "aeionnhhhn",
"output": "NO"
},
{
"input": "h",
"output": "NO"
},
{
"input": "ny",
"output": "NO"
}
] | 1,670,382,145
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 31
| 0
|
VOWELS = ("a", "e", "i", "o", "u")
VOWELS_exc = ("a", "e", "i", "o", "u", "n")
s = input()
s = s.replace("n", "")
for i in range(1, len(s)):
if s[i-1] in VOWELS_exc:
pass
elif s[i-1] not in VOWELS_exc and s[i] in VOWELS:
pass
else:
print("NO")
raise SystemExit(0)
print("YES")
|
Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
|
```python
VOWELS = ("a", "e", "i", "o", "u")
VOWELS_exc = ("a", "e", "i", "o", "u", "n")
s = input()
s = s.replace("n", "")
for i in range(1, len(s)):
if s[i-1] in VOWELS_exc:
pass
elif s[i-1] not in VOWELS_exc and s[i] in VOWELS:
pass
else:
print("NO")
raise SystemExit(0)
print("YES")
```
| 0
|
|
779
|
A
|
Pupils Redistribution
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
In Berland each high school student is characterized by academic performance — integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known — integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
|
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=100) — number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
|
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
|
[
"4\n5 4 4 4\n5 5 4 5\n",
"6\n1 1 1 1 1 1\n5 5 5 5 5 5\n",
"1\n5\n3\n",
"9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n"
] |
[
"1\n",
"3\n",
"-1\n",
"4\n"
] |
none
| 500
|
[
{
"input": "4\n5 4 4 4\n5 5 4 5",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1\n5 5 5 5 5 5",
"output": "3"
},
{
"input": "1\n5\n3",
"output": "-1"
},
{
"input": "9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1",
"output": "4"
},
{
"input": "1\n1\n2",
"output": "-1"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "8\n1 1 2 2 3 3 4 4\n4 4 5 5 1 1 1 1",
"output": "2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1\n2 2 2 2 2 2 2 2 2 2",
"output": "5"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "2\n1 1\n1 1",
"output": "0"
},
{
"input": "2\n1 2\n1 1",
"output": "-1"
},
{
"input": "2\n2 2\n1 1",
"output": "1"
},
{
"input": "2\n1 2\n2 1",
"output": "0"
},
{
"input": "2\n1 1\n2 2",
"output": "1"
},
{
"input": "5\n5 5 5 5 5\n5 5 5 5 5",
"output": "0"
},
{
"input": "5\n5 5 5 3 5\n5 3 5 5 5",
"output": "0"
},
{
"input": "5\n2 3 2 3 3\n2 3 2 2 2",
"output": "1"
},
{
"input": "5\n4 4 1 4 2\n1 2 4 2 2",
"output": "1"
},
{
"input": "50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "50\n1 3 1 3 3 3 1 3 3 3 3 1 1 1 3 3 3 1 3 1 1 1 3 1 3 1 3 3 3 1 3 1 1 3 3 3 1 1 1 1 3 3 1 1 1 3 3 1 1 1\n1 3 1 3 3 1 1 3 1 3 3 1 1 1 1 3 3 1 3 1 1 3 1 1 3 1 1 1 1 3 3 1 3 3 3 3 1 3 3 3 3 3 1 1 3 3 1 1 3 1",
"output": "0"
},
{
"input": "50\n1 1 1 4 1 1 4 1 4 1 1 4 1 1 4 1 1 4 1 1 4 1 4 4 4 1 1 4 1 4 4 4 4 4 4 4 1 4 1 1 1 1 4 1 4 4 1 1 1 4\n1 4 4 1 1 4 1 4 4 1 1 4 1 4 1 1 4 1 1 1 4 4 1 1 4 1 4 1 1 4 4 4 4 1 1 4 4 1 1 1 4 1 4 1 4 1 1 1 4 4",
"output": "0"
},
{
"input": "50\n3 5 1 3 3 4 3 4 2 5 2 1 2 2 5 5 4 5 4 2 1 3 4 2 3 3 3 2 4 3 5 5 5 5 5 5 2 5 2 2 5 4 4 1 5 3 4 2 1 3\n3 5 3 2 5 3 4 4 5 2 3 4 4 4 2 2 4 4 4 3 3 5 5 4 3 1 4 4 5 5 4 1 2 5 5 4 1 2 3 4 5 5 3 2 3 4 3 5 1 1",
"output": "3"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n1 1 3 1 3 1 1 3 1 1 3 1 3 1 1 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 1 1 1 3 1 1 1 3 1 1 3 3 1 3 3 1 3 1 3 3 3 3 1 1 3 3 3 1 1 3 1 3 3 3 1 3 3 3 3 3 1 3 3 3 3 1 3 1 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 1 1 3 1 1 1\n1 1 1 3 3 3 3 3 3 3 1 3 3 3 1 3 3 3 3 3 3 1 3 3 1 3 3 1 1 1 3 3 3 3 3 3 3 1 1 3 3 3 1 1 3 3 1 1 1 3 3 3 1 1 3 1 1 3 3 1 1 3 3 3 3 3 3 1 3 3 3 1 1 3 3 3 1 1 3 3 1 3 1 3 3 1 1 3 3 1 1 3 1 3 3 3 1 3 1 3",
"output": "0"
},
{
"input": "100\n2 4 5 2 5 5 4 4 5 4 4 5 2 5 5 4 5 2 5 2 2 4 5 4 4 4 2 4 2 2 4 2 4 2 2 2 4 5 5 5 4 2 4 5 4 4 2 5 4 2 5 4 5 4 5 4 5 5 5 4 2 2 4 5 2 5 5 2 5 2 4 4 4 5 5 2 2 2 4 4 2 2 2 5 5 2 2 4 5 4 2 4 4 2 5 2 4 4 4 4\n4 4 2 5 2 2 4 2 5 2 5 4 4 5 2 4 5 4 5 2 2 2 2 5 4 5 2 4 2 2 5 2 5 2 4 5 5 5 2 5 4 4 4 4 5 2 2 4 2 4 2 4 5 5 5 4 5 4 5 5 5 2 5 4 4 4 4 4 2 5 5 4 2 4 4 5 5 2 4 4 4 2 2 2 5 4 2 2 4 5 4 4 4 4 2 2 4 5 5 2",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 4 3 1 4 2 1 3 1 1 2 4 4 4 4 1 1 4 1 4 4 1 1 2 3 3 3 2 4 2 3 3 3 1 3 4 2 2 1 3 4 4 3 2 2 2 4 2 1 2 1 2 2 1 1 4 2 1 3 2 4 4 4 2 3 1 3 1 3 2 2 2 2 4 4 1 3 1 1 4 2 3 3 4 4 2 4 4 2 4 3 3 1 3 2 4\n3 1 4 4 2 1 1 1 1 1 1 3 1 1 3 4 3 2 2 4 2 1 4 4 4 4 1 2 3 4 2 3 3 4 3 3 2 4 2 2 2 1 2 4 4 4 2 1 3 4 3 3 4 2 4 4 3 2 4 2 4 2 4 4 1 4 3 1 4 3 3 3 3 1 2 2 2 2 4 1 2 1 3 4 3 1 3 3 4 2 3 3 2 1 3 4 2 1 1 2",
"output": "0"
},
{
"input": "100\n2 4 5 2 1 5 5 2 1 5 1 5 1 1 1 3 4 5 1 1 2 3 3 1 5 5 4 4 4 1 1 1 5 2 3 5 1 2 2 1 1 1 2 2 1 2 4 4 5 1 3 2 5 3 5 5 3 2 2 2 1 3 4 4 4 4 4 5 3 1 4 1 5 4 4 5 4 5 2 4 4 3 1 2 1 4 5 3 3 3 3 2 2 2 3 5 3 1 3 4\n3 2 5 1 5 4 4 3 5 5 5 2 1 4 4 3 2 3 3 5 5 4 5 5 2 1 2 4 4 3 5 1 1 5 1 3 2 5 2 4 4 2 4 2 4 2 3 2 5 1 4 4 1 1 1 5 3 5 1 1 4 5 1 1 2 2 5 3 5 1 1 1 2 3 3 2 3 2 4 4 5 4 2 1 3 4 1 1 2 4 1 5 3 1 2 1 3 4 1 3",
"output": "0"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n1 4 4 1 4 4 1 1 4 1 1 1 1 4 4 4 4 1 1 1 1 1 1 4 4 4 1 1 4 4 1 1 1 1 4 4 4 4 4 1 1 4 4 1 1 1 4 1 1 1 1 4 4 4 4 4 4 1 4 4 4 4 1 1 1 4 1 4 1 1 1 1 4 1 1 1 4 4 4 1 4 4 1 4 4 4 4 4 1 4 1 1 4 1 4 1 1 1 4 4\n4 1 1 4 4 4 1 4 4 4 1 1 4 1 1 4 1 4 4 4 1 1 4 1 4 1 1 1 4 4 1 4 1 4 1 4 4 1 1 4 1 4 1 1 1 4 1 4 4 4 1 4 1 4 4 4 4 1 4 1 1 4 1 1 4 4 4 1 4 1 4 1 4 4 4 1 1 4 1 4 4 4 4 1 1 1 1 1 4 4 1 4 1 4 1 1 1 4 4 1",
"output": "1"
},
{
"input": "100\n5 2 5 2 2 3 3 2 5 3 2 5 3 3 3 5 2 2 5 5 3 3 5 3 2 2 2 3 2 2 2 2 3 5 3 3 2 3 2 5 3 3 5 3 2 2 5 5 5 5 5 2 3 2 2 2 2 3 2 5 2 2 2 3 5 5 5 3 2 2 2 3 5 3 2 5 5 3 5 5 5 3 2 5 2 3 5 3 2 5 5 3 5 2 3 3 2 2 2 2\n5 3 5 3 3 5 2 5 3 2 3 3 5 2 5 2 2 5 2 5 2 5 3 3 5 3 2 2 2 3 5 3 2 2 3 2 2 5 5 2 3 2 3 3 5 3 2 5 2 2 2 3 3 5 3 3 5 2 2 2 3 3 2 2 3 5 3 5 5 3 3 2 5 3 5 2 3 2 5 5 3 2 5 5 2 2 2 2 3 2 2 5 2 5 2 2 3 3 2 5",
"output": "1"
},
{
"input": "100\n4 4 5 4 3 5 5 2 4 5 5 5 3 4 4 2 5 2 5 3 3 3 3 5 3 2 2 2 4 4 4 4 3 3 4 5 3 2 2 2 4 4 5 3 4 5 4 5 5 2 4 2 5 2 3 4 4 5 2 2 4 4 5 5 5 3 5 4 5 5 5 4 3 3 2 4 3 5 5 5 2 4 2 5 4 3 5 3 2 3 5 2 5 2 2 5 4 5 4 3\n5 4 2 4 3 5 2 5 5 3 4 5 4 5 3 3 5 5 2 3 4 2 3 5 2 2 2 4 2 5 2 4 4 5 2 2 4 4 5 5 2 3 4 2 4 5 2 5 2 2 4 5 5 3 5 5 5 4 3 4 4 3 5 5 3 4 5 3 2 3 4 3 4 4 2 5 3 4 5 5 3 5 3 3 4 3 5 3 2 2 4 5 4 5 5 2 3 4 3 5",
"output": "1"
},
{
"input": "100\n1 4 2 2 2 1 4 5 5 5 4 4 5 5 1 3 2 1 4 5 2 3 4 4 5 4 4 4 4 5 1 3 5 5 3 3 3 3 5 1 4 3 5 1 2 4 1 3 5 5 1 3 3 3 1 3 5 4 4 2 2 5 5 5 2 3 2 5 1 3 5 4 5 3 2 2 3 2 3 3 2 5 2 4 2 3 4 1 3 1 3 1 5 1 5 2 3 5 4 5\n1 2 5 3 2 3 4 2 5 1 2 5 3 4 3 3 4 1 5 5 1 3 3 1 1 4 1 4 2 5 4 1 3 4 5 3 2 2 1 4 5 5 2 3 3 5 5 4 2 3 3 5 3 3 5 4 4 5 3 5 1 1 4 4 4 1 3 5 5 5 4 2 4 5 3 2 2 2 5 5 5 1 4 3 1 3 1 2 2 4 5 1 3 2 4 5 1 5 2 5",
"output": "1"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n5 2 2 2 5 2 5 5 5 2 5 2 5 5 5 5 5 5 2 2 2 5 5 2 5 2 2 5 2 5 5 2 5 2 5 2 5 5 5 5 5 2 2 2 2 5 5 2 5 5 5 2 5 5 5 2 5 5 5 2 2 2 5 2 2 2 5 5 2 5 5 5 2 5 2 2 5 2 2 2 5 5 5 5 2 5 2 5 2 2 5 2 5 2 2 2 2 5 5 2\n5 5 2 2 5 5 2 5 2 2 5 5 5 5 2 5 5 2 5 2 2 5 2 2 5 2 5 2 2 5 2 5 2 5 5 2 2 5 5 5 2 5 5 2 5 5 5 2 2 5 5 5 2 5 5 5 2 2 2 5 5 5 2 2 5 5 2 2 2 5 2 5 5 2 5 2 5 2 2 5 5 2 2 5 5 2 2 5 2 2 5 2 2 2 5 5 2 2 2 5",
"output": "1"
},
{
"input": "100\n3 3 2 2 1 2 3 3 2 2 1 1 3 3 1 1 1 2 1 2 3 2 3 3 3 1 2 3 1 2 1 2 3 3 2 1 1 1 1 1 2 2 3 2 1 1 3 3 1 3 3 1 3 1 3 3 3 2 1 2 3 1 3 2 2 2 2 2 2 3 1 3 1 2 2 1 2 3 2 3 3 1 2 1 1 3 1 1 1 2 1 2 2 2 3 2 3 2 1 1\n1 3 1 2 1 1 1 1 1 2 1 2 1 3 2 2 3 2 1 1 2 2 2 1 1 3 2 3 2 1 2 2 3 2 3 1 3 1 1 2 3 1 2 1 3 2 1 2 3 2 3 3 3 2 2 2 3 1 3 1 1 2 1 3 1 3 1 3 3 3 1 3 3 2 1 3 3 3 3 3 2 1 2 2 3 3 2 1 2 2 1 3 3 1 3 2 2 1 1 3",
"output": "1"
},
{
"input": "100\n5 3 3 2 5 3 2 4 2 3 3 5 3 4 5 4 3 3 4 3 2 3 3 4 5 4 2 4 2 4 5 3 3 4 5 3 5 3 5 3 3 2 5 3 4 5 2 5 2 2 4 2 2 2 2 5 4 5 4 3 5 4 2 5 5 3 4 5 2 3 2 2 2 5 3 2 2 2 3 3 5 2 3 2 4 5 3 3 3 5 2 3 3 3 5 4 5 5 5 2\n4 4 4 5 5 3 5 5 4 3 5 4 3 4 3 3 5 3 5 5 3 3 3 5 5 4 4 3 2 5 4 3 3 4 5 3 5 2 4 2 2 2 5 3 5 2 5 5 3 3 2 3 3 4 2 5 2 5 2 4 2 4 2 3 3 4 2 2 2 4 4 3 3 3 4 3 3 3 5 5 3 4 2 2 3 5 5 2 3 4 5 4 5 3 4 2 5 3 2 4",
"output": "3"
},
{
"input": "100\n5 3 4 4 2 5 1 1 4 4 3 5 5 1 4 4 2 5 3 2 1 1 3 2 4 4 4 2 5 2 2 3 1 4 1 4 4 5 3 5 1 4 1 4 1 5 5 3 5 5 1 5 3 5 1 3 3 4 5 3 2 2 4 5 2 5 4 2 4 4 1 1 4 2 4 1 2 2 4 3 4 1 1 1 4 3 5 1 2 1 4 5 4 4 2 1 4 1 3 2\n1 1 1 1 4 2 1 4 1 1 3 5 4 3 5 2 2 4 2 2 4 1 3 4 4 5 1 1 2 2 2 1 4 1 4 4 1 5 5 2 3 5 1 5 4 2 3 2 2 5 4 1 1 4 5 2 4 5 4 4 3 3 2 4 3 4 5 5 4 2 4 2 1 2 3 2 2 5 5 3 1 3 4 3 4 4 5 3 1 1 3 5 1 4 4 2 2 1 4 5",
"output": "2"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 4 3 3 4 4 3 3 3 4 3 4 3 4 4 3 3 3 3 3 3 4 3 3 4 3 3 3 3 4 3 3 3 4 4 4 3 3 4 4 4 3 4 4 3 3 4 3 3 3 4 4 4 3 4 3 3 3 3 3 3 3 4 4 3 3 3 3 4 3 3 3 3 3 4 4 3 3 3 3 3 4 3 4 4 4 4 3 4 3 4 4 4 4 3 3\n4 3 3 3 3 4 4 3 4 4 4 3 3 4 4 3 4 4 4 4 3 4 3 3 3 4 4 4 3 4 3 4 4 3 3 4 3 3 3 3 3 4 3 3 3 3 4 4 4 3 3 4 3 4 4 4 4 3 4 4 3 3 4 3 3 4 3 4 3 4 4 4 4 3 3 4 3 4 4 4 3 3 4 4 4 4 4 3 3 3 4 3 3 4 3 3 3 3 3 3",
"output": "5"
},
{
"input": "100\n4 2 5 2 5 4 2 5 5 4 4 2 4 4 2 4 4 5 2 5 5 2 2 4 4 5 4 5 5 5 2 2 2 2 4 4 5 2 4 4 4 2 2 5 5 4 5 4 4 2 4 5 4 2 4 5 4 2 4 5 4 4 4 4 4 5 4 2 5 2 5 5 5 5 4 2 5 5 4 4 2 5 2 5 2 5 4 2 4 2 4 5 2 5 2 4 2 4 2 4\n5 4 5 4 5 2 2 4 5 2 5 5 5 5 5 4 4 4 4 5 4 5 5 2 4 4 4 4 5 2 4 4 5 5 2 5 2 5 5 4 4 5 2 5 2 5 2 5 4 5 2 5 2 5 2 4 4 5 4 2 5 5 4 2 2 2 5 4 2 2 4 4 4 5 5 2 5 2 2 4 4 4 2 5 4 5 2 2 5 4 4 5 5 4 5 5 4 5 2 5",
"output": "5"
},
{
"input": "100\n3 4 5 3 5 4 5 4 4 4 2 4 5 4 3 2 3 4 3 5 2 5 2 5 4 3 4 2 5 2 5 3 4 5 2 5 4 2 4 5 4 3 2 4 4 5 2 5 5 3 3 5 2 4 4 2 3 3 2 5 5 5 2 4 5 5 4 2 2 5 3 3 2 4 4 2 4 5 5 2 5 5 3 2 5 2 4 4 3 3 5 4 5 5 2 5 4 5 4 3\n4 3 5 5 2 4 2 4 5 5 5 2 3 3 3 3 5 5 5 5 3 5 2 3 5 2 3 2 2 5 5 3 5 3 4 2 2 5 3 3 3 3 5 2 4 5 3 5 3 4 4 4 5 5 3 4 4 2 2 4 4 5 3 2 4 5 5 4 5 2 2 3 5 4 5 5 2 5 4 3 2 3 2 5 4 5 3 4 5 5 3 5 2 2 4 4 3 2 5 2",
"output": "4"
},
{
"input": "100\n4 1 1 2 1 4 4 1 4 5 5 5 2 2 1 3 5 2 1 5 2 1 2 4 4 2 1 2 2 2 4 3 1 4 2 2 3 1 1 4 4 5 4 4 4 5 1 4 1 4 3 1 2 1 2 4 1 2 5 2 1 4 3 4 1 4 2 1 1 1 5 3 3 1 4 1 3 1 4 1 1 2 2 2 3 1 4 3 4 4 5 2 5 4 3 3 3 2 2 1\n5 1 4 4 3 4 4 5 2 3 3 4 4 2 3 2 3 1 3 1 1 4 1 5 4 3 2 4 3 3 3 2 3 4 1 5 4 2 4 2 2 2 5 3 1 2 5 3 2 2 1 1 2 2 3 5 1 2 5 3 2 1 1 2 1 2 4 3 5 4 5 3 2 4 1 3 4 1 4 4 5 4 4 5 4 2 5 3 4 1 4 2 4 2 4 5 4 5 4 2",
"output": "6"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1 4 4 4 4 4 4 4 4 4 4\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "1"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 1 3 1 3 3 3 3 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 4 3 3 3 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 1 3 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3",
"output": "1"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
},
{
"input": "100\n3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5\n3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1",
"output": "25"
},
{
"input": "100\n3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5\n2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4",
"output": "50"
},
{
"input": "100\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "40"
},
{
"input": "100\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5\n2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3",
"output": "30"
},
{
"input": "5\n4 4 4 4 5\n4 5 5 5 5",
"output": "-1"
},
{
"input": "4\n1 1 1 1\n3 3 3 3",
"output": "2"
},
{
"input": "6\n1 1 2 2 3 4\n1 2 3 3 4 4",
"output": "-1"
},
{
"input": "4\n1 1 1 2\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n2 2 2\n4 4 4",
"output": "-1"
},
{
"input": "2\n1 2\n3 4",
"output": "-1"
},
{
"input": "6\n1 1 1 3 3 3\n2 2 2 4 4 4",
"output": "-1"
},
{
"input": "5\n1 2 2 2 2\n1 1 1 1 3",
"output": "-1"
},
{
"input": "2\n1 3\n2 2",
"output": "-1"
},
{
"input": "2\n1 3\n4 5",
"output": "-1"
},
{
"input": "4\n1 2 3 4\n5 5 5 5",
"output": "-1"
},
{
"input": "2\n1 3\n2 4",
"output": "-1"
},
{
"input": "2\n1 2\n4 4",
"output": "-1"
},
{
"input": "2\n1 2\n3 3",
"output": "-1"
},
{
"input": "10\n4 4 4 4 2 3 3 3 3 1\n2 2 2 2 4 1 1 1 1 3",
"output": "-1"
},
{
"input": "6\n1 2 3 3 4 4\n1 1 2 2 3 4",
"output": "-1"
},
{
"input": "5\n3 3 3 3 1\n1 1 1 1 3",
"output": "-1"
},
{
"input": "2\n1 1\n2 3",
"output": "-1"
},
{
"input": "8\n1 1 2 2 3 3 3 3\n2 2 2 2 1 1 1 1",
"output": "2"
},
{
"input": "5\n1 1 1 3 3\n1 1 1 1 2",
"output": "-1"
},
{
"input": "6\n2 2 3 3 4 4\n2 3 4 5 5 5",
"output": "-1"
},
{
"input": "6\n1 1 2 2 3 4\n3 3 4 4 1 2",
"output": "-1"
},
{
"input": "4\n1 2 3 3\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n3 3 3",
"output": "-1"
},
{
"input": "5\n3 3 3 2 2\n2 2 2 3 3",
"output": "-1"
},
{
"input": "10\n1 2 3 4 1 2 3 4 1 2\n1 2 3 4 1 2 3 4 3 4",
"output": "-1"
},
{
"input": "2\n2 2\n1 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n1 1 4",
"output": "-1"
},
{
"input": "4\n3 4 4 4\n3 3 4 4",
"output": "-1"
}
] | 1,657,310,179
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
def main():
n = int(input())
groupA = [int(num) for num in input().split()]
groupB = [int(num) for num in input().split()]
count = [0 for i in range(5)]
for i in range(n):
count[groupA[i] - 1] += 1
if count[groupB[i] - 1] > 0:
count[groupB[i] - 1] -= 1
for i in range(5):
if count[i] % 2 == 1:
print(-1)
return
print(sum(count)//2)
main()
|
Title: Pupils Redistribution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland each high school student is characterized by academic performance — integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known — integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
Input Specification:
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=100) — number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
Output Specification:
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
Demo Input:
['4\n5 4 4 4\n5 5 4 5\n', '6\n1 1 1 1 1 1\n5 5 5 5 5 5\n', '1\n5\n3\n', '9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n']
Demo Output:
['1\n', '3\n', '-1\n', '4\n']
Note:
none
|
```python
def main():
n = int(input())
groupA = [int(num) for num in input().split()]
groupB = [int(num) for num in input().split()]
count = [0 for i in range(5)]
for i in range(n):
count[groupA[i] - 1] += 1
if count[groupB[i] - 1] > 0:
count[groupB[i] - 1] -= 1
for i in range(5):
if count[i] % 2 == 1:
print(-1)
return
print(sum(count)//2)
main()
```
| 0
|
|
245
|
E
|
Mishap in Club
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation"
] | null | null |
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen.
On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended.
Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times.
|
The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive.
|
Print the sought minimum number of people
|
[
"+-+-+\n",
"---"
] |
[
"1\n",
"3"
] |
none
| 0
|
[
{
"input": "+-+-+",
"output": "1"
},
{
"input": "---",
"output": "3"
},
{
"input": "-",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "---",
"output": "3"
},
{
"input": "----",
"output": "4"
},
{
"input": "---+",
"output": "3"
},
{
"input": "--+-",
"output": "2"
},
{
"input": "--++",
"output": "2"
},
{
"input": "-+--",
"output": "2"
},
{
"input": "-++",
"output": "2"
},
{
"input": "-++-",
"output": "2"
},
{
"input": "+",
"output": "1"
},
{
"input": "+-",
"output": "1"
},
{
"input": "+--",
"output": "2"
},
{
"input": "+--+",
"output": "2"
},
{
"input": "++--",
"output": "2"
},
{
"input": "-+++--+-++--+-+--+-+",
"output": "3"
},
{
"input": "++-++--+++++-+++++---+++-++-++-",
"output": "12"
},
{
"input": "----+-+--++---++---++-+-----+--",
"output": "11"
},
{
"input": "-+++---+++++++++++++-++-++++++-++-+-+++-",
"output": "22"
},
{
"input": "++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++",
"output": "300"
},
{
"input": "------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------",
"output": "300"
},
{
"input": "++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++",
"output": "298"
},
{
"input": "++++++++++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++",
"output": "296"
},
{
"input": "+++++++++++++++++++++++++++++++++++++++++-++++++++-++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++-+++++++++++++++++++++++++-++++++++++++++++++++++++-++++-+++++++++++++-++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-+++-+++++++++++++++++",
"output": "280"
},
{
"input": "++++-+---+++--++++-++-++++++++-++-++++-++++++-+-+-+++--+-++++++-+++-++-+++-++++-++++-+-+----+++++---++++-+---+++--+++++-+++-+-++++++----+--+++++++++-+--+++-+-+-++++++--+-+-+-+-++--+-+-----++++++-+++-++--+++++++++---+-++++++-++-++++-+--+-++-++++-+-+--++-+--+++-+-++++++++++++-+++-+----++++++++--+-+-++",
"output": "100"
},
{
"input": "+----++-----+----+++--++---+++--+-++++-++---++++++--++++--++-++--++--++----++++---+--+++----++--++--++--+--++++++++--++--+++----+++----++----++-+--+---+--+-++--+--+--+-+--+---++-+-++--+++++-++------+++-++--+--+--+++++++--++-+--+-+--++++-++--+---+-+-++-+-++----+-++++++-+++--+----++-+--++-----+++-++-+",
"output": "15"
},
{
"input": "-+++----+-++--+-+----+--+++++----+---+-++-+---+++--+---++-+-----+----+------+--+----++-++-----+++--+---+-+-----++++------+--+-----++---+---+---+-++------++++--+-+-------------+---+--+-+--------++---+-++---+-----+++--+---+-++-+---+-+---+++--++-----++------+----+---+---+--+-+-++-+---++--------+----++",
"output": "103"
},
{
"input": "----------+-----------------------------------------------------------+-+-------------+--------------------------------------------------------------------------------+--+-----+-+-------------------------------------------+-----------------------------------------------------------------+-----------",
"output": "280"
}
] | 1,651,890,024
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 92
| 0
|
notes = list(input())
y = 0
MAY = 0
MIY = 0
for i in notes:
if i == '+':
y += 1
MAY = max(MAY, y)
else:
y -= 1
MIY = min(MIY, y)
print(MAY-MIY)
|
Title: Mishap in Club
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen.
On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended.
Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times.
Input Specification:
The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive.
Output Specification:
Print the sought minimum number of people
Demo Input:
['+-+-+\n', '---']
Demo Output:
['1\n', '3']
Note:
none
|
```python
notes = list(input())
y = 0
MAY = 0
MIY = 0
for i in notes:
if i == '+':
y += 1
MAY = max(MAY, y)
else:
y -= 1
MIY = min(MIY, y)
print(MAY-MIY)
```
| 3
|
|
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..."
}
] | 1,669,753,523
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 30
| 0
|
t = int(input())
ans = 0
temp = 0
tab = []
if t == 2:
print(2)
elif t == 3:
print(3)
else:
while(True):
ans += 1
temp += 2
tab.append(2)
if temp == t:
break
elif t - temp == 3:
ans += 1
temp += 3
tab.append(3)
break
print(ans)
print(" ".join(str(x) for x in tab))
|
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
t = int(input())
ans = 0
temp = 0
tab = []
if t == 2:
print(2)
elif t == 3:
print(3)
else:
while(True):
ans += 1
temp += 2
tab.append(2)
if temp == t:
break
elif t - temp == 3:
ans += 1
temp += 3
tab.append(3)
break
print(ans)
print(" ".join(str(x) for x in tab))
```
| 0
|
|
1,005
|
A
|
Tanya and Stairways
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
|
The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
|
In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
|
[
"7\n1 2 3 1 2 3 4\n",
"4\n1 1 1 1\n",
"5\n1 2 3 4 5\n",
"5\n1 2 1 2 1\n"
] |
[
"2\n3 4 ",
"4\n1 1 1 1 ",
"1\n5 ",
"3\n2 2 1 "
] |
none
| 0
|
[
{
"input": "7\n1 2 3 1 2 3 4",
"output": "2\n3 4 "
},
{
"input": "4\n1 1 1 1",
"output": "4\n1 1 1 1 "
},
{
"input": "5\n1 2 3 4 5",
"output": "1\n5 "
},
{
"input": "5\n1 2 1 2 1",
"output": "3\n2 2 1 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "48\n1 2 3 4 1 2 3 1 1 2 3 1 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 1 2 1 2 1 1 2 1 2 1 2 3 1 2 1 2 1",
"output": "20\n4 3 1 3 4 1 4 4 4 1 2 2 2 1 2 2 3 2 2 1 "
},
{
"input": "2\n1 2",
"output": "1\n2 "
},
{
"input": "3\n1 1 2",
"output": "2\n1 2 "
},
{
"input": "4\n1 1 2 3",
"output": "2\n1 3 "
},
{
"input": "8\n1 2 3 1 2 3 4 5",
"output": "2\n3 5 "
},
{
"input": "5\n1 1 1 2 3",
"output": "3\n1 1 3 "
}
] | 1,660,979,721
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 16
| 46
| 0
|
n=int(input())
a=input()
a1=a.split()
for i in range(n): a1[i]=int(a1[i])
s=0
for i in range(n):
if a1[i]==1: s=s+1
print(s)
for i in range(n-1):
if a1[i]>=a1[i+1]: print(a1[i],end=" ")
print(a1[-1])
|
Title: Tanya and Stairways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Input Specification:
The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Output Specification:
In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
Demo Input:
['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n']
Demo Output:
['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 ']
Note:
none
|
```python
n=int(input())
a=input()
a1=a.split()
for i in range(n): a1[i]=int(a1[i])
s=0
for i in range(n):
if a1[i]==1: s=s+1
print(s)
for i in range(n-1):
if a1[i]>=a1[i+1]: print(a1[i],end=" ")
print(a1[-1])
```
| 3
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,675,540,918
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 92
| 4,915,200
|
import copy
n=int(input())
Z={}
S=[]
for i in range(n):
x,y=map(str,input().split())
S.append((x,eval(y)))
if x in Z:
Z[x]=Z[x]+eval(y)
else:
Z[x]=eval(y)
maxm=0
for k in Z:
if maxm<Z[k]:
maxm=Z[k]
G={}
for i in Z:
if Z[i]==maxm:
G[i]=0
for k in G:
G[k]=0
for c in S:
if c[0] in G:
G[c[0]]+=c[1]
if G[c[0]]>=maxm:
print(c[0])
break
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
import copy
n=int(input())
Z={}
S=[]
for i in range(n):
x,y=map(str,input().split())
S.append((x,eval(y)))
if x in Z:
Z[x]=Z[x]+eval(y)
else:
Z[x]=eval(y)
maxm=0
for k in Z:
if maxm<Z[k]:
maxm=Z[k]
G={}
for i in Z:
if Z[i]==maxm:
G[i]=0
for k in G:
G[k]=0
for c in S:
if c[0] in G:
G[c[0]]+=c[1]
if G[c[0]]>=maxm:
print(c[0])
break
```
| 3.917379
|
675
|
B
|
Restoring Painting
|
PROGRAMMING
| 1,400
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
- The painting is a square 3<=×<=3, each cell contains a single integer from 1 to *n*, and different cells may contain either different or equal integers. - The sum of integers in each of four squares 2<=×<=2 is equal to the sum of integers in the top left square 2<=×<=2. - Four elements *a*, *b*, *c* and *d* are known and are located as shown on the picture below.
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
|
The first line of the input contains five integers *n*, *a*, *b*, *c* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=*n*) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
|
Print one integer — the number of distinct valid squares.
|
[
"2 1 1 1 2\n",
"3 3 1 2 3\n"
] |
[
"2\n",
"6\n"
] |
Below are all the possible paintings for the first sample. <img class="tex-graphics" src="https://espresso.codeforces.com/c4c53d4e7b6814d8aad7b72604b6089d61dadb48.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/46a6ad6a5d3db202f3779b045b9dc77fc2348cf1.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In the second sample, only paintings displayed below satisfy all the rules. <img class="tex-graphics" src="https://espresso.codeforces.com/776f231305f8ce7c33e79e887722ce46aa8b6e61.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/2fce9e9a31e70f1e46ea26f11d7305b3414e9b6b.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/be084a4d1f7e475be1183f7dff10e9c89eb175ef.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/96afdb4a35ac14f595d29bea2282f621098902f4.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/79ca8d720334a74910514f017ecf1d0166009a03.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/ad3c37e950bf5702d54f05756db35c831da59ad9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "2 1 1 1 2",
"output": "2"
},
{
"input": "3 3 1 2 3",
"output": "6"
},
{
"input": "1 1 1 1 1",
"output": "1"
},
{
"input": "1000 522 575 426 445",
"output": "774000"
},
{
"input": "99000 52853 14347 64237 88869",
"output": "1296306000"
},
{
"input": "100000 2 2 2 2",
"output": "10000000000"
},
{
"input": "2 1 1 2 2",
"output": "0"
},
{
"input": "10 9 10 8 10",
"output": "70"
},
{
"input": "100 19 16 35 83",
"output": "1700"
},
{
"input": "1000 102 583 606 929",
"output": "150000"
},
{
"input": "10000 1816 3333 6908 7766",
"output": "4750000"
},
{
"input": "100000 80015 84290 50777 30497",
"output": "1696900000"
},
{
"input": "100000 64022 49026 55956 88430",
"output": "6866200000"
},
{
"input": "100000 10263 46628 10268 22948",
"output": "5095500000"
},
{
"input": "100000 81311 81584 51625 57276",
"output": "4600600000"
},
{
"input": "100000 77594 3226 21255 8541",
"output": "1291800000"
},
{
"input": "100000 65131 35523 58220 87645",
"output": "5478900000"
},
{
"input": "100000 83958 32567 91083 95317",
"output": "3012500000"
},
{
"input": "100000 36851 54432 21164 85520",
"output": "1806300000"
},
{
"input": "100000 55732 17473 23832 75148",
"output": "7422500000"
},
{
"input": "100000 60789 25296 49585 25237",
"output": "4015900000"
},
{
"input": "100000 92060 77234 58709 36956",
"output": "2637100000"
},
{
"input": "100000 87223 66046 27153 40823",
"output": "1470700000"
},
{
"input": "100000 3809 35468 34556 51158",
"output": "5173900000"
},
{
"input": "100000 35038 37363 95275 88903",
"output": "0"
},
{
"input": "100000 45274 9250 36558 49486",
"output": "6848000000"
},
{
"input": "100000 1 1 1 1",
"output": "10000000000"
},
{
"input": "100000 1 1 1 100000",
"output": "100000"
},
{
"input": "100000 1 1 100000 1",
"output": "100000"
},
{
"input": "100000 1 1 100000 100000",
"output": "0"
},
{
"input": "100000 1 100000 1 1",
"output": "100000"
},
{
"input": "100000 1 100000 1 100000",
"output": "0"
},
{
"input": "100000 1 100000 100000 1",
"output": "10000000000"
},
{
"input": "100000 1 100000 100000 100000",
"output": "100000"
},
{
"input": "100000 100000 1 1 1",
"output": "100000"
},
{
"input": "100000 100000 1 1 100000",
"output": "10000000000"
},
{
"input": "100000 100000 1 100000 1",
"output": "0"
},
{
"input": "100000 100000 1 100000 100000",
"output": "100000"
},
{
"input": "100000 100000 100000 1 1",
"output": "0"
},
{
"input": "100000 100000 100000 1 100000",
"output": "100000"
},
{
"input": "100000 100000 100000 100000 1",
"output": "100000"
},
{
"input": "100000 100000 100000 100000 100000",
"output": "10000000000"
},
{
"input": "3 3 3 1 1",
"output": "0"
},
{
"input": "10 1 2 5 10",
"output": "0"
},
{
"input": "5 1 1 5 5",
"output": "0"
},
{
"input": "4 4 4 1 1",
"output": "0"
},
{
"input": "10 10 10 1 1",
"output": "0"
},
{
"input": "5 5 5 1 1",
"output": "0"
},
{
"input": "100 100 100 1 1",
"output": "0"
},
{
"input": "3 1 1 3 3",
"output": "0"
},
{
"input": "10 2 10 1 10",
"output": "0"
},
{
"input": "7 7 7 1 1",
"output": "0"
},
{
"input": "5 5 3 4 1",
"output": "0"
},
{
"input": "7 1 1 7 7",
"output": "0"
},
{
"input": "100 1 1 100 100",
"output": "0"
},
{
"input": "123 1 2 3 100",
"output": "2829"
},
{
"input": "10 1 1 10 10",
"output": "0"
},
{
"input": "803 525 6 623 8",
"output": "0"
}
] | 1,694,516,810
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 233
| 0
|
n,a,b,c,d = map(int,input().split())
ans=0
for i in range(1,n+1):
maxx=max(a+b+i+1,max(a+c+i+1,max(c+d+i+1,b+d+i+1)))
e=maxx-(a+b+i)
f=maxx-(a+c+i)
g=maxx-(c+d+i)
h=maxx-(b+d+i)
x=max(e,max(f,max(g,h)))
ans+=n+1-x
print(ans if ans>=0 else 0)
|
Title: Restoring Painting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
- The painting is a square 3<=×<=3, each cell contains a single integer from 1 to *n*, and different cells may contain either different or equal integers. - The sum of integers in each of four squares 2<=×<=2 is equal to the sum of integers in the top left square 2<=×<=2. - Four elements *a*, *b*, *c* and *d* are known and are located as shown on the picture below.
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input Specification:
The first line of the input contains five integers *n*, *a*, *b*, *c* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=*n*) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output Specification:
Print one integer — the number of distinct valid squares.
Demo Input:
['2 1 1 1 2\n', '3 3 1 2 3\n']
Demo Output:
['2\n', '6\n']
Note:
Below are all the possible paintings for the first sample. <img class="tex-graphics" src="https://espresso.codeforces.com/c4c53d4e7b6814d8aad7b72604b6089d61dadb48.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/46a6ad6a5d3db202f3779b045b9dc77fc2348cf1.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In the second sample, only paintings displayed below satisfy all the rules. <img class="tex-graphics" src="https://espresso.codeforces.com/776f231305f8ce7c33e79e887722ce46aa8b6e61.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/2fce9e9a31e70f1e46ea26f11d7305b3414e9b6b.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/be084a4d1f7e475be1183f7dff10e9c89eb175ef.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/96afdb4a35ac14f595d29bea2282f621098902f4.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/79ca8d720334a74910514f017ecf1d0166009a03.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/ad3c37e950bf5702d54f05756db35c831da59ad9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n,a,b,c,d = map(int,input().split())
ans=0
for i in range(1,n+1):
maxx=max(a+b+i+1,max(a+c+i+1,max(c+d+i+1,b+d+i+1)))
e=maxx-(a+b+i)
f=maxx-(a+c+i)
g=maxx-(c+d+i)
h=maxx-(b+d+i)
x=max(e,max(f,max(g,h)))
ans+=n+1-x
print(ans if ans>=0 else 0)
```
| 3
|
|
255
|
A
|
Greg's Workout
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
|
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
|
[
"2\n2 8\n",
"3\n5 1 10\n",
"7\n3 3 2 7 9 6 8\n"
] |
[
"biceps\n",
"back\n",
"chest\n"
] |
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
| 500
|
[
{
"input": "2\n2 8",
"output": "biceps"
},
{
"input": "3\n5 1 10",
"output": "back"
},
{
"input": "7\n3 3 2 7 9 6 8",
"output": "chest"
},
{
"input": "4\n5 6 6 2",
"output": "chest"
},
{
"input": "5\n8 2 2 6 3",
"output": "chest"
},
{
"input": "6\n8 7 2 5 3 4",
"output": "chest"
},
{
"input": "8\n7 2 9 10 3 8 10 6",
"output": "chest"
},
{
"input": "9\n5 4 2 3 4 4 5 2 2",
"output": "chest"
},
{
"input": "10\n4 9 8 5 3 8 8 10 4 2",
"output": "biceps"
},
{
"input": "11\n10 9 7 6 1 3 9 7 1 3 5",
"output": "chest"
},
{
"input": "12\n24 22 6 16 5 21 1 7 2 19 24 5",
"output": "chest"
},
{
"input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24",
"output": "chest"
},
{
"input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7",
"output": "back"
},
{
"input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12",
"output": "chest"
},
{
"input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8",
"output": "biceps"
},
{
"input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19",
"output": "chest"
},
{
"input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21",
"output": "back"
},
{
"input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24",
"output": "chest"
},
{
"input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20",
"output": "chest"
},
{
"input": "1\n10",
"output": "chest"
},
{
"input": "2\n15 3",
"output": "chest"
},
{
"input": "3\n21 11 19",
"output": "chest"
},
{
"input": "4\n19 24 13 15",
"output": "chest"
},
{
"input": "5\n4 24 1 9 19",
"output": "biceps"
},
{
"input": "6\n6 22 24 7 15 24",
"output": "back"
},
{
"input": "7\n10 8 23 23 14 18 14",
"output": "chest"
},
{
"input": "8\n5 16 8 9 17 16 14 7",
"output": "biceps"
},
{
"input": "9\n12 3 10 23 6 4 22 13 12",
"output": "chest"
},
{
"input": "10\n1 9 20 18 20 17 7 24 23 2",
"output": "back"
},
{
"input": "11\n22 25 8 2 18 15 1 13 1 11 4",
"output": "biceps"
},
{
"input": "12\n20 12 14 2 15 6 24 3 11 8 11 14",
"output": "chest"
},
{
"input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18",
"output": "back"
},
{
"input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22",
"output": "biceps"
},
{
"input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6",
"output": "back"
},
{
"input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14",
"output": "chest"
},
{
"input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10",
"output": "biceps"
},
{
"input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17",
"output": "chest"
},
{
"input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14",
"output": "back"
},
{
"input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6",
"output": "back"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "2\n1 7",
"output": "biceps"
},
{
"input": "3\n18 18 23",
"output": "back"
},
{
"input": "4\n12 15 1 13",
"output": "chest"
},
{
"input": "5\n11 14 25 21 21",
"output": "biceps"
},
{
"input": "6\n11 9 12 11 22 18",
"output": "biceps"
},
{
"input": "7\n11 1 16 20 21 25 20",
"output": "chest"
},
{
"input": "8\n1 2 20 9 3 22 17 4",
"output": "back"
},
{
"input": "9\n19 2 10 19 15 20 3 1 13",
"output": "back"
},
{
"input": "10\n11 2 11 8 21 16 2 3 19 9",
"output": "back"
},
{
"input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24",
"output": "chest"
},
{
"input": "12\n4 24 21 3 13 24 22 13 12 21 1 15",
"output": "back"
},
{
"input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3",
"output": "biceps"
},
{
"input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11",
"output": "biceps"
},
{
"input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17",
"output": "back"
},
{
"input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17",
"output": "chest"
},
{
"input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8",
"output": "chest"
},
{
"input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22",
"output": "chest"
},
{
"input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24",
"output": "chest"
},
{
"input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23",
"output": "biceps"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22",
"output": "biceps"
}
] | 1,595,952,007
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 61
| 280
| 20,172,800
|
n = int(input())
a = list(map(int, input().split()))
x = 0
y = 0
z = 0
for i in range(0,n,3):
x += a[i]
for j in range(1,n,3):
y += a[j]
for k in range(2,n,3):
z += a[k]
if max(x,y,z) == x:
print('chest')
elif max(x,y,z) == y:
print('biceps')
else:
print('back')
|
Title: Greg's Workout
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Output Specification:
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
Demo Input:
['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n']
Demo Output:
['biceps\n', 'back\n', 'chest\n']
Note:
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
|
```python
n = int(input())
a = list(map(int, input().split()))
x = 0
y = 0
z = 0
for i in range(0,n,3):
x += a[i]
for j in range(1,n,3):
y += a[j]
for k in range(2,n,3):
z += a[k]
if max(x,y,z) == x:
print('chest')
elif max(x,y,z) == y:
print('biceps')
else:
print('back')
```
| 3
|
|
769
|
B
|
News About Credit
|
PROGRAMMING
| 1,200
|
[
"*special",
"greedy",
"two pointers"
] | null | null |
Polycarp studies at the university in the group which consists of *n* students (including himself). All they are registrated in the social net "TheContacnt!".
Not all students are equally sociable. About each student you know the value *a**i* — the maximum number of messages which the *i*-th student is agree to send per day. The student can't send messages to himself.
In early morning Polycarp knew important news that the programming credit will be tomorrow. For this reason it is necessary to urgently inform all groupmates about this news using private messages.
Your task is to make a plan of using private messages, so that:
- the student *i* sends no more than *a**i* messages (for all *i* from 1 to *n*); - all students knew the news about the credit (initially only Polycarp knew it); - the student can inform the other student only if he knows it himself.
Let's consider that all students are numerated by distinct numbers from 1 to *n*, and Polycarp always has the number 1.
In that task you shouldn't minimize the number of messages, the moment of time, when all knew about credit or some other parameters. Find any way how to use private messages which satisfies requirements above.
|
The first line contains the positive integer *n* (2<=≤<=*n*<=≤<=100) — the number of students.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* equals to the maximum number of messages which can the *i*-th student agree to send. Consider that Polycarp always has the number 1.
|
Print -1 to the first line if it is impossible to inform all students about credit.
Otherwise, in the first line print the integer *k* — the number of messages which will be sent. In each of the next *k* lines print two distinct integers *f* and *t*, meaning that the student number *f* sent the message with news to the student number *t*. All messages should be printed in chronological order. It means that the student, who is sending the message, must already know this news. It is assumed that students can receive repeated messages with news of the credit.
If there are several answers, it is acceptable to print any of them.
|
[
"4\n1 2 1 0\n",
"6\n2 0 1 3 2 0\n",
"3\n0 2 2\n"
] |
[
"3\n1 2\n2 4\n2 3\n",
"6\n1 3\n3 4\n1 2\n4 5\n5 6\n4 6\n",
"-1\n"
] |
In the first test Polycarp (the student number 1) can send the message to the student number 2, who after that can send the message to students number 3 and 4. Thus, all students knew about the credit.
| 1,000
|
[
{
"input": "4\n1 2 1 0",
"output": "3\n1 2\n2 3\n2 4"
},
{
"input": "6\n2 0 1 3 2 0",
"output": "5\n1 4\n1 5\n4 3\n4 2\n4 6"
},
{
"input": "3\n0 2 2",
"output": "-1"
},
{
"input": "2\n0 0",
"output": "-1"
},
{
"input": "2\n1 0",
"output": "1\n1 2"
},
{
"input": "2\n0 1",
"output": "-1"
},
{
"input": "2\n1 1",
"output": "1\n1 2"
},
{
"input": "3\n1 1 0",
"output": "2\n1 2\n2 3"
},
{
"input": "3\n0 1 1",
"output": "-1"
},
{
"input": "3\n1 0 0",
"output": "-1"
},
{
"input": "3\n2 0 0",
"output": "2\n1 2\n1 3"
},
{
"input": "3\n1 0 1",
"output": "2\n1 3\n3 2"
},
{
"input": "3\n1 1 1",
"output": "2\n1 2\n2 3"
},
{
"input": "40\n3 3 2 1 0 0 0 4 5 4 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 2 3 2 0 1 0 0 2 0 3 0 1 0",
"output": "-1"
},
{
"input": "100\n1 0 0 2 0 2 0 0 2 0 0 2 0 0 2 2 2 1 1 2 1 2 2 2 1 2 0 1 0 1 0 2 2 2 0 1 2 0 0 2 0 2 0 1 1 0 1 0 2 0 0 2 1 2 1 2 2 2 2 1 0 2 0 0 1 0 2 0 0 2 0 1 0 2 1 1 2 2 2 2 0 0 2 0 2 1 0 0 0 1 0 2 2 2 0 1 0 1 1 0",
"output": "99\n1 4\n4 6\n4 9\n6 12\n6 15\n9 16\n9 17\n12 20\n12 22\n15 23\n15 24\n16 26\n16 32\n17 33\n17 34\n20 37\n20 40\n22 42\n22 49\n23 52\n23 54\n24 56\n24 57\n26 58\n26 59\n32 62\n32 67\n33 70\n33 74\n34 77\n34 78\n37 79\n37 80\n40 83\n40 85\n42 92\n42 93\n49 94\n49 18\n52 19\n52 21\n54 25\n54 28\n56 30\n56 36\n57 44\n57 45\n58 47\n58 53\n59 55\n59 60\n62 65\n62 72\n67 75\n67 76\n70 86\n70 90\n74 96\n74 98\n77 99\n77 2\n78 3\n78 5\n79 7\n79 8\n80 10\n80 11\n83 13\n83 14\n85 27\n85 29\n92 31\n92 35\n93 38\n93 3..."
},
{
"input": "4\n2 0 0 0",
"output": "-1"
},
{
"input": "4\n2 0 0 1",
"output": "3\n1 4\n1 2\n4 3"
},
{
"input": "4\n2 0 1 0",
"output": "3\n1 3\n1 2\n3 4"
},
{
"input": "4\n2 1 0 0",
"output": "3\n1 2\n1 3\n2 4"
},
{
"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": "-1"
},
{
"input": "100\n99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87\n..."
},
{
"input": "100\n98 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "100\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 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n76 77\n..."
},
{
"input": "100\n1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "99\n1 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n76 7..."
},
{
"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 0",
"output": "99\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n..."
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87\n..."
},
{
"input": "100\n1 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "100\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 98 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99\n1 55\n55 2\n55 3\n55 4\n55 5\n55 6\n55 7\n55 8\n55 9\n55 10\n55 11\n55 12\n55 13\n55 14\n55 15\n55 16\n55 17\n55 18\n55 19\n55 20\n55 21\n55 22\n55 23\n55 24\n55 25\n55 26\n55 27\n55 28\n55 29\n55 30\n55 31\n55 32\n55 33\n55 34\n55 35\n55 36\n55 37\n55 38\n55 39\n55 40\n55 41\n55 42\n55 43\n55 44\n55 45\n55 46\n55 47\n55 48\n55 49\n55 50\n55 51\n55 52\n55 53\n55 54\n55 56\n55 57\n55 58\n55 59\n55 60\n55 61\n55 62\n55 63\n55 64\n55 65\n55 66\n55 67\n55 68\n55 69\n55 70\n55 71\n55 72\n55 73\n55 74\n55 75..."
},
{
"input": "2\n0 1",
"output": "-1"
},
{
"input": "5\n0 0 1 1 2",
"output": "-1"
},
{
"input": "7\n2 0 0 0 1 0 3",
"output": "6\n1 7\n1 5\n7 2\n7 3\n7 4\n5 6"
},
{
"input": "10\n3 0 0 0 0 2 0 1 0 3",
"output": "9\n1 10\n1 6\n1 8\n10 2\n10 3\n10 4\n6 5\n6 7\n8 9"
},
{
"input": "20\n0 2 0 0 2 0 0 2 2 0 0 2 0 2 1 0 1 3 1 1",
"output": "-1"
},
{
"input": "30\n2 0 2 2 0 2 2 0 0 0 3 0 1 1 2 0 0 2 2 0 1 0 3 0 1 0 2 0 0 1",
"output": "29\n1 11\n1 23\n11 3\n11 4\n11 6\n23 7\n23 15\n23 18\n3 19\n3 27\n4 13\n4 14\n6 21\n6 25\n7 30\n7 2\n15 5\n15 8\n18 9\n18 10\n19 12\n19 16\n27 17\n27 20\n13 22\n14 24\n21 26\n25 28\n30 29"
},
{
"input": "31\n2 0 0 4 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 0 3 0 4 3 0 2 0 0 0 3 4",
"output": "30\n1 4\n1 23\n4 31\n4 14\n4 21\n4 24\n23 30\n23 15\n23 26\n23 2\n31 3\n31 5\n31 6\n31 7\n14 8\n14 9\n14 10\n21 11\n21 12\n21 13\n24 16\n24 17\n24 18\n30 19\n30 20\n30 22\n15 25\n15 27\n26 28\n26 29"
},
{
"input": "39\n2 0 3 0 0 2 0 0 2 1 1 0 0 3 3 0 2 0 2 3 0 0 3 0 3 2 0 0 3 0 0 0 3 0 0 0 0 0 0",
"output": "38\n1 3\n1 14\n3 15\n3 20\n3 23\n14 25\n14 29\n14 33\n15 6\n15 9\n15 17\n20 19\n20 26\n20 10\n23 11\n23 2\n23 4\n25 5\n25 7\n25 8\n29 12\n29 13\n29 16\n33 18\n33 21\n33 22\n6 24\n6 27\n9 28\n9 30\n17 31\n17 32\n19 34\n19 35\n26 36\n26 37\n10 38\n11 39"
},
{
"input": "58\n4 2 1 3 5 3 0 0 1 0 3 0 2 1 0 0 0 4 0 0 0 0 0 1 2 3 4 0 1 1 0 0 1 0 0 0 2 0 0 0 0 2 2 0 2 0 0 4 0 2 0 0 0 0 0 1 0 0",
"output": "57\n1 5\n1 18\n1 27\n1 48\n5 4\n5 6\n5 11\n5 26\n5 2\n18 13\n18 25\n18 37\n18 42\n27 43\n27 45\n27 50\n27 3\n48 9\n48 14\n48 24\n48 29\n4 30\n4 33\n4 56\n6 7\n6 8\n6 10\n11 12\n11 15\n11 16\n26 17\n26 19\n26 20\n2 21\n2 22\n13 23\n13 28\n25 31\n25 32\n37 34\n37 35\n42 36\n42 38\n43 39\n43 40\n45 41\n45 44\n50 46\n50 47\n3 49\n9 51\n14 52\n24 53\n29 54\n30 55\n33 57\n56 58"
},
{
"input": "65\n3 0 0 0 0 3 0 0 0 0 0 4 2 0 0 0 0 0 0 0 0 8 0 0 0 0 0 6 7 0 3 0 0 0 0 4 0 3 0 0 0 0 1 0 0 5 0 0 0 0 3 0 0 4 0 0 0 0 0 1 0 0 0 0 7",
"output": "64\n1 22\n1 29\n1 65\n22 28\n22 46\n22 12\n22 36\n22 54\n22 6\n22 31\n22 38\n29 51\n29 13\n29 43\n29 60\n29 2\n29 3\n29 4\n65 5\n65 7\n65 8\n65 9\n65 10\n65 11\n65 14\n28 15\n28 16\n28 17\n28 18\n28 19\n28 20\n46 21\n46 23\n46 24\n46 25\n46 26\n12 27\n12 30\n12 32\n12 33\n36 34\n36 35\n36 37\n36 39\n54 40\n54 41\n54 42\n54 44\n6 45\n6 47\n6 48\n31 49\n31 50\n31 52\n38 53\n38 55\n38 56\n51 57\n51 58\n51 59\n13 61\n13 62\n43 63\n60 64"
},
{
"input": "77\n7 0 0 0 0 0 0 0 0 8 0 0 0 0 3 0 0 0 0 9 0 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 8 0 0 0 0 0 2 6 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 7 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 0 8",
"output": "76\n1 20\n1 63\n1 10\n1 37\n1 77\n1 24\n1 60\n20 44\n20 15\n20 43\n20 55\n20 2\n20 3\n20 4\n20 5\n20 6\n63 7\n63 8\n63 9\n63 11\n63 12\n63 13\n63 14\n63 16\n63 17\n10 18\n10 19\n10 21\n10 22\n10 23\n10 25\n10 26\n10 27\n37 28\n37 29\n37 30\n37 31\n37 32\n37 33\n37 34\n37 35\n77 36\n77 38\n77 39\n77 40\n77 41\n77 42\n77 45\n77 46\n24 47\n24 48\n24 49\n24 50\n24 51\n24 52\n24 53\n60 54\n60 56\n60 57\n60 58\n60 59\n60 61\n60 62\n44 64\n44 65\n44 66\n44 67\n44 68\n44 69\n15 70\n15 71\n15 72\n43 73\n43 74\n55 7..."
},
{
"input": "80\n2 3 0 2 2 1 3 3 3 0 0 0 1 0 1 0 3 1 0 2 0 2 3 0 2 3 0 3 0 0 0 3 0 0 0 2 3 0 0 2 0 0 0 0 0 3 2 0 0 3 0 3 0 3 0 3 1 2 0 0 0 0 0 0 0 1 0 3 0 0 0 1 0 2 0 2 0 0 0 0",
"output": "79\n1 2\n1 7\n2 8\n2 9\n2 17\n7 23\n7 26\n7 28\n8 32\n8 37\n8 46\n9 50\n9 52\n9 54\n17 56\n17 68\n17 4\n23 5\n23 20\n23 22\n26 25\n26 36\n26 40\n28 47\n28 58\n28 74\n32 76\n32 6\n32 13\n37 15\n37 18\n37 57\n46 66\n46 72\n46 3\n50 10\n50 11\n50 12\n52 14\n52 16\n52 19\n54 21\n54 24\n54 27\n56 29\n56 30\n56 31\n68 33\n68 34\n68 35\n4 38\n4 39\n5 41\n5 42\n20 43\n20 44\n22 45\n22 48\n25 49\n25 51\n36 53\n36 55\n40 59\n40 60\n47 61\n47 62\n58 63\n58 64\n74 65\n74 67\n76 69\n76 70\n6 71\n13 73\n15 75\n18 77\n57..."
},
{
"input": "90\n2 0 0 0 0 1 2 0 1 1 0 1 0 4 0 1 1 0 1 0 1 0 1 1 2 0 0 1 2 3 0 1 1 0 0 1 1 0 0 2 0 2 2 1 0 1 0 0 2 0 1 4 2 0 1 2 2 0 1 0 0 5 0 0 3 0 1 2 0 0 0 0 2 3 0 0 3 3 0 3 3 0 0 0 1 0 1 2 2 2",
"output": "89\n1 62\n1 14\n62 52\n62 30\n62 65\n62 74\n62 77\n14 78\n14 80\n14 81\n14 7\n52 25\n52 29\n52 40\n52 42\n30 43\n30 49\n30 53\n65 56\n65 57\n65 68\n74 73\n74 88\n74 89\n77 90\n77 6\n77 9\n78 10\n78 12\n78 16\n80 17\n80 19\n80 21\n81 23\n81 24\n81 28\n7 32\n7 33\n25 36\n25 37\n29 44\n29 46\n40 51\n40 55\n42 59\n42 67\n43 85\n43 87\n49 2\n49 3\n53 4\n53 5\n56 8\n56 11\n57 13\n57 15\n68 18\n68 20\n73 22\n73 26\n88 27\n88 31\n89 34\n89 35\n90 38\n90 39\n6 41\n9 45\n10 47\n12 48\n16 50\n17 54\n19 58\n21 60\n23 ..."
},
{
"input": "99\n1 2 1 0 2 1 2 0 1 2 1 1 2 1 0 2 0 0 1 2 0 1 1 1 0 0 1 1 2 3 1 0 0 0 1 1 0 0 1 2 1 0 2 0 0 2 0 1 0 2 1 1 0 0 3 1 0 2 2 2 2 1 0 2 0 1 1 0 3 2 0 0 1 1 2 0 0 2 0 0 1 2 3 0 3 0 0 3 0 3 0 1 2 1 0 1 0 1 1",
"output": "98\n1 30\n30 55\n30 69\n30 83\n55 85\n55 88\n55 90\n69 2\n69 5\n69 7\n83 10\n83 13\n83 16\n85 20\n85 29\n85 40\n88 43\n88 46\n88 50\n90 58\n90 59\n90 60\n2 61\n2 64\n5 70\n5 75\n7 78\n7 82\n10 93\n10 3\n13 6\n13 9\n16 11\n16 12\n20 14\n20 19\n29 22\n29 23\n40 24\n40 27\n43 28\n43 31\n46 35\n46 36\n50 39\n50 41\n58 48\n58 51\n59 52\n59 56\n60 62\n60 66\n61 67\n61 73\n64 74\n64 81\n70 92\n70 94\n75 96\n75 98\n78 99\n78 4\n82 8\n82 15\n93 17\n93 18\n3 21\n6 25\n9 26\n11 32\n12 33\n14 34\n19 37\n22 38\n23 42\n..."
},
{
"input": "100\n18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 16 0 0 0 0 0 13 0 0 0 0 0 15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 15 0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 16 0",
"output": "99\n1 26\n1 99\n1 38\n1 62\n1 32\n1 71\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n26 14\n26 15\n26 16\n26 17\n26 18\n26 19\n26 20\n26 21\n26 22\n26 23\n26 24\n26 25\n26 27\n26 28\n26 29\n26 30\n99 31\n99 33\n99 34\n99 35\n99 36\n99 37\n99 39\n99 40\n99 41\n99 42\n99 43\n99 44\n99 45\n99 46\n99 47\n99 48\n38 49\n38 50\n38 51\n38 52\n38 53\n38 54\n38 55\n38 56\n38 57\n38 58\n38 59\n38 60\n38 61\n38 63\n38 64\n62 65\n62 66\n62 67\n62 68\n62 69\n62 70\n62 72\n62 73\n62 74\n62 75\n62 76\n6..."
},
{
"input": "100\n3 2 1 0 1 0 0 0 2 3 2 1 0 0 0 3 1 1 3 0 1 1 1 3 0 2 2 2 0 1 1 1 0 0 1 0 2 1 1 2 1 0 0 0 1 1 0 3 0 0 0 4 1 2 0 0 0 1 0 3 2 0 0 2 0 3 0 1 4 2 1 0 0 1 1 0 1 0 0 4 0 1 0 1 0 1 1 0 0 0 0 1 2 0 0 0 3 3 0 2",
"output": "99\n1 52\n1 69\n1 80\n52 10\n52 16\n52 19\n52 24\n69 48\n69 60\n69 66\n69 97\n80 98\n80 2\n80 9\n80 11\n10 26\n10 27\n10 28\n16 37\n16 40\n16 54\n19 61\n19 64\n19 70\n24 93\n24 100\n24 3\n48 5\n48 12\n48 17\n60 18\n60 21\n60 22\n66 23\n66 30\n66 31\n97 32\n97 35\n97 38\n98 39\n98 41\n98 45\n2 46\n2 53\n9 58\n9 68\n11 71\n11 74\n26 75\n26 77\n27 82\n27 84\n28 86\n28 87\n37 92\n37 4\n40 6\n40 7\n54 8\n54 13\n61 14\n61 15\n64 20\n64 25\n70 29\n70 33\n93 34\n93 36\n100 42\n100 43\n3 44\n5 47\n12 49\n17 50\n18 ..."
},
{
"input": "100\n66 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0",
"output": "99\n1 97\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n97 67\n97 68\n97 69\n97 70\n97 71\n97 72\n97 73\n97 74\n97 75\n97 76\n97 77\n97 78\n97 79\n97 80\n97 81\n97 82\n97 83\n..."
},
{
"input": "20\n0 0 3 0 0 0 3 4 2 0 2 0 0 0 0 1 0 1 0 1",
"output": "-1"
},
{
"input": "60\n3 0 0 1 0 0 0 0 3 1 3 4 0 0 0 3 0 0 0 2 0 3 4 1 3 3 0 2 0 4 1 5 3 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 1 1 0 1 0 3 0 0",
"output": "-1"
},
{
"input": "80\n4 0 0 0 0 0 0 3 0 3 0 0 0 4 3 0 1 0 2 0 0 0 5 0 5 0 0 0 0 4 0 3 0 0 0 1 0 0 2 0 5 2 0 0 4 4 0 3 0 0 0 0 0 0 0 2 5 0 2 0 0 0 0 0 0 0 0 0 0 3 0 0 3 5 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "100\n2 0 0 2 0 0 0 0 0 2 0 0 0 5 0 0 0 0 0 0 0 1 0 7 0 0 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 6 0 7 4 0 0 0 0 5 0 0 0 0 0 0 7 4 0 0 0 0 0 0 7 7 0 0 0 0 0 2 0 0 0 0 0 0 0 0 4 7 7 0 0 0",
"output": "-1"
},
{
"input": "100\n1 0 0 0 1 2 3 2 1 0 1 2 3 1 3 1 0 0 1 1 0 0 2 1 2 1 3 3 1 0 0 1 0 2 2 0 3 0 1 1 1 2 0 2 0 1 0 2 0 1 2 2 0 0 0 0 1 0 0 0 1 1 4 0 2 0 1 0 2 0 2 2 2 1 1 0 0 2 0 3 1 0 0 1 1 0 1 0 2 3 2 0 1 2 0 0 0 0 0 1",
"output": "99\n1 63\n63 7\n63 13\n63 15\n63 27\n7 28\n7 37\n7 80\n13 90\n13 6\n13 8\n15 12\n15 23\n15 25\n27 34\n27 35\n27 42\n28 44\n28 48\n28 51\n37 52\n37 65\n37 69\n80 71\n80 72\n80 73\n90 78\n90 89\n90 91\n6 94\n6 5\n8 9\n8 11\n12 14\n12 16\n23 19\n23 20\n25 24\n25 26\n34 29\n34 32\n35 39\n35 40\n42 41\n42 46\n44 50\n44 57\n48 61\n48 62\n51 67\n51 74\n52 75\n52 81\n65 84\n65 85\n69 87\n69 93\n71 100\n71 2\n72 3\n72 4\n73 10\n73 17\n78 18\n78 21\n89 22\n89 30\n91 31\n91 33\n94 36\n94 38\n5 43\n9 45\n11 47\n14 49\n..."
},
{
"input": "100\n5 0 0 1 1 0 0 0 1 0 0 0 0 0 5 0 2 1 0 1 0 6 0 0 0 3 0 0 0 0 0 0 0 0 1 0 3 0 0 0 0 0 0 4 0 0 3 1 1 0 0 4 0 0 0 0 0 0 3 2 3 3 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 2 1 0 5 1 0 8 0 1 0 0 10 4 0 0 0 6 4 0 0 0 0 1",
"output": "99\n1 89\n1 84\n1 22\n1 94\n1 15\n89 81\n89 44\n89 52\n89 90\n89 95\n89 26\n89 37\n89 47\n89 59\n89 61\n84 62\n84 17\n84 60\n84 75\n84 78\n84 4\n84 5\n84 9\n22 18\n22 20\n22 35\n22 48\n22 49\n22 79\n94 82\n94 86\n94 100\n94 2\n94 3\n94 6\n15 7\n15 8\n15 10\n15 11\n15 12\n81 13\n81 14\n81 16\n81 19\n81 21\n44 23\n44 24\n44 25\n44 27\n52 28\n52 29\n52 30\n52 31\n90 32\n90 33\n90 34\n90 36\n95 38\n95 39\n95 40\n95 41\n26 42\n26 43\n26 45\n37 46\n37 50\n37 51\n47 53\n47 54\n47 55\n59 56\n59 57\n59 58\n61 63\n6..."
},
{
"input": "100\n47 0 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 4 0 1 0 2 0 0 1 0 0 0 0 0 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0",
"output": "99\n1 47\n1 12\n1 66\n1 33\n1 93\n1 37\n1 81\n1 25\n1 35\n1 40\n1 46\n1 61\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 34\n1 36\n1 38\n1 39\n1 41\n1 42\n47 43\n47 44\n47 45\n47 48\n47 49\n47 50\n47 51\n47 52\n47 53\n47 54\n47 55\n47 56\n47 57\n47 58\n47 59\n47 60\n47 62\n47 63\n12 64\n12 65\n12 67\n12 68\n12 69\n12 70\n12 71\n12 72\n12 73\n66 74\n66 75\n66 76\n66 77\n66 78\n66 79\n66..."
},
{
"input": "100\n1 0 2 1 1 1 0 0 0 3 2 1 1 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 2 1 0 2 1 1 1 0 0 1 2 1 3 1 1 0 0 2 1 0 1 1 1 2 1 2 0 3 1 2 0 1 1 2 2 1 1 1 1 1 2 0 0 2 1 1 0 1 2 1 1 1 1 1 0 1 1 1 0 3 0 0 2 2 0 0 0 2 1 2 0",
"output": "99\n1 10\n10 42\n10 57\n10 89\n42 3\n42 11\n42 30\n57 33\n57 40\n57 47\n89 53\n89 55\n89 59\n3 63\n3 64\n11 70\n11 73\n30 78\n30 92\n33 93\n33 97\n40 99\n40 4\n47 5\n47 6\n53 12\n53 13\n55 14\n55 16\n59 18\n59 19\n63 20\n63 22\n64 23\n64 24\n70 26\n70 27\n73 28\n73 29\n78 31\n78 34\n92 35\n92 36\n93 39\n93 41\n97 43\n97 44\n99 48\n99 50\n4 51\n5 52\n6 54\n12 58\n13 61\n14 62\n16 65\n18 66\n19 67\n20 68\n22 69\n23 74\n24 75\n26 77\n27 79\n28 80\n29 81\n31 82\n34 83\n35 85\n36 86\n39 87\n41 98\n43 2\n44 7\n4..."
},
{
"input": "100\n83 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 11 0",
"output": "99\n1 99\n1 83\n1 22\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n99 84\n99 85\n99 ..."
},
{
"input": "100\n1 1 0 1 0 1 1 1 2 1 0 1 1 0 1 2 1 1 1 1 2 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 2 2 2 2 1 1 0 1 2 1 1 0 1 1 0 1 0 1 1 0 0 1 1 1 1 1 1 2 0 2 1 2 1 1 0 0 1 1 1 1 0 2 4 2 1 1 1 0 1 2 1 1 1 0 2 1 2",
"output": "99\n1 86\n86 9\n86 16\n86 21\n86 44\n9 45\n9 46\n16 47\n16 52\n21 71\n21 73\n44 75\n44 85\n45 87\n45 93\n46 98\n46 100\n47 2\n47 4\n52 6\n52 7\n71 8\n71 10\n73 12\n73 13\n75 15\n75 17\n85 18\n85 19\n87 20\n87 22\n93 23\n93 25\n98 27\n98 28\n100 29\n100 30\n2 31\n4 32\n6 33\n7 34\n8 35\n10 36\n12 37\n13 38\n15 39\n17 41\n18 42\n19 48\n20 49\n22 51\n23 53\n25 54\n27 56\n28 57\n29 59\n30 61\n31 62\n32 65\n33 66\n34 67\n35 68\n36 69\n37 70\n38 74\n39 76\n41 77\n42 80\n48 81\n49 82\n51 83\n53 88\n54 89\n56 90\n..."
},
{
"input": "100\n99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87\n..."
},
{
"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 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n76 77\n..."
},
{
"input": "2\n0 100",
"output": "-1"
},
{
"input": "2\n100 0",
"output": "1\n1 2"
},
{
"input": "2\n100 100",
"output": "1\n1 2"
}
] | 1,488,633,222
| 4,422
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 46
| 4,505,600
|
n = int(input())
a = list(map(lambda x: int(x), input().split(' ')))
a[1:] = sorted(enumerate(a[1:]), key=lambda x: x[1], reverse=1)
a[0] = (0, a[0])
for i in range(1, len(a)): a[i] = (a[i][0] + 1, a[i][1])
#print(a)
received = [0]*n
p = 1
pair = []
for i in range(n):
if i >= p or p >= n: break
#print('-', a[i][0])
for j in range(a[i][1]):
#print(p, a[p])
pair.append((a[i][0], a[p][0]))
p += 1
if p == n: break
if p+1 < n: print(-1)
else: print(str.join('\n', map(lambda p: '{} {}'.format(p[0]+1, p[1]+1), pair)))
'''
4
1 2 1 0
6
2 0 1 3 2 0
6
2 3 2 0 1 0
'''
'''
6
2 0 1 3 2 0
0 3
0 4
3 1
3 2
3 5
2 0 1 3 2 0
2 3 2 1 0 0
'''
|
Title: News About Credit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp studies at the university in the group which consists of *n* students (including himself). All they are registrated in the social net "TheContacnt!".
Not all students are equally sociable. About each student you know the value *a**i* — the maximum number of messages which the *i*-th student is agree to send per day. The student can't send messages to himself.
In early morning Polycarp knew important news that the programming credit will be tomorrow. For this reason it is necessary to urgently inform all groupmates about this news using private messages.
Your task is to make a plan of using private messages, so that:
- the student *i* sends no more than *a**i* messages (for all *i* from 1 to *n*); - all students knew the news about the credit (initially only Polycarp knew it); - the student can inform the other student only if he knows it himself.
Let's consider that all students are numerated by distinct numbers from 1 to *n*, and Polycarp always has the number 1.
In that task you shouldn't minimize the number of messages, the moment of time, when all knew about credit or some other parameters. Find any way how to use private messages which satisfies requirements above.
Input Specification:
The first line contains the positive integer *n* (2<=≤<=*n*<=≤<=100) — the number of students.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* equals to the maximum number of messages which can the *i*-th student agree to send. Consider that Polycarp always has the number 1.
Output Specification:
Print -1 to the first line if it is impossible to inform all students about credit.
Otherwise, in the first line print the integer *k* — the number of messages which will be sent. In each of the next *k* lines print two distinct integers *f* and *t*, meaning that the student number *f* sent the message with news to the student number *t*. All messages should be printed in chronological order. It means that the student, who is sending the message, must already know this news. It is assumed that students can receive repeated messages with news of the credit.
If there are several answers, it is acceptable to print any of them.
Demo Input:
['4\n1 2 1 0\n', '6\n2 0 1 3 2 0\n', '3\n0 2 2\n']
Demo Output:
['3\n1 2\n2 4\n2 3\n', '6\n1 3\n3 4\n1 2\n4 5\n5 6\n4 6\n', '-1\n']
Note:
In the first test Polycarp (the student number 1) can send the message to the student number 2, who after that can send the message to students number 3 and 4. Thus, all students knew about the credit.
|
```python
n = int(input())
a = list(map(lambda x: int(x), input().split(' ')))
a[1:] = sorted(enumerate(a[1:]), key=lambda x: x[1], reverse=1)
a[0] = (0, a[0])
for i in range(1, len(a)): a[i] = (a[i][0] + 1, a[i][1])
#print(a)
received = [0]*n
p = 1
pair = []
for i in range(n):
if i >= p or p >= n: break
#print('-', a[i][0])
for j in range(a[i][1]):
#print(p, a[p])
pair.append((a[i][0], a[p][0]))
p += 1
if p == n: break
if p+1 < n: print(-1)
else: print(str.join('\n', map(lambda p: '{} {}'.format(p[0]+1, p[1]+1), pair)))
'''
4
1 2 1 0
6
2 0 1 3 2 0
6
2 3 2 0 1 0
'''
'''
6
2 0 1 3 2 0
0 3
0 4
3 1
3 2
3 5
2 0 1 3 2 0
2 3 2 1 0 0
'''
```
| 0
|
|
3
|
B
|
Lorry
|
PROGRAMMING
| 1,900
|
[
"greedy",
"sortings"
] |
B. Lorry
|
2
|
64
|
A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.
|
The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.
|
In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.
|
[
"3 2\n1 2\n2 7\n1 3\n"
] |
[
"7\n2\n"
] |
none
| 0
|
[
{
"input": "3 2\n1 2\n2 7\n1 3",
"output": "7\n2"
},
{
"input": "5 3\n1 9\n2 9\n1 9\n2 10\n1 6",
"output": "24\n3 1 5"
},
{
"input": "10 10\n1 14\n2 15\n2 11\n2 12\n2 9\n1 14\n2 15\n1 9\n2 11\n2 6",
"output": "81\n6 1 7 2 4 9"
},
{
"input": "20 19\n2 47\n1 37\n1 48\n2 42\n2 48\n1 38\n2 47\n1 48\n2 47\n1 41\n2 46\n1 28\n1 49\n1 45\n2 34\n1 43\n2 29\n1 46\n2 45\n2 18",
"output": "630\n13 8 3 18 14 16 10 6 2 5 9 7 1 11"
},
{
"input": "50 27\n2 93\n1 98\n2 62\n1 56\n1 86\n1 42\n2 67\n2 97\n2 59\n1 73\n1 83\n2 96\n1 20\n1 66\n1 84\n1 83\n1 91\n2 97\n1 81\n2 88\n2 63\n1 99\n2 57\n1 39\n1 74\n2 88\n1 30\n2 68\n1 100\n2 57\n1 87\n1 93\n1 83\n1 100\n1 91\n1 14\n1 38\n2 98\n2 85\n2 61\n1 44\n2 93\n2 66\n2 55\n2 74\n1 67\n2 67\n1 85\n2 59\n1 83",
"output": "2055\n34 29 22 2 32 35 17 31 5 48 15 50 33 16 11 19 25 10 46 14 4 38 18 8"
},
{
"input": "1 1\n1 600",
"output": "600\n1"
},
{
"input": "10 14\n2 230\n2 516\n2 527\n2 172\n2 854\n2 61\n1 52\n2 154\n2 832\n2 774",
"output": "3905\n5 9 10 3 2 1 4"
},
{
"input": "8 8\n1 1\n1 1\n1 1\n1 1\n2 100\n2 100\n2 100\n2 100",
"output": "400\n8 7 6 5"
},
{
"input": "8 4\n1 100\n1 100\n1 100\n1 100\n2 1\n2 1\n2 1\n2 1",
"output": "400\n4 3 2 1"
}
] | 1,609,767,973
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 1,528
| 10,547,200
|
r=lambda:map(int,input().split())
[n,v]=r()
t=[]
o=[]
for i in range(n):
[tp,p]=r()
if tp==1: o+=[[p,i]]
else: t+=[[p,i]]
t.sort(reverse=True)
o.sort(reverse=True)
nt=len(t)
no=len(o)
for i in range(1,nt): t[i][0]+=t[i-1][0]
for i in range(1,no): o[i][0]+=o[i-1][0]
a=[0,0,0]
for i in range(v+1):
if i>no: break
j=min(nt, (v-i)//2)
r=0
if i: r+=o[i-1][0]
if j: r+=t[j-1][0]
a=max(a,[r,i,j])
print (a[0])
for i in range(a[1]): print (o[i][1]+1)
for i in range(a[2]): print (t[i][1]+1)
|
Title: Lorry
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.
Input Specification:
The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.
Output Specification:
In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.
Demo Input:
['3 2\n1 2\n2 7\n1 3\n']
Demo Output:
['7\n2\n']
Note:
none
|
```python
r=lambda:map(int,input().split())
[n,v]=r()
t=[]
o=[]
for i in range(n):
[tp,p]=r()
if tp==1: o+=[[p,i]]
else: t+=[[p,i]]
t.sort(reverse=True)
o.sort(reverse=True)
nt=len(t)
no=len(o)
for i in range(1,nt): t[i][0]+=t[i-1][0]
for i in range(1,no): o[i][0]+=o[i-1][0]
a=[0,0,0]
for i in range(v+1):
if i>no: break
j=min(nt, (v-i)//2)
r=0
if i: r+=o[i-1][0]
if j: r+=t[j-1][0]
a=max(a,[r,i,j])
print (a[0])
for i in range(a[1]): print (o[i][1]+1)
for i in range(a[2]): print (t[i][1]+1)
```
| 3.539417
|
232
|
B
|
Table
|
PROGRAMMING
| 1,900
|
[
"bitmasks",
"combinatorics",
"dp",
"math"
] | null | null |
John Doe has an *n*<=×<=*m* table. John Doe can paint points in some table cells, not more than one point in one table cell. John Doe wants to use such operations to make each square subtable of size *n*<=×<=*n* have exactly *k* points.
John Doe wondered, how many distinct ways to fill the table with points are there, provided that the condition must hold. As this number can be rather large, John Doe asks to find its remainder after dividing by 1000000007 (109<=+<=7).
You should assume that John always paints a point exactly in the center of some cell. Two ways to fill a table are considered distinct, if there exists a table cell, that has a point in one way and doesn't have it in the other.
|
A single line contains space-separated integers *n*, *m*, *k* (1<=≤<=*n*<=≤<=100; *n*<=≤<=*m*<=≤<=1018; 0<=≤<=*k*<=≤<=*n*2) — the number of rows of the table, the number of columns of the table and the number of points each square must contain.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
In a single line print a single integer — the remainder from dividing the described number of ways by 1000000007 (109<=+<=7).
|
[
"5 6 1\n"
] |
[
"45"
] |
Let's consider the first test case:
| 1,000
|
[
{
"input": "5 6 1",
"output": "45"
},
{
"input": "1 1000000000000000000 0",
"output": "1"
},
{
"input": "100 1000000 5000",
"output": "13662512"
},
{
"input": "100 1000000000000000000 10000",
"output": "1"
},
{
"input": "2 1791938441 1",
"output": "216278738"
},
{
"input": "3 783497007 1",
"output": "83405186"
},
{
"input": "4 1484499480 1",
"output": "484459942"
},
{
"input": "5 559703685 1",
"output": "841924438"
},
{
"input": "6 26236608 1",
"output": "320953373"
},
{
"input": "7 167374275 1",
"output": "845164424"
},
{
"input": "8 703528038 1",
"output": "322849207"
},
{
"input": "9 402260913 1",
"output": "895045202"
},
{
"input": "10 360430679 1",
"output": "769907582"
},
{
"input": "11 1726671165 1",
"output": "56626752"
},
{
"input": "12 534773931 1",
"output": "534937763"
},
{
"input": "13 685597394 1",
"output": "718244179"
},
{
"input": "14 922147616 1",
"output": "761962772"
},
{
"input": "15 1033532424 1",
"output": "694740913"
},
{
"input": "16 1812819014 1",
"output": "373277664"
},
{
"input": "17 65579254 1",
"output": "319329231"
},
{
"input": "18 1679162402 1",
"output": "156808889"
},
{
"input": "19 1146762197 1",
"output": "61583889"
},
{
"input": "20 1486959259 1",
"output": "835319650"
},
{
"input": "21 1609925409 1",
"output": "301675407"
},
{
"input": "22 285780757 1",
"output": "46653057"
},
{
"input": "23 740166386 1",
"output": "25188003"
},
{
"input": "24 1050116263 1",
"output": "697818450"
},
{
"input": "25 155354114 1",
"output": "324981149"
},
{
"input": "26 591654786 1",
"output": "131698277"
},
{
"input": "27 705741648 1",
"output": "937918901"
},
{
"input": "28 82742333 1",
"output": "135556923"
},
{
"input": "29 1239133968 1",
"output": "452197425"
},
{
"input": "30 452387544 1",
"output": "16600985"
},
{
"input": "31 1369710639 1",
"output": "16208445"
},
{
"input": "32 504681835 1",
"output": "358188883"
},
{
"input": "33 581065536 1",
"output": "266009369"
},
{
"input": "34 1083266459 1",
"output": "983560631"
},
{
"input": "35 359145576 1",
"output": "530694730"
},
{
"input": "36 290905083 1",
"output": "563027265"
},
{
"input": "37 1877886816 1",
"output": "589422305"
},
{
"input": "38 1150965180 1",
"output": "66410021"
},
{
"input": "39 27777879 1",
"output": "754316484"
},
{
"input": "40 1910205205 1",
"output": "824953571"
},
{
"input": "41 1950702099 1",
"output": "169527953"
},
{
"input": "2 1936883231 4",
"output": "1"
},
{
"input": "3 765857968 9",
"output": "1"
},
{
"input": "4 427306495 10",
"output": "333983125"
},
{
"input": "5 1715975476 10",
"output": "501939193"
},
{
"input": "6 657924949 10",
"output": "510906867"
},
{
"input": "7 116917520 10",
"output": "317300520"
},
{
"input": "8 307047272 10",
"output": "128349211"
},
{
"input": "9 1859135821 10",
"output": "975542783"
},
{
"input": "10 1018604235 10",
"output": "865916063"
},
{
"input": "5 37 1",
"output": "1015625"
},
{
"input": "5 37 2",
"output": "233903772"
},
{
"input": "5 37 3",
"output": "644646209"
},
{
"input": "5 37 4",
"output": "767122993"
},
{
"input": "5 37 5",
"output": "9140856"
},
{
"input": "5 37 6",
"output": "374658095"
},
{
"input": "5 37 7",
"output": "484718996"
},
{
"input": "5 37 8",
"output": "142285080"
},
{
"input": "5 37 9",
"output": "324758465"
},
{
"input": "5 37 10",
"output": "186553763"
},
{
"input": "5 37 11",
"output": "117291874"
},
{
"input": "5 37 12",
"output": "328884166"
},
{
"input": "5 37 13",
"output": "328884166"
},
{
"input": "5 37 14",
"output": "117291874"
},
{
"input": "5 37 15",
"output": "186553763"
},
{
"input": "5 37 16",
"output": "324758465"
},
{
"input": "5 37 17",
"output": "142285080"
},
{
"input": "5 37 18",
"output": "484718996"
},
{
"input": "5 37 19",
"output": "374658095"
},
{
"input": "5 37 20",
"output": "9140856"
},
{
"input": "5 37 21",
"output": "767122993"
},
{
"input": "5 37 22",
"output": "644646209"
},
{
"input": "5 37 23",
"output": "233903772"
},
{
"input": "5 37 24",
"output": "1015625"
},
{
"input": "5 37 25",
"output": "1"
},
{
"input": "44 1103339866 549",
"output": "526084438"
},
{
"input": "80 2000000000 3200",
"output": "703575100"
},
{
"input": "80 2000000000 6398",
"output": "218525968"
},
{
"input": "80 2000000000 6400",
"output": "1"
},
{
"input": "81 1229669603 3200",
"output": "641340015"
},
{
"input": "82 1223376193 3200",
"output": "202027569"
},
{
"input": "83 253746842 3200",
"output": "274764243"
},
{
"input": "84 927691094 3200",
"output": "149168097"
},
{
"input": "85 1801239280 3200",
"output": "310862287"
},
{
"input": "86 641728038 3200",
"output": "501969615"
},
{
"input": "87 434070509 3200",
"output": "762701365"
},
{
"input": "88 439646755 3200",
"output": "285691888"
},
{
"input": "89 286041042 3200",
"output": "104408464"
},
{
"input": "90 1091567971 3200",
"output": "165310445"
},
{
"input": "91 1675398540 3200",
"output": "734035315"
},
{
"input": "92 548977630 3200",
"output": "684032743"
},
{
"input": "93 1816430427 3200",
"output": "501670378"
},
{
"input": "94 1634867304 3200",
"output": "583832610"
},
{
"input": "95 1404439427 3200",
"output": "755298335"
},
{
"input": "96 649014433 3200",
"output": "92151836"
},
{
"input": "97 1244188575 3200",
"output": "847426943"
},
{
"input": "98 220789222 3200",
"output": "584032379"
},
{
"input": "99 281272480 3200",
"output": "183722548"
},
{
"input": "100 1785754808 3200",
"output": "118701841"
},
{
"input": "80 500000000000000000 3200",
"output": "722478695"
},
{
"input": "79 1000000000000000000 3107",
"output": "631080535"
},
{
"input": "2 500000000000000000 2",
"output": "262146"
},
{
"input": "100 1000000000000000000 5000",
"output": "824224847"
},
{
"input": "3 1000000000000000000 2",
"output": "341903757"
},
{
"input": "1 1000000000000000000 1",
"output": "1"
},
{
"input": "100 1000000000000000000 1221",
"output": "43180268"
},
{
"input": "5 1000000000000000000 10",
"output": "317580731"
},
{
"input": "100 12345678901234567 3123",
"output": "75657938"
},
{
"input": "20 24 20",
"output": "228455439"
}
] | 1,622,902,798
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 4,000
| 9,523,200
|
N = 101
# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)
# array to precompute inverse of 1! to N!
naturalNumInverse = [None] * (N + 1)
# array to store factorial of
# first N numbers
fact = [None] * (N + 1)
# Function to precompute inverse of numbers
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
# precompute inverse of natural numbers
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
# Function to calculate factorial of 1 to N
def factorial(p):
fact[0] = 1
# precompute factorials
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
# Function to return nCr % p in O(1) time
def Binomial(N, R, p):
# n C r = n!*inverse(r!)*inverse((n-r)!)
ans = ((fact[N] * factorialNumInverse[R]) % p *
factorialNumInverse[N - R]) % p
return ans
p=10**9+7
InverseofNumber(p)
InverseofFactorial(p)
factorial(p)
def power(x, y, p):
res = 1 # Initialize result
# Update x if it is more
# than or equal to p
x = x % p
if (x == 0):
return 0
while (y > 0):
# If y is odd, multiply
# x with result
if ((y & 1) == 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
n,m,k=map(int,input().split())
dp=[[0 for j in range(k+1)]for u in range(n+1)]
dp[0][0]=1
r=m%n
q=m//n
for i in range(1,n+1):
tot=q
if i<=r:
tot+=1
for s in range(k+1):
for l in range(min(n+1,s+1)):
dp[i][s]=(dp[i][s]+(dp[i-1][s-l])*(power(Binomial(n,l,p),tot,p)))
dp[i][s]=dp[i][s]%p
print(dp[n][k]%p)
|
Title: Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
John Doe has an *n*<=×<=*m* table. John Doe can paint points in some table cells, not more than one point in one table cell. John Doe wants to use such operations to make each square subtable of size *n*<=×<=*n* have exactly *k* points.
John Doe wondered, how many distinct ways to fill the table with points are there, provided that the condition must hold. As this number can be rather large, John Doe asks to find its remainder after dividing by 1000000007 (109<=+<=7).
You should assume that John always paints a point exactly in the center of some cell. Two ways to fill a table are considered distinct, if there exists a table cell, that has a point in one way and doesn't have it in the other.
Input Specification:
A single line contains space-separated integers *n*, *m*, *k* (1<=≤<=*n*<=≤<=100; *n*<=≤<=*m*<=≤<=1018; 0<=≤<=*k*<=≤<=*n*2) — the number of rows of the table, the number of columns of the table and the number of points each square must contain.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
In a single line print a single integer — the remainder from dividing the described number of ways by 1000000007 (109<=+<=7).
Demo Input:
['5 6 1\n']
Demo Output:
['45']
Note:
Let's consider the first test case:
|
```python
N = 101
# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)
# array to precompute inverse of 1! to N!
naturalNumInverse = [None] * (N + 1)
# array to store factorial of
# first N numbers
fact = [None] * (N + 1)
# Function to precompute inverse of numbers
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
# precompute inverse of natural numbers
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
# Function to calculate factorial of 1 to N
def factorial(p):
fact[0] = 1
# precompute factorials
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
# Function to return nCr % p in O(1) time
def Binomial(N, R, p):
# n C r = n!*inverse(r!)*inverse((n-r)!)
ans = ((fact[N] * factorialNumInverse[R]) % p *
factorialNumInverse[N - R]) % p
return ans
p=10**9+7
InverseofNumber(p)
InverseofFactorial(p)
factorial(p)
def power(x, y, p):
res = 1 # Initialize result
# Update x if it is more
# than or equal to p
x = x % p
if (x == 0):
return 0
while (y > 0):
# If y is odd, multiply
# x with result
if ((y & 1) == 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
n,m,k=map(int,input().split())
dp=[[0 for j in range(k+1)]for u in range(n+1)]
dp[0][0]=1
r=m%n
q=m//n
for i in range(1,n+1):
tot=q
if i<=r:
tot+=1
for s in range(k+1):
for l in range(min(n+1,s+1)):
dp[i][s]=(dp[i][s]+(dp[i-1][s-l])*(power(Binomial(n,l,p),tot,p)))
dp[i][s]=dp[i][s]%p
print(dp[n][k]%p)
```
| 0
|
|
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"
}
] | 1,675,769,625
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
a=int(input())
for i in range (a):
b=input()
if len(b)<=10:
print(b)
else:
print(b[0],len(b)-2,b[1],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
a=int(input())
for i in range (a):
b=input()
if len(b)<=10:
print(b)
else:
print(b[0],len(b)-2,b[1],sep=" ")
```
| 0
|
556
|
A
|
Case of the Zeros and Ones
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
|
Output the minimum length of the string that may remain after applying the described operations several times.
|
[
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] |
[
"0\n",
"1\n",
"6\n"
] |
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 250
|
[
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "2\n01",
"output": "0"
},
{
"input": "2\n10",
"output": "0"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "3\n001",
"output": "1"
},
{
"input": "6\n110110",
"output": "2"
},
{
"input": "7\n0000011",
"output": "3"
},
{
"input": "6\n110010",
"output": "0"
},
{
"input": "6\n110100",
"output": "0"
},
{
"input": "3\n100",
"output": "1"
},
{
"input": "6\n010111",
"output": "2"
},
{
"input": "8\n01011100",
"output": "0"
},
{
"input": "6\n001011",
"output": "0"
},
{
"input": "7\n1110000",
"output": "1"
},
{
"input": "9\n011111101",
"output": "5"
}
] | 1,610,119,539
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
n=int(input())
s=input()
try:
s=s.replace("01","")
s=s.replace("10","")
except:
pass
print(len(s))
|
Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
n=int(input())
s=input()
try:
s=s.replace("01","")
s=s.replace("10","")
except:
pass
print(len(s))
```
| 0
|
|
35
|
A
|
Shell Game
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Shell Game
|
2
|
64
|
Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too?
|
The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3.
|
In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles.
|
[
"1\n1 2\n2 1\n2 1\n",
"1\n2 1\n3 1\n1 3\n"
] |
[
"2\n",
"2\n"
] |
none
| 500
|
[
{
"input": "1\n1 2\n2 1\n2 1",
"output": "2"
},
{
"input": "1\n2 1\n3 1\n1 3",
"output": "2"
},
{
"input": "3\n3 1\n2 1\n1 2",
"output": "1"
},
{
"input": "1\n1 3\n1 2\n2 3",
"output": "2"
},
{
"input": "3\n3 2\n3 1\n3 1",
"output": "2"
},
{
"input": "1\n2 1\n1 3\n1 3",
"output": "2"
},
{
"input": "3\n3 1\n2 3\n3 2",
"output": "1"
},
{
"input": "2\n1 3\n1 2\n2 1",
"output": "2"
},
{
"input": "1\n1 3\n3 2\n1 2",
"output": "1"
},
{
"input": "1\n1 3\n1 3\n2 3",
"output": "1"
},
{
"input": "2\n1 2\n2 3\n2 1",
"output": "2"
},
{
"input": "3\n1 3\n3 2\n2 1",
"output": "2"
},
{
"input": "1\n1 2\n2 1\n2 3",
"output": "1"
},
{
"input": "1\n2 3\n1 3\n1 2",
"output": "3"
},
{
"input": "2\n3 1\n3 2\n2 3",
"output": "2"
},
{
"input": "2\n1 3\n3 1\n3 1",
"output": "2"
},
{
"input": "1\n3 2\n1 3\n3 1",
"output": "1"
},
{
"input": "3\n1 3\n1 2\n1 3",
"output": "2"
},
{
"input": "1\n3 2\n3 1\n1 2",
"output": "3"
},
{
"input": "2\n2 3\n1 3\n1 3",
"output": "3"
}
] | 1,633,993,552
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n=int(input())
for i in range(3):
a,b = map(int,input().split())
if a == n:
n=b
elif b=n:
n = a
print(n)
|
Title: Shell Game
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too?
Input Specification:
The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3.
Output Specification:
In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles.
Demo Input:
['1\n1 2\n2 1\n2 1\n', '1\n2 1\n3 1\n1 3\n']
Demo Output:
['2\n', '2\n']
Note:
none
|
```python
n=int(input())
for i in range(3):
a,b = map(int,input().split())
if a == n:
n=b
elif b=n:
n = a
print(n)
```
| -1
|
469
|
A
|
I Wanna Be the Guy
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
|
The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100).
The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.
|
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
|
[
"4\n3 1 2 3\n2 2 4\n",
"4\n3 1 2 3\n2 2 3\n"
] |
[
"I become the guy.\n",
"Oh, my keyboard!\n"
] |
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
| 500
|
[
{
"input": "4\n3 1 2 3\n2 2 4",
"output": "I become the guy."
},
{
"input": "4\n3 1 2 3\n2 2 3",
"output": "Oh, my keyboard!"
},
{
"input": "10\n5 8 6 1 5 4\n6 1 3 2 9 4 6",
"output": "Oh, my keyboard!"
},
{
"input": "10\n8 8 10 7 3 1 4 2 6\n8 9 5 10 3 7 2 4 8",
"output": "I become the guy."
},
{
"input": "10\n9 6 1 8 3 9 7 5 10 4\n7 1 3 2 7 6 9 5",
"output": "I become the guy."
},
{
"input": "100\n75 83 69 73 30 76 37 48 14 41 42 21 35 15 50 61 86 85 46 3 31 13 78 10 2 44 80 95 56 82 38 75 77 4 99 9 84 53 12 11 36 74 39 72 43 89 57 28 54 1 51 66 27 22 93 59 68 88 91 29 7 20 63 8 52 23 64 58 100 79 65 49 96 71 33 45\n83 50 89 73 34 28 99 67 77 44 19 60 68 42 8 27 94 85 14 39 17 78 24 21 29 63 92 32 86 22 71 81 31 82 65 48 80 59 98 3 70 55 37 12 15 72 47 9 11 33 16 7 91 74 13 64 38 84 6 61 93 90 45 69 1 54 52 100 57 10 35 49 53 75 76 43 62 5 4 18 36 96 79 23",
"output": "Oh, my keyboard!"
},
{
"input": "1\n1 1\n1 1",
"output": "I become the guy."
},
{
"input": "1\n0\n1 1",
"output": "I become the guy."
},
{
"input": "1\n1 1\n0",
"output": "I become the guy."
},
{
"input": "1\n0\n0",
"output": "Oh, my keyboard!"
},
{
"input": "100\n0\n0",
"output": "Oh, my keyboard!"
},
{
"input": "100\n44 71 70 55 49 43 16 53 7 95 58 56 38 76 67 94 20 73 29 90 25 30 8 84 5 14 77 52 99 91 66 24 39 37 22 44 78 12 63 59 32 51 15 82 34\n56 17 10 96 80 69 13 81 31 57 4 48 68 89 50 45 3 33 36 2 72 100 64 87 21 75 54 74 92 65 23 40 97 61 18 28 98 93 35 83 9 79 46 27 41 62 88 6 47 60 86 26 42 85 19 1 11",
"output": "I become the guy."
},
{
"input": "100\n78 63 59 39 11 58 4 2 80 69 22 95 90 26 65 16 30 100 66 99 67 79 54 12 23 28 45 56 70 74 60 82 73 91 68 43 92 75 51 21 17 97 86 44 62 47 85 78 72 64 50 81 71 5 57 13 31 76 87 9 49 96 25 42 19 35 88 53 7 83 38 27 29 41 89 93 10 84 18\n78 1 16 53 72 99 9 36 59 49 75 77 94 79 35 4 92 42 82 83 76 97 20 68 55 47 65 50 14 30 13 67 98 8 7 40 64 32 87 10 33 90 93 18 26 71 17 46 24 28 89 58 37 91 39 34 25 48 84 31 96 95 80 88 3 51 62 52 85 61 12 15 27 6 45 38 2 22 60",
"output": "I become the guy."
},
{
"input": "2\n2 2 1\n0",
"output": "I become the guy."
},
{
"input": "2\n1 2\n2 1 2",
"output": "I become the guy."
},
{
"input": "80\n57 40 1 47 36 69 24 76 5 72 26 4 29 62 6 60 3 70 8 64 18 37 16 14 13 21 25 7 66 68 44 74 61 39 38 33 15 63 34 65 10 23 56 51 80 58 49 75 71 12 50 57 2 30 54 27 17 52\n61 22 67 15 28 41 26 1 80 44 3 38 18 37 79 57 11 7 65 34 9 36 40 5 48 29 64 31 51 63 27 4 50 13 24 32 58 23 19 46 8 73 39 2 21 56 77 53 59 78 43 12 55 45 30 74 33 68 42 47 17 54",
"output": "Oh, my keyboard!"
},
{
"input": "100\n78 87 96 18 73 32 38 44 29 64 40 70 47 91 60 69 24 1 5 34 92 94 99 22 83 65 14 68 15 20 74 31 39 100 42 4 97 46 25 6 8 56 79 9 71 35 54 19 59 93 58 62 10 85 57 45 33 7 86 81 30 98 26 61 84 41 23 28 88 36 66 51 80 53 37 63 43 95 75\n76 81 53 15 26 37 31 62 24 87 41 39 75 86 46 76 34 4 51 5 45 65 67 48 68 23 71 27 94 47 16 17 9 96 84 89 88 100 18 52 69 42 6 92 7 64 49 12 98 28 21 99 25 55 44 40 82 19 36 30 77 90 14 43 50 3 13 95 78 35 20 54 58 11 2 1 33",
"output": "Oh, my keyboard!"
},
{
"input": "100\n77 55 26 98 13 91 78 60 23 76 12 11 36 62 84 80 18 1 68 92 81 67 19 4 2 10 17 77 96 63 15 69 46 97 82 42 83 59 50 72 14 40 89 9 52 29 56 31 74 39 45 85 22 99 44 65 95 6 90 38 54 32 49 34 3 70 75 33 94 53 21 71 5 66 73 41 100 24\n69 76 93 5 24 57 59 6 81 4 30 12 44 15 67 45 73 3 16 8 47 95 20 64 68 85 54 17 90 86 66 58 13 37 42 51 35 32 1 28 43 80 7 14 48 19 62 55 2 91 25 49 27 26 38 79 89 99 22 60 75 53 88 82 34 21 87 71 72 61",
"output": "I become the guy."
},
{
"input": "100\n74 96 32 63 12 69 72 99 15 22 1 41 79 77 71 31 20 28 75 73 85 37 38 59 42 100 86 89 55 87 68 4 24 57 52 8 92 27 56 98 95 58 34 9 45 14 11 36 66 76 61 19 25 23 78 49 90 26 80 43 70 13 65 10 5 74 81 21 44 60 97 3 47 93 6\n64 68 21 27 16 91 23 22 33 12 71 88 90 50 62 43 28 29 57 59 5 74 10 95 35 1 67 93 36 32 86 40 6 64 78 46 89 15 84 53 18 30 17 85 2 3 47 92 25 48 76 51 20 82 52 83 99 63 80 11 94 54 39 7 58",
"output": "I become the guy."
},
{
"input": "100\n75 11 98 44 47 88 94 23 78 59 70 2 43 39 34 63 71 19 42 61 30 74 14 77 97 53 92 60 67 36 37 13 6 86 62 46 41 3 25 93 7 12 27 48 55 49 31 35 51 10 57 54 95 82 28 90 73 26 17 50 81 56 20 87 40 85 72 64 99 29 91 5 80 18 24 52\n72 93 59 5 88 47 9 58 48 1 43 50 100 87 61 91 45 98 99 56 25 84 53 73 78 54 63 38 37 2 77 95 89 85 4 90 10 33 12 22 74 32 34 70 71 52 96 57 15 66 31 27 75 8 21 39 62 44 67 94 81 68 14 19 36 28 11 79 16 65 46 83 76",
"output": "Oh, my keyboard!"
},
{
"input": "3\n1 2\n2 2 3",
"output": "Oh, my keyboard!"
},
{
"input": "4\n1 2\n3 1 3 4",
"output": "I become the guy."
},
{
"input": "6\n2 1 2\n3 4 5 6",
"output": "Oh, my keyboard!"
}
] | 1,699,041,536
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 26
| 46
| 0
|
n,x,y=int(input()),set(list(map(int,input().split()))),set(list(map(int,input().split())))
lx = 0 if x=={0} else len(x)
ly= 0 if y=={0} else len(y)
print("I become the guy." if n == (lx+ly-len([i for i in x if i in y])) else "Oh, my keyboard!")
|
Title: I Wanna Be the Guy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input Specification:
The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100).
The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.
Output Specification:
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Demo Input:
['4\n3 1 2 3\n2 2 4\n', '4\n3 1 2 3\n2 2 3\n']
Demo Output:
['I become the guy.\n', 'Oh, my keyboard!\n']
Note:
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
|
```python
n,x,y=int(input()),set(list(map(int,input().split()))),set(list(map(int,input().split())))
lx = 0 if x=={0} else len(x)
ly= 0 if y=={0} else len(y)
print("I become the guy." if n == (lx+ly-len([i for i in x if i in y])) else "Oh, my keyboard!")
```
| 0
|
|
615
|
A
|
Bulbs
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
|
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
|
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
|
[
"3 4\n2 1 4\n3 1 3 1\n1 2\n",
"3 3\n1 1\n1 2\n1 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
| 500
|
[
{
"input": "3 4\n2 1 4\n3 1 3 1\n1 2",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n1 1",
"output": "NO"
},
{
"input": "3 4\n1 1\n1 2\n1 3",
"output": "NO"
},
{
"input": "1 5\n5 1 2 3 4 5",
"output": "YES"
},
{
"input": "1 5\n5 4 4 1 2 3",
"output": "NO"
},
{
"input": "1 5\n5 1 1 1 1 5",
"output": "NO"
},
{
"input": "2 5\n4 3 1 4 2\n4 2 3 4 5",
"output": "YES"
},
{
"input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1",
"output": "YES"
},
{
"input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6",
"output": "YES"
},
{
"input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 4\n3 1 2 3",
"output": "NO"
},
{
"input": "1 4\n3 2 3 4",
"output": "NO"
},
{
"input": "2 4\n3 2 3 4\n1 1",
"output": "YES"
},
{
"input": "2 4\n3 1 2 3\n1 4",
"output": "YES"
},
{
"input": "5 1\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "1 1\n0",
"output": "NO"
},
{
"input": "1 10\n10 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "1 1\n1 1",
"output": "YES"
},
{
"input": "1 100\n99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "NO"
},
{
"input": "1 3\n3 1 2 1",
"output": "NO"
},
{
"input": "1 100\n100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "YES"
}
] | 1,675,505,669
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 45
| 62
| 32,358,400
|
n, m = map(int, input().split())
a = []
for i in range(n):
*b, = map(int, input().split())
del b[0]
for j in b:
if j not in a:
a.append(j)
if len(a) == m:
print("YES")
else:
print("NO")
|
Title: Bulbs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input Specification:
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
Output Specification:
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
Demo Input:
['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
|
```python
n, m = map(int, input().split())
a = []
for i in range(n):
*b, = map(int, input().split())
del b[0]
for j in b:
if j not in a:
a.append(j)
if len(a) == m:
print("YES")
else:
print("NO")
```
| 3
|
|
821
|
A
|
Okabe and Future Gadget Laboratory
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an *n* by *n* square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every *x*,<=*y* such that 1<=≤<=*x*,<=*y*<=≤<=*n* and *a**x*,<=*y*<=≠<=1, there should exist two indices *s* and *t* so that *a**x*,<=*y*<==<=*a**x*,<=*s*<=+<=*a**t*,<=*y*, where *a**i*,<=*j* denotes the integer in *i*-th row and *j*-th column.
Help Okabe determine whether a given lab is good!
|
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=50) — the size of the lab.
The next *n* lines contain *n* space-separated integers denoting a row of the grid. The *j*-th integer in the *i*-th row is *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=105).
|
Print "Yes" if the given lab is good and "No" otherwise.
You can output each letter in upper or lower case.
|
[
"3\n1 1 2\n2 3 1\n6 4 1\n",
"3\n1 5 2\n1 1 1\n1 2 3\n"
] |
[
"Yes\n",
"No\n"
] |
In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes".
In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
| 500
|
[
{
"input": "3\n1 1 2\n2 3 1\n6 4 1",
"output": "Yes"
},
{
"input": "3\n1 5 2\n1 1 1\n1 2 3",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 11 1 2\n2 5 1 4\n3 9 4 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 7 1 1\n1 3 1 2\n2 6 3 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 12 1 2\n4 4 1 3\n5 10 6 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 13 1 2\n4 5 1 3\n5 11 6 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 13 1 2\n4 5 1 3\n7 11 6 1",
"output": "No"
},
{
"input": "4\n1 1 1 1\n1 14 1 6\n4 5 1 5\n3 8 2 1",
"output": "Yes"
},
{
"input": "10\n3 2 6 5 1 2 1 1 2 1\n1 2 2 3 1 4 1 1 1 1\n1 1 3 1 2 2 2 1 1 1\n1 3 1 2 1 2 1 1 1 1\n12 1 4 3 3 1 1 1 2 1\n8 1 1 1 1 3 6 5 1 3\n1 1 1 2 4 1 3 2 1 1\n1 1 1 1 1 1 1 2 1 1\n1 2 1 1 1 1 1 1 1 1\n2 1 1 1 1 1 1 1 1 2",
"output": "Yes"
},
{
"input": "4\n1 2 3 5\n1 1 1 1\n1 1 1 1\n1 1 1 1",
"output": "No"
},
{
"input": "1\n2",
"output": "No"
},
{
"input": "3\n1 1 4\n1 1 2\n1 1 2",
"output": "No"
},
{
"input": "1\n5",
"output": "No"
},
{
"input": "3\n1 2 1\n2 2 3\n1 3 1",
"output": "No"
},
{
"input": "3\n2 1 1\n2 1 1\n2 1 1",
"output": "No"
},
{
"input": "1\n60000",
"output": "No"
}
] | 1,500,610,073
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 77
| 4,608,000
|
n = int(input())
lab = []
def is_valid_cell(i,j):
row = lab[i]
column = [lab[i][j] - lab[x][j] for x in range(n)]
for each in column:
if each in row:
return True
return False
for i in range(n):
line = [int(x) for x in input().split(' ')]
lab.append(line)
result = 'YES'
for i in range(n):
for j in range(n):
if (lab[i][j] != 1):
if (not is_valid_cell(i,j)):
result = 'NO'
break
else:
continue
break
print(result)
|
Title: Okabe and Future Gadget Laboratory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an *n* by *n* square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every *x*,<=*y* such that 1<=≤<=*x*,<=*y*<=≤<=*n* and *a**x*,<=*y*<=≠<=1, there should exist two indices *s* and *t* so that *a**x*,<=*y*<==<=*a**x*,<=*s*<=+<=*a**t*,<=*y*, where *a**i*,<=*j* denotes the integer in *i*-th row and *j*-th column.
Help Okabe determine whether a given lab is good!
Input Specification:
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=50) — the size of the lab.
The next *n* lines contain *n* space-separated integers denoting a row of the grid. The *j*-th integer in the *i*-th row is *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=105).
Output Specification:
Print "Yes" if the given lab is good and "No" otherwise.
You can output each letter in upper or lower case.
Demo Input:
['3\n1 1 2\n2 3 1\n6 4 1\n', '3\n1 5 2\n1 1 1\n1 2 3\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes".
In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
|
```python
n = int(input())
lab = []
def is_valid_cell(i,j):
row = lab[i]
column = [lab[i][j] - lab[x][j] for x in range(n)]
for each in column:
if each in row:
return True
return False
for i in range(n):
line = [int(x) for x in input().split(' ')]
lab.append(line)
result = 'YES'
for i in range(n):
for j in range(n):
if (lab[i][j] != 1):
if (not is_valid_cell(i,j)):
result = 'NO'
break
else:
continue
break
print(result)
```
| 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"
}
] | 1,685,848,707
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 1,000
| 0
|
n,m,a=map(int,input().split())
qt=n*m
if qt<=(a*a):
print('1')
else:
result=2
st=a*a*2
while st<qt:
result+=2
st+=a*a*2
print(result)
|
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
n,m,a=map(int,input().split())
qt=n*m
if qt<=(a*a):
print('1')
else:
result=2
st=a*a*2
while st<qt:
result+=2
st+=a*a*2
print(result)
```
| 0
|
195
|
B
|
After Training
|
PROGRAMMING
| 1,300
|
[
"data structures",
"implementation",
"math"
] | null | null |
After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.
|
The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly.
|
Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball.
|
[
"4 3\n",
"3 1\n"
] |
[
"2\n1\n3\n2\n",
"1\n1\n1\n"
] |
none
| 1,000
|
[
{
"input": "4 3",
"output": "2\n1\n3\n2"
},
{
"input": "3 1",
"output": "1\n1\n1"
},
{
"input": "10 3",
"output": "2\n1\n3\n2\n1\n3\n2\n1\n3\n2"
},
{
"input": "6 5",
"output": "3\n2\n4\n1\n5\n3"
},
{
"input": "2 6",
"output": "3\n4"
},
{
"input": "5 2",
"output": "1\n2\n1\n2\n1"
},
{
"input": "85702 100000",
"output": "50000\n50001\n49999\n50002\n49998\n50003\n49997\n50004\n49996\n50005\n49995\n50006\n49994\n50007\n49993\n50008\n49992\n50009\n49991\n50010\n49990\n50011\n49989\n50012\n49988\n50013\n49987\n50014\n49986\n50015\n49985\n50016\n49984\n50017\n49983\n50018\n49982\n50019\n49981\n50020\n49980\n50021\n49979\n50022\n49978\n50023\n49977\n50024\n49976\n50025\n49975\n50026\n49974\n50027\n49973\n50028\n49972\n50029\n49971\n50030\n49970\n50031\n49969\n50032\n49968\n50033\n49967\n50034\n49966\n50035\n49965\n50036\n49964\n..."
},
{
"input": "9 2",
"output": "1\n2\n1\n2\n1\n2\n1\n2\n1"
},
{
"input": "45 88",
"output": "44\n45\n43\n46\n42\n47\n41\n48\n40\n49\n39\n50\n38\n51\n37\n52\n36\n53\n35\n54\n34\n55\n33\n56\n32\n57\n31\n58\n30\n59\n29\n60\n28\n61\n27\n62\n26\n63\n25\n64\n24\n65\n23\n66\n22"
},
{
"input": "61 51",
"output": "26\n25\n27\n24\n28\n23\n29\n22\n30\n21\n31\n20\n32\n19\n33\n18\n34\n17\n35\n16\n36\n15\n37\n14\n38\n13\n39\n12\n40\n11\n41\n10\n42\n9\n43\n8\n44\n7\n45\n6\n46\n5\n47\n4\n48\n3\n49\n2\n50\n1\n51\n26\n25\n27\n24\n28\n23\n29\n22\n30\n21"
},
{
"input": "21 57",
"output": "29\n28\n30\n27\n31\n26\n32\n25\n33\n24\n34\n23\n35\n22\n36\n21\n37\n20\n38\n19\n39"
},
{
"input": "677 787",
"output": "394\n393\n395\n392\n396\n391\n397\n390\n398\n389\n399\n388\n400\n387\n401\n386\n402\n385\n403\n384\n404\n383\n405\n382\n406\n381\n407\n380\n408\n379\n409\n378\n410\n377\n411\n376\n412\n375\n413\n374\n414\n373\n415\n372\n416\n371\n417\n370\n418\n369\n419\n368\n420\n367\n421\n366\n422\n365\n423\n364\n424\n363\n425\n362\n426\n361\n427\n360\n428\n359\n429\n358\n430\n357\n431\n356\n432\n355\n433\n354\n434\n353\n435\n352\n436\n351\n437\n350\n438\n349\n439\n348\n440\n347\n441\n346\n442\n345\n443\n344\n444\n343\n4..."
},
{
"input": "37 849",
"output": "425\n424\n426\n423\n427\n422\n428\n421\n429\n420\n430\n419\n431\n418\n432\n417\n433\n416\n434\n415\n435\n414\n436\n413\n437\n412\n438\n411\n439\n410\n440\n409\n441\n408\n442\n407\n443"
},
{
"input": "453 855",
"output": "428\n427\n429\n426\n430\n425\n431\n424\n432\n423\n433\n422\n434\n421\n435\n420\n436\n419\n437\n418\n438\n417\n439\n416\n440\n415\n441\n414\n442\n413\n443\n412\n444\n411\n445\n410\n446\n409\n447\n408\n448\n407\n449\n406\n450\n405\n451\n404\n452\n403\n453\n402\n454\n401\n455\n400\n456\n399\n457\n398\n458\n397\n459\n396\n460\n395\n461\n394\n462\n393\n463\n392\n464\n391\n465\n390\n466\n389\n467\n388\n468\n387\n469\n386\n470\n385\n471\n384\n472\n383\n473\n382\n474\n381\n475\n380\n476\n379\n477\n378\n478\n377\n4..."
},
{
"input": "165 374",
"output": "187\n188\n186\n189\n185\n190\n184\n191\n183\n192\n182\n193\n181\n194\n180\n195\n179\n196\n178\n197\n177\n198\n176\n199\n175\n200\n174\n201\n173\n202\n172\n203\n171\n204\n170\n205\n169\n206\n168\n207\n167\n208\n166\n209\n165\n210\n164\n211\n163\n212\n162\n213\n161\n214\n160\n215\n159\n216\n158\n217\n157\n218\n156\n219\n155\n220\n154\n221\n153\n222\n152\n223\n151\n224\n150\n225\n149\n226\n148\n227\n147\n228\n146\n229\n145\n230\n144\n231\n143\n232\n142\n233\n141\n234\n140\n235\n139\n236\n138\n237\n137\n238\n1..."
},
{
"input": "328 3",
"output": "2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3..."
},
{
"input": "8 80",
"output": "40\n41\n39\n42\n38\n43\n37\n44"
},
{
"input": "90 544",
"output": "272\n273\n271\n274\n270\n275\n269\n276\n268\n277\n267\n278\n266\n279\n265\n280\n264\n281\n263\n282\n262\n283\n261\n284\n260\n285\n259\n286\n258\n287\n257\n288\n256\n289\n255\n290\n254\n291\n253\n292\n252\n293\n251\n294\n250\n295\n249\n296\n248\n297\n247\n298\n246\n299\n245\n300\n244\n301\n243\n302\n242\n303\n241\n304\n240\n305\n239\n306\n238\n307\n237\n308\n236\n309\n235\n310\n234\n311\n233\n312\n232\n313\n231\n314\n230\n315\n229\n316\n228\n317"
},
{
"input": "85 60",
"output": "30\n31\n29\n32\n28\n33\n27\n34\n26\n35\n25\n36\n24\n37\n23\n38\n22\n39\n21\n40\n20\n41\n19\n42\n18\n43\n17\n44\n16\n45\n15\n46\n14\n47\n13\n48\n12\n49\n11\n50\n10\n51\n9\n52\n8\n53\n7\n54\n6\n55\n5\n56\n4\n57\n3\n58\n2\n59\n1\n60\n30\n31\n29\n32\n28\n33\n27\n34\n26\n35\n25\n36\n24\n37\n23\n38\n22\n39\n21\n40\n20\n41\n19\n42\n18"
},
{
"input": "392 5",
"output": "3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3..."
},
{
"input": "8 87",
"output": "44\n43\n45\n42\n46\n41\n47\n40"
},
{
"input": "6 358",
"output": "179\n180\n178\n181\n177\n182"
},
{
"input": "501 70",
"output": "35\n36\n34\n37\n33\n38\n32\n39\n31\n40\n30\n41\n29\n42\n28\n43\n27\n44\n26\n45\n25\n46\n24\n47\n23\n48\n22\n49\n21\n50\n20\n51\n19\n52\n18\n53\n17\n54\n16\n55\n15\n56\n14\n57\n13\n58\n12\n59\n11\n60\n10\n61\n9\n62\n8\n63\n7\n64\n6\n65\n5\n66\n4\n67\n3\n68\n2\n69\n1\n70\n35\n36\n34\n37\n33\n38\n32\n39\n31\n40\n30\n41\n29\n42\n28\n43\n27\n44\n26\n45\n25\n46\n24\n47\n23\n48\n22\n49\n21\n50\n20\n51\n19\n52\n18\n53\n17\n54\n16\n55\n15\n56\n14\n57\n13\n58\n12\n59\n11\n60\n10\n61\n9\n62\n8\n63\n7\n64\n6\n65\n5\n6..."
},
{
"input": "3834 1",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1..."
},
{
"input": "1 8828",
"output": "4414"
},
{
"input": "69230 89906",
"output": "44953\n44954\n44952\n44955\n44951\n44956\n44950\n44957\n44949\n44958\n44948\n44959\n44947\n44960\n44946\n44961\n44945\n44962\n44944\n44963\n44943\n44964\n44942\n44965\n44941\n44966\n44940\n44967\n44939\n44968\n44938\n44969\n44937\n44970\n44936\n44971\n44935\n44972\n44934\n44973\n44933\n44974\n44932\n44975\n44931\n44976\n44930\n44977\n44929\n44978\n44928\n44979\n44927\n44980\n44926\n44981\n44925\n44982\n44924\n44983\n44923\n44984\n44922\n44985\n44921\n44986\n44920\n44987\n44919\n44988\n44918\n44989\n44917\n..."
},
{
"input": "27646 59913",
"output": "29957\n29956\n29958\n29955\n29959\n29954\n29960\n29953\n29961\n29952\n29962\n29951\n29963\n29950\n29964\n29949\n29965\n29948\n29966\n29947\n29967\n29946\n29968\n29945\n29969\n29944\n29970\n29943\n29971\n29942\n29972\n29941\n29973\n29940\n29974\n29939\n29975\n29938\n29976\n29937\n29977\n29936\n29978\n29935\n29979\n29934\n29980\n29933\n29981\n29932\n29982\n29931\n29983\n29930\n29984\n29929\n29985\n29928\n29986\n29927\n29987\n29926\n29988\n29925\n29989\n29924\n29990\n29923\n29991\n29922\n29992\n29921\n29993\n..."
},
{
"input": "37006 54783",
"output": "27392\n27391\n27393\n27390\n27394\n27389\n27395\n27388\n27396\n27387\n27397\n27386\n27398\n27385\n27399\n27384\n27400\n27383\n27401\n27382\n27402\n27381\n27403\n27380\n27404\n27379\n27405\n27378\n27406\n27377\n27407\n27376\n27408\n27375\n27409\n27374\n27410\n27373\n27411\n27372\n27412\n27371\n27413\n27370\n27414\n27369\n27415\n27368\n27416\n27367\n27417\n27366\n27418\n27365\n27419\n27364\n27420\n27363\n27421\n27362\n27422\n27361\n27423\n27360\n27424\n27359\n27425\n27358\n27426\n27357\n27427\n27356\n27428\n..."
},
{
"input": "1 100000",
"output": "50000"
},
{
"input": "100000 1",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1..."
},
{
"input": "100000 100000",
"output": "50000\n50001\n49999\n50002\n49998\n50003\n49997\n50004\n49996\n50005\n49995\n50006\n49994\n50007\n49993\n50008\n49992\n50009\n49991\n50010\n49990\n50011\n49989\n50012\n49988\n50013\n49987\n50014\n49986\n50015\n49985\n50016\n49984\n50017\n49983\n50018\n49982\n50019\n49981\n50020\n49980\n50021\n49979\n50022\n49978\n50023\n49977\n50024\n49976\n50025\n49975\n50026\n49974\n50027\n49973\n50028\n49972\n50029\n49971\n50030\n49970\n50031\n49969\n50032\n49968\n50033\n49967\n50034\n49966\n50035\n49965\n50036\n49964\n..."
},
{
"input": "100000 13",
"output": "7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n..."
},
{
"input": "100000 44",
"output": "22\n23\n21\n24\n20\n25\n19\n26\n18\n27\n17\n28\n16\n29\n15\n30\n14\n31\n13\n32\n12\n33\n11\n34\n10\n35\n9\n36\n8\n37\n7\n38\n6\n39\n5\n40\n4\n41\n3\n42\n2\n43\n1\n44\n22\n23\n21\n24\n20\n25\n19\n26\n18\n27\n17\n28\n16\n29\n15\n30\n14\n31\n13\n32\n12\n33\n11\n34\n10\n35\n9\n36\n8\n37\n7\n38\n6\n39\n5\n40\n4\n41\n3\n42\n2\n43\n1\n44\n22\n23\n21\n24\n20\n25\n19\n26\n18\n27\n17\n28\n16\n29\n15\n30\n14\n31\n13\n32\n12\n33\n11\n34\n10\n35\n9\n36\n8\n37\n7\n38\n6\n39\n5\n40\n4\n41\n3\n42\n2\n43\n1\n44\n22\n23\n21..."
},
{
"input": "100000 37820",
"output": "18910\n18911\n18909\n18912\n18908\n18913\n18907\n18914\n18906\n18915\n18905\n18916\n18904\n18917\n18903\n18918\n18902\n18919\n18901\n18920\n18900\n18921\n18899\n18922\n18898\n18923\n18897\n18924\n18896\n18925\n18895\n18926\n18894\n18927\n18893\n18928\n18892\n18929\n18891\n18930\n18890\n18931\n18889\n18932\n18888\n18933\n18887\n18934\n18886\n18935\n18885\n18936\n18884\n18937\n18883\n18938\n18882\n18939\n18881\n18940\n18880\n18941\n18879\n18942\n18878\n18943\n18877\n18944\n18876\n18945\n18875\n18946\n18874\n..."
},
{
"input": "99999 77777",
"output": "38889\n38888\n38890\n38887\n38891\n38886\n38892\n38885\n38893\n38884\n38894\n38883\n38895\n38882\n38896\n38881\n38897\n38880\n38898\n38879\n38899\n38878\n38900\n38877\n38901\n38876\n38902\n38875\n38903\n38874\n38904\n38873\n38905\n38872\n38906\n38871\n38907\n38870\n38908\n38869\n38909\n38868\n38910\n38867\n38911\n38866\n38912\n38865\n38913\n38864\n38914\n38863\n38915\n38862\n38916\n38861\n38917\n38860\n38918\n38859\n38919\n38858\n38920\n38857\n38921\n38856\n38922\n38855\n38923\n38854\n38924\n38853\n38925\n..."
},
{
"input": "1991 1935",
"output": "968\n967\n969\n966\n970\n965\n971\n964\n972\n963\n973\n962\n974\n961\n975\n960\n976\n959\n977\n958\n978\n957\n979\n956\n980\n955\n981\n954\n982\n953\n983\n952\n984\n951\n985\n950\n986\n949\n987\n948\n988\n947\n989\n946\n990\n945\n991\n944\n992\n943\n993\n942\n994\n941\n995\n940\n996\n939\n997\n938\n998\n937\n999\n936\n1000\n935\n1001\n934\n1002\n933\n1003\n932\n1004\n931\n1005\n930\n1006\n929\n1007\n928\n1008\n927\n1009\n926\n1010\n925\n1011\n924\n1012\n923\n1013\n922\n1014\n921\n1015\n920\n1016\n919\n1017..."
},
{
"input": "17 812",
"output": "406\n407\n405\n408\n404\n409\n403\n410\n402\n411\n401\n412\n400\n413\n399\n414\n398"
},
{
"input": "30078 300",
"output": "150\n151\n149\n152\n148\n153\n147\n154\n146\n155\n145\n156\n144\n157\n143\n158\n142\n159\n141\n160\n140\n161\n139\n162\n138\n163\n137\n164\n136\n165\n135\n166\n134\n167\n133\n168\n132\n169\n131\n170\n130\n171\n129\n172\n128\n173\n127\n174\n126\n175\n125\n176\n124\n177\n123\n178\n122\n179\n121\n180\n120\n181\n119\n182\n118\n183\n117\n184\n116\n185\n115\n186\n114\n187\n113\n188\n112\n189\n111\n190\n110\n191\n109\n192\n108\n193\n107\n194\n106\n195\n105\n196\n104\n197\n103\n198\n102\n199\n101\n200\n100\n201\n9..."
},
{
"input": "10500 5",
"output": "3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3..."
},
{
"input": "90091 322",
"output": "161\n162\n160\n163\n159\n164\n158\n165\n157\n166\n156\n167\n155\n168\n154\n169\n153\n170\n152\n171\n151\n172\n150\n173\n149\n174\n148\n175\n147\n176\n146\n177\n145\n178\n144\n179\n143\n180\n142\n181\n141\n182\n140\n183\n139\n184\n138\n185\n137\n186\n136\n187\n135\n188\n134\n189\n133\n190\n132\n191\n131\n192\n130\n193\n129\n194\n128\n195\n127\n196\n126\n197\n125\n198\n124\n199\n123\n200\n122\n201\n121\n202\n120\n203\n119\n204\n118\n205\n117\n206\n116\n207\n115\n208\n114\n209\n113\n210\n112\n211\n111\n212\n1..."
},
{
"input": "8471 92356",
"output": "46178\n46179\n46177\n46180\n46176\n46181\n46175\n46182\n46174\n46183\n46173\n46184\n46172\n46185\n46171\n46186\n46170\n46187\n46169\n46188\n46168\n46189\n46167\n46190\n46166\n46191\n46165\n46192\n46164\n46193\n46163\n46194\n46162\n46195\n46161\n46196\n46160\n46197\n46159\n46198\n46158\n46199\n46157\n46200\n46156\n46201\n46155\n46202\n46154\n46203\n46153\n46204\n46152\n46205\n46151\n46206\n46150\n46207\n46149\n46208\n46148\n46209\n46147\n46210\n46146\n46211\n46145\n46212\n46144\n46213\n46143\n46214\n46142\n..."
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1\n1"
},
{
"input": "52097 88310",
"output": "44155\n44156\n44154\n44157\n44153\n44158\n44152\n44159\n44151\n44160\n44150\n44161\n44149\n44162\n44148\n44163\n44147\n44164\n44146\n44165\n44145\n44166\n44144\n44167\n44143\n44168\n44142\n44169\n44141\n44170\n44140\n44171\n44139\n44172\n44138\n44173\n44137\n44174\n44136\n44175\n44135\n44176\n44134\n44177\n44133\n44178\n44132\n44179\n44131\n44180\n44130\n44181\n44129\n44182\n44128\n44183\n44127\n44184\n44126\n44185\n44125\n44186\n44124\n44187\n44123\n44188\n44122\n44189\n44121\n44190\n44120\n44191\n44119\n..."
}
] | 1,620,290,672
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 6,963,200
|
n,m = map(int,input().split())
mid = (m+1)//2
lst = [0]*m
lst[0] = mid
p = 1
for i in range(1,mid):
lst[p] = mid-i
lst[p+1] = mid+1
p += 2
if m%2==0:
lst[n-1] = m
x = n//m
v = n-x*m
lst *= x
lst.extend(lst[:v])
print(*lst)
|
Title: After Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.
Input Specification:
The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly.
Output Specification:
Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball.
Demo Input:
['4 3\n', '3 1\n']
Demo Output:
['2\n1\n3\n2\n', '1\n1\n1\n']
Note:
none
|
```python
n,m = map(int,input().split())
mid = (m+1)//2
lst = [0]*m
lst[0] = mid
p = 1
for i in range(1,mid):
lst[p] = mid-i
lst[p+1] = mid+1
p += 2
if m%2==0:
lst[n-1] = m
x = n//m
v = n-x*m
lst *= x
lst.extend(lst[:v])
print(*lst)
```
| 0
|
|
584
|
A
|
Olesya and Rodion
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
|
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
|
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
|
[
"3 2\n"
] |
[
"712"
] |
none
| 500
|
[
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input": "18 8",
"output": "888888888888888888"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "100 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "10 2",
"output": "2222222222"
},
{
"input": "18 10",
"output": "111111111111111110"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "7 6",
"output": "6666666"
},
{
"input": "4 4",
"output": "4444"
},
{
"input": "14 7",
"output": "77777777777777"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "2 8",
"output": "88"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 9",
"output": "99999"
},
{
"input": "4 8",
"output": "8888"
},
{
"input": "3 4",
"output": "444"
},
{
"input": "9 4",
"output": "444444444"
},
{
"input": "8 10",
"output": "11111110"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "20 3",
"output": "33333333333333333333"
},
{
"input": "15 10",
"output": "111111111111110"
},
{
"input": "31 4",
"output": "4444444444444444444444444444444"
},
{
"input": "18 9",
"output": "999999999999999999"
},
{
"input": "72 4",
"output": "444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "76 8",
"output": "8888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "12 5",
"output": "555555555555"
},
{
"input": "54 5",
"output": "555555555555555555555555555555555555555555555555555555"
},
{
"input": "96 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "15 9",
"output": "999999999999999"
},
{
"input": "100 2",
"output": "2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222"
},
{
"input": "99 3",
"output": "333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333"
},
{
"input": "98 4",
"output": "44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "97 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "100 6",
"output": "6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "98 8",
"output": "88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "97 9",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "3 10",
"output": "110"
},
{
"input": "2 4",
"output": "44"
},
{
"input": "2 10",
"output": "10"
},
{
"input": "10 10",
"output": "1111111110"
},
{
"input": "4 10",
"output": "1110"
},
{
"input": "25 10",
"output": "1111111111111111111111110"
},
{
"input": "50 10",
"output": "11111111111111111111111111111111111111111111111110"
},
{
"input": "5 10",
"output": "11110"
},
{
"input": "99 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "20 10",
"output": "11111111111111111110"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "89 5",
"output": "55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "6 10",
"output": "111110"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "100 7",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
}
] | 1,666,114,708
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 46
| 0
|
list1 = list(input().split(" "))
x = int(list1[0])
y = int(list1[1])
for n in range(10**(x-1), 10**x):
if n % y == 0:
print(n)
break
|
Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none
|
```python
list1 = list(input().split(" "))
x = int(list1[0])
y = int(list1[1])
for n in range(10**(x-1), 10**x):
if n % y == 0:
print(n)
break
```
| 0
|
|
690
|
D2
|
The Wall (medium)
|
PROGRAMMING
| 1,800
|
[
"combinatorics"
] | null | null |
Heidi the Cow is aghast: cracks in the northern Wall? Zombies gathering outside, forming groups, preparing their assault? This must not happen! Quickly, she fetches her HC2 (Handbook of Crazy Constructions) and looks for the right chapter:
How to build a wall:
1. Take a set of bricks. 1. Select one of the possible wall designs. Computing the number of possible designs is left as an exercise to the reader. 1. Place bricks on top of each other, according to the chosen design.
This seems easy enough. But Heidi is a Coding Cow, not a Constructing Cow. Her mind keeps coming back to point 2b. Despite the imminent danger of a zombie onslaught, she wonders just how many possible walls she could build with up to *n* bricks.
A wall is a set of wall segments as defined in the easy version. How many different walls can be constructed such that the wall consists of at least 1 and at most *n* bricks? Two walls are different if there exist a column *c* and a row *r* such that one wall has a brick in this spot, and the other does not.
Along with *n*, you will be given *C*, the width of the wall (as defined in the easy version). Return the number of different walls modulo 106<=+<=3.
|
The first line contains two space-separated integers *n* and *C*, 1<=≤<=*n*<=≤<=500000, 1<=≤<=*C*<=≤<=200000.
|
Print the number of different walls that Heidi could build, modulo 106<=+<=3.
|
[
"5 1\n",
"2 2\n",
"3 2\n",
"11 5\n",
"37 63\n"
] |
[
"5\n",
"5\n",
"9\n",
"4367\n",
"230574\n"
] |
The number 10<sup class="upper-index">6</sup> + 3 is prime.
In the second sample case, the five walls are:
In the third sample case, the nine walls are the five as in the second sample case and in addition the following four:
| 0
|
[
{
"input": "5 1",
"output": "5"
},
{
"input": "2 2",
"output": "5"
},
{
"input": "3 2",
"output": "9"
},
{
"input": "11 5",
"output": "4367"
},
{
"input": "37 63",
"output": "230574"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "350000 140000",
"output": "453366"
},
{
"input": "350000 160000",
"output": "155549"
},
{
"input": "350000 180000",
"output": "708073"
},
{
"input": "350000 200000",
"output": "504934"
},
{
"input": "400000 140000",
"output": "956370"
},
{
"input": "400000 160000",
"output": "480365"
},
{
"input": "400000 180000",
"output": "376155"
},
{
"input": "400000 200000",
"output": "388234"
},
{
"input": "450000 140000",
"output": "175993"
},
{
"input": "450000 160000",
"output": "926957"
},
{
"input": "450000 180000",
"output": "135727"
},
{
"input": "450000 200000",
"output": "997315"
},
{
"input": "500000 140000",
"output": "775486"
},
{
"input": "500000 160000",
"output": "298591"
},
{
"input": "500000 180000",
"output": "901135"
},
{
"input": "500000 200000",
"output": "781209"
}
] | 1,490,347,933
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 22
| 935
| 18,124,800
|
n, c = map(int, input().split())
mod = 10 ** 6 +3
inv = [0, 1]
for i in range(2, max(n, c) + 1):
inv.append(inv[mod % i] * (mod - mod // i) % mod)
ans = 1
for i in range(1, n + c + 1):
ans = ans * i % mod
for i in range(1, c + 1):
ans = ans * inv[i] % mod
for i in range(1, n + 1):
ans = ans * inv[i] % mod
ans += mod - 1
ans %= mod
print(ans)
|
Title: The Wall (medium)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Heidi the Cow is aghast: cracks in the northern Wall? Zombies gathering outside, forming groups, preparing their assault? This must not happen! Quickly, she fetches her HC2 (Handbook of Crazy Constructions) and looks for the right chapter:
How to build a wall:
1. Take a set of bricks. 1. Select one of the possible wall designs. Computing the number of possible designs is left as an exercise to the reader. 1. Place bricks on top of each other, according to the chosen design.
This seems easy enough. But Heidi is a Coding Cow, not a Constructing Cow. Her mind keeps coming back to point 2b. Despite the imminent danger of a zombie onslaught, she wonders just how many possible walls she could build with up to *n* bricks.
A wall is a set of wall segments as defined in the easy version. How many different walls can be constructed such that the wall consists of at least 1 and at most *n* bricks? Two walls are different if there exist a column *c* and a row *r* such that one wall has a brick in this spot, and the other does not.
Along with *n*, you will be given *C*, the width of the wall (as defined in the easy version). Return the number of different walls modulo 106<=+<=3.
Input Specification:
The first line contains two space-separated integers *n* and *C*, 1<=≤<=*n*<=≤<=500000, 1<=≤<=*C*<=≤<=200000.
Output Specification:
Print the number of different walls that Heidi could build, modulo 106<=+<=3.
Demo Input:
['5 1\n', '2 2\n', '3 2\n', '11 5\n', '37 63\n']
Demo Output:
['5\n', '5\n', '9\n', '4367\n', '230574\n']
Note:
The number 10<sup class="upper-index">6</sup> + 3 is prime.
In the second sample case, the five walls are:
In the third sample case, the nine walls are the five as in the second sample case and in addition the following four:
|
```python
n, c = map(int, input().split())
mod = 10 ** 6 +3
inv = [0, 1]
for i in range(2, max(n, c) + 1):
inv.append(inv[mod % i] * (mod - mod // i) % mod)
ans = 1
for i in range(1, n + c + 1):
ans = ans * i % mod
for i in range(1, c + 1):
ans = ans * inv[i] % mod
for i in range(1, n + 1):
ans = ans * inv[i] % mod
ans += mod - 1
ans %= mod
print(ans)
```
| 3
|
|
761
|
D
|
Dasha and Very Difficult Problem
|
PROGRAMMING
| 1,700
|
[
"binary search",
"brute force",
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Dasha logged into the system and began to solve problems. One of them is as follows:
Given two sequences *a* and *b* of length *n* each you need to write a sequence *c* of length *n*, the *i*-th element of which is calculated as follows: *c**i*<==<=*b**i*<=-<=*a**i*.
About sequences *a* and *b* we know that their elements are in the range from *l* to *r*. More formally, elements satisfy the following conditions: *l*<=≤<=*a**i*<=≤<=*r* and *l*<=≤<=*b**i*<=≤<=*r*. About sequence *c* we know that all its elements are distinct.
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence *a* and the compressed sequence of the sequence *c* were known from that test.
Let's give the definition to a compressed sequence. A compressed sequence of sequence *c* of length *n* is a sequence *p* of length *n*, so that *p**i* equals to the number of integers which are less than or equal to *c**i* in the sequence *c*. For example, for the sequence *c*<==<=[250,<=200,<=300,<=100,<=50] the compressed sequence will be *p*<==<=[4,<=3,<=5,<=2,<=1]. Pay attention that in *c* all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to *n* inclusively.
Help Dasha to find any sequence *b* for which the calculated compressed sequence of sequence *c* is correct.
|
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*l*<=≤<=*r*<=≤<=109) — the length of the sequence and boundaries of the segment where the elements of sequences *a* and *b* are.
The next line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (*l*<=≤<=*a**i*<=≤<=*r*) — the elements of the sequence *a*.
The next line contains *n* distinct integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the compressed sequence of the sequence *c*.
|
If there is no the suitable sequence *b*, then in the only line print "-1".
Otherwise, in the only line print *n* integers — the elements of any suitable sequence *b*.
|
[
"5 1 5\n1 1 1 1 1\n3 1 5 4 2\n",
"4 2 9\n3 4 8 9\n3 2 1 4\n",
"6 1 5\n1 1 1 1 1 1\n2 3 5 4 1 6\n"
] |
[
"3 1 5 4 2 ",
"2 2 2 9 ",
"-1\n"
] |
Sequence *b* which was found in the second sample is suitable, because calculated sequence *c* = [2 - 3, 2 - 4, 2 - 8, 9 - 9] = [ - 1, - 2, - 6, 0] (note that *c*<sub class="lower-index">*i*</sub> = *b*<sub class="lower-index">*i*</sub> - *a*<sub class="lower-index">*i*</sub>) has compressed sequence equals to *p* = [3, 2, 1, 4].
| 2,000
|
[
{
"input": "5 1 5\n1 1 1 1 1\n3 1 5 4 2",
"output": "3 1 5 4 2 "
},
{
"input": "4 2 9\n3 4 8 9\n3 2 1 4",
"output": "2 2 2 9 "
},
{
"input": "6 1 5\n1 1 1 1 1 1\n2 3 5 4 1 6",
"output": "-1"
},
{
"input": "5 1 7\n1 4 4 6 5\n5 2 1 4 3",
"output": "2 2 1 6 4 "
},
{
"input": "5 10 100\n12 14 15 11 13\n4 2 1 5 3",
"output": "10 10 10 10 10 "
},
{
"input": "2 1 1000000000\n1000000000 1\n2 1",
"output": "-1"
},
{
"input": "2 1 1000000000\n1000000000 1\n1 2",
"output": "1 1 "
},
{
"input": "5 1 5\n1 1 1 1 1\n1 2 3 4 5",
"output": "1 2 3 4 5 "
},
{
"input": "5 1 5\n1 1 1 1 1\n2 3 1 5 4",
"output": "2 3 1 5 4 "
},
{
"input": "1 1000000000 1000000000\n1000000000\n1",
"output": "1000000000 "
},
{
"input": "6 3 7\n6 7 5 5 5 5\n2 1 4 3 5 6",
"output": "3 3 4 3 5 6 "
},
{
"input": "3 5 100\n10 50 100\n3 2 1",
"output": "5 5 5 "
},
{
"input": "10 1 10\n9 2 9 5 5 2 6 8 2 8\n2 10 1 6 7 8 5 3 9 4",
"output": "2 3 1 2 3 1 2 2 2 3 "
},
{
"input": "30 100 200\n102 108 122 116 107 145 195 145 119 110 187 196 140 174 104 190 193 181 118 127 157 111 139 175 173 191 181 105 142 166\n30 26 20 23 27 15 2 14 21 25 6 1 17 10 29 5 3 7 22 19 13 24 18 9 11 4 8 28 16 12",
"output": "100 100 100 100 100 101 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 101 100 100 100 "
},
{
"input": "50 10 15\n13 14 12 14 12 15 13 10 11 11 15 10 14 11 14 12 11 10 10 13 11 13 12 15 14 15 13 15 13 10 12 10 15 15 10 14 11 11 10 14 11 12 13 12 10 11 13 15 14 11\n20 10 25 13 29 6 24 47 37 39 2 44 12 33 9 26 35 49 46 19 38 23 30 5 16 4 21 7 18 45 28 43 1 8 48 15 36 40 50 11 32 27 22 31 42 41 17 3 14 34",
"output": "-1"
},
{
"input": "10 8 10\n8 10 10 9 8 10 10 10 10 10\n9 5 6 8 10 2 7 3 1 4",
"output": "-1"
},
{
"input": "50 1 50\n6 20 27 26 46 35 41 41 4 12 48 13 20 5 35 7 37 3 18 17 22 42 28 1 30 46 38 38 42 29 33 42 8 45 22 40 11 33 12 15 16 49 30 19 29 40 5 12 32 22\n47 27 35 40 31 26 42 19 11 49 37 3 41 34 36 30 16 44 12 4 46 20 14 39 32 25 18 1 21 6 2 17 48 7 5 24 29 33 50 43 8 15 28 23 22 13 45 38 10 9",
"output": "-1"
},
{
"input": "11 5 11\n9 8 7 5 7 5 9 5 10 5 7\n3 4 6 9 5 11 2 10 1 8 7",
"output": "6 6 7 8 6 10 5 9 5 7 8 "
}
] | 1,689,602,726
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1689602726.6624253")# 1689602726.6624458
|
Title: Dasha and Very Difficult Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dasha logged into the system and began to solve problems. One of them is as follows:
Given two sequences *a* and *b* of length *n* each you need to write a sequence *c* of length *n*, the *i*-th element of which is calculated as follows: *c**i*<==<=*b**i*<=-<=*a**i*.
About sequences *a* and *b* we know that their elements are in the range from *l* to *r*. More formally, elements satisfy the following conditions: *l*<=≤<=*a**i*<=≤<=*r* and *l*<=≤<=*b**i*<=≤<=*r*. About sequence *c* we know that all its elements are distinct.
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence *a* and the compressed sequence of the sequence *c* were known from that test.
Let's give the definition to a compressed sequence. A compressed sequence of sequence *c* of length *n* is a sequence *p* of length *n*, so that *p**i* equals to the number of integers which are less than or equal to *c**i* in the sequence *c*. For example, for the sequence *c*<==<=[250,<=200,<=300,<=100,<=50] the compressed sequence will be *p*<==<=[4,<=3,<=5,<=2,<=1]. Pay attention that in *c* all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to *n* inclusively.
Help Dasha to find any sequence *b* for which the calculated compressed sequence of sequence *c* is correct.
Input Specification:
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*l*<=≤<=*r*<=≤<=109) — the length of the sequence and boundaries of the segment where the elements of sequences *a* and *b* are.
The next line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (*l*<=≤<=*a**i*<=≤<=*r*) — the elements of the sequence *a*.
The next line contains *n* distinct integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the compressed sequence of the sequence *c*.
Output Specification:
If there is no the suitable sequence *b*, then in the only line print "-1".
Otherwise, in the only line print *n* integers — the elements of any suitable sequence *b*.
Demo Input:
['5 1 5\n1 1 1 1 1\n3 1 5 4 2\n', '4 2 9\n3 4 8 9\n3 2 1 4\n', '6 1 5\n1 1 1 1 1 1\n2 3 5 4 1 6\n']
Demo Output:
['3 1 5 4 2 ', '2 2 2 9 ', '-1\n']
Note:
Sequence *b* which was found in the second sample is suitable, because calculated sequence *c* = [2 - 3, 2 - 4, 2 - 8, 9 - 9] = [ - 1, - 2, - 6, 0] (note that *c*<sub class="lower-index">*i*</sub> = *b*<sub class="lower-index">*i*</sub> - *a*<sub class="lower-index">*i*</sub>) has compressed sequence equals to *p* = [3, 2, 1, 4].
|
```python
print("_RANDOM_GUESS_1689602726.6624253")# 1689602726.6624458
```
| 0
|
|
334
|
A
|
Candy Bags
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
|
The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.
|
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
|
[
"2\n"
] |
[
"1 4\n2 3\n"
] |
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
| 500
|
[
{
"input": "2",
"output": "1 4\n2 3"
},
{
"input": "4",
"output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9"
},
{
"input": "6",
"output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19"
},
{
"input": "8",
"output": "1 64 2 63 3 62 4 61\n5 60 6 59 7 58 8 57\n9 56 10 55 11 54 12 53\n13 52 14 51 15 50 16 49\n17 48 18 47 19 46 20 45\n21 44 22 43 23 42 24 41\n25 40 26 39 27 38 28 37\n29 36 30 35 31 34 32 33"
},
{
"input": "10",
"output": "1 100 2 99 3 98 4 97 5 96\n6 95 7 94 8 93 9 92 10 91\n11 90 12 89 13 88 14 87 15 86\n16 85 17 84 18 83 19 82 20 81\n21 80 22 79 23 78 24 77 25 76\n26 75 27 74 28 73 29 72 30 71\n31 70 32 69 33 68 34 67 35 66\n36 65 37 64 38 63 39 62 40 61\n41 60 42 59 43 58 44 57 45 56\n46 55 47 54 48 53 49 52 50 51"
},
{
"input": "100",
"output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951\n51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 993..."
},
{
"input": "62",
"output": "1 3844 2 3843 3 3842 4 3841 5 3840 6 3839 7 3838 8 3837 9 3836 10 3835 11 3834 12 3833 13 3832 14 3831 15 3830 16 3829 17 3828 18 3827 19 3826 20 3825 21 3824 22 3823 23 3822 24 3821 25 3820 26 3819 27 3818 28 3817 29 3816 30 3815 31 3814\n32 3813 33 3812 34 3811 35 3810 36 3809 37 3808 38 3807 39 3806 40 3805 41 3804 42 3803 43 3802 44 3801 45 3800 46 3799 47 3798 48 3797 49 3796 50 3795 51 3794 52 3793 53 3792 54 3791 55 3790 56 3789 57 3788 58 3787 59 3786 60 3785 61 3784 62 3783\n63 3782 64 3781 65 378..."
},
{
"input": "66",
"output": "1 4356 2 4355 3 4354 4 4353 5 4352 6 4351 7 4350 8 4349 9 4348 10 4347 11 4346 12 4345 13 4344 14 4343 15 4342 16 4341 17 4340 18 4339 19 4338 20 4337 21 4336 22 4335 23 4334 24 4333 25 4332 26 4331 27 4330 28 4329 29 4328 30 4327 31 4326 32 4325 33 4324\n34 4323 35 4322 36 4321 37 4320 38 4319 39 4318 40 4317 41 4316 42 4315 43 4314 44 4313 45 4312 46 4311 47 4310 48 4309 49 4308 50 4307 51 4306 52 4305 53 4304 54 4303 55 4302 56 4301 57 4300 58 4299 59 4298 60 4297 61 4296 62 4295 63 4294 64 4293 65 4292..."
},
{
"input": "18",
"output": "1 324 2 323 3 322 4 321 5 320 6 319 7 318 8 317 9 316\n10 315 11 314 12 313 13 312 14 311 15 310 16 309 17 308 18 307\n19 306 20 305 21 304 22 303 23 302 24 301 25 300 26 299 27 298\n28 297 29 296 30 295 31 294 32 293 33 292 34 291 35 290 36 289\n37 288 38 287 39 286 40 285 41 284 42 283 43 282 44 281 45 280\n46 279 47 278 48 277 49 276 50 275 51 274 52 273 53 272 54 271\n55 270 56 269 57 268 58 267 59 266 60 265 61 264 62 263 63 262\n64 261 65 260 66 259 67 258 68 257 69 256 70 255 71 254 72 253\n73 252 7..."
},
{
"input": "68",
"output": "1 4624 2 4623 3 4622 4 4621 5 4620 6 4619 7 4618 8 4617 9 4616 10 4615 11 4614 12 4613 13 4612 14 4611 15 4610 16 4609 17 4608 18 4607 19 4606 20 4605 21 4604 22 4603 23 4602 24 4601 25 4600 26 4599 27 4598 28 4597 29 4596 30 4595 31 4594 32 4593 33 4592 34 4591\n35 4590 36 4589 37 4588 38 4587 39 4586 40 4585 41 4584 42 4583 43 4582 44 4581 45 4580 46 4579 47 4578 48 4577 49 4576 50 4575 51 4574 52 4573 53 4572 54 4571 55 4570 56 4569 57 4568 58 4567 59 4566 60 4565 61 4564 62 4563 63 4562 64 4561 65 4560..."
},
{
"input": "86",
"output": "1 7396 2 7395 3 7394 4 7393 5 7392 6 7391 7 7390 8 7389 9 7388 10 7387 11 7386 12 7385 13 7384 14 7383 15 7382 16 7381 17 7380 18 7379 19 7378 20 7377 21 7376 22 7375 23 7374 24 7373 25 7372 26 7371 27 7370 28 7369 29 7368 30 7367 31 7366 32 7365 33 7364 34 7363 35 7362 36 7361 37 7360 38 7359 39 7358 40 7357 41 7356 42 7355 43 7354\n44 7353 45 7352 46 7351 47 7350 48 7349 49 7348 50 7347 51 7346 52 7345 53 7344 54 7343 55 7342 56 7341 57 7340 58 7339 59 7338 60 7337 61 7336 62 7335 63 7334 64 7333 65 7332..."
},
{
"input": "96",
"output": "1 9216 2 9215 3 9214 4 9213 5 9212 6 9211 7 9210 8 9209 9 9208 10 9207 11 9206 12 9205 13 9204 14 9203 15 9202 16 9201 17 9200 18 9199 19 9198 20 9197 21 9196 22 9195 23 9194 24 9193 25 9192 26 9191 27 9190 28 9189 29 9188 30 9187 31 9186 32 9185 33 9184 34 9183 35 9182 36 9181 37 9180 38 9179 39 9178 40 9177 41 9176 42 9175 43 9174 44 9173 45 9172 46 9171 47 9170 48 9169\n49 9168 50 9167 51 9166 52 9165 53 9164 54 9163 55 9162 56 9161 57 9160 58 9159 59 9158 60 9157 61 9156 62 9155 63 9154 64 9153 65 9152..."
},
{
"input": "12",
"output": "1 144 2 143 3 142 4 141 5 140 6 139\n7 138 8 137 9 136 10 135 11 134 12 133\n13 132 14 131 15 130 16 129 17 128 18 127\n19 126 20 125 21 124 22 123 23 122 24 121\n25 120 26 119 27 118 28 117 29 116 30 115\n31 114 32 113 33 112 34 111 35 110 36 109\n37 108 38 107 39 106 40 105 41 104 42 103\n43 102 44 101 45 100 46 99 47 98 48 97\n49 96 50 95 51 94 52 93 53 92 54 91\n55 90 56 89 57 88 58 87 59 86 60 85\n61 84 62 83 63 82 64 81 65 80 66 79\n67 78 68 77 69 76 70 75 71 74 72 73"
},
{
"input": "88",
"output": "1 7744 2 7743 3 7742 4 7741 5 7740 6 7739 7 7738 8 7737 9 7736 10 7735 11 7734 12 7733 13 7732 14 7731 15 7730 16 7729 17 7728 18 7727 19 7726 20 7725 21 7724 22 7723 23 7722 24 7721 25 7720 26 7719 27 7718 28 7717 29 7716 30 7715 31 7714 32 7713 33 7712 34 7711 35 7710 36 7709 37 7708 38 7707 39 7706 40 7705 41 7704 42 7703 43 7702 44 7701\n45 7700 46 7699 47 7698 48 7697 49 7696 50 7695 51 7694 52 7693 53 7692 54 7691 55 7690 56 7689 57 7688 58 7687 59 7686 60 7685 61 7684 62 7683 63 7682 64 7681 65 7680..."
},
{
"input": "28",
"output": "1 784 2 783 3 782 4 781 5 780 6 779 7 778 8 777 9 776 10 775 11 774 12 773 13 772 14 771\n15 770 16 769 17 768 18 767 19 766 20 765 21 764 22 763 23 762 24 761 25 760 26 759 27 758 28 757\n29 756 30 755 31 754 32 753 33 752 34 751 35 750 36 749 37 748 38 747 39 746 40 745 41 744 42 743\n43 742 44 741 45 740 46 739 47 738 48 737 49 736 50 735 51 734 52 733 53 732 54 731 55 730 56 729\n57 728 58 727 59 726 60 725 61 724 62 723 63 722 64 721 65 720 66 719 67 718 68 717 69 716 70 715\n71 714 72 713 73 712 74 7..."
},
{
"input": "80",
"output": "1 6400 2 6399 3 6398 4 6397 5 6396 6 6395 7 6394 8 6393 9 6392 10 6391 11 6390 12 6389 13 6388 14 6387 15 6386 16 6385 17 6384 18 6383 19 6382 20 6381 21 6380 22 6379 23 6378 24 6377 25 6376 26 6375 27 6374 28 6373 29 6372 30 6371 31 6370 32 6369 33 6368 34 6367 35 6366 36 6365 37 6364 38 6363 39 6362 40 6361\n41 6360 42 6359 43 6358 44 6357 45 6356 46 6355 47 6354 48 6353 49 6352 50 6351 51 6350 52 6349 53 6348 54 6347 55 6346 56 6345 57 6344 58 6343 59 6342 60 6341 61 6340 62 6339 63 6338 64 6337 65 6336..."
},
{
"input": "48",
"output": "1 2304 2 2303 3 2302 4 2301 5 2300 6 2299 7 2298 8 2297 9 2296 10 2295 11 2294 12 2293 13 2292 14 2291 15 2290 16 2289 17 2288 18 2287 19 2286 20 2285 21 2284 22 2283 23 2282 24 2281\n25 2280 26 2279 27 2278 28 2277 29 2276 30 2275 31 2274 32 2273 33 2272 34 2271 35 2270 36 2269 37 2268 38 2267 39 2266 40 2265 41 2264 42 2263 43 2262 44 2261 45 2260 46 2259 47 2258 48 2257\n49 2256 50 2255 51 2254 52 2253 53 2252 54 2251 55 2250 56 2249 57 2248 58 2247 59 2246 60 2245 61 2244 62 2243 63 2242 64 2241 65 224..."
},
{
"input": "54",
"output": "1 2916 2 2915 3 2914 4 2913 5 2912 6 2911 7 2910 8 2909 9 2908 10 2907 11 2906 12 2905 13 2904 14 2903 15 2902 16 2901 17 2900 18 2899 19 2898 20 2897 21 2896 22 2895 23 2894 24 2893 25 2892 26 2891 27 2890\n28 2889 29 2888 30 2887 31 2886 32 2885 33 2884 34 2883 35 2882 36 2881 37 2880 38 2879 39 2878 40 2877 41 2876 42 2875 43 2874 44 2873 45 2872 46 2871 47 2870 48 2869 49 2868 50 2867 51 2866 52 2865 53 2864 54 2863\n55 2862 56 2861 57 2860 58 2859 59 2858 60 2857 61 2856 62 2855 63 2854 64 2853 65 285..."
},
{
"input": "58",
"output": "1 3364 2 3363 3 3362 4 3361 5 3360 6 3359 7 3358 8 3357 9 3356 10 3355 11 3354 12 3353 13 3352 14 3351 15 3350 16 3349 17 3348 18 3347 19 3346 20 3345 21 3344 22 3343 23 3342 24 3341 25 3340 26 3339 27 3338 28 3337 29 3336\n30 3335 31 3334 32 3333 33 3332 34 3331 35 3330 36 3329 37 3328 38 3327 39 3326 40 3325 41 3324 42 3323 43 3322 44 3321 45 3320 46 3319 47 3318 48 3317 49 3316 50 3315 51 3314 52 3313 53 3312 54 3311 55 3310 56 3309 57 3308 58 3307\n59 3306 60 3305 61 3304 62 3303 63 3302 64 3301 65 330..."
},
{
"input": "64",
"output": "1 4096 2 4095 3 4094 4 4093 5 4092 6 4091 7 4090 8 4089 9 4088 10 4087 11 4086 12 4085 13 4084 14 4083 15 4082 16 4081 17 4080 18 4079 19 4078 20 4077 21 4076 22 4075 23 4074 24 4073 25 4072 26 4071 27 4070 28 4069 29 4068 30 4067 31 4066 32 4065\n33 4064 34 4063 35 4062 36 4061 37 4060 38 4059 39 4058 40 4057 41 4056 42 4055 43 4054 44 4053 45 4052 46 4051 47 4050 48 4049 49 4048 50 4047 51 4046 52 4045 53 4044 54 4043 55 4042 56 4041 57 4040 58 4039 59 4038 60 4037 61 4036 62 4035 63 4034 64 4033\n65 403..."
},
{
"input": "44",
"output": "1 1936 2 1935 3 1934 4 1933 5 1932 6 1931 7 1930 8 1929 9 1928 10 1927 11 1926 12 1925 13 1924 14 1923 15 1922 16 1921 17 1920 18 1919 19 1918 20 1917 21 1916 22 1915\n23 1914 24 1913 25 1912 26 1911 27 1910 28 1909 29 1908 30 1907 31 1906 32 1905 33 1904 34 1903 35 1902 36 1901 37 1900 38 1899 39 1898 40 1897 41 1896 42 1895 43 1894 44 1893\n45 1892 46 1891 47 1890 48 1889 49 1888 50 1887 51 1886 52 1885 53 1884 54 1883 55 1882 56 1881 57 1880 58 1879 59 1878 60 1877 61 1876 62 1875 63 1874 64 1873 65 187..."
}
] | 1,418,982,699
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 154
| 307,200
|
n = int(input())
r = [str(i) for i in range(1,n*n+1)]
rr = [r.pop() for i in range(n*n//2)]
for i in range(n):
s = ""
for ii in range(n//2):
s += r.pop() +" "+ rr.pop()+" "
print(s)
|
Title: Candy Bags
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
Input Specification:
The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.
Output Specification:
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Demo Input:
['2\n']
Demo Output:
['1 4\n2 3\n']
Note:
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
|
```python
n = int(input())
r = [str(i) for i in range(1,n*n+1)]
rr = [r.pop() for i in range(n*n//2)]
for i in range(n):
s = ""
for ii in range(n//2):
s += r.pop() +" "+ rr.pop()+" "
print(s)
```
| 3
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,686,146,540
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 27
| 93
| 0
|
s=input()
s=s[1:-1]
if not s:
count=0
else:
s_l=s.split(", ")
count=len(set(s_l))
print(count)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
s=input()
s=s[1:-1]
if not s:
count=0
else:
s_l=s.split(", ")
count=len(set(s_l))
print(count)
```
| 3
|
|
237
|
A
|
Free Cash
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
|
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
|
[
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] |
[
"2\n",
"1\n"
] |
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
| 500
|
[
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6",
"output": "3"
},
{
"input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32",
"output": "6"
},
{
"input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30",
"output": "10"
},
{
"input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20",
"output": "8"
},
{
"input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49",
"output": "1"
},
{
"input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16",
"output": "5"
},
{
"input": "2\n0 24\n1 0",
"output": "1"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "1\n1 5",
"output": "1"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "3"
},
{
"input": "1\n5 0",
"output": "1"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "5"
},
{
"input": "1\n10 10",
"output": "1"
},
{
"input": "1\n8 0",
"output": "1"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "10"
},
{
"input": "2\n0 0\n0 1",
"output": "1"
},
{
"input": "2\n8 5\n8 5",
"output": "2"
}
] | 1,670,666,413
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 67
| 1,434
| 9,830,400
|
n = int(input())
d = {}
for i in range(n):
h, m = map(int, input().split())
if (h*60+m) in d:
d[h*60+m]+=1
else:
d[h*60+m]=1
print(max(list(d.values())))
|
Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
|
```python
n = int(input())
d = {}
for i in range(n):
h, m = map(int, input().split())
if (h*60+m) in d:
d[h*60+m]+=1
else:
d[h*60+m]=1
print(max(list(d.values())))
```
| 3
|
|
952
|
C
|
Ravioli Sort
|
PROGRAMMING
| 1,600
|
[
"implementation"
] | null | null |
Everybody knows of [spaghetti sort](https://en.wikipedia.org/wiki/Spaghetti_sort). You decided to implement an analog sorting algorithm yourself, but as you survey your pantry you realize you're out of spaghetti! The only type of pasta you have is ravioli, but you are not going to let this stop you...
You come up with the following algorithm. For each number in the array *a**i*, build a stack of *a**i* ravioli. The image shows the stack for *a**i*<==<=4.
Arrange the stacks in one row in the order in which the corresponding numbers appear in the input array. Find the tallest one (if there are several stacks of maximal height, use the leftmost one). Remove it and add its height to the end of the output array. Shift the stacks in the row so that there is no gap between them. Repeat the procedure until all stacks have been removed.
At first you are very happy with your algorithm, but as you try it on more inputs you realize that it doesn't always produce the right sorted array. Turns out when two stacks of ravioli are next to each other (at any step of the process) and differ in height by two or more, the top ravioli of the taller stack slides down on top of the lower stack.
Given an input array, figure out whether the described algorithm will sort it correctly.
|
The first line of input contains a single number *n* (1<=≤<=*n*<=≤<=10) — the size of the array.
The second line of input contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the elements of the array.
|
Output "YES" if the array can be sorted using the described procedure and "NO" if it can not.
|
[
"3\n1 2 3\n",
"3\n3 1 2\n"
] |
[
"YES\n",
"NO\n"
] |
In the second example the array will change even before the tallest stack is chosen for the first time: ravioli from stack of height 3 will slide on the stack of height 1, and the algorithm will output an array {2, 2, 2}.
| 0
|
[
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "3\n3 1 2",
"output": "NO"
},
{
"input": "1\n13",
"output": "YES"
},
{
"input": "10\n67 67 67 67 67 67 67 67 67 67",
"output": "YES"
},
{
"input": "10\n16 17 16 15 14 15 16 17 16 15",
"output": "YES"
},
{
"input": "4\n54 54 54 55",
"output": "YES"
},
{
"input": "3\n68 67 67",
"output": "YES"
},
{
"input": "5\n46 46 47 46 45",
"output": "YES"
},
{
"input": "4\n14 15 15 16",
"output": "YES"
},
{
"input": "6\n59 59 60 60 59 58",
"output": "YES"
},
{
"input": "3\n40 40 40",
"output": "YES"
},
{
"input": "4\n90 91 90 91",
"output": "YES"
},
{
"input": "10\n9 9 9 10 10 9 8 8 9 9",
"output": "YES"
},
{
"input": "3\n22 23 24",
"output": "YES"
},
{
"input": "9\n71 71 70 70 71 70 69 70 71",
"output": "YES"
},
{
"input": "9\n15 14 14 13 13 12 13 13 14",
"output": "YES"
},
{
"input": "4\n61 60 60 60",
"output": "YES"
},
{
"input": "4\n16 17 17 18",
"output": "YES"
},
{
"input": "6\n87 86 86 86 85 86",
"output": "YES"
},
{
"input": "5\n64 63 63 62 61",
"output": "YES"
},
{
"input": "9\n13 80 13 38 98 85 11 73 74",
"output": "NO"
},
{
"input": "10\n2 83 18 65 58 95 37 51 86 47",
"output": "NO"
},
{
"input": "6\n47 100 96 2 96 43",
"output": "NO"
},
{
"input": "10\n28 61 23 73 61 33 45 55 18 43",
"output": "NO"
},
{
"input": "10\n95 51 52 8 44 39 77 17 96 88",
"output": "NO"
},
{
"input": "5\n14 91 91 91 84",
"output": "NO"
},
{
"input": "4\n92 18 29 93",
"output": "NO"
},
{
"input": "7\n23 37 39 8 72 31 85",
"output": "NO"
},
{
"input": "4\n61 28 3 81",
"output": "NO"
},
{
"input": "4\n83 100 81 75",
"output": "NO"
},
{
"input": "9\n95 7 97 61 90 7 30 65 39",
"output": "NO"
},
{
"input": "3\n90 39 98",
"output": "NO"
},
{
"input": "3\n76 9 12",
"output": "NO"
},
{
"input": "3\n69 26 73",
"output": "NO"
},
{
"input": "10\n55 39 93 42 97 40 36 38 11 97",
"output": "NO"
},
{
"input": "5\n21 57 40 94 17",
"output": "NO"
},
{
"input": "7\n35 91 87 78 17 71 63",
"output": "NO"
},
{
"input": "7\n20 21 95 73 49 98 53",
"output": "NO"
},
{
"input": "8\n46 4 30 85 52 6 84 13",
"output": "NO"
},
{
"input": "10\n79 84 22 38 23 22 33 42 13 96",
"output": "NO"
}
] | 1,588,427,949
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 202
| 2,048,000
|
n=int(input())
a=list(map(int,input().split()))
b=[abs(a[i]-a[i-1]) for i in range(1,n)]
print("YNEOS"[max(b)>1::2])
|
Title: Ravioli Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Everybody knows of [spaghetti sort](https://en.wikipedia.org/wiki/Spaghetti_sort). You decided to implement an analog sorting algorithm yourself, but as you survey your pantry you realize you're out of spaghetti! The only type of pasta you have is ravioli, but you are not going to let this stop you...
You come up with the following algorithm. For each number in the array *a**i*, build a stack of *a**i* ravioli. The image shows the stack for *a**i*<==<=4.
Arrange the stacks in one row in the order in which the corresponding numbers appear in the input array. Find the tallest one (if there are several stacks of maximal height, use the leftmost one). Remove it and add its height to the end of the output array. Shift the stacks in the row so that there is no gap between them. Repeat the procedure until all stacks have been removed.
At first you are very happy with your algorithm, but as you try it on more inputs you realize that it doesn't always produce the right sorted array. Turns out when two stacks of ravioli are next to each other (at any step of the process) and differ in height by two or more, the top ravioli of the taller stack slides down on top of the lower stack.
Given an input array, figure out whether the described algorithm will sort it correctly.
Input Specification:
The first line of input contains a single number *n* (1<=≤<=*n*<=≤<=10) — the size of the array.
The second line of input contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the elements of the array.
Output Specification:
Output "YES" if the array can be sorted using the described procedure and "NO" if it can not.
Demo Input:
['3\n1 2 3\n', '3\n3 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the second example the array will change even before the tallest stack is chosen for the first time: ravioli from stack of height 3 will slide on the stack of height 1, and the algorithm will output an array {2, 2, 2}.
|
```python
n=int(input())
a=list(map(int,input().split()))
b=[abs(a[i]-a[i-1]) for i in range(1,n)]
print("YNEOS"[max(b)>1::2])
```
| -1
|
|
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"
}
] | 1,640,623,758
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 124
| 0
|
board = input()
rboard = board.split()
m,n = rboard
print((int(m)*int(n))//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
board = input()
rboard = board.split()
m,n = rboard
print((int(m)*int(n))//2)
```
| 3.969
|
500
|
A
|
New Year Transportation
|
PROGRAMMING
| 1,000
|
[
"dfs and similar",
"graphs",
"implementation"
] | null | null |
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
|
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
|
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
|
[
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
| 500
|
[
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1",
"output": "YES"
},
{
"input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1",
"output": "NO"
},
{
"input": "10 3\n8 3 5 4 2 3 2 2 1",
"output": "NO"
},
{
"input": "10 9\n8 3 5 4 2 3 2 2 1",
"output": "YES"
},
{
"input": "3 2\n1 1",
"output": "YES"
},
{
"input": "3 2\n2 1",
"output": "NO"
},
{
"input": "4 2\n2 1 1",
"output": "NO"
},
{
"input": "4 4\n2 2 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 1 1",
"output": "YES"
},
{
"input": "3 3\n2 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1",
"output": "YES"
},
{
"input": "8 8\n1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5 5\n1 1 1 1",
"output": "YES"
}
] | 1,684,767,383
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 61
| 3,584,000
|
import sys
input = lambda: sys.stdin.readline().strip()
n,m=map(int,input().split())
x=list(map(int,input().split()))
l=1
k='NO'
for i in range(n):
if l>n-1:
break
elif x[l-1]+l==m:
k='YES'
break
l+=x[l-1]
print(k)
|
Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
|
```python
import sys
input = lambda: sys.stdin.readline().strip()
n,m=map(int,input().split())
x=list(map(int,input().split()))
l=1
k='NO'
for i in range(n):
if l>n-1:
break
elif x[l-1]+l==m:
k='YES'
break
l+=x[l-1]
print(k)
```
| 3
|
|
362
|
A
|
Two Semiknights Meet
|
PROGRAMMING
| 1,500
|
[
"greedy",
"math"
] | null | null |
A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.
Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square. After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights can move through these squares but their meetings in these squares don't count.
Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.
Please see the test case analysis.
|
The first line contains number *t* (1<=≤<=*t*<=≤<=50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights. The semiknight's squares are considered good for the meeting. The tests are separated by empty line.
|
For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise.
|
[
"2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#\n"
] |
[
"YES\nNO\n"
] |
Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from square (8, 1) to square (6, 3). Then both semiknights go to (4, 5) but this square is bad, so they move together to square (2, 7).
On the second board the semiknights will never meet.
| 1,000
|
[
{
"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#",
"output": "YES\nNO"
},
{
"input": "3\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n####K#K#\n\n........\nK......K\n........\n#......#\n.#....#.\n..####..\n........\n........\n\n.#..#...\n.##.##..\n..###...\n..#K###.\n..####..\n......K.\n..#####.\n..#####.",
"output": "NO\nNO\nNO"
},
{
"input": "1\nK.#....#\n...#..#.\n..#.....\n..#.###.\n..#.....\n...#....\n.#.....#\n.#...##K",
"output": "NO"
},
{
"input": "2\n....#..K\n...#....\n..##.#..\n.#.#.#..\n.#.....#\n.#......\n###.....\nK#.#....\n\nK.#.....\n..#...#.\n#.....#.\n..#.#..#\n#.......\n..#..#..\n....#...\nK..##.##",
"output": "NO\nNO"
},
{
"input": "5\n........\n...KK...\n..####..\n...##...\n........\n..####..\n.######.\n#......#\n\n........\n.K......\n..#.....\n...#....\n....#...\n.....#..\n......#.\n.......K\n\n........\n...K....\n##...##.\n#.#.#..#\n.##.###.\n#..K#..#\n.##..##.\n........\n\n........\n.K..K...\n..##....\n..####..\n.#....#.\n.#.....#\n..#####.\n........\n\nK.......\n........\n........\n........\n........\n........\n........\n.......K",
"output": "NO\nNO\nYES\nNO\nNO"
}
] | 1,653,586,176
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 3,379,200
|
for _ in range(int(input())):
n = int(input())
for i in range(2,30):
if n % (2**i-1) == 0:
print(n//(2**i-1))
break
|
Title: Two Semiknights Meet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.
Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square. After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights can move through these squares but their meetings in these squares don't count.
Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.
Please see the test case analysis.
Input Specification:
The first line contains number *t* (1<=≤<=*t*<=≤<=50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights. The semiknight's squares are considered good for the meeting. The tests are separated by empty line.
Output Specification:
For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise.
Demo Input:
['2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#\n']
Demo Output:
['YES\nNO\n']
Note:
Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from square (8, 1) to square (6, 3). Then both semiknights go to (4, 5) but this square is bad, so they move together to square (2, 7).
On the second board the semiknights will never meet.
|
```python
for _ in range(int(input())):
n = int(input())
for i in range(2,30):
if n % (2**i-1) == 0:
print(n//(2**i-1))
break
```
| -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"
}
] | 1,649,157,660
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 9
| 140
| 3,584,000
|
import math
n = int(input())
ne = []
for i in range(int(math.sqrt(n) + 1)):
ne.append(str(i**2))
ans = 100
n = str(n)
intn = int(n)
for i in range(1 << len(n)):
s = ""
cnt = 0
for j in range(len(n)):
if (i>>j)&1:
s += n[j]
else :
cnt += 1
if s in ne:
ans = min(ans, cnt)
print(ans if ans != 100 else -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 math
n = int(input())
ne = []
for i in range(int(math.sqrt(n) + 1)):
ne.append(str(i**2))
ans = 100
n = str(n)
intn = int(n)
for i in range(1 << len(n)):
s = ""
cnt = 0
for j in range(len(n)):
if (i>>j)&1:
s += n[j]
else :
cnt += 1
if s in ne:
ans = min(ans, cnt)
print(ans if ans != 100 else -1)
```
| 0
|
|
540
|
A
|
Combination Lock
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
|
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
|
[
"5\n82195\n64723\n"
] |
[
"13\n"
] |
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
"input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746",
"output": "245"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "3\n607\n684",
"output": "5"
},
{
"input": "4\n0809\n0636",
"output": "8"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762280548\n9519431339078678836940020",
"output": "72"
},
{
"input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024",
"output": "305"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762285484\n9519431339078678836940202",
"output": "74"
},
{
"input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831",
"output": "357"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8378525987476228048406972\n9194339078883694020217816",
"output": "55"
},
{
"input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599",
"output": "274"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "1\n7\n7",
"output": "0"
},
{
"input": "1\n0\n5",
"output": "5"
},
{
"input": "1\n2\n7",
"output": "5"
},
{
"input": "1\n7\n9",
"output": "2"
},
{
"input": "1\n9\n7",
"output": "2"
},
{
"input": "1\n2\n9",
"output": "3"
},
{
"input": "1\n9\n2",
"output": "3"
},
{
"input": "25\n3164978461316464614169874\n9413979197249127496597357",
"output": "66"
},
{
"input": "4\n9999\n9999",
"output": "0"
},
{
"input": "2\n11\n11",
"output": "0"
}
] | 1,624,977,136
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 77
| 0
|
n= int(input())
a = str(input())
b = str(input())
lock = 0
for i in range(n):
if int(a[i]) > int(b[i]):
large = int(a[i])
small = int(b[i])
else:
large = int(b[i])
small = int(a[i])
temp = large - 1
count1 = large - small
count2 = small + (9 - temp)
if count1 > count2:
lock += count2
else:
lock += count1
print(lock)
|
Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n= int(input())
a = str(input())
b = str(input())
lock = 0
for i in range(n):
if int(a[i]) > int(b[i]):
large = int(a[i])
small = int(b[i])
else:
large = int(b[i])
small = int(a[i])
temp = large - 1
count1 = large - small
count2 = small + (9 - temp)
if count1 > count2:
lock += count2
else:
lock += count1
print(lock)
```
| 3
|
|
929
|
A
|
Прокат велосипедов
|
PROGRAMMING
| 1,400
|
[
"*special",
"greedy",
"implementation"
] | null | null |
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
|
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
|
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
|
[
"4 4\n3 6 8 10\n",
"2 9\n10 20\n",
"12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n"
] |
[
"2\n",
"-1\n",
"6\n"
] |
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
| 500
|
[
{
"input": "4 4\n3 6 8 10",
"output": "2"
},
{
"input": "2 9\n10 20",
"output": "-1"
},
{
"input": "12 3\n4 6 7 9 10 11 13 15 17 18 20 21",
"output": "6"
},
{
"input": "2 1\n11164 11165",
"output": "1"
},
{
"input": "3 7\n45823 45825 45829",
"output": "1"
},
{
"input": "2 100000\n0 100000",
"output": "1"
},
{
"input": "50 15001\n1269 1580 5431 5916 6642 14145 15760 19922 20589 22062 24138 33454 33505 35916 37012 42577 43140 49457 54720 55188 56205 56639 56870 57997 58379 59088 59297 61805 61861 63005 64013 68848 71182 71497 72715 73008 75616 76042 76614 80690 83089 85033 86408 92392 92763 93833 95085 95815 97152 99379",
"output": "8"
},
{
"input": "5 7\n6219 6222 6229 6231 6236",
"output": "3"
},
{
"input": "10 448\n78449 78573 78599 78742 78748 78759 78853 79091 79298 79324",
"output": "3"
},
{
"input": "20 19191\n11272 12386 14108 24663 24932 26547 29656 44677 45436 45654 48562 62367 71049 71238 78182 84042 88736 92026 96836 99343",
"output": "6"
},
{
"input": "2 100000\n1 100000",
"output": "1"
},
{
"input": "2 99999\n0 100000",
"output": "-1"
},
{
"input": "2 2\n1 3",
"output": "1"
},
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "2 2\n1 4",
"output": "-1"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
}
] | 1,520,006,525
| 1,625
|
Python 3
|
OK
|
TESTS
| 52
| 62
| 5,734,400
|
n, k = (int(s) for s in input().split())
x = [int(s) for s in input().split()]
cur = x[0]
res = 1
for i in range(n - 1):
if x[i + 1] > cur + k:
cur = x[i]
res += 1
if x[i + 1] > cur + k:
res = -1
break
print(res)
|
Title: Прокат велосипедов
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
Input Specification:
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
Output Specification:
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
Demo Input:
['4 4\n3 6 8 10\n', '2 9\n10 20\n', '12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
|
```python
n, k = (int(s) for s in input().split())
x = [int(s) for s in input().split()]
cur = x[0]
res = 1
for i in range(n - 1):
if x[i + 1] > cur + k:
cur = x[i]
res += 1
if x[i + 1] > cur + k:
res = -1
break
print(res)
```
| 3
|
|
115
|
A
|
Party
|
PROGRAMMING
| 900
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
|
Print a single integer denoting the minimum number of groups that will be formed in the party.
|
[
"5\n-1\n1\n2\n1\n-1\n"
] |
[
"3\n"
] |
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
| 500
|
[
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": "2"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n2\n-1",
"output": "2"
},
{
"input": "2\n-1\n-1",
"output": "1"
},
{
"input": "3\n2\n-1\n1",
"output": "3"
},
{
"input": "3\n-1\n-1\n-1",
"output": "1"
},
{
"input": "5\n4\n5\n1\n-1\n4",
"output": "3"
},
{
"input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7",
"output": "4"
},
{
"input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4",
"output": "5"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1",
"output": "2"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1",
"output": "1"
},
{
"input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11",
"output": "12"
},
{
"input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2",
"output": "11"
},
{
"input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6",
"output": "6"
},
{
"input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1",
"output": "3"
},
{
"input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1",
"output": "2"
},
{
"input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1",
"output": "3"
},
{
"input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8",
"output": "5"
},
{
"input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10",
"output": "6"
},
{
"input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1",
"output": "4"
},
{
"input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9",
"output": "4"
},
{
"input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8",
"output": "3"
},
{
"input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11",
"output": "7"
},
{
"input": "7\n7\n6\n-1\n-1\n4\n4\n-1",
"output": "3"
},
{
"input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34",
"output": "6"
},
{
"input": "5\n2\n3\n4\n5\n-1",
"output": "5"
}
] | 1,678,038,157
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
def getseniors(graph,node):
listSeniors=[]
for j in range(len(graph[node])):
if graph[node][j]==1:
listSeniors.append(j)
return listSeniors
def maxdepth(node):
l=getseniors(managerof,node)
if l==[]:
depth=1
else :
depth=max(maxdepth(j) for j in l)+1
return depth
n=int(input())
managerof=[[0 for _ in range(n+1)] for _ in range(n+1)]
for i in range(1,n+1):
x=int(input())
if (x!=-1):
managerof[x][i]=1
startnode=1
for i in range(1,n+1):
if managerof[i].count(1)>=managerof[startnode].count(1):
startnode=i
print(startnode)
print(maxdepth(startnode))
|
Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
|
```python
def getseniors(graph,node):
listSeniors=[]
for j in range(len(graph[node])):
if graph[node][j]==1:
listSeniors.append(j)
return listSeniors
def maxdepth(node):
l=getseniors(managerof,node)
if l==[]:
depth=1
else :
depth=max(maxdepth(j) for j in l)+1
return depth
n=int(input())
managerof=[[0 for _ in range(n+1)] for _ in range(n+1)]
for i in range(1,n+1):
x=int(input())
if (x!=-1):
managerof[x][i]=1
startnode=1
for i in range(1,n+1):
if managerof[i].count(1)>=managerof[startnode].count(1):
startnode=i
print(startnode)
print(maxdepth(startnode))
```
| 0
|
|
547
|
A
|
Mike and Frog
|
PROGRAMMING
| 2,200
|
[
"brute force",
"greedy",
"implementation",
"math"
] | null | null |
Mike has a frog and a flower. His frog is named Xaniar and his flower is named Abol. Initially(at time 0), height of Xaniar is *h*1 and height of Abol is *h*2. Each second, Mike waters Abol and Xaniar.
So, if height of Xaniar is *h*1 and height of Abol is *h*2, after one second height of Xaniar will become and height of Abol will become where *x*1,<=*y*1,<=*x*2 and *y*2 are some integer numbers and denotes the remainder of *a* modulo *b*.
Mike is a competitive programmer fan. He wants to know the minimum time it takes until height of Xania is *a*1 and height of Abol is *a*2.
Mike has asked you for your help. Calculate the minimum time or say it will never happen.
|
The first line of input contains integer *m* (2<=≤<=*m*<=≤<=106).
The second line of input contains integers *h*1 and *a*1 (0<=≤<=*h*1,<=*a*1<=<<=*m*).
The third line of input contains integers *x*1 and *y*1 (0<=≤<=*x*1,<=*y*1<=<<=*m*).
The fourth line of input contains integers *h*2 and *a*2 (0<=≤<=*h*2,<=*a*2<=<<=*m*).
The fifth line of input contains integers *x*2 and *y*2 (0<=≤<=*x*2,<=*y*2<=<<=*m*).
It is guaranteed that *h*1<=≠<=*a*1 and *h*2<=≠<=*a*2.
|
Print the minimum number of seconds until Xaniar reaches height *a*1 and Abol reaches height *a*2 or print -1 otherwise.
|
[
"5\n4 2\n1 1\n0 1\n2 3\n",
"1023\n1 2\n1 0\n1 2\n1 1\n"
] |
[
"3\n",
"-1\n"
] |
In the first sample, heights sequences are following:
Xaniar: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/85da629b05969e7a8a6636d995b8fe7a0494e8f4.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Abol: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea95da14490864ae8b8bfcd4a8b7c02ad3a666b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 750
|
[
{
"input": "5\n4 2\n1 1\n0 1\n2 3",
"output": "3"
},
{
"input": "1023\n1 2\n1 0\n1 2\n1 1",
"output": "-1"
},
{
"input": "1023\n1 2\n1 2\n1 2\n1 2",
"output": "512"
},
{
"input": "2\n0 1\n1 0\n1 0\n0 1",
"output": "-1"
},
{
"input": "17\n15 12\n15 12\n12 14\n1 11",
"output": "-1"
},
{
"input": "29\n4 0\n1 1\n25 20\n16 0",
"output": "170"
},
{
"input": "91\n9 64\n75 32\n60 81\n35 46",
"output": "5"
},
{
"input": "91\n38 74\n66 10\n40 76\n17 13",
"output": "-1"
},
{
"input": "100\n11 20\n99 31\n60 44\n45 64",
"output": "3"
},
{
"input": "9999\n4879 6224\n63 7313\n4279 6583\n438 1627",
"output": "4"
},
{
"input": "10000\n8681 4319\n9740 5980\n24 137\n462 7971",
"output": "-1"
},
{
"input": "100000\n76036 94415\n34870 43365\n56647 26095\n88580 30995",
"output": "5"
},
{
"input": "100000\n90861 77058\n96282 30306\n45940 25601\n17117 48287",
"output": "-1"
},
{
"input": "1000000\n220036 846131\n698020 485511\n656298 242999\n766802 905433",
"output": "5297"
},
{
"input": "1000000\n536586 435396\n748740 34356\n135075 790803\n547356 534911",
"output": "9958"
},
{
"input": "1000000\n661647 690400\n864868 326304\n581148 452012\n327910 197092",
"output": "1021"
},
{
"input": "1000000\n233404 949288\n893747 751429\n692094 57207\n674400 583468",
"output": "-1"
},
{
"input": "1000000\n358465 242431\n977171 267570\n170871 616951\n711850 180241",
"output": "-1"
},
{
"input": "1000000\n707719 502871\n60595 816414\n649648 143990\n525107 66615",
"output": "-1"
},
{
"input": "999983\n192005 690428\n971158 641039\n974183 1882\n127579 312317",
"output": "470479"
},
{
"input": "999983\n420528 808305\n387096 497121\n596163 353326\n47177 758204",
"output": "548500"
},
{
"input": "999983\n651224 992349\n803017 393514\n258455 402487\n888310 244420",
"output": "126531"
},
{
"input": "999983\n151890 906425\n851007 9094\n696594 968184\n867017 157783",
"output": "-1"
},
{
"input": "999983\n380412 325756\n266945 907644\n318575 83081\n786616 603671",
"output": "-1"
},
{
"input": "999983\n570797 704759\n723177 763726\n978676 238272\n708387 89886",
"output": "-1"
},
{
"input": "999983\n408725 408721\n1 1\n378562 294895\n984270 0",
"output": "499981500166"
},
{
"input": "999983\n639420 639416\n1 1\n507684 954997\n466316 0",
"output": "499981500166"
},
{
"input": "999983\n867942 867939\n1 1\n963840 536667\n899441 0",
"output": "999964000320"
},
{
"input": "999961\n664221 931770\n530542 936292\n885122 515424\n868560 472225",
"output": "-1"
},
{
"input": "999961\n744938 661980\n845908 76370\n237399 381935\n418010 938769",
"output": "203332"
},
{
"input": "999961\n89288 89284\n1 1\n764559 727291\n999322 0",
"output": "999920001595"
},
{
"input": "1000000\n661703 661699\n1 1\n425192 823944\n854093 0",
"output": "-1"
},
{
"input": "100019\n98811 98807\n1 1\n91322 14787\n72253 0",
"output": "10003600319"
},
{
"input": "524288\n199980 199978\n1 1\n236260 325076\n81773 0",
"output": "-1"
},
{
"input": "524288\n47283 489031\n305624 183135\n141146 335913\n519614 150715",
"output": "19"
},
{
"input": "524288\n83398 33987\n158854 211502\n36433 18758\n218812 517001",
"output": "-1"
},
{
"input": "912488\n681639 518634\n168348 212018\n255428 4970\n31726 664998",
"output": "34838"
},
{
"input": "129081\n128454 36771\n116353 2940\n95311 22200\n579 118683",
"output": "68409"
},
{
"input": "129081\n45717 106320\n121816 69841\n5161 4872\n102076 100020",
"output": "-1"
},
{
"input": "4\n1 2\n1 1\n0 1\n2 0",
"output": "-1"
},
{
"input": "3\n1 0\n1 1\n1 2\n2 0",
"output": "5"
},
{
"input": "3\n0 2\n1 0\n2 0\n2 1",
"output": "-1"
},
{
"input": "2\n0 1\n0 1\n0 1\n0 1",
"output": "1"
},
{
"input": "2\n0 1\n1 0\n0 1\n1 0",
"output": "-1"
},
{
"input": "2\n0 1\n1 1\n0 1\n1 1",
"output": "1"
},
{
"input": "2\n0 1\n1 1\n0 1\n1 0",
"output": "-1"
},
{
"input": "2\n0 1\n1 0\n0 1\n1 1",
"output": "-1"
},
{
"input": "1000000\n1 0\n1 1\n1 0\n1 1",
"output": "999999"
},
{
"input": "1000000\n2 1\n1 1\n2 0\n1 2",
"output": "999999"
},
{
"input": "6\n1 2\n3 5\n0 2\n4 2",
"output": "1"
},
{
"input": "545\n26 40\n477 97\n454 394\n15 264",
"output": "90"
},
{
"input": "3\n1 0\n0 1\n0 2\n1 0",
"output": "-1"
},
{
"input": "1376\n1227 1349\n313 193\n1113 361\n1314 23",
"output": "338"
},
{
"input": "1376\n1322 1320\n1 1\n776 495\n38 0",
"output": "-1"
},
{
"input": "1376\n152 405\n1083 1328\n76 856\n49 629",
"output": "-1"
},
{
"input": "1392\n1060 796\n512 242\n1386 1346\n1310 1199",
"output": "-1"
},
{
"input": "100000\n5827 41281\n41285 70821\n99199 42807\n65667 94952",
"output": "13770"
},
{
"input": "100000\n51157 27741\n40564 90740\n45270 52367\n31585 92150",
"output": "-1"
},
{
"input": "100000\n70525 70522\n1 1\n89465 30265\n33279 0",
"output": "-1"
},
{
"input": "10\n1 6\n7 9\n1 4\n4 0",
"output": "1"
},
{
"input": "10\n9 6\n0 8\n3 0\n2 7",
"output": "-1"
},
{
"input": "10\n4 2\n1 1\n7 3\n9 0",
"output": "-1"
},
{
"input": "6\n5 1\n1 1\n3 1\n3 0",
"output": "-1"
},
{
"input": "999983\n3 1\n1 1\n8 1\n2 0",
"output": "499981500168"
},
{
"input": "18\n3 9\n3 0\n1 3\n3 0",
"output": "1"
},
{
"input": "18\n1 3\n3 0\n3 9\n3 0",
"output": "1"
},
{
"input": "16\n1 0\n2 0\n1 2\n2 0",
"output": "-1"
},
{
"input": "16\n8 0\n2 0\n1 4\n2 0",
"output": "2"
},
{
"input": "999983\n2 1\n2 0\n1 0\n1 1",
"output": "499982500152"
},
{
"input": "324\n2 54\n3 0\n27 108\n2 0",
"output": "-1"
},
{
"input": "999993\n499997 1\n2 3\n1 4\n1 1",
"output": "39325724721"
},
{
"input": "999983\n1 37827\n1 1\n2 192083\n3 0",
"output": "404303164556"
},
{
"input": "41222\n30759 26408\n31332 39118\n5026 25812\n1 9030",
"output": "58900566"
},
{
"input": "100007\n2 1\n2 0\n3 1\n1 1",
"output": "434330399"
},
{
"input": "8\n0 4\n4 4\n1 4\n2 0",
"output": "2"
}
] | 1,508,584,802
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 17
| 1,000
| 5,529,600
|
from math import gcd
f = lambda: map(int, input().split())
m = int(input())
def g(u, v):
h, a = u
x, y = v
s = 0
while h != a:
h = (h * x + y) % m
s += 1
if s > m: break
h = (a * x + y) % m
d = 1
while h != a:
h = (h * x + y) % m
d += 1
if d > m: break
return s, d
(u, x), (v, y) = g(f(), f()), g(f(), f())
if u > m or v > m: q = -1
elif u == v: q = u
elif x > m: q = -1 if u < v or (v - u) % y else u
elif y > m: q = -1 if v < u or (v - u) % x else v
elif (v - u) % gcd(x, y): q = -1
else:
while u != v:
if u < v: u += x
else: v += y
q = u
print(q)
|
Title: Mike and Frog
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a frog and a flower. His frog is named Xaniar and his flower is named Abol. Initially(at time 0), height of Xaniar is *h*1 and height of Abol is *h*2. Each second, Mike waters Abol and Xaniar.
So, if height of Xaniar is *h*1 and height of Abol is *h*2, after one second height of Xaniar will become and height of Abol will become where *x*1,<=*y*1,<=*x*2 and *y*2 are some integer numbers and denotes the remainder of *a* modulo *b*.
Mike is a competitive programmer fan. He wants to know the minimum time it takes until height of Xania is *a*1 and height of Abol is *a*2.
Mike has asked you for your help. Calculate the minimum time or say it will never happen.
Input Specification:
The first line of input contains integer *m* (2<=≤<=*m*<=≤<=106).
The second line of input contains integers *h*1 and *a*1 (0<=≤<=*h*1,<=*a*1<=<<=*m*).
The third line of input contains integers *x*1 and *y*1 (0<=≤<=*x*1,<=*y*1<=<<=*m*).
The fourth line of input contains integers *h*2 and *a*2 (0<=≤<=*h*2,<=*a*2<=<<=*m*).
The fifth line of input contains integers *x*2 and *y*2 (0<=≤<=*x*2,<=*y*2<=<<=*m*).
It is guaranteed that *h*1<=≠<=*a*1 and *h*2<=≠<=*a*2.
Output Specification:
Print the minimum number of seconds until Xaniar reaches height *a*1 and Abol reaches height *a*2 or print -1 otherwise.
Demo Input:
['5\n4 2\n1 1\n0 1\n2 3\n', '1023\n1 2\n1 0\n1 2\n1 1\n']
Demo Output:
['3\n', '-1\n']
Note:
In the first sample, heights sequences are following:
Xaniar: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/85da629b05969e7a8a6636d995b8fe7a0494e8f4.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Abol: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea95da14490864ae8b8bfcd4a8b7c02ad3a666b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
from math import gcd
f = lambda: map(int, input().split())
m = int(input())
def g(u, v):
h, a = u
x, y = v
s = 0
while h != a:
h = (h * x + y) % m
s += 1
if s > m: break
h = (a * x + y) % m
d = 1
while h != a:
h = (h * x + y) % m
d += 1
if d > m: break
return s, d
(u, x), (v, y) = g(f(), f()), g(f(), f())
if u > m or v > m: q = -1
elif u == v: q = u
elif x > m: q = -1 if u < v or (v - u) % y else u
elif y > m: q = -1 if v < u or (v - u) % x else v
elif (v - u) % gcd(x, y): q = -1
else:
while u != v:
if u < v: u += x
else: v += y
q = u
print(q)
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,624,591,518
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 124
| 0
|
str = list(input().split(' ')) #number of participants
m = int(str[0]) #number of participants
n = int(str[-1]) #k-th finisher score
print((m*n)//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
str = list(input().split(' ')) #number of participants
m = int(str[0]) #number of participants
n = int(str[-1]) #k-th finisher score
print((m*n)//2)
```
| 3.969
|
306
|
A
|
Candies
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
|
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
|
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
|
[
"12 3\n",
"15 4\n",
"18 7\n"
] |
[
"4 4 4 ",
"3 4 4 4 ",
"2 2 2 3 3 3 3 "
] |
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
| 500
|
[
{
"input": "12 4",
"output": "3 3 3 3 "
},
{
"input": "15 6",
"output": "2 2 2 3 3 3 "
},
{
"input": "18 8",
"output": "2 2 2 2 2 2 3 3 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "100 1",
"output": "100 "
},
{
"input": "100 100",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100 10",
"output": "10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "64 8",
"output": "8 8 8 8 8 8 8 8 "
},
{
"input": "81 27",
"output": "3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100 99",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 "
},
{
"input": "100 51",
"output": "1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100 49",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 "
},
{
"input": "100 13",
"output": "7 7 7 7 8 8 8 8 8 8 8 8 8 "
},
{
"input": "97 11",
"output": "8 8 9 9 9 9 9 9 9 9 9 "
},
{
"input": "19 3",
"output": "6 6 7 "
},
{
"input": "99 10",
"output": "9 10 10 10 10 10 10 10 10 10 "
},
{
"input": "88 5",
"output": "17 17 18 18 18 "
},
{
"input": "87 34",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "97 55",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
}
] | 1,638,896,164
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 154
| 0
|
a,b=map(int,input().split())
l=[a//b]*b
a%=b
for i in range(a):
l[i]+=1
for i in l:
print(i, end=" ")
|
Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
Input Specification:
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
Output Specification:
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
Demo Input:
['12 3\n', '15 4\n', '18 7\n']
Demo Output:
['4 4 4 ', '3 4 4 4 ', '2 2 2 3 3 3 3 ']
Note:
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
|
```python
a,b=map(int,input().split())
l=[a//b]*b
a%=b
for i in range(a):
l[i]+=1
for i in l:
print(i, end=" ")
```
| 3
|
|
907
|
A
|
Masha and Bears
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation"
] | null | null |
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
|
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
|
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
|
[
"50 30 10 10\n",
"100 50 10 21\n"
] |
[
"50\n30\n10\n",
"-1\n"
] |
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
| 500
|
[
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"input": "100 99 98 100",
"output": "-1"
},
{
"input": "100 40 30 40",
"output": "-1"
},
{
"input": "100 50 19 25",
"output": "100\n51\n25"
},
{
"input": "100 50 19 30",
"output": "100\n61\n30"
},
{
"input": "49 48 25 49",
"output": "-1"
},
{
"input": "48 47 23 46",
"output": "94\n93\n46"
},
{
"input": "37 23 16 20",
"output": "42\n41\n20"
},
{
"input": "98 2 1 1",
"output": "98\n3\n1"
},
{
"input": "99 22 13 14",
"output": "99\n29\n14"
},
{
"input": "97 95 3 2",
"output": "97\n95\n3"
},
{
"input": "27 3 2 3",
"output": "-1"
},
{
"input": "13 7 6 2",
"output": "-1"
},
{
"input": "19 17 11 6",
"output": "19\n17\n11"
},
{
"input": "15 6 4 5",
"output": "15\n11\n5"
},
{
"input": "21 3 1 3",
"output": "-1"
},
{
"input": "100 99 98 97",
"output": "196\n195\n98"
},
{
"input": "99 32 13 5",
"output": "-1"
},
{
"input": "5 4 3 1",
"output": "-1"
},
{
"input": "50 20 4 8",
"output": "50\n20\n8"
},
{
"input": "60 30 10 1",
"output": "-1"
},
{
"input": "100 50 10 1",
"output": "-1"
},
{
"input": "40 30 20 10",
"output": "40\n30\n20"
},
{
"input": "50 49 5 10",
"output": "50\n49\n10"
},
{
"input": "99 98 50 97",
"output": "196\n195\n97"
},
{
"input": "50 30 10 5",
"output": "50\n30\n10"
},
{
"input": "4 3 2 1",
"output": "4\n3\n2"
},
{
"input": "100 50 3 1",
"output": "-1"
}
] | 1,689,636,898
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1689636898.6921825")# 1689636898.6922042
|
Title: Masha and Bears
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
|
```python
print("_RANDOM_GUESS_1689636898.6921825")# 1689636898.6922042
```
| 0
|
|
258
|
A
|
Little Elephant and Bits
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
|
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
|
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
|
[
"101\n",
"110010\n"
] |
[
"11\n",
"11010\n"
] |
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
| 500
|
[
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,597,818,359
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 218
| 409,600
|
def littleElephant():
s=input()
if s[0]=='0':
s=s[:i]+s[i+1:]
print(s)
return
for i in range(len(s)):
if s[i]=='0':
s=s[:i]+s[i+1:]
print(s)
return
s=s[:0]+s[1:]
print(s)
return
littleElephant()
|
Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
|
```python
def littleElephant():
s=input()
if s[0]=='0':
s=s[:i]+s[i+1:]
print(s)
return
for i in range(len(s)):
if s[i]=='0':
s=s[:i]+s[i+1:]
print(s)
return
s=s[:0]+s[1:]
print(s)
return
littleElephant()
```
| 3
|
|
4
|
A
|
Watermelon
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Watermelon
|
1
|
64
|
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
|
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
|
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
|
[
"8\n"
] |
[
"YES\n"
] |
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
| 0
|
[
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,692,612,528
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
def is_even(number):
return number % 2 == 0
def can_divide_watermelon(weight):
if is_even(weight):
return "YES"
else:
return "NO"
def main():
weight = int(input())
print(can_divide_watermelon(weight))
if __name__ == "__main__":
main()
|
Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
|
```python
def is_even(number):
return number % 2 == 0
def can_divide_watermelon(weight):
if is_even(weight):
return "YES"
else:
return "NO"
def main():
weight = int(input())
print(can_divide_watermelon(weight))
if __name__ == "__main__":
main()
```
| 0
|
112
|
A
|
Petya and Strings
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Petya and Strings
|
2
|
256
|
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
|
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
|
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
|
[
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] |
[
"0\n",
"-1\n",
"1\n"
] |
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order
| 500
|
[
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp",
"output": "0"
},
{
"input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH",
"output": "-1"
},
{
"input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt",
"output": "1"
},
{
"input": "nCeNVIzHqPceNhjHeHvJvgBsNFiXBATRrjSTXJzhLMDMxiJztphxBRlDlqwDFImWeEPkggZCXSRwelOdpNrYnTepiOqpvkr\nHJbjJFtlvNxIbkKlxQUwmZHJFVNMwPAPDRslIoXISBYHHfymyIaQHLgECPxAmqnOCizwXnIUBRmpYUBVPenoUKhCobKdOjL",
"output": "1"
},
{
"input": "ttXjenUAlfixytHEOrPkgXmkKTSGYuyVXGIHYmWWYGlBYpHkujueqBSgjLguSgiMGJWATIGEUjjAjKXdMiVbHozZUmqQtFrT\nJziDBFBDmDJCcGqFsQwDFBYdOidLxxhBCtScznnDgnsiStlWFnEXQrJxqTXKPxZyIGfLIToETKWZBPUIBmLeImrlSBWCkTNo",
"output": "1"
},
{
"input": "AjQhPqSVhwQQjcgCycjKorWBgFCRuQBwgdVuAPSMJAvTyxGVuFHjfJzkKfsmfhFbKqFrFIohSZBbpjgEHebezmVlGLTPSCTMf\nXhxWuSnMmKFrCUOwkTUmvKAfbTbHWzzOTzxJatLLCdlGnHVaBUnxDlsqpvjLHMThOPAFBggVKDyKBrZAmjnjrhHlrnSkyzBja",
"output": "-1"
},
{
"input": "HCIgYtnqcMyjVngziNflxKHtdTmcRJhzMAjFAsNdWXFJYEhiTzsQUtFNkAbdrFBRmvLirkuirqTDvIpEfyiIqkrwsjvpPWTEdI\nErqiiWKsmIjyZuzgTlTqxYZwlrpvRyaVhRTOYUqtPMVGGtWOkDCOOQRKrkkRzPftyQCkYkzKkzTPqqXmeZhvvEEiEhkdOmoMvy",
"output": "1"
},
{
"input": "mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR",
"output": "1"
},
{
"input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd",
"output": "1"
},
{
"input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI",
"output": "-1"
},
{
"input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC",
"output": "-1"
},
{
"input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl",
"output": "-1"
},
{
"input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo",
"output": "-1"
},
{
"input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV",
"output": "-1"
},
{
"input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ",
"output": "-1"
},
{
"input": "hJDANKUNBisOOINDsTixJmYgHNogtpwswwcvVMptfGwIjvqgwTYFcqTdyAqaqlnhOCMtsnWXQqtjFwQlEcBtMFAtSqnqthVb\nrNquIcjNWESjpPVWmzUJFrelpUZeGDmSvCurCqVmKHKVAAPkaHksniOlzjiKYIJtvbuQWZRufMebpTFPqyxIWWjfPaWYiNlK",
"output": "-1"
},
{
"input": "ycLoapxsfsDTHMSfAAPIUpiEhQKUIXUcXEiopMBuuZLHtfPpLmCHwNMNQUwsEXxCEmKHTBSnKhtQhGWUvppUFZUgSpbeChX\ndCZhgVXofkGousCzObxZSJwXcHIaqUDSCPKzXntcVmPxtNcXmVcjsetZYxedmgQzXTZHMvzjoaXCMKsncGciSDqQWIIRlys",
"output": "1"
},
{
"input": "nvUbnrywIePXcoukIhwTfUVcHUEgXcsMyNQhmMlTltZiCooyZiIKRIGVHMCnTKgzXXIuvoNDEZswKoACOBGSyVNqTNQqMhAG\nplxuGSsyyJjdvpddrSebOARSAYcZKEaKjqbCwvjhNykuaECoQVHTVFMKXwvrQXRaqXsHsBaGVhCxGRxNyGUbMlxOarMZNXxy",
"output": "-1"
},
{
"input": "EncmXtAblQzcVRzMQqdDqXfAhXbtJKQwZVWyHoWUckohnZqfoCmNJDzexFgFJYrwNHGgzCJTzQQFnxGlhmvQTpicTkEeVICKac\nNIUNZoMLFMyAjVgQLITELJSodIXcGSDWfhFypRoGYuogJpnqGTotWxVqpvBHjFOWcDRDtARsaHarHaOkeNWEHGTaGOFCOFEwvK",
"output": "-1"
},
{
"input": "UG\nak",
"output": "1"
},
{
"input": "JZR\nVae",
"output": "-1"
},
{
"input": "a\nZ",
"output": "-1"
},
{
"input": "rk\nkv",
"output": "1"
},
{
"input": "RvuT\nbJzE",
"output": "1"
},
{
"input": "PPS\nydq",
"output": "-1"
},
{
"input": "q\nq",
"output": "0"
},
{
"input": "peOw\nIgSJ",
"output": "1"
},
{
"input": "PyK\noKN",
"output": "1"
},
{
"input": "O\ni",
"output": "1"
},
{
"input": "NmGY\npDlP",
"output": "-1"
},
{
"input": "nG\nZf",
"output": "-1"
},
{
"input": "m\na",
"output": "1"
},
{
"input": "MWyB\nWZEV",
"output": "-1"
},
{
"input": "Gre\nfxc",
"output": "1"
},
{
"input": "Ooq\nwap",
"output": "-1"
},
{
"input": "XId\nlbB",
"output": "1"
},
{
"input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF",
"output": "0"
},
{
"input": "DQBdtSEDtFGiNRUeJNbOIfDZnsryUlzJHGTXGFXnwsVyxNtLgmklmFvRCzYETBVdmkpJJIvIOkMDgCFHZOTODiYrkwXd\nDQbDtsEdTFginRUEJNBOIfdZnsryulZJHGtxGFxnwSvYxnTLgmKlmFVRCzyEtBVdmKpJjiVioKMDgCFhzoTODiYrKwXD",
"output": "0"
},
{
"input": "tYWRijFQSzHBpCjUzqBtNvBKyzZRnIdWEuyqnORBQTLyOQglIGfYJIRjuxnbLvkqZakNqPiGDvgpWYkfxYNXsdoKXZtRkSasfa\nTYwRiJfqsZHBPcJuZQBTnVbkyZZRnidwEuYQnorbQTLYOqGligFyjirJUxnblVKqZaknQpigDVGPwyKfxyNXSDoKxztRKSaSFA",
"output": "0"
},
{
"input": "KhScXYiErQIUtmVhNTCXSLAviefIeHIIdiGhsYnPkSBaDTvMkyanfMLBOvDWgRybLtDqvXVdVjccNunDyijhhZEAKBrdz\nkHsCXyiErqIuTMVHNTCxSLaViEFIEhIIDiGHsYNpKsBAdTvMKyANFMLBovdwGRYbLtdQVxvDVJCcNUndYiJHhzeakBrdZ",
"output": "0"
},
{
"input": "cpPQMpjRQJKQVXjWDYECXbagSmNcVfOuBWNZxihdERraVuiOpSVDCPgTGuSQALNoVjySceHcKXwOEpSzXrEqWwwrYeppNiWhDVg\nCPPqmPjRqJkQvxJwdyECXBAGsMNcVfOuBWNzxIhderRavUiOpSvDCpGTgusqAlNovjyScEhCKXwoePSZxrEQwWwryEPPniWHDvG",
"output": "0"
},
{
"input": "SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ",
"output": "0"
},
{
"input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ",
"output": "0"
},
{
"input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi",
"output": "0"
},
{
"input": "abac\nadaa",
"output": "-1"
},
{
"input": "Bbc\nabc",
"output": "1"
},
{
"input": "aaaba\naaaab",
"output": "1"
}
] | 1,695,803,849
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 53
| 124
| 28,774,400
|
l1=str.lower(input())
l2=str.lower(input())
if l1==l2:
print(0)
elif l1>l2:
print(1)
else:
print(-1)
|
Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order
|
```python
l1=str.lower(input())
l2=str.lower(input())
if l1==l2:
print(0)
elif l1>l2:
print(1)
else:
print(-1)
```
| 3.915404
|
118
|
A
|
String Task
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
|
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
|
Print the resulting string. It is guaranteed that this string is not empty.
|
[
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] |
[
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] |
none
| 500
|
[
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": ".g.g.d.v.q"
},
{
"input": "pumesz",
"output": ".p.m.s.z"
},
{
"input": "g",
"output": ".g"
},
{
"input": "zjuotps",
"output": ".z.j.t.p.s"
},
{
"input": "jzbwuehe",
"output": ".j.z.b.w.h"
},
{
"input": "tnkgwuugu",
"output": ".t.n.k.g.w.g"
},
{
"input": "kincenvizh",
"output": ".k.n.c.n.v.z.h"
},
{
"input": "xattxjenual",
"output": ".x.t.t.x.j.n.l"
},
{
"input": "ktajqhpqsvhw",
"output": ".k.t.j.q.h.p.q.s.v.h.w"
},
{
"input": "xnhcigytnqcmy",
"output": ".x.n.h.c.g.t.n.q.c.m"
},
{
"input": "jfmtbejyilxcec",
"output": ".j.f.m.t.b.j.l.x.c.c"
},
{
"input": "D",
"output": ".d"
},
{
"input": "ab",
"output": ".b"
},
{
"input": "Ab",
"output": ".b"
},
{
"input": "aB",
"output": ".b"
},
{
"input": "AB",
"output": ".b"
},
{
"input": "ba",
"output": ".b"
},
{
"input": "bA",
"output": ".b"
},
{
"input": "Ba",
"output": ".b"
},
{
"input": "BA",
"output": ".b"
},
{
"input": "aab",
"output": ".b"
},
{
"input": "baa",
"output": ".b"
},
{
"input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM",
"output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m"
},
{
"input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN",
"output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n"
},
{
"input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ",
"output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j"
},
{
"input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea",
"output": ".w.z.t.c.v.v.v.v.z.v"
},
{
"input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl",
"output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l"
},
{
"input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg",
"output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g"
},
{
"input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ",
"output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q"
},
{
"input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg",
"output": ".l.t.g.m.g"
},
{
"input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy",
"output": ".k.z.d.x.f"
},
{
"input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ",
"output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q"
},
{
"input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY",
"output": ".m.r.c.g.p"
},
{
"input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx",
"output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x"
},
{
"input": "YB",
"output": ".b"
},
{
"input": "fly",
"output": ".f.l"
},
{
"input": "YyyYYYyyYxdwdawdDAWDdaddYYYY",
"output": ".x.d.w.d.w.d.d.w.d.d.d.d"
}
] | 1,699,083,789
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 14
| 92
| 0
|
str = input()
strArray = []
for i in range(len(str)):
if str[i].lower() == "a" or str[i].lower() == 'e' or str[i].lower() == 'i' or str[i].lower() == 'o' or str[i].lower() == 'u':
continue
else:
strArray.append('.')
strArray.append(str[i].lower())
print("".join(strArray))
|
Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none
|
```python
str = input()
strArray = []
for i in range(len(str)):
if str[i].lower() == "a" or str[i].lower() == 'e' or str[i].lower() == 'i' or str[i].lower() == 'o' or str[i].lower() == 'u':
continue
else:
strArray.append('.')
strArray.append(str[i].lower())
print("".join(strArray))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types:
- replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*.
Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task.
|
The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
|
Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*.
|
[
"47\n74\n",
"774\n744\n",
"777\n444\n"
] |
[
"1\n",
"1\n",
"3\n"
] |
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
| 0
|
[
{
"input": "47\n74",
"output": "1"
},
{
"input": "774\n744",
"output": "1"
},
{
"input": "777\n444",
"output": "3"
},
{
"input": "74747474\n77777777",
"output": "4"
},
{
"input": "444444444444\n777777777777",
"output": "12"
},
{
"input": "4744744447774474447474774\n4477774777444444444777447",
"output": "8"
},
{
"input": "7\n4",
"output": "1"
},
{
"input": "4\n7",
"output": "1"
},
{
"input": "7777777777\n7777777774",
"output": "1"
},
{
"input": "47777777777\n77777777774",
"output": "1"
},
{
"input": "47747477747744447774774444444777444747474747777774\n44777444774477447777444774477777477774444477447777",
"output": "14"
},
{
"input": "44447777447744444777777747477444777444447744444\n47444747774774744474747744447744477747777777447",
"output": "13"
},
{
"input": "4447744774744774744747744774474474444447477477444747477444\n7477477444744774744744774774744474744447744774744477744477",
"output": "14"
},
{
"input": "44747744777777444\n47774747747744777",
"output": "6"
},
{
"input": "44447774444474477747774774477777474774744744477444447777477477744747477774744444744777777777747777477447744774744444747477744744\n77777474477477747774777777474474477444474777477747747777477747747744474474747774747747444777474444744744444477477777747744747477",
"output": "37"
},
{
"input": "774774747744474477447477777447477747477474777477744744747444774474477477747474477447774444774744777\n744477444747477447477777774477447444447747477747477747774477474447474477477474444777444444447474747",
"output": "27"
},
{
"input": "4747447477\n4747444744",
"output": "3"
},
{
"input": "47744447444\n74477447744",
"output": "4"
},
{
"input": "447444777744\n777747744477",
"output": "6"
},
{
"input": "474777477774444\n774747777774477",
"output": "4"
},
{
"input": "47744474447747744777777447\n44744747477474777744777477",
"output": "7"
},
{
"input": "77447447444777777744744747744747774747477774777774447447777474477477774774777\n74777777444744447447474474477747747444444447447774444444747777444747474777447",
"output": "28"
},
{
"input": "7\n7",
"output": "0"
},
{
"input": "444\n444",
"output": "0"
},
{
"input": "77747\n47474",
"output": "3"
}
] | 1,644,981,094
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 124
| 307,200
|
s = input()
t = input()
count_7, count_4, count = 0, 0, 0
for i in range(len(s)):
if s[i] == t[i]:
continue
if s[i] == '7':
count_7 += 1
else:
count_4 += 1
print(count_7 if count_7>=count_4 else count_4)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types:
- replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*.
Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task.
Input Specification:
The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output Specification:
Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*.
Demo Input:
['47\n74\n', '774\n744\n', '777\n444\n']
Demo Output:
['1\n', '1\n', '3\n']
Note:
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
|
```python
s = input()
t = input()
count_7, count_4, count = 0, 0, 0
for i in range(len(s)):
if s[i] == t[i]:
continue
if s[i] == '7':
count_7 += 1
else:
count_4 += 1
print(count_7 if count_7>=count_4 else count_4)
```
| 3
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,652,299,870
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
st = input() # hello
l = list(st)
for i in l :
if l.count(i) > 2 :
l.remove(i)
s = "".join(l)
if "hello" in s :
print("YES")
else:
print("NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
st = input() # hello
l = list(st)
for i in l :
if l.count(i) > 2 :
l.remove(i)
s = "".join(l)
if "hello" in s :
print("YES")
else:
print("NO")
```
| 0
|
954
|
A
|
Diagonal Walking
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.
In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.
Your problem is to print the minimum possible length of the sequence of moves after the replacements.
|
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
|
Print the minimum possible length of the sequence of moves after all replacements are done.
|
[
"5\nRUURU\n",
"17\nUUURRRRRUUURURUUU\n"
] |
[
"3\n",
"13\n"
] |
In the first test the shortened sequence of moves may be DUD (its length is 3).
In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
| 0
|
[
{
"input": "5\nRUURU",
"output": "3"
},
{
"input": "17\nUUURRRRRUUURURUUU",
"output": "13"
},
{
"input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU",
"output": "100"
},
{
"input": "100\nRRURRUUUURURRRURRRRURRRRRRURRUURRRUUURUURURRURUURUURRUURUURRURURUUUUURUUUUUURRUUURRRURRURRRUURRUUUUR",
"output": "67"
},
{
"input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUURUUUUUUUUUUUUUUUUUUUUU",
"output": "99"
},
{
"input": "3\nRUR",
"output": "2"
},
{
"input": "1\nR",
"output": "1"
},
{
"input": "5\nRURUU",
"output": "3"
},
{
"input": "1\nU",
"output": "1"
},
{
"input": "2\nUR",
"output": "1"
},
{
"input": "23\nUUUUUUUUUUUUUUUUUUUUUUU",
"output": "23"
}
] | 1,638,185,993
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n = int(input())
s=input()
c = 6
i = 0
while i<n-1:
if x[i] l = s(i + 1):
c = c+1
i = 1+2
else:
i = 1+1
t = n-c
print (t)
|
Title: Diagonal Walking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.
In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.
Your problem is to print the minimum possible length of the sequence of moves after the replacements.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
Output Specification:
Print the minimum possible length of the sequence of moves after all replacements are done.
Demo Input:
['5\nRUURU\n', '17\nUUURRRRRUUURURUUU\n']
Demo Output:
['3\n', '13\n']
Note:
In the first test the shortened sequence of moves may be DUD (its length is 3).
In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
|
```python
n = int(input())
s=input()
c = 6
i = 0
while i<n-1:
if x[i] l = s(i + 1):
c = c+1
i = 1+2
else:
i = 1+1
t = n-c
print (t)
```
| -1
|
|
299
|
A
|
Ksusha and Array
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory",
"sortings"
] | null | null |
Ksusha is a beginner coder. Today she starts studying arrays. She has array *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers.
Her university teacher gave her a task. Find such number in the array, that all array elements are divisible by it. Help her and find the number!
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105), showing how many numbers the array has. The next line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the array elements.
|
Print a single integer — the number from the array, such that all array elements are divisible by it. If such number doesn't exist, print -1.
If there are multiple answers, you are allowed to print any of them.
|
[
"3\n2 2 4\n",
"5\n2 1 3 1 6\n",
"3\n2 3 5\n"
] |
[
"2\n",
"1\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "3\n2 2 4",
"output": "2"
},
{
"input": "5\n2 1 3 1 6",
"output": "1"
},
{
"input": "3\n2 3 5",
"output": "-1"
},
{
"input": "1\n331358794",
"output": "331358794"
},
{
"input": "5\n506904227 214303304 136194869 838256937 183952885",
"output": "-1"
},
{
"input": "2\n500000000 1000000000",
"output": "500000000"
},
{
"input": "2\n4 6",
"output": "-1"
},
{
"input": "5\n10 8 6 4 2",
"output": "2"
},
{
"input": "2\n6 10",
"output": "-1"
},
{
"input": "1\n1000000000",
"output": "1000000000"
},
{
"input": "2\n6 8",
"output": "-1"
},
{
"input": "5\n2 2 2 2 1000000000",
"output": "2"
},
{
"input": "2\n6 4",
"output": "-1"
}
] | 1,675,302,628
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 7,884,800
|
n=int(input())
x=list(map(int,input().split()))
c=0
for i in x:
if i%min(x)==0: c+=1
else: break
if c==len(x): print(min(x))
else: print(-1)
|
Title: Ksusha and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksusha is a beginner coder. Today she starts studying arrays. She has array *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers.
Her university teacher gave her a task. Find such number in the array, that all array elements are divisible by it. Help her and find the number!
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105), showing how many numbers the array has. The next line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the array elements.
Output Specification:
Print a single integer — the number from the array, such that all array elements are divisible by it. If such number doesn't exist, print -1.
If there are multiple answers, you are allowed to print any of them.
Demo Input:
['3\n2 2 4\n', '5\n2 1 3 1 6\n', '3\n2 3 5\n']
Demo Output:
['2\n', '1\n', '-1\n']
Note:
none
|
```python
n=int(input())
x=list(map(int,input().split()))
c=0
for i in x:
if i%min(x)==0: c+=1
else: break
if c==len(x): print(min(x))
else: print(-1)
```
| 0
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,678,170,271
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 2,764,800
|
k = input()
l = input()
i = 1
while True:
if (k*i)%10 == l:
break
i+=1
print(i)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
k = input()
l = input()
i = 1
while True:
if (k*i)%10 == l:
break
i+=1
print(i)
```
| -1
|
|
148
|
A
|
Insomnia cure
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
|
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
|
Output the number of damaged dragons.
|
[
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] |
[
"12\n",
"17\n"
] |
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
| 1,000
|
[
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"input": "8\n4\n1\n10\n59392",
"output": "59392"
},
{
"input": "4\n1\n8\n7\n44835",
"output": "44835"
},
{
"input": "6\n1\n7\n2\n62982",
"output": "62982"
},
{
"input": "2\n7\n4\n9\n56937",
"output": "35246"
},
{
"input": "2\n9\n8\n1\n75083",
"output": "75083"
},
{
"input": "8\n7\n7\n6\n69038",
"output": "24656"
},
{
"input": "4\n4\n2\n3\n54481",
"output": "36320"
},
{
"input": "6\n4\n9\n8\n72628",
"output": "28244"
},
{
"input": "9\n7\n8\n10\n42357",
"output": "16540"
},
{
"input": "5\n6\n4\n3\n60504",
"output": "36302"
},
{
"input": "7\n2\n3\n8\n21754",
"output": "15539"
},
{
"input": "1\n2\n10\n4\n39901",
"output": "39901"
},
{
"input": "3\n4\n7\n1\n58048",
"output": "58048"
},
{
"input": "9\n10\n4\n6\n52003",
"output": "21956"
},
{
"input": "5\n10\n9\n3\n70149",
"output": "32736"
},
{
"input": "5\n5\n5\n10\n55592",
"output": "11118"
},
{
"input": "1\n5\n2\n6\n49547",
"output": "49547"
},
{
"input": "3\n7\n7\n7\n84046",
"output": "36019"
},
{
"input": "10\n2\n1\n8\n63537",
"output": "63537"
},
{
"input": "7\n2\n10\n5\n81684",
"output": "53678"
},
{
"input": "7\n1\n6\n8\n99831",
"output": "99831"
},
{
"input": "3\n9\n3\n3\n61082",
"output": "20360"
},
{
"input": "5\n5\n8\n9\n79228",
"output": "29931"
},
{
"input": "1\n5\n5\n6\n89535",
"output": "89535"
},
{
"input": "5\n5\n2\n1\n7682",
"output": "7682"
},
{
"input": "7\n4\n8\n8\n25829",
"output": "9224"
},
{
"input": "3\n4\n3\n3\n87079",
"output": "43539"
},
{
"input": "5\n8\n4\n9\n5226",
"output": "2438"
},
{
"input": "4\n9\n6\n6\n1070",
"output": "415"
},
{
"input": "10\n10\n10\n10\n100000",
"output": "10000"
},
{
"input": "1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "10\n10\n10\n10\n1",
"output": "0"
},
{
"input": "10\n10\n1\n10\n10",
"output": "10"
},
{
"input": "10\n10\n1\n10\n100",
"output": "100"
},
{
"input": "2\n2\n2\n2\n1",
"output": "0"
}
] | 1,680,634,348
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 280
| 5,017,600
|
a = [int(input()) for _ in range(5)]
s = set()
for b in a[:-1]:
c = b
while c <= a[-1]:
s.add(c)
c += b
print(len(s))
|
Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
|
```python
a = [int(input()) for _ in range(5)]
s = set()
for b in a[:-1]:
c = b
while c <= a[-1]:
s.add(c)
c += b
print(len(s))
```
| 3
|
|
939
|
A
|
Love Triangle
|
PROGRAMMING
| 800
|
[
"graphs"
] | null | null |
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
|
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
|
[
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] |
[
"YES\n",
"NO\n"
] |
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles.
| 500
|
[
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"input": "100\n50 40 60 87 39 58 44 84 46 68 16 57 77 87 92 95 42 31 74 15 36 84 30 3 47 15 87 90 76 66 6 63 74 19 40 49 6 84 41 9 77 34 7 12 11 73 58 24 81 14 81 29 65 100 1 85 64 32 38 4 54 67 32 81 80 7 100 71 29 80 4 52 47 7 78 56 52 75 81 37 16 41 27 28 58 60 62 47 29 40 37 14 59 91 12 54 25 58 12 43",
"output": "NO"
},
{
"input": "100\n25 6 46 37 87 99 70 31 46 12 94 40 87 56 28 8 94 39 13 12 67 13 71 39 83 48 40 14 62 41 16 71 20 41 83 41 68 98 23 82 62 83 62 35 49 22 31 21 66 98 54 39 34 52 11 28 47 89 25 44 68 36 91 46 82 86 88 48 27 93 7 9 53 36 16 100 84 84 44 25 58 66 16 46 72 21 91 78 4 17 44 17 47 67 93 89 75 44 56 50",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "YES"
},
{
"input": "5\n2 1 4 5 3",
"output": "YES"
},
{
"input": "5\n5 4 5 5 2",
"output": "YES"
}
] | 1,695,568,858
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
n = int(input())
arr = list(map(int, input().split()))
f = True
for i in range (len(arr)):
if arr[arr[i]-1] == i+1:
f = False
#print(i, f)
print("YES") if f else print("NO")
|
Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles.
|
```python
n = int(input())
arr = list(map(int, input().split()))
f = True
for i in range (len(arr)):
if arr[arr[i]-1] == i+1:
f = False
#print(i, f)
print("YES") if f else print("NO")
```
| 0
|
|
313
|
A
|
Ilya and Bank Account
|
PROGRAMMING
| 900
|
[
"implementation",
"number theory"
] | null | null |
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
|
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
|
In a single line print an integer — the maximum state of the bank account that Ilya can get.
|
[
"2230\n",
"-10\n",
"-100003\n"
] |
[
"2230\n",
"0\n",
"-10000\n"
] |
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
| 500
|
[
{
"input": "2230",
"output": "2230"
},
{
"input": "-10",
"output": "0"
},
{
"input": "-100003",
"output": "-10000"
},
{
"input": "544883178",
"output": "544883178"
},
{
"input": "-847251738",
"output": "-84725173"
},
{
"input": "423654797",
"output": "423654797"
},
{
"input": "-623563697",
"output": "-62356367"
},
{
"input": "645894116",
"output": "645894116"
},
{
"input": "-384381709",
"output": "-38438170"
},
{
"input": "437587210",
"output": "437587210"
},
{
"input": "-297534606",
"output": "-29753460"
},
{
"input": "891773002",
"output": "891773002"
},
{
"input": "-56712976",
"output": "-5671296"
},
{
"input": "963662765",
"output": "963662765"
},
{
"input": "-272656295",
"output": "-27265625"
},
{
"input": "383441522",
"output": "383441522"
},
{
"input": "-477665112",
"output": "-47766511"
},
{
"input": "791725034",
"output": "791725034"
},
{
"input": "-812168727",
"output": "-81216872"
},
{
"input": "528894922",
"output": "528894922"
},
{
"input": "-479977172",
"output": "-47997712"
},
{
"input": "568044564",
"output": "568044564"
},
{
"input": "-392784794",
"output": "-39278474"
},
{
"input": "925596634",
"output": "925596634"
},
{
"input": "-836078769",
"output": "-83607876"
},
{
"input": "71036059",
"output": "71036059"
},
{
"input": "-337396162",
"output": "-33739612"
},
{
"input": "87129297",
"output": "87129297"
},
{
"input": "-648171877",
"output": "-64817187"
},
{
"input": "20218400",
"output": "20218400"
},
{
"input": "10",
"output": "10"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "-1000000000",
"output": "-100000000"
},
{
"input": "-102",
"output": "-10"
},
{
"input": "-120",
"output": "-10"
},
{
"input": "-20",
"output": "0"
},
{
"input": "-15",
"output": "-1"
},
{
"input": "-5575533",
"output": "-557553"
},
{
"input": "-50",
"output": "0"
},
{
"input": "-31",
"output": "-1"
},
{
"input": "-55",
"output": "-5"
},
{
"input": "-81",
"output": "-1"
},
{
"input": "-99",
"output": "-9"
},
{
"input": "-23",
"output": "-2"
},
{
"input": "-41",
"output": "-1"
},
{
"input": "-24",
"output": "-2"
},
{
"input": "46",
"output": "46"
}
] | 1,698,090,892
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
n = int(input())
if n >= 0:
print(n)
else:
option1 = n // 10
option2 = (n // 100) * 10 + n % 10
print(max(option1, option2))
|
Title: Ilya and Bank Account
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
Input Specification:
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
Output Specification:
In a single line print an integer — the maximum state of the bank account that Ilya can get.
Demo Input:
['2230\n', '-10\n', '-100003\n']
Demo Output:
['2230\n', '0\n', '-10000\n']
Note:
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
|
```python
n = int(input())
if n >= 0:
print(n)
else:
option1 = n // 10
option2 = (n // 100) * 10 + n % 10
print(max(option1, option2))
```
| 0
|
|
14
|
A
|
Letter
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Letter
|
1
|
64
|
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
|
The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
|
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
|
[
"6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n",
"3 3\n***\n*.*\n***\n"
] |
[
"***\n*..\n***\n*..\n***\n",
"***\n*.*\n***\n"
] |
none
| 0
|
[
{
"input": "6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..",
"output": "***\n*..\n***\n*..\n***"
},
{
"input": "3 3\n***\n*.*\n***",
"output": "***\n*.*\n***"
},
{
"input": "1 1\n*",
"output": "*"
},
{
"input": "2 1\n*\n*",
"output": "*\n*"
},
{
"input": "5 1\n.\n*\n.\n.\n.",
"output": "*"
},
{
"input": "1 6\n*****.",
"output": "*****"
},
{
"input": "2 2\n..\n*.",
"output": "*"
},
{
"input": "8 2\n..\n..\n..\n..\n..\n*.\n..\n..",
"output": "*"
},
{
"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": "*\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n*"
},
{
"input": "2 1\n*\n.",
"output": "*"
},
{
"input": "5 1\n*\n*\n*\n*\n*",
"output": "*\n*\n*\n*\n*"
},
{
"input": "2 2\n*.\n..",
"output": "*"
},
{
"input": "3 4\n...*\n*...\n..*.",
"output": "...*\n*...\n..*."
},
{
"input": "8 2\n**\n**\n**\n**\n**\n**\n**\n**",
"output": "**\n**\n**\n**\n**\n**\n**\n**"
},
{
"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": "*\n*\n*\n.\n.\n*\n.\n*\n*\n*\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n.\n.\n.\n.\n*\n*\n.\n*\n*\n.\n.\n.\n.\n*\n*"
}
] | 1,684,616,920
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
n,m=map(int,input().split(" "))
s,e,c ,s_1,e_1=100,0,0,-1,0 # start ,end , count ,start_rectangle,end_rectangle
l=[]
for i in range(n):
temp=input()
if ("*" in temp ):
if (s_1==-1 ):s_1=i
if(temp.index("*")<s):
s=temp.index("*")
if(temp.rindex("*")>e ):
e=temp.rindex("*")
if ( e_1<i ):e_1=i
l.append(list(temp))
for i in range(s_1,e_1+1):
print("".join(l[i][s:e+1]))
|
Title: Letter
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input Specification:
The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output Specification:
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Demo Input:
['6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n', '3 3\n***\n*.*\n***\n']
Demo Output:
['***\n*..\n***\n*..\n***\n', '***\n*.*\n***\n']
Note:
none
|
```python
n,m=map(int,input().split(" "))
s,e,c ,s_1,e_1=100,0,0,-1,0 # start ,end , count ,start_rectangle,end_rectangle
l=[]
for i in range(n):
temp=input()
if ("*" in temp ):
if (s_1==-1 ):s_1=i
if(temp.index("*")<s):
s=temp.index("*")
if(temp.rindex("*")>e ):
e=temp.rindex("*")
if ( e_1<i ):e_1=i
l.append(list(temp))
for i in range(s_1,e_1+1):
print("".join(l[i][s:e+1]))
```
| 3.954
|
804
|
B
|
Minimum number of steps
|
PROGRAMMING
| 1,400
|
[
"combinatorics",
"greedy",
"implementation",
"math"
] | null | null |
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
|
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
|
Print the minimum number of steps modulo 109<=+<=7.
|
[
"ab\n",
"aab\n"
] |
[
"1\n",
"3\n"
] |
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
| 1,000
|
[
{
"input": "ab",
"output": "1"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaaaabaabababaaaaaba",
"output": "17307"
},
{
"input": "abaabaaabbabaabab",
"output": "1795"
},
{
"input": "abbaa",
"output": "2"
},
{
"input": "abbaaabaabaaaaabbbbaababaaaaabaabbaaaaabbaabbaaaabbbabbbabb",
"output": "690283580"
},
{
"input": "aababbaaaabbaabbbbbbbbabbababbbaaabbaaabbabbba",
"output": "2183418"
},
{
"input": "aabbaababbabbbaabbaababaaaabbaaaabaaaaaababbaaaabaababbabbbb",
"output": "436420225"
},
{
"input": "aaabaaaabbababbaabbababbbbaaaaaaabbabbba",
"output": "8431094"
},
{
"input": "abbbbababbabbbbbabaabbbaabbbbbbbaaab",
"output": "8180"
},
{
"input": "bbababbbaabaaaaaaaabbabbbb",
"output": "40979"
},
{
"input": "abbbaaabbbbbabaabbaaabbbababbbaabaabababababa",
"output": "2065758"
},
{
"input": "abaaaaaabaaaabbabbaaabbbbabababaaaaabbaabbaaaaabbbaababaaaaaaabbbbbaaaaabaababbabababbabbbbaabbaabbabbbabaabbaabbaaaaaab",
"output": "235606597"
},
{
"input": "abbbbbbbbbbbbbbbbbbbbbbbbbbaababaaaaaaabaabaaababaabaababaaabababaababab",
"output": "7"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaaaaaaabaabaaababaabaababaaabababaabbbbbbb",
"output": "557763786"
}
] | 1,682,901,339
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 202
| 59,904,000
|
new_seq = list(input())
new_bs = 0
new_mod = 1000000007
new_plays = 0
new_seq.reverse()
for char in new_seq:
if char == 'b':
new_bs = (new_bs + 1) % new_mod
else:
new_plays = (new_plays + new_bs) % new_mod
new_bs = (new_bs * 2) % new_mod
print(new_plays)
|
Title: Minimum number of steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input Specification:
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output Specification:
Print the minimum number of steps modulo 109<=+<=7.
Demo Input:
['ab\n', 'aab\n']
Demo Output:
['1\n', '3\n']
Note:
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
|
```python
new_seq = list(input())
new_bs = 0
new_mod = 1000000007
new_plays = 0
new_seq.reverse()
for char in new_seq:
if char == 'b':
new_bs = (new_bs + 1) % new_mod
else:
new_plays = (new_plays + new_bs) % new_mod
new_bs = (new_bs * 2) % new_mod
print(new_plays)
```
| 3
|
|
486
|
A
|
Calculating Function
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
|
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
|
Print *f*(*n*) in a single line.
|
[
"4\n",
"5\n"
] |
[
"2\n",
"-3\n"
] |
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3
| 500
|
[
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
"output": "50"
},
{
"input": "101",
"output": "-51"
},
{
"input": "102",
"output": "51"
},
{
"input": "103",
"output": "-52"
},
{
"input": "104",
"output": "52"
},
{
"input": "105",
"output": "-53"
},
{
"input": "106",
"output": "53"
},
{
"input": "107",
"output": "-54"
},
{
"input": "108",
"output": "54"
},
{
"input": "109",
"output": "-55"
},
{
"input": "208170109961052",
"output": "104085054980526"
},
{
"input": "46017661651072",
"output": "23008830825536"
},
{
"input": "4018154546667",
"output": "-2009077273334"
},
{
"input": "288565475053",
"output": "-144282737527"
},
{
"input": "3052460231",
"output": "-1526230116"
},
{
"input": "29906716",
"output": "14953358"
},
{
"input": "87897701693326",
"output": "43948850846663"
},
{
"input": "8240",
"output": "4120"
},
{
"input": "577935",
"output": "-288968"
},
{
"input": "62",
"output": "31"
},
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "1"
},
{
"input": "9999999999999",
"output": "-5000000000000"
},
{
"input": "1000000000000",
"output": "500000000000"
},
{
"input": "99999999999999",
"output": "-50000000000000"
},
{
"input": "999999999999999",
"output": "-500000000000000"
},
{
"input": "42191359342",
"output": "21095679671"
},
{
"input": "100000000000000",
"output": "50000000000000"
},
{
"input": "145645214654154",
"output": "72822607327077"
},
{
"input": "4294967296",
"output": "2147483648"
},
{
"input": "3037000499",
"output": "-1518500250"
},
{
"input": "10000000000001",
"output": "-5000000000001"
},
{
"input": "100000017040846",
"output": "50000008520423"
},
{
"input": "98979894985999",
"output": "-49489947493000"
}
] | 1,697,918,268
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
def f(n):
if n % 2:
return -(n + 1) // 2
return n // 2
# Sat Oct 21 2023 22:57:46 GMT+0300 (Moscow Standard Time)
|
Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3
|
```python
def f(n):
if n % 2:
return -(n + 1) // 2
return n // 2
# Sat Oct 21 2023 22:57:46 GMT+0300 (Moscow Standard Time)
```
| 0
|
|
156
|
B
|
Suspects
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"data structures",
"implementation"
] | null | null |
As Sherlock Holmes was investigating a crime, he identified *n* suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to *n*. After that, he asked each one: "Which one committed the crime?". Suspect number *i* answered either "The crime was committed by suspect number *a**i*", or "Suspect number *a**i* didn't commit the crime". Also, the suspect could say so about himself (*a**i*<==<=*i*).
Sherlock Holmes understood for sure that exactly *m* answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=*n*) — the total number of suspects and the number of suspects who told the truth. Next *n* lines contain the suspects' answers. The *i*-th line contains either "+*a**i*" (without the quotes), if the suspect number *i* says that the crime was committed by suspect number *a**i*, or "-*a**i*" (without the quotes), if the suspect number *i* says that the suspect number *a**i* didn't commit the crime (*a**i* is an integer, 1<=≤<=*a**i*<=≤<=*n*).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly *m* people told the truth.
|
Print *n* lines. Line number *i* should contain "Truth" if suspect number *i* has told the truth for sure. Print "Lie" if the suspect number *i* lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
|
[
"1 1\n+1\n",
"3 2\n-1\n-2\n-3\n",
"4 1\n+2\n-3\n+4\n-1\n"
] |
[
"Truth\n",
"Not defined\nNot defined\nNot defined\n",
"Lie\nNot defined\nLie\nNot defined\n"
] |
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
| 1,000
|
[
{
"input": "1 1\n+1",
"output": "Truth"
},
{
"input": "3 2\n-1\n-2\n-3",
"output": "Not defined\nNot defined\nNot defined"
},
{
"input": "4 1\n+2\n-3\n+4\n-1",
"output": "Lie\nNot defined\nLie\nNot defined"
},
{
"input": "1 0\n-1",
"output": "Lie"
},
{
"input": "2 2\n+1\n+1",
"output": "Truth\nTruth"
},
{
"input": "2 1\n+2\n+1",
"output": "Not defined\nNot defined"
},
{
"input": "2 0\n-2\n-2",
"output": "Lie\nLie"
},
{
"input": "3 1\n+2\n+3\n+3",
"output": "Truth\nLie\nLie"
},
{
"input": "6 3\n+5\n+5\n+5\n+1\n+1\n+1",
"output": "Not defined\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined"
},
{
"input": "4 3\n-4\n-3\n-1\n-3",
"output": "Not defined\nTruth\nNot defined\nTruth"
},
{
"input": "10 4\n-8\n+1\n-6\n-10\n+5\n-6\n-8\n-8\n-4\n-8",
"output": "Lie\nLie\nTruth\nTruth\nLie\nTruth\nLie\nLie\nTruth\nLie"
},
{
"input": "10 5\n-4\n+4\n+4\n-9\n-9\n-4\n-4\n+2\n-9\n-4",
"output": "Lie\nTruth\nTruth\nTruth\nTruth\nLie\nLie\nLie\nTruth\nLie"
},
{
"input": "7 2\n+5\n+5\n+5\n-2\n+1\n-5\n-6",
"output": "Lie\nLie\nLie\nNot defined\nLie\nTruth\nNot defined"
},
{
"input": "7 4\n+7\n-3\n-3\n-4\n+3\n+3\n+3",
"output": "Not defined\nNot defined\nNot defined\nTruth\nNot defined\nNot defined\nNot defined"
},
{
"input": "6 3\n-6\n-1\n+5\n+1\n+6\n+1",
"output": "Truth\nNot defined\nNot defined\nNot defined\nLie\nNot defined"
},
{
"input": "5 3\n-2\n+2\n+2\n-3\n+5",
"output": "Not defined\nNot defined\nNot defined\nTruth\nNot defined"
},
{
"input": "3 0\n-2\n-2\n-2",
"output": "Lie\nLie\nLie"
},
{
"input": "5 3\n-1\n-1\n-4\n+1\n-4",
"output": "Lie\nLie\nTruth\nTruth\nTruth"
},
{
"input": "9 6\n+2\n+7\n+7\n-1\n-4\n+7\n-7\n+7\n+5",
"output": "Lie\nTruth\nTruth\nTruth\nTruth\nTruth\nLie\nTruth\nLie"
},
{
"input": "64 28\n+54\n+44\n+55\n-3\n+33\n-54\n-54\n-7\n+33\n+54\n+54\n+26\n-54\n+14\n-54\n-47\n+25\n-54\n-54\n-54\n-52\n+54\n+54\n+54\n+54\n+20\n+7\n+54\n+4\n+32\n+46\n-54\n-47\n+15\n+32\n-54\n+7\n+62\n-16\n-54\n+3\n+54\n+54\n+54\n+54\n-54\n+54\n-54\n+54\n-52\n+27\n-7\n+54\n-5\n-54\n-18\n+1\n+58\n+28\n-46\n+61\n-54\n-49\n-43",
"output": "Not defined\nNot defined\nNot defined\nTruth\nLie\nNot defined\nNot defined\nTruth\nLie\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nTruth\nNot defined\nNot defined\nNot defined\nNot defined\nTruth\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nLie\nNot defined\nNot defined\nLie\nLie\nNot defined\nTruth\nNot defined\nLie\nNot defined\nLie\nNot defined\nTruth\nNot defined\nLie\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\n..."
},
{
"input": "43 18\n-14\n-28\n+16\n+10\n+25\n-30\n+25\n+30\n+25\n+25\n+25\n+25\n-25\n+22\n+3\n-17\n+16\n-25\n+10\n+14\n+41\n+25\n-25\n+33\n+24\n-23\n-25\n+25\n-22\n+29\n+28\n-25\n-25\n-29\n+11\n+26\n-25\n+25\n+10\n+1\n-20\n-17\n+23",
"output": "Truth\nTruth\nNot defined\nLie\nNot defined\nTruth\nNot defined\nLie\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nLie\nLie\nTruth\nNot defined\nNot defined\nLie\nLie\nLie\nNot defined\nNot defined\nLie\nLie\nTruth\nNot defined\nNot defined\nTruth\nLie\nLie\nNot defined\nNot defined\nTruth\nLie\nLie\nNot defined\nNot defined\nLie\nLie\nTruth\nTruth\nLie"
},
{
"input": "28 12\n+10\n-7\n+17\n-20\n+7\n-7\n+13\n-21\n-7\n-7\n-18\n+7\n+7\n+3\n+6\n+14\n+7\n-24\n-21\n-7\n-7\n+4\n+7\n-7\n+21\n-7\n-26\n+7",
"output": "Lie\nLie\nLie\nTruth\nTruth\nLie\nLie\nTruth\nLie\nLie\nTruth\nTruth\nTruth\nLie\nLie\nLie\nTruth\nTruth\nTruth\nLie\nLie\nLie\nTruth\nLie\nLie\nLie\nTruth\nTruth"
},
{
"input": "17 9\n-6\n+16\n+5\n+16\n-17\n+17\n-11\n+5\n+14\n+5\n-8\n-5\n+6\n-2\n-11\n+4\n+17",
"output": "Truth\nNot defined\nNot defined\nNot defined\nTruth\nLie\nTruth\nNot defined\nLie\nNot defined\nTruth\nNot defined\nLie\nTruth\nTruth\nLie\nLie"
},
{
"input": "14 3\n+14\n+12\n-9\n+9\n-9\n-9\n+8\n+9\n+2\n+1\n-13\n-9\n+13\n+3",
"output": "Lie\nLie\nLie\nTruth\nLie\nLie\nLie\nTruth\nLie\nLie\nTruth\nLie\nLie\nLie"
},
{
"input": "10 4\n-9\n-8\n-5\n-9\n-7\n-9\n-9\n-9\n-4\n-9",
"output": "Lie\nTruth\nTruth\nLie\nTruth\nLie\nLie\nLie\nTruth\nLie"
},
{
"input": "10 5\n-10\n-10\n-10\n-5\n-1\n+10\n-3\n-10\n-9\n-10",
"output": "Lie\nLie\nLie\nTruth\nTruth\nTruth\nTruth\nLie\nTruth\nLie"
},
{
"input": "10 4\n-3\n-3\n-3\n-3\n-3\n-2\n-2\n-6\n-7\n-3",
"output": "Lie\nLie\nLie\nLie\nLie\nTruth\nTruth\nTruth\nTruth\nLie"
},
{
"input": "10 6\n-9\n-7\n-5\n-5\n-4\n-2\n-8\n-5\n-5\n-9",
"output": "Truth\nTruth\nLie\nLie\nTruth\nTruth\nTruth\nLie\nLie\nTruth"
},
{
"input": "10 4\n-8\n-2\n-8\n+1\n-4\n-8\n-2\n-8\n-8\n-1",
"output": "Lie\nTruth\nLie\nLie\nTruth\nLie\nTruth\nLie\nLie\nTruth"
},
{
"input": "10 2\n-8\n+10\n+1\n+8\n+4\n+8\n+6\n-8\n+10\n+1",
"output": "Not defined\nLie\nLie\nNot defined\nLie\nNot defined\nLie\nNot defined\nLie\nLie"
},
{
"input": "10 3\n+9\n+3\n+8\n+3\n+6\n-3\n+6\n+8\n+3\n+7",
"output": "Lie\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nNot defined\nLie"
},
{
"input": "10 8\n-2\n+9\n+9\n-4\n+9\n+9\n+4\n-9\n-3\n+9",
"output": "Truth\nTruth\nTruth\nTruth\nTruth\nTruth\nLie\nLie\nTruth\nTruth"
},
{
"input": "10 7\n-4\n+6\n+4\n+9\n+6\n+6\n+6\n+6\n+6\n+2",
"output": "Truth\nTruth\nLie\nLie\nTruth\nTruth\nTruth\nTruth\nTruth\nLie"
},
{
"input": "10 4\n+3\n+5\n+6\n+10\n+5\n+5\n+6\n+8\n+5\n+6",
"output": "Lie\nTruth\nLie\nLie\nTruth\nTruth\nLie\nLie\nTruth\nLie"
},
{
"input": "10 7\n-6\n-10\n-3\n-1\n-3\n-7\n-2\n-7\n-7\n-3",
"output": "Truth\nTruth\nNot defined\nTruth\nNot defined\nNot defined\nTruth\nNot defined\nNot defined\nNot defined"
},
{
"input": "10 5\n-4\n-4\n-1\n-5\n-7\n-4\n-4\n-4\n-1\n-7",
"output": "Lie\nLie\nTruth\nTruth\nTruth\nLie\nLie\nLie\nTruth\nTruth"
},
{
"input": "10 5\n-9\n-7\n-6\n-3\n-10\n-10\n-10\n-10\n-10\n-2",
"output": "Truth\nTruth\nTruth\nTruth\nLie\nLie\nLie\nLie\nLie\nTruth"
},
{
"input": "10 3\n-10\n-10\n-10\n-3\n-10\n-10\n-10\n-8\n-4\n-10",
"output": "Lie\nLie\nLie\nTruth\nLie\nLie\nLie\nTruth\nTruth\nLie"
},
{
"input": "10 5\n-8\n-8\n-4\n-9\n-10\n-2\n-9\n-8\n-8\n-8",
"output": "Lie\nLie\nTruth\nTruth\nTruth\nTruth\nTruth\nLie\nLie\nLie"
},
{
"input": "10 5\n+7\n+8\n+9\n+1\n+7\n+7\n+7\n+6\n+6\n+7",
"output": "Truth\nLie\nLie\nLie\nTruth\nTruth\nTruth\nLie\nLie\nTruth"
},
{
"input": "10 5\n+2\n+2\n+2\n+2\n+9\n+10\n+8\n+7\n+4\n+2",
"output": "Truth\nTruth\nTruth\nTruth\nLie\nLie\nLie\nLie\nLie\nTruth"
},
{
"input": "10 9\n+7\n+7\n+7\n+7\n+7\n+7\n+5\n+7\n+7\n+7",
"output": "Truth\nTruth\nTruth\nTruth\nTruth\nTruth\nLie\nTruth\nTruth\nTruth"
},
{
"input": "10 3\n+10\n+2\n+10\n+9\n+1\n+9\n+4\n+9\n+3\n+2",
"output": "Lie\nLie\nLie\nTruth\nLie\nTruth\nLie\nTruth\nLie\nLie"
},
{
"input": "10 6\n+10\n+10\n+10\n+3\n+10\n+10\n+6\n+6\n+10\n+8",
"output": "Truth\nTruth\nTruth\nLie\nTruth\nTruth\nLie\nLie\nTruth\nLie"
},
{
"input": "3 2\n-1\n+2\n+3",
"output": "Truth\nNot defined\nNot defined"
}
] | 1,655,210,681
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 97
| 902
| 4,710,400
|
[n, m] = input().split()
n = int(n)
m = int(m)
arr = []
ps = []
ns = []
suspects = []
for i in range(0, n):
a = input()
a = int(a)
arr.append(a)
ps.append(0)
ns.append(0)
suspects.append(0)
sum_ps = 0
sum_ns = 0
for i in arr:
if i > 0:
ps[i - 1] += 1
sum_ps += 1
else:
ns[-i - 1] += 1
sum_ns += 1
num_suspects = 0
for i in range(0, n):
if ps[i] + sum_ns - ns[i] == m:
suspects[i] = 1
num_suspects += 1
for i in arr:
if i > 0:
if suspects[i - 1]:
if num_suspects == 1:
print("Truth")
else:
print("Not defined")
else:
print("Lie")
else:
if suspects[-i - 1]:
if num_suspects == 1:
print("Lie")
else:
print("Not defined")
else:
print("Truth")
|
Title: Suspects
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As Sherlock Holmes was investigating a crime, he identified *n* suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to *n*. After that, he asked each one: "Which one committed the crime?". Suspect number *i* answered either "The crime was committed by suspect number *a**i*", or "Suspect number *a**i* didn't commit the crime". Also, the suspect could say so about himself (*a**i*<==<=*i*).
Sherlock Holmes understood for sure that exactly *m* answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=*n*) — the total number of suspects and the number of suspects who told the truth. Next *n* lines contain the suspects' answers. The *i*-th line contains either "+*a**i*" (without the quotes), if the suspect number *i* says that the crime was committed by suspect number *a**i*, or "-*a**i*" (without the quotes), if the suspect number *i* says that the suspect number *a**i* didn't commit the crime (*a**i* is an integer, 1<=≤<=*a**i*<=≤<=*n*).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly *m* people told the truth.
Output Specification:
Print *n* lines. Line number *i* should contain "Truth" if suspect number *i* has told the truth for sure. Print "Lie" if the suspect number *i* lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Demo Input:
['1 1\n+1\n', '3 2\n-1\n-2\n-3\n', '4 1\n+2\n-3\n+4\n-1\n']
Demo Output:
['Truth\n', 'Not defined\nNot defined\nNot defined\n', 'Lie\nNot defined\nLie\nNot defined\n']
Note:
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
|
```python
[n, m] = input().split()
n = int(n)
m = int(m)
arr = []
ps = []
ns = []
suspects = []
for i in range(0, n):
a = input()
a = int(a)
arr.append(a)
ps.append(0)
ns.append(0)
suspects.append(0)
sum_ps = 0
sum_ns = 0
for i in arr:
if i > 0:
ps[i - 1] += 1
sum_ps += 1
else:
ns[-i - 1] += 1
sum_ns += 1
num_suspects = 0
for i in range(0, n):
if ps[i] + sum_ns - ns[i] == m:
suspects[i] = 1
num_suspects += 1
for i in arr:
if i > 0:
if suspects[i - 1]:
if num_suspects == 1:
print("Truth")
else:
print("Not defined")
else:
print("Lie")
else:
if suspects[-i - 1]:
if num_suspects == 1:
print("Lie")
else:
print("Not defined")
else:
print("Truth")
```
| 3
|
|
327
|
A
|
Flipping Game
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dp",
"implementation"
] | null | null |
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
|
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
|
[
"5\n1 0 0 1 0\n",
"4\n1 0 0 1\n"
] |
[
"4\n",
"4\n"
] |
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.
| 500
|
[
{
"input": "5\n1 0 0 1 0",
"output": "4"
},
{
"input": "4\n1 0 0 1",
"output": "4"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "8\n1 0 0 0 1 0 0 0",
"output": "7"
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "18"
},
{
"input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "22"
},
{
"input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1",
"output": "70"
},
{
"input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1",
"output": "60"
},
{
"input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0",
"output": "11"
},
{
"input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1",
"output": "18"
},
{
"input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1",
"output": "36"
},
{
"input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "44"
},
{
"input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1",
"output": "61"
},
{
"input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0",
"output": "61"
},
{
"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": "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": "99"
},
{
"input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0",
"output": "61"
},
{
"input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1",
"output": "59"
},
{
"input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1",
"output": "61"
},
{
"input": "2\n1 1",
"output": "1"
}
] | 1,676,119,899
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 0
|
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
#from math import gcd
mod = int(1e9 + 7)
inf = int(1e20)
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
ii = lambda :int(input())
py = lambda :print("YES")
pn = lambda :print("NO")
n=ii()
arr=li()
que=deque(arr)
total=sum(arr)
res=cnt=0
while que:
while que and que[0]==1:
cnt+=que.popleft()
total-=1
while que and que[-1]==1:
cnt+=que.pop()
total-=1
res=max(res,cnt+max(total,len(que)-total))
while que and que[0]==0:
que.popleft()
while que and que[-1]==0:
que.pop()
print(res)
|
Title: Flipping Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Output Specification:
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Demo Input:
['5\n1 0 0 1 0\n', '4\n1 0 0 1\n']
Demo Output:
['4\n', '4\n']
Note:
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.
|
```python
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
#from math import gcd
mod = int(1e9 + 7)
inf = int(1e20)
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
ii = lambda :int(input())
py = lambda :print("YES")
pn = lambda :print("NO")
n=ii()
arr=li()
que=deque(arr)
total=sum(arr)
res=cnt=0
while que:
while que and que[0]==1:
cnt+=que.popleft()
total-=1
while que and que[-1]==1:
cnt+=que.pop()
total-=1
res=max(res,cnt+max(total,len(que)-total))
while que and que[0]==0:
que.popleft()
while que and que[-1]==0:
que.pop()
print(res)
```
| 0
|
|
587
|
A
|
Duff and Weight Lifting
|
PROGRAMMING
| 1,500
|
[
"greedy"
] | null | null |
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of *i*-th of them is 2*w**i* pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2*a*1,<=...,<=2*a**k* if and only if there exists a non-negative integer *x* such that 2*a*1<=+<=2*a*2<=+<=...<=+<=2*a**k*<==<=2*x*, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
|
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=106), the number of weights.
The second line contains *n* integers *w*1,<=...,<=*w**n* separated by spaces (0<=≤<=*w**i*<=≤<=106 for each 1<=≤<=*i*<=≤<=*n*), the powers of two forming the weights values.
|
Print the minimum number of steps in a single line.
|
[
"5\n1 1 2 3 3\n",
"4\n0 1 2 3\n"
] |
[
"2\n",
"4\n"
] |
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
| 500
|
[
{
"input": "5\n1 1 2 3 3",
"output": "2"
},
{
"input": "4\n0 1 2 3",
"output": "4"
},
{
"input": "1\n120287",
"output": "1"
},
{
"input": "2\n28288 0",
"output": "2"
},
{
"input": "2\n95745 95745",
"output": "1"
},
{
"input": "13\n92 194 580495 0 10855 41704 13 96429 33 213 0 92 140599",
"output": "11"
},
{
"input": "13\n688743 688743 1975 688743 688743 688743 688743 688743 688743 0 0 688743 688743",
"output": "4"
},
{
"input": "35\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",
"output": "3"
},
{
"input": "35\n130212 3176 77075 8071 18 1369 7539 1683 80757 1847 0 1374 122 8524 4 2 21333 270264 4 9254 151921 0 1 33596 73002 54382 0 1 29233 75952 15 38892 1877 6167 4",
"output": "31"
},
{
"input": "35\n0 0 298 0 0 0 0 0 689063 65442 0 984598 2054 43668 0 369 0 2054 0 996220 0 16327 369 0 996220 0 0 0 4693 2054 348 0 118 0 0",
"output": "16"
},
{
"input": "100\n196 1681 196 0 61 93 196 196 196 196 196 0 0 96 18 1576 0 93 666463 18 93 1 1278 8939 93 196 196 1278 3 0 67416 869956 10 56489 196 745 39 783 196 8939 196 81 69634 4552 39 3 14 20 25 8 10 4 7302 0 19579 20 1140 15990 7302 0 19579 4142 11 1354 75252 93 311 1278 0 79475 10 75252 93 7302 0 81 408441 19579 10 39 19 37748 4364 31135 47700 105818 47700 10 4142 543356 3 30647 45917 60714 8939 18 22925 7302 93 75252",
"output": "59"
}
] | 1,608,157,033
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 7,782,400
|
a = input()
b = input()
a_1 = ""
l_b = []
conter = 0
dct = {}
for i in range(len(b)):
if b[i] != " ":
a_1 = a_1 + b[i]
else:
dct[int(a_1)] = dct.get(int(a_1), 0) + 1
a_1 = ""
dct[int(a_1)] = dct.get(int(a_1), 0) + 1
def mainfn2(dct):
x = 0
for item in dct:
if dct[item] > 1:
dct[item + 1] = dct.get(item + 1, 0) + (dct[item] // 2)
dct[item] = dct[item] % 2
x = 0
break
else:
x = x + 1
if x > 0:
return dct
mainfn2(dct)
mainfn2(dct)
for item in dct:
if dct[item] != 0:
conter = conter +1
print(conter)
|
Title: Duff and Weight Lifting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of *i*-th of them is 2*w**i* pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2*a*1,<=...,<=2*a**k* if and only if there exists a non-negative integer *x* such that 2*a*1<=+<=2*a*2<=+<=...<=+<=2*a**k*<==<=2*x*, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=106), the number of weights.
The second line contains *n* integers *w*1,<=...,<=*w**n* separated by spaces (0<=≤<=*w**i*<=≤<=106 for each 1<=≤<=*i*<=≤<=*n*), the powers of two forming the weights values.
Output Specification:
Print the minimum number of steps in a single line.
Demo Input:
['5\n1 1 2 3 3\n', '4\n0 1 2 3\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
|
```python
a = input()
b = input()
a_1 = ""
l_b = []
conter = 0
dct = {}
for i in range(len(b)):
if b[i] != " ":
a_1 = a_1 + b[i]
else:
dct[int(a_1)] = dct.get(int(a_1), 0) + 1
a_1 = ""
dct[int(a_1)] = dct.get(int(a_1), 0) + 1
def mainfn2(dct):
x = 0
for item in dct:
if dct[item] > 1:
dct[item + 1] = dct.get(item + 1, 0) + (dct[item] // 2)
dct[item] = dct[item] % 2
x = 0
break
else:
x = x + 1
if x > 0:
return dct
mainfn2(dct)
mainfn2(dct)
for item in dct:
if dct[item] != 0:
conter = conter +1
print(conter)
```
| 0
|
|
725
|
A
|
Jumping Ball
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
|
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
|
[
"4\n<<><\n",
"5\n>>>>>\n",
"4\n>><<\n"
] |
[
"2",
"5",
"0"
] |
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field.
| 500
|
[
{
"input": "4\n<<><",
"output": "2"
},
{
"input": "5\n>>>>>",
"output": "5"
},
{
"input": "4\n>><<",
"output": "0"
},
{
"input": "3\n<<>",
"output": "3"
},
{
"input": "3\n<<<",
"output": "3"
},
{
"input": "3\n><<",
"output": "0"
},
{
"input": "1\n<",
"output": "1"
},
{
"input": "2\n<>",
"output": "2"
},
{
"input": "3\n<>>",
"output": "3"
},
{
"input": "3\n><>",
"output": "1"
},
{
"input": "2\n><",
"output": "0"
},
{
"input": "2\n>>",
"output": "2"
},
{
"input": "2\n<<",
"output": "2"
},
{
"input": "1\n>",
"output": "1"
},
{
"input": "3\n>><",
"output": "0"
},
{
"input": "3\n>>>",
"output": "3"
},
{
"input": "3\n<><",
"output": "1"
},
{
"input": "10\n<<<><<<>>>",
"output": "6"
},
{
"input": "20\n><><<><<<>>>>>>>>>>>",
"output": "11"
},
{
"input": "20\n<<<<<<<<<<><<<<>>>>>",
"output": "15"
},
{
"input": "50\n<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<>><<>><<<<<>><>><<<>><><<>>><<>>><<<<><><><<><<<<><>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "49"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>><<>><>><>><<><><><><>>>><><<<>>>><<<>>>>>>><><",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<",
"output": "100"
},
{
"input": "100\n>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "100"
},
{
"input": "12\n<<>><<>><<>>",
"output": "4"
},
{
"input": "6\n<<><>>",
"output": "4"
},
{
"input": "6\n><>>>>",
"output": "4"
},
{
"input": "8\n>>>><<<>",
"output": "1"
},
{
"input": "4\n<><>",
"output": "2"
},
{
"input": "4\n><><",
"output": "0"
},
{
"input": "7\n<<>>><>",
"output": "3"
},
{
"input": "10\n><><>>>>>>",
"output": "6"
},
{
"input": "5\n<><>>",
"output": "3"
},
{
"input": "12\n<><<<<>>>>>>",
"output": "7"
},
{
"input": "6\n<>><<>",
"output": "2"
},
{
"input": "6\n>>><>>",
"output": "2"
},
{
"input": "10\n><><>>>><>",
"output": "1"
},
{
"input": "5\n><>>>",
"output": "3"
},
{
"input": "5\n<<><>",
"output": "3"
},
{
"input": "5\n<><<<",
"output": "1"
},
{
"input": "4\n<><<",
"output": "1"
},
{
"input": "8\n<<>><<>>",
"output": "4"
},
{
"input": "7\n<<><>>>",
"output": "5"
},
{
"input": "5\n><<>>",
"output": "2"
},
{
"input": "10\n<<<<<>>>>>",
"output": "10"
},
{
"input": "6\n><<<<<",
"output": "0"
},
{
"input": "8\n<<><><>>",
"output": "4"
},
{
"input": "10\n<<<<><<<><",
"output": "4"
},
{
"input": "12\n<<<>>>><<>>>",
"output": "6"
},
{
"input": "4\n><>>",
"output": "2"
},
{
"input": "11\n<<><<>><<>>",
"output": "4"
}
] | 1,663,313,074
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 409,600
|
input()
string = input()
print(abs(string.count(">") - string.count("<")))
|
Title: Jumping Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
Output Specification:
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
Demo Input:
['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n']
Demo Output:
['2', '5', '0']
Note:
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field.
|
```python
input()
string = input()
print(abs(string.count(">") - string.count("<")))
```
| 0
|
|
869
|
B
|
The Eternal Immortality
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
|
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
|
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
|
[
"2 4\n",
"0 10\n",
"107 109\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
| 1,000
|
[
{
"input": "2 4",
"output": "2"
},
{
"input": "0 10",
"output": "0"
},
{
"input": "107 109",
"output": "2"
},
{
"input": "10 13",
"output": "6"
},
{
"input": "998244355 998244359",
"output": "4"
},
{
"input": "999999999000000000 1000000000000000000",
"output": "0"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 15",
"output": "0"
},
{
"input": "24 26",
"output": "0"
},
{
"input": "14 60",
"output": "0"
},
{
"input": "11 79",
"output": "0"
},
{
"input": "1230 1232",
"output": "2"
},
{
"input": "2633 2634",
"output": "4"
},
{
"input": "535 536",
"output": "6"
},
{
"input": "344319135 396746843",
"output": "0"
},
{
"input": "696667767 696667767",
"output": "1"
},
{
"input": "419530302 610096911",
"output": "0"
},
{
"input": "238965115 821731161",
"output": "0"
},
{
"input": "414626436 728903812",
"output": "0"
},
{
"input": "274410639 293308324",
"output": "0"
},
{
"input": "650636673091305697 650636673091305702",
"output": "0"
},
{
"input": "651240548333620923 651240548333620924",
"output": "4"
},
{
"input": "500000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "0"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1"
},
{
"input": "0 4",
"output": "4"
},
{
"input": "50000000062000007 50000000062000011",
"output": "0"
},
{
"input": "0 0",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "0 2",
"output": "2"
},
{
"input": "10000000000012 10000000000015",
"output": "0"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "12 23",
"output": "0"
},
{
"input": "0 11",
"output": "0"
},
{
"input": "11111234567890 11111234567898",
"output": "0"
},
{
"input": "0 3",
"output": "6"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999999999999999997 999999999999999999",
"output": "2"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "0 1",
"output": "1"
},
{
"input": "101 1002",
"output": "0"
},
{
"input": "0 100000000000000001",
"output": "0"
},
{
"input": "99999999999999997 99999999999999999",
"output": "2"
},
{
"input": "14 15",
"output": "5"
},
{
"input": "8 19",
"output": "0"
},
{
"input": "12 22",
"output": "0"
},
{
"input": "999999999999996 999999999999999",
"output": "4"
},
{
"input": "1 3",
"output": "6"
},
{
"input": "124 125",
"output": "5"
},
{
"input": "11 32",
"output": "0"
},
{
"input": "0 5",
"output": "0"
},
{
"input": "0 999999",
"output": "0"
},
{
"input": "151151151515 151151151526",
"output": "0"
},
{
"input": "6 107",
"output": "0"
},
{
"input": "5 16",
"output": "0"
},
{
"input": "7 16",
"output": "0"
},
{
"input": "6 19",
"output": "0"
},
{
"input": "11113111111111 13111111111111",
"output": "0"
},
{
"input": "1 1000",
"output": "0"
},
{
"input": "24 25",
"output": "5"
},
{
"input": "0 100000000000",
"output": "0"
},
{
"input": "1 22",
"output": "0"
},
{
"input": "999999999999999996 999999999999999999",
"output": "4"
}
] | 1,589,610,150
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 1,000
| 8,294,400
|
from math import factorial
x1,x2=map(int,input().split())
y = str(factorial(x2) // factorial(x1))
print(y[-1])
|
Title: The Eternal Immortality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
Input Specification:
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output Specification:
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
Demo Input:
['2 4\n', '0 10\n', '107 109\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
|
```python
from math import factorial
x1,x2=map(int,input().split())
y = str(factorial(x2) // factorial(x1))
print(y[-1])
```
| 0
|
|
202
|
A
|
LLPS
|
PROGRAMMING
| 800
|
[
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null |
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
|
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
|
Print the lexicographically largest palindromic subsequence of string *s*.
|
[
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] |
[
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] |
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
| 500
|
[
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,601,097,151
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 218
| 0
|
s=input()
c=max(s)
k=s.count(c)
print(c*k)
|
Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
```python
s=input()
c=max(s)
k=s.count(c)
print(c*k)
```
| 3
|
|
574
|
A
|
Bear and Elections
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
|
The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
|
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
|
[
"5\n5 1 11 2 8\n",
"4\n1 8 8 8\n",
"2\n7 6\n"
] |
[
"4\n",
"6\n",
"0\n"
] |
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.
In the third sample Limak is a winner without bribing any citizen.
| 500
|
[
{
"input": "5\n5 1 11 2 8",
"output": "4"
},
{
"input": "4\n1 8 8 8",
"output": "6"
},
{
"input": "2\n7 6",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "10\n100 200 57 99 1 1000 200 200 200 500",
"output": "451"
},
{
"input": "16\n7 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "932"
},
{
"input": "100\n47 64 68 61 68 66 69 61 69 65 69 63 62 60 68 65 64 65 65 62 63 68 60 70 63 63 65 67 70 69 68 69 61 65 63 60 60 65 61 60 70 66 66 65 62 60 65 68 61 62 67 64 66 65 67 68 60 69 70 63 65 62 64 65 67 67 69 68 66 69 70 67 65 70 60 66 70 67 67 64 69 69 66 68 60 64 62 62 68 69 67 69 60 70 69 68 62 63 68 66",
"output": "23"
},
{
"input": "2\n96 97",
"output": "1"
},
{
"input": "2\n1000 1000",
"output": "1"
},
{
"input": "3\n999 1000 1000",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "7\n10 940 926 990 946 980 985",
"output": "817"
},
{
"input": "10\n5 3 4 5 5 2 1 8 4 1",
"output": "2"
},
{
"input": "15\n17 15 17 16 13 17 13 16 14 14 17 17 13 15 17",
"output": "1"
},
{
"input": "20\n90 5 62 9 50 7 14 43 44 44 56 13 71 22 43 35 52 60 73 54",
"output": "0"
},
{
"input": "30\n27 85 49 7 77 38 4 68 23 28 81 100 40 9 78 38 1 60 60 49 98 44 45 92 46 39 98 24 37 39",
"output": "58"
},
{
"input": "51\n90 47 100 12 21 96 2 68 84 60 2 9 33 8 45 13 59 50 100 93 22 97 4 81 51 2 3 78 19 16 25 63 52 34 79 32 34 87 7 42 96 93 30 33 33 43 69 8 63 58 57",
"output": "8"
},
{
"input": "77\n1000 2 2 3 1 1 1 3 3 2 1 1 3 2 2 2 3 2 3 1 3 1 1 2 2 2 3 1 1 2 2 2 3 2 1 3 3 1 2 3 3 3 2 1 3 2 1 3 3 2 3 3 2 1 3 1 1 1 2 3 2 3 1 3 1 2 1 2 2 2 1 2 2 3 2 2 2",
"output": "0"
},
{
"input": "91\n3 92 89 83 85 80 91 94 95 82 92 95 80 88 90 85 81 90 87 86 94 88 90 87 88 82 95 84 84 93 83 95 91 85 89 88 88 85 87 90 93 80 89 95 94 92 93 86 83 82 86 84 91 80 90 95 84 86 84 85 84 92 82 84 83 91 87 95 94 95 90 95 86 92 86 80 95 86 88 80 82 87 84 83 91 93 81 81 91 89 88",
"output": "89"
},
{
"input": "100\n1 3 71 47 64 82 58 61 61 35 52 36 57 62 63 54 52 21 78 100 24 94 4 80 99 62 43 72 21 70 90 4 23 14 72 4 76 49 71 96 96 99 78 7 32 11 14 61 19 69 1 68 100 77 86 54 14 86 47 53 30 88 67 66 61 70 17 63 40 5 99 53 38 31 91 18 41 5 77 61 53 30 87 21 23 54 52 17 23 75 58 99 99 63 20 1 78 72 28 11",
"output": "90"
},
{
"input": "100\n1 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99"
},
{
"input": "94\n3 100 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 100 99 100 99 100 100 100 99 100 99 100 99 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 99 99 100 99 100 99 99 99 100 100 99 100 100 99 99 100 100 100 99 100 99 99 99 99 99 100 100 100 100 100 100 100 100 100 99 99 99 99 100 99 100 99 100 100",
"output": "97"
},
{
"input": "97\n99 99 98 98 100 98 99 99 98 100 100 100 99 99 100 99 99 98 99 99 98 98 98 100 100 99 98 99 100 98 99 98 98 100 98 99 100 98 98 99 98 98 99 98 100 99 99 99 99 98 98 98 100 99 100 100 99 99 100 99 99 98 98 98 100 100 98 100 100 99 98 99 100 98 98 98 98 99 99 98 98 99 100 100 98 98 99 98 99 100 98 99 100 98 99 99 100",
"output": "2"
},
{
"input": "100\n100 55 70 81 73 51 6 75 45 85 33 61 98 63 11 59 1 8 14 28 78 74 44 80 7 69 7 5 90 73 43 78 64 64 43 92 59 70 80 19 33 39 31 70 38 85 24 23 86 79 98 56 92 63 92 4 36 8 79 74 2 81 54 13 69 44 49 63 17 76 78 99 42 36 47 71 19 90 9 58 83 53 27 2 35 51 65 59 90 51 74 87 84 48 98 44 84 100 84 93",
"output": "1"
},
{
"input": "100\n100 637 498 246 615 901 724 673 793 33 282 908 477 185 185 969 34 859 90 70 107 492 227 918 919 131 620 182 802 703 779 184 403 891 448 499 628 553 905 392 70 396 8 575 66 908 992 496 792 174 667 355 836 610 855 377 244 827 836 808 667 354 800 114 746 556 75 894 162 367 99 718 394 273 833 776 151 433 315 470 759 12 552 613 85 793 775 649 225 86 296 624 557 201 209 595 697 527 282 168",
"output": "749"
},
{
"input": "100\n107 172 549 883 564 56 399 970 173 990 224 217 601 381 948 631 159 958 512 136 61 584 633 202 652 355 26 723 663 237 410 721 688 552 699 24 748 186 461 88 34 243 872 205 471 298 654 693 244 33 359 533 471 116 386 653 654 887 531 303 335 829 319 340 827 89 602 191 422 289 361 200 593 421 592 402 256 813 606 589 741 9 148 893 3 142 50 169 219 360 642 45 810 818 507 624 561 743 303 111",
"output": "729"
},
{
"input": "90\n670 694 651 729 579 539 568 551 707 638 604 544 502 531 775 805 558 655 506 729 802 778 653 737 591 770 594 535 588 604 658 713 779 705 504 563 513 651 529 572 505 553 515 750 621 574 727 774 714 725 665 798 670 747 751 635 755 798 635 717 583 682 517 546 740 802 743 507 658 700 645 671 533 594 506 633 768 584 672 666 703 522 530 501 592 528 678 708 619 786",
"output": "111"
},
{
"input": "90\n10 265 429 431 343 305 806 746 284 313 503 221 594 351 83 653 232 431 427 610 458 88 255 215 529 205 492 549 55 694 535 104 45 327 816 432 595 549 454 141 216 557 250 415 531 494 190 749 718 380 78 447 784 347 196 814 16 780 262 462 776 315 160 307 593 694 692 41 528 725 376 777 337 44 438 630 345 502 384 184 742 429 570 361 394 267 820 778 662 377",
"output": "714"
},
{
"input": "95\n800 280 176 472 587 763 588 838 760 378 667 231 566 278 713 305 354 815 140 220 188 409 109 180 251 268 474 590 853 143 235 691 313 785 386 92 783 471 43 342 718 592 678 404 256 362 239 504 163 85 521 81 356 73 754 589 380 159 196 862 838 509 149 42 366 630 467 292 698 123 187 796 576 37 689 800 186 518 488 432 159 860 349 799 282 304 880 283 23 312 55 507 734 370 490",
"output": "52"
},
{
"input": "100\n95 88 84 85 74 97 100 52 91 94 62 66 90 56 86 66 95 73 79 68 54 67 99 52 82 62 81 71 93 85 72 72 95 52 72 63 57 90 92 89 88 77 84 78 95 59 72 86 98 64 89 64 80 70 54 93 88 86 79 78 94 64 89 66 50 90 54 82 52 96 99 54 81 66 83 79 69 80 51 73 81 69 93 82 76 52 58 87 93 92 52 67 78 63 63 87 77 95 58 78",
"output": "4"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "100\n999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "2"
},
{
"input": "100\n901 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "99"
},
{
"input": "4\n8 5 11 3",
"output": "2"
},
{
"input": "3\n3 10 1",
"output": "4"
}
] | 1,441,035,860
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 46
| 0
|
import math
n = int(input())
v = list(map(int , input().split()))
v[1:]= sorted(v[1:])
ans=0
while(v[0]<=v[-1]):
#print(v , "last is" , v[-1])
v[-1]-=1
v[0]+=1
ans+=1
v[1:]= sorted(v[1:])
print(ans)
|
Title: Bear and Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
Output Specification:
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
Demo Input:
['5\n5 1 11 2 8\n', '4\n1 8 8 8\n', '2\n7 6\n']
Demo Output:
['4\n', '6\n', '0\n']
Note:
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.
In the third sample Limak is a winner without bribing any citizen.
|
```python
import math
n = int(input())
v = list(map(int , input().split()))
v[1:]= sorted(v[1:])
ans=0
while(v[0]<=v[-1]):
#print(v , "last is" , v[-1])
v[-1]-=1
v[0]+=1
ans+=1
v[1:]= sorted(v[1:])
print(ans)
```
| 3
|
|
1
|
C
|
Ancient Berland Circus
|
PROGRAMMING
| 2,100
|
[
"geometry",
"math"
] |
C. Ancient Berland Circus
|
2
|
64
|
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
|
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
|
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
|
[
"0.000000 0.000000\n1.000000 1.000000\n0.000000 1.000000\n"
] |
[
"1.00000000\n"
] | 0
|
[
{
"input": "0.000000 0.000000\n1.000000 1.000000\n0.000000 1.000000",
"output": "1.00000000"
},
{
"input": "71.756151 7.532275\n-48.634784 100.159986\n91.778633 158.107739",
"output": "9991.27897663"
},
{
"input": "18.716839 40.852752\n66.147248 -4.083161\n111.083161 43.347248",
"output": "4268.87997505"
},
{
"input": "-13.242302 -45.014124\n-33.825369 51.083964\n84.512928 -55.134407",
"output": "16617.24002771"
},
{
"input": "115.715093 141.583620\n136.158119 -23.780834\n173.673212 64.802787",
"output": "24043.74046813"
},
{
"input": "17.288379 68.223317\n48.776683 71.688379\n23.170559 106.572762",
"output": "1505.27997374"
},
{
"input": "76.820252 66.709341\n61.392328 82.684207\n44.267775 -2.378694",
"output": "6503.44762335"
},
{
"input": "-46.482632 -31.161247\n19.689679 -70.646972\n-17.902656 -58.455808",
"output": "23949.55226823"
},
{
"input": "34.236058 108.163949\n28.639345 104.566515\n25.610069 86.002927",
"output": "780.93431702"
},
{
"input": "25.428124 39.407248\n17.868098 39.785933\n11.028461 43.028890",
"output": "1152.21351717"
},
{
"input": "36.856072 121.845502\n46.453956 109.898647\n-30.047767 77.590282",
"output": "5339.35578947"
},
{
"input": "-18.643272 56.008305\n9.107608 -22.094058\n-6.456146 70.308320",
"output": "9009.25177521"
},
{
"input": "88.653021 18.024220\n51.942488 -2.527850\n76.164701 24.553012",
"output": "1452.52866331"
},
{
"input": "80.181999 -38.076894\n23.381778 122.535736\n47.118815 140.734014",
"output": "28242.17663744"
},
{
"input": "1.514204 81.400629\n32.168797 100.161401\n7.778734 46.010993",
"output": "3149.43107333"
},
{
"input": "84.409605 38.496141\n77.788313 39.553807\n75.248391 59.413884",
"output": "438.85760782"
},
{
"input": "12.272903 101.825792\n-51.240438 -12.708472\n-29.729299 77.882032",
"output": "24908.67540438"
},
{
"input": "35.661751 27.283571\n96.513550 51.518022\n97.605986 131.258287",
"output": "13324.78113326"
},
{
"input": "-20.003518 -4.671086\n93.588632 6.362759\n-24.748109 24.792124",
"output": "11191.04493104"
},
{
"input": "93.583067 132.858352\n63.834975 19.353720\n33.677824 102.529376",
"output": "10866.49390021"
},
{
"input": "-7.347450 36.971423\n84.498728 89.423536\n75.469963 98.022482",
"output": "8977.83404724"
},
{
"input": "51.679280 56.072393\n-35.819256 73.390532\n-10.661374 129.756454",
"output": "7441.86549199"
},
{
"input": "97.326813 61.492460\n100.982131 57.717635\n68.385216 22.538372",
"output": "1840.59945324"
},
{
"input": "-16.356805 109.310423\n124.529388 25.066276\n-37.892043 80.604904",
"output": "22719.36404168"
},
{
"input": "103.967164 63.475916\n86.466163 59.341930\n69.260229 73.258917",
"output": "1621.96700296"
},
{
"input": "122.381894 -48.763263\n163.634346 -22.427845\n26.099674 73.681862",
"output": "22182.51901824"
},
{
"input": "119.209229 133.905087\n132.001535 22.179509\n96.096673 0.539763",
"output": "16459.52899209"
},
{
"input": "77.145533 85.041789\n67.452820 52.513188\n80.503843 85.000149",
"output": "1034.70898496"
},
{
"input": "28.718442 36.116251\n36.734593 35.617015\n76.193973 99.136077",
"output": "6271.48941610"
},
{
"input": "0.376916 17.054676\n100.187614 85.602831\n1.425829 132.750915",
"output": "13947.47744984"
},
{
"input": "46.172435 -22.819705\n17.485134 -1.663888\n101.027565 111.619705",
"output": "16483.23337238"
},
{
"input": "55.957968 -72.765994\n39.787413 -75.942282\n24.837014 128.144762",
"output": "32799.66697178"
},
{
"input": "40.562163 -47.610606\n10.073051 -54.490068\n54.625875 -40.685797",
"output": "31224.34817875"
},
{
"input": "20.965151 74.716562\n167.264364 81.864800\n5.931644 48.813212",
"output": "30115.26346791"
},
{
"input": "105.530943 80.920069\n40.206723 125.323331\n40.502256 -85.455877",
"output": "36574.64621711"
},
{
"input": "104.636703 49.583778\n85.940583 95.426299\n69.375168 93.234795",
"output": "2632.68754075"
},
{
"input": "72.873708 -59.083734\n110.911118 -6.206576\n-44.292395 13.106202",
"output": "19244.42781859"
},
{
"input": "49.320630 48.119616\n65.888396 93.514980\n27.342377 97.600590",
"output": "2437.50897386"
},
{
"input": "6.949504 69.606390\n26.139268 72.136945\n24.032442 57.407195",
"output": "372.09309018"
},
{
"input": "-21.925928 -24.623076\n-33.673619 -11.677794\n4.692348 52.266292",
"output": "5669.99444283"
},
{
"input": "109.515505 37.575315\n5.377080 101.729711\n17.501630 103.324931",
"output": "25142.85604936"
},
{
"input": "-56.880888 172.997993\n81.126977 42.144034\n-51.413417 17.057807",
"output": "29051.57171313"
},
{
"input": "80.895061 94.491414\n42.361631 65.191687\n77.556800 76.694829",
"output": "2386.01792476"
},
{
"input": "165.094169 94.574129\n46.867578 147.178855\n174.685774 62.705213",
"output": "32087.47120554"
},
{
"input": "146.604506 -3.502359\n24.935572 44.589981\n106.160918 -51.162271",
"output": "13799.61044048"
},
{
"input": "139.847022 19.153937\n104.096879 75.379874\n49.164271 46.404632",
"output": "7083.26303902"
},
{
"input": "31.312532 151.532355\n182.646053 56.534075\n15.953947 127.065925",
"output": "25712.80766033"
},
{
"input": "42.147045 64.165917\n70.260284 4.962470\n10.532991 76.277713",
"output": "14261.92257159"
},
{
"input": "129.400249 -44.695226\n122.278798 -53.696996\n44.828427 -83.507917",
"output": "26227.47891833"
},
{
"input": "28.420253 0.619862\n10.966628 21.724132\n14.618862 10.754642",
"output": "1760.14006648"
}
] | 1,693,844,595
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
import math
epsilon = 1e-6
x1, y1 = map(float, input().split())
x2, y2 = map(float, input().split())
x3, y3 = map(float, input().split())
d12_2 = (x1 - x2) ** 2 + (y1 - y2) ** 2
d23_2 = (x3 - x2) ** 2 + (y3 - y2) ** 2
if abs(d12_2 - d23_2) > epsilon:
if d23_2 > d12_2:
x1, y1, x2, y2 = x2, y2, x1, y1
else:
x3, y3, x2, y2 = x2, y2, x3, y3
# maintenant, d12 = d23
v12 = [x2 - x1, y2 - y1]
v12_norm = (v12[0] ** 2 + v12[1] ** 2) ** 0.5
v23 = [x3 - x2, y3 - y2]
v23_norm = (v23[0] ** 2 + v23[1] ** 2) ** 0.5
v12_v23 = v12[0] * v23[0] + v12[1] * v23[1]
cos_angle = v12_v23/(v12_norm * v23_norm)
angle = math.acos(cos_angle)
n = 1/ (0.5-angle/(2*math.pi))
d12_2 = (x1 - x2) ** 2 + (y1 - y2) ** 2
d23_2 = (x3 - x2) ** 2 + (y3 - y2) ** 2
L = (d12_2 + d23_2)/2
A = n*L**2/(4*math.tan(math.pi/n))
print(A)
|
Title: Ancient Berland Circus
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input Specification:
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output Specification:
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Demo Input:
['0.000000 0.000000\n1.000000 1.000000\n0.000000 1.000000\n']
Demo Output:
['1.00000000\n']
|
```python
import math
epsilon = 1e-6
x1, y1 = map(float, input().split())
x2, y2 = map(float, input().split())
x3, y3 = map(float, input().split())
d12_2 = (x1 - x2) ** 2 + (y1 - y2) ** 2
d23_2 = (x3 - x2) ** 2 + (y3 - y2) ** 2
if abs(d12_2 - d23_2) > epsilon:
if d23_2 > d12_2:
x1, y1, x2, y2 = x2, y2, x1, y1
else:
x3, y3, x2, y2 = x2, y2, x3, y3
# maintenant, d12 = d23
v12 = [x2 - x1, y2 - y1]
v12_norm = (v12[0] ** 2 + v12[1] ** 2) ** 0.5
v23 = [x3 - x2, y3 - y2]
v23_norm = (v23[0] ** 2 + v23[1] ** 2) ** 0.5
v12_v23 = v12[0] * v23[0] + v12[1] * v23[1]
cos_angle = v12_v23/(v12_norm * v23_norm)
angle = math.acos(cos_angle)
n = 1/ (0.5-angle/(2*math.pi))
d12_2 = (x1 - x2) ** 2 + (y1 - y2) ** 2
d23_2 = (x3 - x2) ** 2 + (y3 - y2) ** 2
L = (d12_2 + d23_2)/2
A = n*L**2/(4*math.tan(math.pi/n))
print(A)
```
| 0
|
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,697,303,477
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 122
| 0
|
word = input()
print(word[0].capitalize() + word[1::])
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
word = input()
print(word[0].capitalize() + word[1::])
```
| 3
|
|
755
|
A
|
PolandBall and Hypothesis
|
PROGRAMMING
| 800
|
[
"brute force",
"graphs",
"math",
"number theory"
] | null | null |
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
|
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
|
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
|
[
"3\n",
"4\n"
] |
[
"1",
"2"
] |
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
| 500
|
[
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "153",
"output": "1"
},
{
"input": "1000",
"output": "1"
},
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "4"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "11",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "996",
"output": "3"
},
{
"input": "36",
"output": "4"
},
{
"input": "210",
"output": "4"
},
{
"input": "270",
"output": "4"
},
{
"input": "306",
"output": "4"
},
{
"input": "330",
"output": "5"
},
{
"input": "336",
"output": "4"
},
{
"input": "600",
"output": "4"
},
{
"input": "726",
"output": "4"
},
{
"input": "988",
"output": "1"
},
{
"input": "12",
"output": "2"
},
{
"input": "987",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "986",
"output": "1"
},
{
"input": "14",
"output": "1"
},
{
"input": "985",
"output": "1"
},
{
"input": "15",
"output": "1"
},
{
"input": "984",
"output": "1"
},
{
"input": "16",
"output": "2"
},
{
"input": "983",
"output": "1"
},
{
"input": "17",
"output": "1"
},
{
"input": "982",
"output": "2"
},
{
"input": "18",
"output": "3"
},
{
"input": "981",
"output": "1"
},
{
"input": "19",
"output": "1"
},
{
"input": "980",
"output": "1"
},
{
"input": "20",
"output": "1"
},
{
"input": "979",
"output": "1"
},
{
"input": "21",
"output": "1"
},
{
"input": "978",
"output": "1"
},
{
"input": "22",
"output": "2"
},
{
"input": "977",
"output": "1"
},
{
"input": "23",
"output": "1"
},
{
"input": "976",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "975",
"output": "1"
},
{
"input": "25",
"output": "1"
},
{
"input": "2",
"output": "4"
},
{
"input": "6",
"output": "4"
},
{
"input": "996",
"output": "3"
},
{
"input": "8",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "330",
"output": "5"
},
{
"input": "18",
"output": "3"
},
{
"input": "12",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "999",
"output": "1"
},
{
"input": "726",
"output": "4"
},
{
"input": "876",
"output": "3"
}
] | 1,664,266,204
| 304
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 46
| 0
|
n = int(input())
if n % 2 == 1 or n % 3 == 2:
m = 1 # рез-т чётный или дел. на 3
elif n % 3 == 1:
m = 2 # рез-т станет делиться на 3
else:
m = n - 2
print(m)
|
Title: PolandBall and Hypothesis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
Input Specification:
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
Output Specification:
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
Demo Input:
['3\n', '4\n']
Demo Output:
['1', '2']
Note:
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
|
```python
n = int(input())
if n % 2 == 1 or n % 3 == 2:
m = 1 # рез-т чётный или дел. на 3
elif n % 3 == 1:
m = 2 # рез-т станет делиться на 3
else:
m = n - 2
print(m)
```
| 0
|
|
547
|
C
|
Mike and Foam
|
PROGRAMMING
| 2,300
|
[
"bitmasks",
"combinatorics",
"dp",
"math",
"number theory"
] | null | null |
Mike is a bartender at Rico's bar. At Rico's, they put beer glasses in a special shelf. There are *n* kinds of beer at Rico's numbered from 1 to *n*. *i*-th kind of beer has *a**i* milliliters of foam on it.
Maxim is Mike's boss. Today he told Mike to perform *q* queries. Initially the shelf is empty. In each request, Maxim gives him a number *x*. If beer number *x* is already in the shelf, then Mike should remove it from the shelf, otherwise he should put it in the shelf.
After each query, Mike should tell him the score of the shelf. Bears are geeks. So they think that the score of a shelf is the number of pairs (*i*,<=*j*) of glasses in the shelf such that *i*<=<<=*j* and where is the greatest common divisor of numbers *a* and *b*.
Mike is tired. So he asked you to help him in performing these requests.
|
The first line of input contains numbers *n* and *q* (1<=≤<=*n*,<=*q*<=≤<=2<=×<=105), the number of different kinds of beer and number of queries.
The next line contains *n* space separated integers, *a*1,<=*a*2,<=... ,<=*a**n* (1<=≤<=*a**i*<=≤<=5<=×<=105), the height of foam in top of each kind of beer.
The next *q* lines contain the queries. Each query consists of a single integer integer *x* (1<=≤<=*x*<=≤<=*n*), the index of a beer that should be added or removed from the shelf.
|
For each query, print the answer for that query in one line.
|
[
"5 6\n1 2 3 4 6\n1\n2\n3\n4\n5\n1\n"
] |
[
"0\n1\n3\n5\n6\n2\n"
] |
none
| 1,750
|
[
{
"input": "5 6\n1 2 3 4 6\n1\n2\n3\n4\n5\n1",
"output": "0\n1\n3\n5\n6\n2"
},
{
"input": "3 3\n151790 360570 1\n2\n3\n3",
"output": "0\n1\n0"
},
{
"input": "1 1\n1\n1",
"output": "0"
},
{
"input": "5 10\n1 1 1 1 1\n1\n2\n3\n4\n5\n5\n4\n3\n2\n1",
"output": "0\n1\n3\n6\n10\n6\n3\n1\n0\n0"
},
{
"input": "1 2\n499590\n1\n1",
"output": "0\n0"
}
] | 1,606,032,727
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 2,000
| 1,945,600
|
import math
from sys import stdin,stdout
nq = list(map(int,stdin.readline().split(' ')))
num,q = nq[0],nq[1]
beers = list(map(int,stdin.readline().split(' ')))
mask = [0 for _ in range(num)]
resultados = list()
def comprobar(a,b):
if math.gcd(a,b) == 1:
return True
else:
return False
def Calc(cheves,mascara):
score = 0
for m in range(num):
if mascara[num-m-1] ==1:
for k in range(num-m-1):
chevea,cheveb = beers[num-m-1],beers[k]
if (mascara[k] ==1) and (comprobar(chevea,cheveb)):
score +=1
return score
for n in range(q):
operacion = int(stdin.readline())
if mask[operacion-1] == 1:
mask[operacion-1] = 0
else:
mask[operacion-1] = 1
resultados.append(Calc(beers,mask))
for result in resultados:
print(result)
|
Title: Mike and Foam
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is a bartender at Rico's bar. At Rico's, they put beer glasses in a special shelf. There are *n* kinds of beer at Rico's numbered from 1 to *n*. *i*-th kind of beer has *a**i* milliliters of foam on it.
Maxim is Mike's boss. Today he told Mike to perform *q* queries. Initially the shelf is empty. In each request, Maxim gives him a number *x*. If beer number *x* is already in the shelf, then Mike should remove it from the shelf, otherwise he should put it in the shelf.
After each query, Mike should tell him the score of the shelf. Bears are geeks. So they think that the score of a shelf is the number of pairs (*i*,<=*j*) of glasses in the shelf such that *i*<=<<=*j* and where is the greatest common divisor of numbers *a* and *b*.
Mike is tired. So he asked you to help him in performing these requests.
Input Specification:
The first line of input contains numbers *n* and *q* (1<=≤<=*n*,<=*q*<=≤<=2<=×<=105), the number of different kinds of beer and number of queries.
The next line contains *n* space separated integers, *a*1,<=*a*2,<=... ,<=*a**n* (1<=≤<=*a**i*<=≤<=5<=×<=105), the height of foam in top of each kind of beer.
The next *q* lines contain the queries. Each query consists of a single integer integer *x* (1<=≤<=*x*<=≤<=*n*), the index of a beer that should be added or removed from the shelf.
Output Specification:
For each query, print the answer for that query in one line.
Demo Input:
['5 6\n1 2 3 4 6\n1\n2\n3\n4\n5\n1\n']
Demo Output:
['0\n1\n3\n5\n6\n2\n']
Note:
none
|
```python
import math
from sys import stdin,stdout
nq = list(map(int,stdin.readline().split(' ')))
num,q = nq[0],nq[1]
beers = list(map(int,stdin.readline().split(' ')))
mask = [0 for _ in range(num)]
resultados = list()
def comprobar(a,b):
if math.gcd(a,b) == 1:
return True
else:
return False
def Calc(cheves,mascara):
score = 0
for m in range(num):
if mascara[num-m-1] ==1:
for k in range(num-m-1):
chevea,cheveb = beers[num-m-1],beers[k]
if (mascara[k] ==1) and (comprobar(chevea,cheveb)):
score +=1
return score
for n in range(q):
operacion = int(stdin.readline())
if mask[operacion-1] == 1:
mask[operacion-1] = 0
else:
mask[operacion-1] = 1
resultados.append(Calc(beers,mask))
for result in resultados:
print(result)
```
| 0
|
|
222
|
A
|
Shooshuns and Sequence
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
|
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
|
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
|
[
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] |
[
"1\n",
"-1\n"
] |
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
| 500
|
[
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "4 3\n1 2 3 4",
"output": "-1"
},
{
"input": "5 3\n2 1 1 1 1",
"output": "1"
},
{
"input": "5 3\n1 1 1 1 1",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 1\n2 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 2\n2 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 2\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 3\n3 2 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "-1"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "20 10\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 11\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "20 9\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20 20\n20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10",
"output": "19"
}
] | 1,665,401,301
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 92
| 0
|
n , k = map(int,input().split())
a = list(map(int,input().split()))
l = a[k-1]
for i in a[k:]:
if i != l:
print(-1)
break
else:
for i,val in enumerate(reversed(a[:k-1])):
if val != l:
print(i+1)
break
else:
print(0)
|
Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
|
```python
n , k = map(int,input().split())
a = list(map(int,input().split()))
l = a[k-1]
for i in a[k:]:
if i != l:
print(-1)
break
else:
for i,val in enumerate(reversed(a[:k-1])):
if val != l:
print(i+1)
break
else:
print(0)
```
| 0
|
|
833
|
B
|
The Bakery
|
PROGRAMMING
| 2,200
|
[
"binary search",
"data structures",
"divide and conquer",
"dp",
"two pointers"
] | null | null |
Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven which can bake several types of cakes, and opened the bakery.
Soon the expenses started to overcome the income, so Slastyona decided to study the sweets market. She learned it's profitable to pack cakes in boxes, and that the more distinct cake types a box contains (let's denote this number as the value of the box), the higher price it has.
She needs to change the production technology! The problem is that the oven chooses the cake types on its own and Slastyona can't affect it. However, she knows the types and order of *n* cakes the oven is going to bake today. Slastyona has to pack exactly *k* boxes with cakes today, and she has to put in each box several (at least one) cakes the oven produced one right after another (in other words, she has to put in a box a continuous segment of cakes).
Slastyona wants to maximize the total value of all boxes with cakes. Help her determine this maximum possible total value.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=35000, 1<=≤<=*k*<=≤<=*min*(*n*,<=50)) – the number of cakes and the number of boxes, respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) – the types of cakes in the order the oven bakes them.
|
Print the only integer – the maximum total value of all boxes with cakes.
|
[
"4 1\n1 2 2 1\n",
"7 2\n1 3 3 1 4 4 4\n",
"8 3\n7 7 8 7 7 8 1 7\n"
] |
[
"2\n",
"5\n",
"6\n"
] |
In the first example Slastyona has only one box. She has to put all cakes in it, so that there are two types of cakes in the box, so the value is equal to 2.
In the second example it is profitable to put the first two cakes in the first box, and all the rest in the second. There are two distinct types in the first box, and three in the second box then, so the total value is 5.
| 1,250
|
[
{
"input": "4 1\n1 2 2 1",
"output": "2"
},
{
"input": "7 2\n1 3 3 1 4 4 4",
"output": "5"
},
{
"input": "8 3\n7 7 8 7 7 8 1 7",
"output": "6"
},
{
"input": "50 5\n2 39 16 20 32 36 29 42 14 23 9 4 28 17 45 16 46 1 39 33 47 15 34 43 6 6 12 15 46 11 32 16 31 12 41 33 47 35 22 3 10 4 9 36 22 22 17 33 15 46",
"output": "49"
},
{
"input": "50 6\n41 23 41 16 33 6 25 15 15 11 14 12 21 2 16 42 25 3 5 17 37 11 10 50 33 28 1 35 9 50 40 8 33 14 43 34 8 42 6 31 21 11 39 34 2 30 33 44 38 45",
"output": "50"
},
{
"input": "50 7\n34 49 9 4 29 13 17 46 12 12 14 20 11 36 33 19 46 5 28 48 35 7 31 7 21 47 3 5 26 42 6 50 27 15 45 39 20 48 1 5 37 30 11 36 28 35 8 13 19 45",
"output": "50"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "42 12\n18 1 14 10 6 40 25 10 39 7 36 25 36 11 11 24 38 27 22 26 41 38 24 35 1 6 15 40 7 14 41 11 38 5 41 32 28 2 38 22 12 2",
"output": "42"
},
{
"input": "2 2\n1 1",
"output": "2"
}
] | 1,564,010,199
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
#define xx first
#define yy second
using namespace std;
typedef pair <int, int> pii;
const int inf = 1 << 30;
const int N = 35e4 + 1;
const int K = 50;
int dp[K][N];
int arr[N];
int cnt[N];
int n, k;
int res;
int pre;
int ql = 1;
int qr = 0;
bool was[N];
void input() {
cin >> n >> k;
for(int i = 1; i <= n; ++i) {
cin >> arr[i];
}
}
void add(int key, int add) {
if(cnt[key] > 0) {
res--;
}
cnt[key] += add;
if(cnt[key] > 0) {
res++;
}
}
void update(int l, int r) {
while(ql < l) {
add(arr[ql++], -1);
}
while(ql > l) {
add(arr[--ql], 1);
}
while(qr < r) {
add(arr[++qr], 1);
}
while(qr > r) {
add(arr[qr--], -1);
}
}
void divide_and_conquer(int i, int l, int r, int optl, int optr) {
if(l > r) {
return;
}
pii opt = {-inf, 0};
int q = l + r >> 1;
for(int j = optl; j <= min(q, optr); ++j) {
update(j, q);
opt = max(opt, pii(dp[i - 1][j - 1] + res, j));
}
dp[i][q] = opt.xx;
divide_and_conquer(i, l, q - 1, optl, opt.yy);
divide_and_conquer(i, q + 1, r, opt.yy, optr);
}
void solve() {
for(int i = 1; i <= n; ++i) {
if(!was[arr[i]]) {
was[arr[i]] = 1;
pre++;
}
dp[0][i] = pre;
}
for(int i = 1; i < k; ++i) {
divide_and_conquer(i, 1, n, 1, n);
}
cout << dp[k - 1][n] << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
input();
solve();
}
|
Title: The Bakery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven which can bake several types of cakes, and opened the bakery.
Soon the expenses started to overcome the income, so Slastyona decided to study the sweets market. She learned it's profitable to pack cakes in boxes, and that the more distinct cake types a box contains (let's denote this number as the value of the box), the higher price it has.
She needs to change the production technology! The problem is that the oven chooses the cake types on its own and Slastyona can't affect it. However, she knows the types and order of *n* cakes the oven is going to bake today. Slastyona has to pack exactly *k* boxes with cakes today, and she has to put in each box several (at least one) cakes the oven produced one right after another (in other words, she has to put in a box a continuous segment of cakes).
Slastyona wants to maximize the total value of all boxes with cakes. Help her determine this maximum possible total value.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=35000, 1<=≤<=*k*<=≤<=*min*(*n*,<=50)) – the number of cakes and the number of boxes, respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) – the types of cakes in the order the oven bakes them.
Output Specification:
Print the only integer – the maximum total value of all boxes with cakes.
Demo Input:
['4 1\n1 2 2 1\n', '7 2\n1 3 3 1 4 4 4\n', '8 3\n7 7 8 7 7 8 1 7\n']
Demo Output:
['2\n', '5\n', '6\n']
Note:
In the first example Slastyona has only one box. She has to put all cakes in it, so that there are two types of cakes in the box, so the value is equal to 2.
In the second example it is profitable to put the first two cakes in the first box, and all the rest in the second. There are two distinct types in the first box, and three in the second box then, so the total value is 5.
|
```python
#include <bits/stdc++.h>
#define xx first
#define yy second
using namespace std;
typedef pair <int, int> pii;
const int inf = 1 << 30;
const int N = 35e4 + 1;
const int K = 50;
int dp[K][N];
int arr[N];
int cnt[N];
int n, k;
int res;
int pre;
int ql = 1;
int qr = 0;
bool was[N];
void input() {
cin >> n >> k;
for(int i = 1; i <= n; ++i) {
cin >> arr[i];
}
}
void add(int key, int add) {
if(cnt[key] > 0) {
res--;
}
cnt[key] += add;
if(cnt[key] > 0) {
res++;
}
}
void update(int l, int r) {
while(ql < l) {
add(arr[ql++], -1);
}
while(ql > l) {
add(arr[--ql], 1);
}
while(qr < r) {
add(arr[++qr], 1);
}
while(qr > r) {
add(arr[qr--], -1);
}
}
void divide_and_conquer(int i, int l, int r, int optl, int optr) {
if(l > r) {
return;
}
pii opt = {-inf, 0};
int q = l + r >> 1;
for(int j = optl; j <= min(q, optr); ++j) {
update(j, q);
opt = max(opt, pii(dp[i - 1][j - 1] + res, j));
}
dp[i][q] = opt.xx;
divide_and_conquer(i, l, q - 1, optl, opt.yy);
divide_and_conquer(i, q + 1, r, opt.yy, optr);
}
void solve() {
for(int i = 1; i <= n; ++i) {
if(!was[arr[i]]) {
was[arr[i]] = 1;
pre++;
}
dp[0][i] = pre;
}
for(int i = 1; i < k; ++i) {
divide_and_conquer(i, 1, n, 1, n);
}
cout << dp[k - 1][n] << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
input();
solve();
}
```
| -1
|
|
801
|
B
|
Valued Keys
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
|
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
|
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
|
[
"ab\naa\n",
"nzwzl\nniwel\n",
"ab\nba\n"
] |
[
"ba\n",
"xiyez\n",
"-1\n"
] |
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
| 1,000
|
[
{
"input": "ab\naa",
"output": "ba"
},
{
"input": "nzwzl\nniwel",
"output": "xiyez"
},
{
"input": "ab\nba",
"output": "-1"
},
{
"input": "r\nl",
"output": "l"
},
{
"input": "d\ny",
"output": "-1"
},
{
"input": "yvowz\ncajav",
"output": "cajav"
},
{
"input": "lwzjp\ninjit",
"output": "-1"
},
{
"input": "epqnlxmiicdidyscjaxqznwur\neodnlemiicdedmkcgavqbnqmm",
"output": "eodnlemiicdedmkcgavqbnqmm"
},
{
"input": "qqdabbsxiibnnjgsgxllfvdqj\nuxmypqtwfdezewdxfgplannrs",
"output": "-1"
},
{
"input": "aanerbaqslfmqmuciqbxyznkevukvznpkmxlcorpmrenwxhzfgbmlfpxtkqpxdrmcqcmbf\naanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf",
"output": "aanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf"
},
{
"input": "mbyrkhjctrcrayisflptgfudwgrtegidhqicsjqafvdloritbjhciyxuwavxknezwwudnk\nvvixsutlbdewqoabqhpuerfkzrddcqptfwmxdlxwbvsaqfjoxztlddvwgflcteqbwaiaen",
"output": "-1"
},
{
"input": "eufycwztywhbjrpqobvknwfqmnboqcfdiahkagykeibbsqpljcghhmsgfmswwsanzyiwtvuirwmppfivtekaywkzskyydfvkjgxb\necfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb",
"output": "ecfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb"
},
{
"input": "qvpltcffyeghtbdhjyhfteojezyzziardduzrbwuxmzzkkoehfnxecafizxglboauhynfbawlfxenmykquyhrxswhjuovvogntok\nchvkcvzxptbcepdjfezcpuvtehewbnvqeoezlcnzhpfwujbmhafoeqmjhtwisnobauinkzyigrvahpuetkgpdjfgbzficsmuqnym",
"output": "-1"
},
{
"input": "nmuwjdihouqrnsuahimssnrbxdpwvxiyqtenahtrlshjkmnfuttnpqhgcagoptinnaptxaccptparldzrhpgbyrzedghudtsswxi\nnilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib",
"output": "nilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib"
},
{
"input": "dyxgwupoauwqtcfoyfjdotzirwztdfrueqiypxoqvkmhiehdppwtdoxrbfvtairdbuvlqohjflznggjpifhwjrshcrfbjtklpykx\ngzqlnoizhxolnditjdhlhptjsbczehicudoybzilwnshmywozwnwuipcgirgzldtvtowdsokfeafggwserzdazkxyddjttiopeew",
"output": "-1"
},
{
"input": "hbgwuqzougqzlxemvyjpeizjfwhgugrfnhbrlxkmkdalikfyunppwgdzmalbwewybnjzqsohwhjkdcyhhzmysflambvhpsjilsyv\nfbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv",
"output": "fbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv"
},
{
"input": "xnjjhjfuhgyxqhpzmvgbaohqarugdoaczcfecofltwemieyxolswkcwhlfagfrgmoiqrgftokbqwtxgxzweozzlikrvafiabivlk\npjfosalbsitcnqiazhmepfifjxvmazvdgffcnozmnqubhonwjldmpdsjagmamniylzjdbklcyrzivjyzgnogahobpkwpwpvraqns",
"output": "-1"
},
{
"input": "zrvzedssbsrfldqvjpgmsefrmsatspzoitwvymahiptphiystjlsauzquzqqbmljobdhijcpdvatorwmyojqgnezvzlgjibxepcf\npesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf",
"output": "pesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf"
},
{
"input": "pdvkuwyzntzfqpblzmbynknyhlnqbxijuqaincviugxohcsrofozrrsategwkbwxcvkyzxhurokefpbdnmcfogfhsojayysqbrow\nbvxruombdrywlcjkrltyayaazwpauuhbtgwfzdrmfwwucgffucwelzvpsdgtapogchblzahsrfymjlaghkbmbssghrpxalkslcvp",
"output": "-1"
},
{
"input": "tgharsjyihroiiahwgbjezlxvlterxivdhtzjcqegzmtigqmrehvhiyjeywegxaseoyoacouijudbiruoghgxvxadwzgdxtnxlds\ntghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp",
"output": "tghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp"
},
{
"input": "jsinejpfwhzloulxndzvzftgogfdagrsscxmatldssqsgaknnbkcvhptebjjpkjhrjegrotzwcdosezkedzxeoyibmyzunkguoqj\nkfmvybobocdpipiripysioruqvloopvbggpjksgmwzyqwyxnesmvhsawnbbmntulspvsysfkjqwpvoelliopbaukyagedextzoej",
"output": "-1"
},
{
"input": "nttdcfceptruiomtmwzestrfchnqpgqeztpcvthzelfyggjgqadylzubpvbrlgndrcsursczpxlnoyoadxezncqalupfzmjeqihe\nkttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe",
"output": "kttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe"
},
{
"input": "diuopwglduasnaxgduwslbzoyayoypzznqspljcyqehweydhlwifcvnjmaowuvyqfwynjghecqvxdvuquuwpvwrjljozocaxnktv\ntrdydprdzmjhgbhzytelrfjpgsebijicsigmwhynmcyjtqrvojcndodchzxfcvyqjxqzwibccdvsjqhsnectdjyrrhzkeamukang",
"output": "-1"
},
{
"input": "ftfr\nftfr",
"output": "ftfr"
},
{
"input": "ftr\nftr",
"output": "ftr"
},
{
"input": "shftr\nshftr",
"output": "shftr"
},
{
"input": "vkvkkv\nvkvkkv",
"output": "vkvkkv"
},
{
"input": "ftrd\nftrd",
"output": "ftrd"
},
{
"input": "fztr\nfztr",
"output": "fztr"
},
{
"input": "frtr\nfrtr",
"output": "frtr"
},
{
"input": "shdftr\nshdftr",
"output": "shdftr"
},
{
"input": "zzz\nzzz",
"output": "zzz"
},
{
"input": "shtr\nshtr",
"output": "shtr"
},
{
"input": "aaaaa\nzzzzz",
"output": "-1"
},
{
"input": "efr\nefr",
"output": "efr"
}
] | 1,677,085,044
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
x,y=input(''),input('')
z=[]
for i in range(len(x)):
if x[i]==y[i]:
z.append('z')
else:
z.append(y[i])
print(''.join(z))
|
Title: Valued Keys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
Input Specification:
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
Output Specification:
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
Demo Input:
['ab\naa\n', 'nzwzl\nniwel\n', 'ab\nba\n']
Demo Output:
['ba\n', 'xiyez\n', '-1\n']
Note:
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
|
```python
x,y=input(''),input('')
z=[]
for i in range(len(x)):
if x[i]==y[i]:
z.append('z')
else:
z.append(y[i])
print(''.join(z))
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,660,734,174
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
# your code goes here
a=int(input())
if a%2!=0 and a==0:
print("NO")
else:
print("YES")
|
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
# your code goes here
a=int(input())
if a%2!=0 and a==0:
print("NO")
else:
print("YES")
```
| -1
|
635
|
A
|
Orchestra
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Paul is at the orchestra. The string section is arranged in an *r*<=×<=*c* rectangular grid and is filled with violinists with the exception of *n* violists. Paul really likes violas, so he would like to take a picture including at least *k* of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.
Two pictures are considered to be different if the coordinates of corresponding rectangles are different.
|
The first line of input contains four space-separated integers *r*, *c*, *n*, *k* (1<=≤<=*r*,<=*c*,<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=*n*) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.
The next *n* lines each contain two integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=*r*, 1<=≤<=*y**i*<=≤<=*c*): the position of the *i*-th viola. It is guaranteed that no location appears more than once in the input.
|
Print a single integer — the number of photographs Paul can take which include at least *k* violas.
|
[
"2 2 1 1\n1 2\n",
"3 2 3 3\n1 1\n3 1\n2 2\n",
"3 2 3 2\n1 1\n3 1\n2 2\n"
] |
[
"4\n",
"1\n",
"4\n"
] |
We will use '*' to denote violinists and '#' to denote violists.
In the first sample, the orchestra looks as follows
In the second sample, the orchestra looks as follows
In the third sample, the orchestra looks the same as in the second sample.
| 500
|
[
{
"input": "2 2 1 1\n1 2",
"output": "4"
},
{
"input": "3 2 3 3\n1 1\n3 1\n2 2",
"output": "1"
},
{
"input": "3 2 3 2\n1 1\n3 1\n2 2",
"output": "4"
},
{
"input": "1 1 1 1\n1 1",
"output": "1"
},
{
"input": "10 10 10 10\n6 1\n3 8\n10 6\n10 3\n10 4\n8 9\n2 3\n5 7\n5 9\n5 1",
"output": "4"
},
{
"input": "10 10 10 1\n8 2\n9 10\n6 8\n10 7\n1 8\n4 4\n6 3\n2 3\n8 8\n7 2",
"output": "2073"
},
{
"input": "5 9 2 2\n4 6\n1 5",
"output": "40"
},
{
"input": "6 4 10 2\n2 3\n2 1\n1 2\n6 1\n1 4\n4 4\n2 4\n1 1\n6 3\n4 2",
"output": "103"
},
{
"input": "8 2 4 4\n3 2\n3 1\n2 2\n7 1",
"output": "4"
},
{
"input": "2 6 2 2\n1 2\n1 5",
"output": "8"
},
{
"input": "7 5 3 1\n5 5\n4 5\n1 4",
"output": "135"
},
{
"input": "10 10 10 6\n3 4\n10 9\n6 5\n4 9\n2 10\n10 10\n9 8\n8 2\n5 6\n1 5",
"output": "78"
},
{
"input": "10 10 10 4\n5 7\n9 7\n5 8\n3 7\n8 9\n6 10\n3 2\n10 8\n4 1\n8 10",
"output": "414"
},
{
"input": "10 10 10 1\n8 10\n2 9\n1 10\n3 1\n8 5\n10 1\n4 10\n10 2\n5 3\n9 3",
"output": "1787"
},
{
"input": "10 10 10 3\n2 7\n6 3\n10 2\n2 4\n7 8\n1 2\n3 1\n7 6\n6 8\n9 7",
"output": "751"
},
{
"input": "10 10 10 2\n4 4\n1 7\n10 5\n2 8\n5 5\n6 9\n7 3\n9 5\n5 3\n6 6",
"output": "1416"
},
{
"input": "10 10 10 10\n8 2\n1 4\n9 9\n5 2\n1 7\n1 5\n3 10\n6 9\n7 8\n3 3",
"output": "4"
},
{
"input": "10 10 10 10\n1 9\n5 2\n5 1\n8 5\n9 10\n10 2\n5 4\n4 3\n3 6\n1 5",
"output": "1"
},
{
"input": "10 10 10 10\n5 6\n4 4\n8 9\n5 7\n9 2\n6 4\n7 3\n6 10\n10 3\n3 8",
"output": "6"
},
{
"input": "10 10 10 1\n9 3\n7 5\n8 2\n3 8\n1 6\n3 9\n7 3\n10 4\n5 3\n1 3",
"output": "1987"
},
{
"input": "10 10 10 1\n1 10\n10 8\n7 4\n7 2\n1 3\n6 6\n10 1\n2 7\n9 3\n3 10",
"output": "1991"
},
{
"input": "10 10 10 1\n4 5\n9 6\n3 6\n6 10\n5 2\n1 7\n4 9\n10 8\n8 1\n2 9",
"output": "2082"
}
] | 1,458,357,381
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 22
| 78
| 0
|
r,c,n,p = map(int,input().split())
resp = 0
m = []
for i in range(r):
aux = []
for j in range(c):
aux.append(False)
m.append(aux[:])
for i in range(n):
x,y = map(int,input().split())
m[x-1][y-1] = True
for i in range(r):
for j in range(c):
for a in range(r,i,-1):
for b in range(c,j,-1):
cont = 0
for k in range(i,a):
for l in range(j,b):
if m[k][l]:
cont += 1
if cont>=p:
resp += 1
print(resp)
|
Title: Orchestra
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Paul is at the orchestra. The string section is arranged in an *r*<=×<=*c* rectangular grid and is filled with violinists with the exception of *n* violists. Paul really likes violas, so he would like to take a picture including at least *k* of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.
Two pictures are considered to be different if the coordinates of corresponding rectangles are different.
Input Specification:
The first line of input contains four space-separated integers *r*, *c*, *n*, *k* (1<=≤<=*r*,<=*c*,<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=*n*) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.
The next *n* lines each contain two integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=*r*, 1<=≤<=*y**i*<=≤<=*c*): the position of the *i*-th viola. It is guaranteed that no location appears more than once in the input.
Output Specification:
Print a single integer — the number of photographs Paul can take which include at least *k* violas.
Demo Input:
['2 2 1 1\n1 2\n', '3 2 3 3\n1 1\n3 1\n2 2\n', '3 2 3 2\n1 1\n3 1\n2 2\n']
Demo Output:
['4\n', '1\n', '4\n']
Note:
We will use '*' to denote violinists and '#' to denote violists.
In the first sample, the orchestra looks as follows
In the second sample, the orchestra looks as follows
In the third sample, the orchestra looks the same as in the second sample.
|
```python
r,c,n,p = map(int,input().split())
resp = 0
m = []
for i in range(r):
aux = []
for j in range(c):
aux.append(False)
m.append(aux[:])
for i in range(n):
x,y = map(int,input().split())
m[x-1][y-1] = True
for i in range(r):
for j in range(c):
for a in range(r,i,-1):
for b in range(c,j,-1):
cont = 0
for k in range(i,a):
for l in range(j,b):
if m[k][l]:
cont += 1
if cont>=p:
resp += 1
print(resp)
```
| 3
|
|
304
|
A
|
Pythagorean Theorem II
|
PROGRAMMING
| 1,200
|
[
"brute force",
"math"
] | null | null |
In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
The theorem can be written as an equation relating the lengths of the sides *a*, *b* and *c*, often called the Pythagorean equation:
where *c* represents the length of the hypotenuse, and *a* and *b* represent the lengths of the other two sides.
Given *n*, your task is to count how many right-angled triangles with side-lengths *a*, *b* and *c* that satisfied an inequality 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*.
|
The only line contains one integer *n* (1<=≤<=*n*<=≤<=104) as we mentioned above.
|
Print a single integer — the answer to the problem.
|
[
"5\n",
"74\n"
] |
[
"1\n",
"35\n"
] |
none
| 500
|
[
{
"input": "5",
"output": "1"
},
{
"input": "74",
"output": "35"
},
{
"input": "1000",
"output": "881"
},
{
"input": "586",
"output": "472"
},
{
"input": "2",
"output": "0"
},
{
"input": "362",
"output": "258"
},
{
"input": "778",
"output": "653"
},
{
"input": "194",
"output": "120"
},
{
"input": "906",
"output": "786"
},
{
"input": "659",
"output": "535"
},
{
"input": "75",
"output": "37"
},
{
"input": "787",
"output": "664"
},
{
"input": "851",
"output": "730"
},
{
"input": "563",
"output": "446"
},
{
"input": "979",
"output": "862"
},
{
"input": "395",
"output": "291"
},
{
"input": "755",
"output": "634"
},
{
"input": "171",
"output": "103"
},
{
"input": "883",
"output": "759"
},
{
"input": "400",
"output": "294"
},
{
"input": "817",
"output": "693"
},
{
"input": "177",
"output": "107"
},
{
"input": "593",
"output": "476"
},
{
"input": "305",
"output": "214"
},
{
"input": "721",
"output": "595"
},
{
"input": "785",
"output": "664"
},
{
"input": "497",
"output": "383"
},
{
"input": "913",
"output": "791"
},
{
"input": "625",
"output": "507"
},
{
"input": "334",
"output": "236"
},
{
"input": "10000",
"output": "12471"
},
{
"input": "9999",
"output": "12467"
}
] | 1,481,969,377
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 30
| 3,000
| 0
|
import math
n=int(input())
cnt=0
for i in range(1,n+1):
for j in range(i,n+1):
m= i*i + j*j
val=math.sqrt(m)
val=int(val)
if val > n:
break
if val*val==m:
cnt=cnt+1
print (cnt)
|
Title: Pythagorean Theorem II
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
The theorem can be written as an equation relating the lengths of the sides *a*, *b* and *c*, often called the Pythagorean equation:
where *c* represents the length of the hypotenuse, and *a* and *b* represent the lengths of the other two sides.
Given *n*, your task is to count how many right-angled triangles with side-lengths *a*, *b* and *c* that satisfied an inequality 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*.
Input Specification:
The only line contains one integer *n* (1<=≤<=*n*<=≤<=104) as we mentioned above.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['5\n', '74\n']
Demo Output:
['1\n', '35\n']
Note:
none
|
```python
import math
n=int(input())
cnt=0
for i in range(1,n+1):
for j in range(i,n+1):
m= i*i + j*j
val=math.sqrt(m)
val=int(val)
if val > n:
break
if val*val==m:
cnt=cnt+1
print (cnt)
```
| 0
|
|
935
|
A
|
Fafa and his Company
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
|
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
|
Print a single integer representing the answer to the problem.
|
[
"2\n",
"10\n"
] |
[
"1\n",
"3\n"
] |
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
| 500
|
[
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
},
{
"input": "1024",
"output": "10"
},
{
"input": "99999",
"output": "11"
},
{
"input": "10007",
"output": "1"
},
{
"input": "4096",
"output": "12"
},
{
"input": "65536",
"output": "16"
},
{
"input": "40320",
"output": "95"
},
{
"input": "30030",
"output": "63"
},
{
"input": "161",
"output": "3"
},
{
"input": "1000",
"output": "15"
},
{
"input": "10000",
"output": "24"
},
{
"input": "777",
"output": "7"
},
{
"input": "121",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "40000",
"output": "34"
},
{
"input": "99990",
"output": "47"
},
{
"input": "98765",
"output": "3"
},
{
"input": "56789",
"output": "3"
},
{
"input": "13579",
"output": "3"
},
{
"input": "97531",
"output": "3"
},
{
"input": "12345",
"output": "7"
},
{
"input": "54321",
"output": "7"
},
{
"input": "83160",
"output": "127"
},
{
"input": "9",
"output": "2"
},
{
"input": "21",
"output": "3"
},
{
"input": "11",
"output": "1"
},
{
"input": "15",
"output": "3"
}
] | 1,667,827,555
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 1,126,400
|
import statistics
n=int(input())
l=[]
for i in range(1,(n//2)+1):
l.append(n//i)
print(statistics.median(l))
|
Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
|
```python
import statistics
n=int(input())
l=[]
for i in range(1,(n//2)+1):
l.append(n//i)
print(statistics.median(l))
```
| 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"
}
] | 1,633,601,923
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 6,758,400
|
s_in = input()
r = [None]
for c in s_in:
if c != r[-1]: r.append(c)
s_out = ''.join(r[1:])
if 'helo' in s_out:
print('YES')
else:
print('NO')
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
s_in = input()
r = [None]
for c in s_in:
if c != r[-1]: r.append(c)
s_out = ''.join(r[1:])
if 'helo' in s_out:
print('YES')
else:
print('NO')
```
| 0
|
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"
}
] | 1,694,255,256
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
word = str(input())
chang = len(word)-2
if chang >= 9:
print(word[0]+f'{chang}'+word[-1])
else:
print(word)
|
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
word = str(input())
chang = len(word)-2
if chang >= 9:
print(word[0]+f'{chang}'+word[-1])
else:
print(word)
```
| 0
|
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