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values | name stringlengths 2 54 | type stringclasses 2
values | rating int64 0 3.4k | tags listlengths 0 7 | title stringclasses 393
values | time-limit stringclasses 7
values | memory-limit stringclasses 6
values | problem-description stringlengths 0 2.97k | input-specification stringlengths 4 1.87k | output-specification stringlengths 4 1.12k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 3.5k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3
values | verdict stringclasses 1
value | testset stringclasses 9
values | passedTestCount int64 1 402 | timeConsumedMillis int64 15 8.06k | memoryConsumedBytes int64 0 514M | code stringlengths 11 61.4k | prompt stringlengths 297 7.35k | response stringlengths 25 61.4k | score float64 2.82 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
128 | A | Statues | PROGRAMMING | 1,500 | [
"dfs and similar"
] | null | null | In this task Anna and Maria play a game with a very unpleasant rival. Anna and Maria are in the opposite squares of a chessboard (8<=×<=8): Anna is in the upper right corner, and Maria is in the lower left one. Apart from them, the board has several statues. Each statue occupies exactly one square. A square that contains a statue cannot have anything or anyone — neither any other statues, nor Anna, nor Maria.
Anna is present on the board as a figurant (she stands still and never moves), and Maria has been actively involved in the game. Her goal is — to come to Anna's square. Maria and statues move in turn, Maria moves first. During one move Maria can go to any adjacent on the side or diagonal cell in which there is no statue, or she can stay in the cell where she is. The statues during their move must go one square down simultaneously, and those statues that were in the bottom row fall from the board and are no longer appeared.
At that moment, when one of the statues is in the cell in which the Maria is, the statues are declared winners. At the moment when Maria comes into the cell where Anna has been waiting, Maria is declared the winner.
Obviously, nothing depends on the statues, so it all depends on Maria. Determine who will win, if Maria does not make a strategic error. | You are given the 8 strings whose length equals 8, describing the initial position on the board. The first line represents the top row of the board, the next one — for the second from the top, and so on, the last line represents the bottom row. Each character string matches a single cell board in the appropriate row, and the characters are in the same manner as that of the corresponding cell. If the cell is empty, the corresponding character is ".". If a cell has Maria, then it is represented by character "M". If a cell has Anna, it is represented by the character "A". If a cell has a statue, then the cell is represented by character "S".
It is guaranteed that the last character of the first row is always "A", the first character of the last line is always "M". The remaining characters are "." or "S". | If Maria wins, print string "WIN". If the statues win, print string "LOSE". | [
".......A\n........\n........\n........\n........\n........\n........\nM.......\n",
".......A\n........\n........\n........\n........\n........\nSS......\nM.......\n",
".......A\n........\n........\n........\n........\n.S......\nS.......\nMS......\n"
] | [
"WIN\n",
"LOSE\n",
"LOSE\n"
] | none | 1,000 | [
{
"input": ".SSSSSSA\n.SSSSSSS\n.SSSSSSS\n.SSSSSSS\n.SSSSSSS\n.SSSSSSS\n.SSSSSSS\nMSSSSSSS",
"output": "WIN"
},
{
"input": "SSSSSSSA\nSSSSSSSS\nSSSSSSSS\nSSSSSSSS\nSSSSSSSS\nSSSSSSSS\nSSSSSSSS\nMSSSSSSS",
"output": "LOSE"
},
{
"input": "SSSSSSSA\n......SS\n.......S\n.......S\n.......S\n.... | 1,455,024,134 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 124 | 0 | r, s = [63], ''.join(input() + 'T' for i in range(8)) + 'T' * 9
for i in range(0, 72, 9):
t = set()
for x in r:
for y in (x, x - 1, x + 1, x - 9, x + 9, x - 10, x - 8, x + 10, x + 8):
if s[y] == 'T': continue
if (y < i or s[y - i] != 'S') and (y < i + 9 or s[y - i - 9] != 'S'): t.add(y)
r = t
print('WIN' if r else 'LOSE') | Title: Statues
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this task Anna and Maria play a game with a very unpleasant rival. Anna and Maria are in the opposite squares of a chessboard (8<=×<=8): Anna is in the upper right corner, and Maria is in the lower left one. Apart from them, the board has several statues. Each statue occupies exactly one square. A square that contains a statue cannot have anything or anyone — neither any other statues, nor Anna, nor Maria.
Anna is present on the board as a figurant (she stands still and never moves), and Maria has been actively involved in the game. Her goal is — to come to Anna's square. Maria and statues move in turn, Maria moves first. During one move Maria can go to any adjacent on the side or diagonal cell in which there is no statue, or she can stay in the cell where she is. The statues during their move must go one square down simultaneously, and those statues that were in the bottom row fall from the board and are no longer appeared.
At that moment, when one of the statues is in the cell in which the Maria is, the statues are declared winners. At the moment when Maria comes into the cell where Anna has been waiting, Maria is declared the winner.
Obviously, nothing depends on the statues, so it all depends on Maria. Determine who will win, if Maria does not make a strategic error.
Input Specification:
You are given the 8 strings whose length equals 8, describing the initial position on the board. The first line represents the top row of the board, the next one — for the second from the top, and so on, the last line represents the bottom row. Each character string matches a single cell board in the appropriate row, and the characters are in the same manner as that of the corresponding cell. If the cell is empty, the corresponding character is ".". If a cell has Maria, then it is represented by character "M". If a cell has Anna, it is represented by the character "A". If a cell has a statue, then the cell is represented by character "S".
It is guaranteed that the last character of the first row is always "A", the first character of the last line is always "M". The remaining characters are "." or "S".
Output Specification:
If Maria wins, print string "WIN". If the statues win, print string "LOSE".
Demo Input:
['.......A\n........\n........\n........\n........\n........\n........\nM.......\n', '.......A\n........\n........\n........\n........\n........\nSS......\nM.......\n', '.......A\n........\n........\n........\n........\n.S......\nS.......\nMS......\n']
Demo Output:
['WIN\n', 'LOSE\n', 'LOSE\n']
Note:
none | ```python
r, s = [63], ''.join(input() + 'T' for i in range(8)) + 'T' * 9
for i in range(0, 72, 9):
t = set()
for x in r:
for y in (x, x - 1, x + 1, x - 9, x + 9, x - 10, x - 8, x + 10, x + 8):
if s[y] == 'T': continue
if (y < i or s[y - i] != 'S') and (y < i + 9 or s[y - i - 9] != 'S'): t.add(y)
r = t
print('WIN' if r else 'LOSE')
``` | 3 | |
509 | A | Maximum in Table | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. | The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. | Print a single line containing a positive integer *m* — the maximum value in the table. | [
"1\n",
"5\n"
] | [
"1",
"70"
] | In the second test the rows of the table look as follows: | 0 | [
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
... | 1,597,246,829 | 2,147,483,647 | PyPy 3 | OK | TESTS | 10 | 124 | 0 | n = int(input())
M = []
for i in range(n):
row = [1]
if i == 0:
for j in range(1, n):
row.append(1)
else:
for j in range(1, n):
row.append(row[j-1] + M[i-1][j])
M.append(row)
print(M[n-1][n-1])
| Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows: | ```python
n = int(input())
M = []
for i in range(n):
row = [1]
if i == 0:
for j in range(1, n):
row.append(1)
else:
for j in range(1, n):
row.append(row[j-1] + M[i-1][j])
M.append(row)
print(M[n-1][n-1])
``` | 3 | |
910 | A | The Way to Home | PROGRAMMING | 800 | [
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null | A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1. | The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one. | If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1. | [
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] | [
"2\n",
"-1\n",
"3\n",
"4\n"
] | In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | 500 | [
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
... | 1,514,404,726 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 62 | 5,529,600 | #10010101
n,e = map(int,input().split())
s = input()
i = 0
d = e
ans = -1
flag = 0
while(i<n-1):
if(n-i-1 < e):
e = n - i - 1
d = e
if(d==0):
flag = 1
print("-1")
break
if(s[i+d]=='1'):
i = i + d
ans+=1
d = e
else:
d-=1
if(flag==0):
print(ans+1)
| Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | ```python
#10010101
n,e = map(int,input().split())
s = input()
i = 0
d = e
ans = -1
flag = 0
while(i<n-1):
if(n-i-1 < e):
e = n - i - 1
d = e
if(d==0):
flag = 1
print("-1")
break
if(s[i+d]=='1'):
i = i + d
ans+=1
d = e
else:
d-=1
if(flag==0):
print(ans+1)
``` | 3 | |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,689,808,656 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 124 | 0 | print(4 - len(set(map(int , input().split()))))
| Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
print(4 - len(set(map(int , input().split()))))
``` | 3 | |
433 | B | Kuriyama Mirai's Stones | PROGRAMMING | 1,200 | [
"dp",
"implementation",
"sortings"
] | null | null | Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one. | Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input. | [
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] | [
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] | Please note that the answers to the questions may overflow 32-bit integer type. | 1,500 | [
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1... | 1,649,915,393 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 436 | 10,649,600 | import sys
import threading
from sys import stdin, stdout
from itertools import accumulate
input = stdin.readline
print = stdout.write
if __name__ == "__main__":
n = int(input().strip())
arr = list(map(int, input().strip().split()))
arr_sort = sorted(arr) #this is the t = O(nlogn) step
cumulative_orig = list(accumulate(arr)) #this is t = O(n)
cumulative_sort = list(accumulate(arr_sort))
m = int(input().strip())
for query in range(m):
type, left_index, right_index = map(int, input().strip().split())
left_index -= 1 #Our code assumes 0-indexing of lists but question expects 1-indexing (nonsense!)
right_index -= 1
if type == 1:
print(f"{cumulative_orig[right_index] - cumulative_orig[left_index] + arr[left_index]}\n")
else:
print(f"{cumulative_sort[right_index] - cumulative_sort[left_index] + arr_sort[left_index]}\n") | Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type. | ```python
import sys
import threading
from sys import stdin, stdout
from itertools import accumulate
input = stdin.readline
print = stdout.write
if __name__ == "__main__":
n = int(input().strip())
arr = list(map(int, input().strip().split()))
arr_sort = sorted(arr) #this is the t = O(nlogn) step
cumulative_orig = list(accumulate(arr)) #this is t = O(n)
cumulative_sort = list(accumulate(arr_sort))
m = int(input().strip())
for query in range(m):
type, left_index, right_index = map(int, input().strip().split())
left_index -= 1 #Our code assumes 0-indexing of lists but question expects 1-indexing (nonsense!)
right_index -= 1
if type == 1:
print(f"{cumulative_orig[right_index] - cumulative_orig[left_index] + arr[left_index]}\n")
else:
print(f"{cumulative_sort[right_index] - cumulative_sort[left_index] + arr_sort[left_index]}\n")
``` | 3 | |
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"
},
{
"inpu... | 1,621,612,819 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 62 | 0 | n=int(input())
a=["Sheldon","Leonard","Penny","Rajesh","Howard"]
R=1
while(R*5<n):
n-=R*5
R*=2
print(a[(n-1)//R])
| 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
n=int(input())
a=["Sheldon","Leonard","Penny","Rajesh","Howard"]
R=1
while(R*5<n):
n-=R*5
R*=2
print(a[(n-1)//R])
``` | 3.969 |
608 | A | Saitama Destroys Hotel | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to *s* and elevator initially starts on floor *s* at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0. | The first line of input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=100, 1<=≤<=*s*<=≤<=1000) — the number of passengers and the number of the top floor respectively.
The next *n* lines each contain two space-separated integers *f**i* and *t**i* (1<=≤<=*f**i*<=≤<=*s*, 1<=≤<=*t**i*<=≤<=1000) — the floor and the time of arrival in seconds for the passenger number *i*. | Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0. | [
"3 7\n2 1\n3 8\n5 2\n",
"5 10\n2 77\n3 33\n8 21\n9 12\n10 64\n"
] | [
"11\n",
"79\n"
] | In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | 500 | [
{
"input": "3 7\n2 1\n3 8\n5 2",
"output": "11"
},
{
"input": "5 10\n2 77\n3 33\n8 21\n9 12\n10 64",
"output": "79"
},
{
"input": "1 1000\n1000 1000",
"output": "2000"
},
{
"input": "1 1\n1 1",
"output": "2"
},
{
"input": "1 1000\n1 1",
"output": "1000"
},
... | 1,619,171,833 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 62 | 204,800 | n,s=(map(int,input().split()))
x=[]
for i in range(n):
a,b=(map(int,input().split()))
x.append([a,b])
x.sort(key=lambda i:i[0],reverse=True)
c=0
p=s
for i in x:
c=max(i[1],c+p-i[0])
p=i[0]
c+=i[0]
print(c) | Title: Saitama Destroys Hotel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to *s* and elevator initially starts on floor *s* at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input Specification:
The first line of input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=100, 1<=≤<=*s*<=≤<=1000) — the number of passengers and the number of the top floor respectively.
The next *n* lines each contain two space-separated integers *f**i* and *t**i* (1<=≤<=*f**i*<=≤<=*s*, 1<=≤<=*t**i*<=≤<=1000) — the floor and the time of arrival in seconds for the passenger number *i*.
Output Specification:
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Demo Input:
['3 7\n2 1\n3 8\n5 2\n', '5 10\n2 77\n3 33\n8 21\n9 12\n10 64\n']
Demo Output:
['11\n', '79\n']
Note:
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | ```python
n,s=(map(int,input().split()))
x=[]
for i in range(n):
a,b=(map(int,input().split()))
x.append([a,b])
x.sort(key=lambda i:i[0],reverse=True)
c=0
p=s
for i in x:
c=max(i[1],c+p-i[0])
p=i[0]
c+=i[0]
print(c)
``` | 3 | |
573 | A | Bear and Poker | PROGRAMMING | 1,300 | [
"implementation",
"math",
"number theory"
] | null | null | Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? | First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. | Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. | [
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | 500 | [
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724... | 1,561,604,379 | 2,147,483,647 | PyPy 3 | OK | TESTS | 70 | 342 | 12,902,400 | def euc(x, y):
if y == 0: return x
return euc(y, x%y)
def judge(x):
m = x
while m % 2 == 0:
m //= 2
while m % 3 == 0:
m //= 3
return m == 1
N = int(input())
A = list(map(int, input().split()))
A.sort()
ans = 'Yes'
for i in range(0, N-1):
X = A[i]*A[N-1]//euc(A[i], A[N-1])
if not judge(X//euc(A[i], A[N-1])):
ans = 'No'
break
print(ans) | Title: Bear and Poker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | ```python
def euc(x, y):
if y == 0: return x
return euc(y, x%y)
def judge(x):
m = x
while m % 2 == 0:
m //= 2
while m % 3 == 0:
m //= 3
return m == 1
N = int(input())
A = list(map(int, input().split()))
A.sort()
ans = 'Yes'
for i in range(0, N-1):
X = A[i]*A[N-1]//euc(A[i], A[N-1])
if not judge(X//euc(A[i], A[N-1])):
ans = 'No'
break
print(ans)
``` | 3 | |
876 | B | Divisiblity of Differences | PROGRAMMING | 1,300 | [
"implementation",
"math",
"number theory"
] | null | null | You are given a multiset of *n* integers. You should select exactly *k* of them in a such way that the difference between any two of them is divisible by *m*, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset. | First line contains three integers *n*, *k* and *m* (2<=≤<=*k*<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the numbers in the multiset. | If it is not possible to select *k* numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print *k* integers *b*1,<=*b*2,<=...,<=*b**k* — the selected numbers. If there are multiple possible solutions, print any of them. | [
"3 2 3\n1 8 4\n",
"3 3 3\n1 8 4\n",
"4 3 5\n2 7 7 7\n"
] | [
"Yes\n1 4 ",
"No",
"Yes\n2 7 7 "
] | none | 1,000 | [
{
"input": "3 2 3\n1 8 4",
"output": "Yes\n1 4 "
},
{
"input": "3 3 3\n1 8 4",
"output": "No"
},
{
"input": "4 3 5\n2 7 7 7",
"output": "Yes\n2 7 7 "
},
{
"input": "9 9 5\n389149775 833127990 969340400 364457730 48649145 316121525 640054660 924273385 973207825",
"output":... | 1,508,176,544 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 233 | 13,619,200 | n,k,m=map(int,input().split())
a=list(map(int,input().split()))
mods=[0]*m
mod=0
for i in range(n):
mod=a[i]%m
mods[mod]+=1
if mods[mod]==k:
break
else:
print('No')
exit()
print('Yes')
results=[None]*k
count=0
for i in range(n):
cur=a[i]
if cur%m==mod:
results[count]=cur
count+=1
if count==k:
print(' '.join(map(str,results)))
break | Title: Divisiblity of Differences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a multiset of *n* integers. You should select exactly *k* of them in a such way that the difference between any two of them is divisible by *m*, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.
Input Specification:
First line contains three integers *n*, *k* and *m* (2<=≤<=*k*<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the numbers in the multiset.
Output Specification:
If it is not possible to select *k* numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print *k* integers *b*1,<=*b*2,<=...,<=*b**k* — the selected numbers. If there are multiple possible solutions, print any of them.
Demo Input:
['3 2 3\n1 8 4\n', '3 3 3\n1 8 4\n', '4 3 5\n2 7 7 7\n']
Demo Output:
['Yes\n1 4 ', 'No', 'Yes\n2 7 7 ']
Note:
none | ```python
n,k,m=map(int,input().split())
a=list(map(int,input().split()))
mods=[0]*m
mod=0
for i in range(n):
mod=a[i]%m
mods[mod]+=1
if mods[mod]==k:
break
else:
print('No')
exit()
print('Yes')
results=[None]*k
count=0
for i in range(n):
cur=a[i]
if cur%m==mod:
results[count]=cur
count+=1
if count==k:
print(' '.join(map(str,results)))
break
``` | 3 | |
803 | A | Maximal Binary Matrix | PROGRAMMING | 1,400 | [
"constructive algorithms"
] | null | null | You are given matrix with *n* rows and *n* columns filled with zeroes. You should put *k* ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1. | The first line consists of two numbers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=106). | If the answer exists then output resulting matrix. Otherwise output -1. | [
"2 1\n",
"3 2\n",
"2 5\n"
] | [
"1 0 \n0 0 \n",
"1 0 0 \n0 1 0 \n0 0 0 \n",
"-1\n"
] | none | 0 | [
{
"input": "2 1",
"output": "1 0 \n0 0 "
},
{
"input": "3 2",
"output": "1 0 0 \n0 1 0 \n0 0 0 "
},
{
"input": "2 5",
"output": "-1"
},
{
"input": "1 0",
"output": "0 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "20 398",
"output": "1 1 1 1 ... | 1,493,714,562 | 2,147,483,647 | Python 3 | OK | TESTS | 168 | 62 | 5,632,000 | # -*- coding: utf-8 -*-
n,k = map(int, input().split(' '))
if k>n**2:
print(-1)
elif k==n**2:
a = [['1']*n for i in range(n)]
for i in range(n):
a[i] = ' '.join(a[i])
print(a[i])
else:
a = [['0']*n for i in range(n)]
c = 0
i = 0
b = True
while b and i<n:
for j in range(n):
if i==j:
if c+1>k:
continue
a[i][j] = '1'
c += 1
else:
if a[j][i]!='1':
if c + 2 >k:
continue
a[i][j] = '1'
a[j][i] = '1'
c += 2
if c==k:
b = False
break
i += 1
for i in range(n):
a[i] = ' '.join(a[i])
print(a[i])
| Title: Maximal Binary Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given matrix with *n* rows and *n* columns filled with zeroes. You should put *k* ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input Specification:
The first line consists of two numbers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=106).
Output Specification:
If the answer exists then output resulting matrix. Otherwise output -1.
Demo Input:
['2 1\n', '3 2\n', '2 5\n']
Demo Output:
['1 0 \n0 0 \n', '1 0 0 \n0 1 0 \n0 0 0 \n', '-1\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
n,k = map(int, input().split(' '))
if k>n**2:
print(-1)
elif k==n**2:
a = [['1']*n for i in range(n)]
for i in range(n):
a[i] = ' '.join(a[i])
print(a[i])
else:
a = [['0']*n for i in range(n)]
c = 0
i = 0
b = True
while b and i<n:
for j in range(n):
if i==j:
if c+1>k:
continue
a[i][j] = '1'
c += 1
else:
if a[j][i]!='1':
if c + 2 >k:
continue
a[i][j] = '1'
a[j][i] = '1'
c += 2
if c==k:
b = False
break
i += 1
for i in range(n):
a[i] = ' '.join(a[i])
print(a[i])
``` | 3 | |
998 | B | Cutting | PROGRAMMING | 1,200 | [
"dp",
"greedy",
"sortings"
] | null | null | There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins. | First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers | Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins. | [
"6 4\n1 2 5 10 15 20\n",
"4 10\n1 3 2 4\n",
"6 100\n1 2 3 4 5 6\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins. | 1,000 | [
{
"input": "6 4\n1 2 5 10 15 20",
"output": "1"
},
{
"input": "4 10\n1 3 2 4",
"output": "0"
},
{
"input": "6 100\n1 2 3 4 5 6",
"output": "2"
},
{
"input": "2 100\n13 78",
"output": "0"
},
{
"input": "10 1\n56 56 98 2 11 64 97 41 95 53",
"output": "0"
},
... | 1,601,402,468 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 514 | 10,752,000 | import math
from decimal import *
import random
mod = int(1e9)+7
n,k = map(int, input().split())
arr = list(map(int, input().split()))
odd, even = 0,0
pos = []
for i in range(n-1):
if(arr[i]%2==0):
even+=1
elif(arr[i]%2==1):
odd+=1
if(odd==even):
pos.append(abs(arr[i+1]-arr[i]))
pos = sorted(pos)
ans= 0
if(pos):
while(k-pos[ans] >=0):
k-=pos[ans]
ans+=1
if(ans==len(pos)):
break
print(ans)
| Title: Cutting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.
Input Specification:
First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers
Output Specification:
Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.
Demo Input:
['6 4\n1 2 5 10 15 20\n', '4 10\n1 3 2 4\n', '6 100\n1 2 3 4 5 6\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins. | ```python
import math
from decimal import *
import random
mod = int(1e9)+7
n,k = map(int, input().split())
arr = list(map(int, input().split()))
odd, even = 0,0
pos = []
for i in range(n-1):
if(arr[i]%2==0):
even+=1
elif(arr[i]%2==1):
odd+=1
if(odd==even):
pos.append(abs(arr[i+1]-arr[i]))
pos = sorted(pos)
ans= 0
if(pos):
while(k-pos[ans] >=0):
k-=pos[ans]
ans+=1
if(ans==len(pos)):
break
print(ans)
``` | 3 | |
385 | B | Bear and Strings | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation",
"math",
"strings"
] | null | null | The bear has a string *s*<==<=*s*1*s*2... *s*|*s*| (record |*s*| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices *i*,<=*j* (1<=≤<=*i*<=≤<=*j*<=≤<=|*s*|), that string *x*(*i*,<=*j*)<==<=*s**i**s**i*<=+<=1... *s**j* contains at least one string "bear" as a substring.
String *x*(*i*,<=*j*) contains string "bear", if there is such index *k* (*i*<=≤<=*k*<=≤<=*j*<=-<=3), that *s**k*<==<=*b*, *s**k*<=+<=1<==<=*e*, *s**k*<=+<=2<==<=*a*, *s**k*<=+<=3<==<=*r*.
Help the bear cope with the given problem. | The first line contains a non-empty string *s* (1<=≤<=|*s*|<=≤<=5000). It is guaranteed that the string only consists of lowercase English letters. | Print a single number — the answer to the problem. | [
"bearbtear\n",
"bearaabearc\n"
] | [
"6\n",
"20\n"
] | In the first sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11). | 1,000 | [
{
"input": "bearbtear",
"output": "6"
},
{
"input": "bearaabearc",
"output": "20"
},
{
"input": "pbearbearhbearzqbearjkterasjhy",
"output": "291"
},
{
"input": "pbearjbearbebearnbabcffbearbearwubearjezpiorrbearbearjbdlbearbearqbearjbearwipmsbearoaftrsebearzsnqb",
"output"... | 1,503,282,772 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 124 | 0 | s = input()
ind = [i for (i, _) in enumerate(s) if s.startswith('bear', i)]
L = len(s)
r, z = 0, 0
for e in ind:
for _ in range(z, e + 1):
z = e + 1
r += L - (e + 3)
print(r) | Title: Bear and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The bear has a string *s*<==<=*s*1*s*2... *s*|*s*| (record |*s*| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices *i*,<=*j* (1<=≤<=*i*<=≤<=*j*<=≤<=|*s*|), that string *x*(*i*,<=*j*)<==<=*s**i**s**i*<=+<=1... *s**j* contains at least one string "bear" as a substring.
String *x*(*i*,<=*j*) contains string "bear", if there is such index *k* (*i*<=≤<=*k*<=≤<=*j*<=-<=3), that *s**k*<==<=*b*, *s**k*<=+<=1<==<=*e*, *s**k*<=+<=2<==<=*a*, *s**k*<=+<=3<==<=*r*.
Help the bear cope with the given problem.
Input Specification:
The first line contains a non-empty string *s* (1<=≤<=|*s*|<=≤<=5000). It is guaranteed that the string only consists of lowercase English letters.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['bearbtear\n', 'bearaabearc\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11). | ```python
s = input()
ind = [i for (i, _) in enumerate(s) if s.startswith('bear', i)]
L = len(s)
r, z = 0, 0
for e in ind:
for _ in range(z, e + 1):
z = e + 1
r += L - (e + 3)
print(r)
``` | 3 | |
151 | A | Soft Drinking | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make? | The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space. | Print a single integer — the number of toasts each friend can make. | [
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] | [
"2\n",
"3\n",
"0\n"
] | A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | 500 | [
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
... | 1,688,359,741 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | a = list(map(int, input().split()))
total_drink = a[1] * a[2]
total_slice = a[3] * a[4]
total_salt = a[5] // a[7]
print(min(total_drink//a[6], total_slice, total_salt) // a[0]) | Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | ```python
a = list(map(int, input().split()))
total_drink = a[1] * a[2]
total_slice = a[3] * a[4]
total_salt = a[5] // a[7]
print(min(total_drink//a[6], total_slice, total_salt) // a[0])
``` | 3 | |
715 | B | Complete The Graph | PROGRAMMING | 2,300 | [
"binary search",
"constructive algorithms",
"graphs",
"shortest paths"
] | null | null | ZS the Coder has drawn an undirected graph of *n* vertices numbered from 0 to *n*<=-<=1 and *m* edges between them. Each edge of the graph is weighted, each weight is a positive integer.
The next day, ZS the Coder realized that some of the weights were erased! So he wants to reassign positive integer weight to each of the edges which weights were erased, so that the length of the shortest path between vertices *s* and *t* in the resulting graph is exactly *L*. Can you help him? | The first line contains five integers *n*,<=*m*,<=*L*,<=*s*,<=*t* (2<=≤<=*n*<=≤<=1000,<=<=1<=≤<=*m*<=≤<=10<=000,<=<=1<=≤<=*L*<=≤<=109,<=<=0<=≤<=*s*,<=*t*<=≤<=*n*<=-<=1,<=<=*s*<=≠<=*t*) — the number of vertices, number of edges, the desired length of shortest path, starting vertex and ending vertex respectively.
Then, *m* lines describing the edges of the graph follow. *i*-th of them contains three integers, *u**i*,<=*v**i*,<=*w**i* (0<=≤<=*u**i*,<=*v**i*<=≤<=*n*<=-<=1,<=<=*u**i*<=≠<=*v**i*,<=<=0<=≤<=*w**i*<=≤<=109). *u**i* and *v**i* denote the endpoints of the edge and *w**i* denotes its weight. If *w**i* is equal to 0 then the weight of the corresponding edge was erased.
It is guaranteed that there is at most one edge between any pair of vertices. | Print "NO" (without quotes) in the only line if it's not possible to assign the weights in a required way.
Otherwise, print "YES" in the first line. Next *m* lines should contain the edges of the resulting graph, with weights assigned to edges which weights were erased. *i*-th of them should contain three integers *u**i*, *v**i* and *w**i*, denoting an edge between vertices *u**i* and *v**i* of weight *w**i*. The edges of the new graph must coincide with the ones in the graph from the input. The weights that were not erased must remain unchanged whereas the new weights can be any positive integer not exceeding 1018.
The order of the edges in the output doesn't matter. The length of the shortest path between *s* and *t* must be equal to *L*.
If there are multiple solutions, print any of them. | [
"5 5 13 0 4\n0 1 5\n2 1 2\n3 2 3\n1 4 0\n4 3 4\n",
"2 1 123456789 0 1\n0 1 0\n",
"2 1 999999999 1 0\n0 1 1000000000\n"
] | [
"YES\n0 1 5\n2 1 2\n3 2 3\n1 4 8\n4 3 4\n",
"YES\n0 1 123456789\n",
"NO\n"
] | Here's how the graph in the first sample case looks like :
In the first sample case, there is only one missing edge weight. Placing the weight of 8 gives a shortest path from 0 to 4 of length 13.
In the second sample case, there is only a single edge. Clearly, the only way is to replace the missing weight with 123456789.
In the last sample case, there is no weights to assign but the length of the shortest path doesn't match the required value, so the answer is "NO". | 1,000 | [
{
"input": "5 5 13 0 4\n0 1 5\n2 1 2\n3 2 3\n1 4 0\n4 3 4",
"output": "YES\n0 1 5\n2 1 2\n3 2 3\n1 4 8\n4 3 4"
},
{
"input": "2 1 123456789 0 1\n0 1 0",
"output": "YES\n0 1 123456789"
},
{
"input": "2 1 999999999 1 0\n0 1 1000000000",
"output": "NO"
},
{
"input": "4 5 10 1 2\... | 1,684,655,110 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 179 | 280 | 13,824,000 | import heapq
import random
import sys
from math import inf
from types import GeneratorType
RANDOM = random.randint(1, 10 ** 9)
class FastIO:
def __init__(self):
return
@staticmethod
def read_str():
return sys.stdin.readline()
def read_int(self):
return int(self.read_str())
def read_float(self):
return float(self.read_str())
def read_ints(self):
return map(int, self.read_str().split())
def read_floats(self):
return map(float, self.read_str().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self.read_str().split())
def read_list_ints(self):
return list(map(int, self.read_str().split()))
def read_list_floats(self):
return list(map(float, self.read_str().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self.read_str().split()))
def read_list_strs(self):
return self.read_str().split()
def read_list_str(self):
return list(self.read_str())
@staticmethod
def st(x):
return print(x)
@staticmethod
def lst(x):
return print(*x)
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
def ask(self, lst):
self.lst(lst)
sys.stdout.flush()
res = self.read_int()
return res
@staticmethod
def accumulate(nums):
n = len(nums)
pre = [0] * (n + 1)
for i in range(n):
pre[i + 1] = pre[i] + nums[i]
return pre
class Wrapper(int):
# 用来规避 py 哈希碰撞的问题和进行加速
def __init__(self, x):
int.__init__(x)
# 原理是异或一个随机种子
def __hash__(self):
# 也可以将数组排序后进行哈希计数
return super(Wrapper, self).__hash__() ^ RANDOM
class Solution:
def __init__(self):
return
@staticmethod
def main(ac=FastIO()):
n, m, target, source, destination = ac.read_ints()
edges = []
dct = [[] for _ in range(n)]
book = [0] * m
for ind in range(m):
i, j, w = ac.read_list_ints()
if w == 0:
w = 1
book[ind] = 1
edges.append([i, j, w])
dct[i].append([ind, j])
dct[j].append([ind, i])
# 第一遍最短路计算最小情况下的距离
dis0 = [inf] * n
stack = [[0, source]]
dis0[source] = 0
while stack:
d, i = heapq.heappop(stack)
if dis0[i] < d:
continue
for ind, j in dct[i]:
dj = edges[ind][2] + d
if dj < dis0[j]:
dis0[j] = dj
heapq.heappush(stack, [dj, j])
if dis0[destination] > target:
ac.st("NO")
return
# 第二遍最短路
dis1 = [inf] * n
stack = [[0, source]]
dis1[source] = 0
while stack:
d, i = heapq.heappop(stack)
if dis1[i] < d:
continue
for ind, j in dct[i]:
if book[ind]:
# 假设 (i, j) 是最短路上的边
if (edges[ind][2] + dis1[i]) + (dis0[destination] - dis0[j]) < target:
# 此时还有一些增长空间即（当前到达 j 的距离）加上（剩余 j 到 destination）的距离仍旧小于 target
x = target - (edges[ind][2] + dis1[i]) - (dis0[destination] - dis0[j])
edges[ind][2] += x
book[ind] = 0
dj = edges[ind][2] + d
if dj < dis1[j]:
dis1[j] = dj
heapq.heappush(stack, [dj, j])
if dis1[destination] == target:
ac.st("YES")
for e in edges:
ac.lst(e)
else:
ac.st("NO")
return
Solution().main()
| Title: Complete The Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder has drawn an undirected graph of *n* vertices numbered from 0 to *n*<=-<=1 and *m* edges between them. Each edge of the graph is weighted, each weight is a positive integer.
The next day, ZS the Coder realized that some of the weights were erased! So he wants to reassign positive integer weight to each of the edges which weights were erased, so that the length of the shortest path between vertices *s* and *t* in the resulting graph is exactly *L*. Can you help him?
Input Specification:
The first line contains five integers *n*,<=*m*,<=*L*,<=*s*,<=*t* (2<=≤<=*n*<=≤<=1000,<=<=1<=≤<=*m*<=≤<=10<=000,<=<=1<=≤<=*L*<=≤<=109,<=<=0<=≤<=*s*,<=*t*<=≤<=*n*<=-<=1,<=<=*s*<=≠<=*t*) — the number of vertices, number of edges, the desired length of shortest path, starting vertex and ending vertex respectively.
Then, *m* lines describing the edges of the graph follow. *i*-th of them contains three integers, *u**i*,<=*v**i*,<=*w**i* (0<=≤<=*u**i*,<=*v**i*<=≤<=*n*<=-<=1,<=<=*u**i*<=≠<=*v**i*,<=<=0<=≤<=*w**i*<=≤<=109). *u**i* and *v**i* denote the endpoints of the edge and *w**i* denotes its weight. If *w**i* is equal to 0 then the weight of the corresponding edge was erased.
It is guaranteed that there is at most one edge between any pair of vertices.
Output Specification:
Print "NO" (without quotes) in the only line if it's not possible to assign the weights in a required way.
Otherwise, print "YES" in the first line. Next *m* lines should contain the edges of the resulting graph, with weights assigned to edges which weights were erased. *i*-th of them should contain three integers *u**i*, *v**i* and *w**i*, denoting an edge between vertices *u**i* and *v**i* of weight *w**i*. The edges of the new graph must coincide with the ones in the graph from the input. The weights that were not erased must remain unchanged whereas the new weights can be any positive integer not exceeding 1018.
The order of the edges in the output doesn't matter. The length of the shortest path between *s* and *t* must be equal to *L*.
If there are multiple solutions, print any of them.
Demo Input:
['5 5 13 0 4\n0 1 5\n2 1 2\n3 2 3\n1 4 0\n4 3 4\n', '2 1 123456789 0 1\n0 1 0\n', '2 1 999999999 1 0\n0 1 1000000000\n']
Demo Output:
['YES\n0 1 5\n2 1 2\n3 2 3\n1 4 8\n4 3 4\n', 'YES\n0 1 123456789\n', 'NO\n']
Note:
Here's how the graph in the first sample case looks like :
In the first sample case, there is only one missing edge weight. Placing the weight of 8 gives a shortest path from 0 to 4 of length 13.
In the second sample case, there is only a single edge. Clearly, the only way is to replace the missing weight with 123456789.
In the last sample case, there is no weights to assign but the length of the shortest path doesn't match the required value, so the answer is "NO". | ```python
import heapq
import random
import sys
from math import inf
from types import GeneratorType
RANDOM = random.randint(1, 10 ** 9)
class FastIO:
def __init__(self):
return
@staticmethod
def read_str():
return sys.stdin.readline()
def read_int(self):
return int(self.read_str())
def read_float(self):
return float(self.read_str())
def read_ints(self):
return map(int, self.read_str().split())
def read_floats(self):
return map(float, self.read_str().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self.read_str().split())
def read_list_ints(self):
return list(map(int, self.read_str().split()))
def read_list_floats(self):
return list(map(float, self.read_str().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self.read_str().split()))
def read_list_strs(self):
return self.read_str().split()
def read_list_str(self):
return list(self.read_str())
@staticmethod
def st(x):
return print(x)
@staticmethod
def lst(x):
return print(*x)
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
def ask(self, lst):
self.lst(lst)
sys.stdout.flush()
res = self.read_int()
return res
@staticmethod
def accumulate(nums):
n = len(nums)
pre = [0] * (n + 1)
for i in range(n):
pre[i + 1] = pre[i] + nums[i]
return pre
class Wrapper(int):
# 用来规避 py 哈希碰撞的问题和进行加速
def __init__(self, x):
int.__init__(x)
# 原理是异或一个随机种子
def __hash__(self):
# 也可以将数组排序后进行哈希计数
return super(Wrapper, self).__hash__() ^ RANDOM
class Solution:
def __init__(self):
return
@staticmethod
def main(ac=FastIO()):
n, m, target, source, destination = ac.read_ints()
edges = []
dct = [[] for _ in range(n)]
book = [0] * m
for ind in range(m):
i, j, w = ac.read_list_ints()
if w == 0:
w = 1
book[ind] = 1
edges.append([i, j, w])
dct[i].append([ind, j])
dct[j].append([ind, i])
# 第一遍最短路计算最小情况下的距离
dis0 = [inf] * n
stack = [[0, source]]
dis0[source] = 0
while stack:
d, i = heapq.heappop(stack)
if dis0[i] < d:
continue
for ind, j in dct[i]:
dj = edges[ind][2] + d
if dj < dis0[j]:
dis0[j] = dj
heapq.heappush(stack, [dj, j])
if dis0[destination] > target:
ac.st("NO")
return
# 第二遍最短路
dis1 = [inf] * n
stack = [[0, source]]
dis1[source] = 0
while stack:
d, i = heapq.heappop(stack)
if dis1[i] < d:
continue
for ind, j in dct[i]:
if book[ind]:
# 假设 (i, j) 是最短路上的边
if (edges[ind][2] + dis1[i]) + (dis0[destination] - dis0[j]) < target:
# 此时还有一些增长空间即（当前到达 j 的距离）加上（剩余 j 到 destination）的距离仍旧小于 target
x = target - (edges[ind][2] + dis1[i]) - (dis0[destination] - dis0[j])
edges[ind][2] += x
book[ind] = 0
dj = edges[ind][2] + d
if dj < dis1[j]:
dis1[j] = dj
heapq.heappush(stack, [dj, j])
if dis1[destination] == target:
ac.st("YES")
for e in edges:
ac.lst(e)
else:
ac.st("NO")
return
Solution().main()
``` | 3 | |
667 | A | Pouring Rain | PROGRAMMING | 1,100 | [
"geometry",
"math"
] | null | null | A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.
Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.
Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom.
You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.
Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds.
Note one milliliter equals to one cubic centimeter. | The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where:
- *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup. | If it is impossible to make the cup empty, print "NO" (without quotes).
Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104. | [
"1 2 3 100\n",
"1 1 1 1\n"
] | [
"NO\n",
"YES\n3.659792366325\n"
] | In the first example the water fills the cup faster than you can drink from it.
In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds. | 500 | [
{
"input": "1 2 3 100",
"output": "NO"
},
{
"input": "1 1 1 1",
"output": "YES\n3.659792366325"
},
{
"input": "48 7946 7992 72",
"output": "NO"
},
{
"input": "72 6791 8546 46",
"output": "NO"
},
{
"input": "100 5635 9099 23",
"output": "NO"
},
{
"input... | 1,578,216,171 | 2,147,483,647 | PyPy 3 | OK | TESTS | 23 | 140 | 0 | from math import pi
X = list(map(int, input().split()))
RainPouring = X[-1] * (X[0] / 2) ** 2 * pi
if RainPouring >= X[-2]: print("NO");exit()
X[-2] -= RainPouring
Volume = X[1] * (X[0] / 2) ** 2 * pi
print("YES", Volume / X[-2], sep='\n')
# Caption: With the help of ARPA
| Title: Pouring Rain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.
Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.
Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom.
You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.
Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds.
Note one milliliter equals to one cubic centimeter.
Input Specification:
The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where:
- *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup.
Output Specification:
If it is impossible to make the cup empty, print "NO" (without quotes).
Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104.
Demo Input:
['1 2 3 100\n', '1 1 1 1\n']
Demo Output:
['NO\n', 'YES\n3.659792366325\n']
Note:
In the first example the water fills the cup faster than you can drink from it.
In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds. | ```python
from math import pi
X = list(map(int, input().split()))
RainPouring = X[-1] * (X[0] / 2) ** 2 * pi
if RainPouring >= X[-2]: print("NO");exit()
X[-2] -= RainPouring
Volume = X[1] * (X[0] / 2) ** 2 * pi
print("YES", Volume / X[-2], sep='\n')
# Caption: With the help of ARPA
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters. | Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. | [
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] | [
"a***b\n",
"***gmg***\n",
"*********\n"
] | The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | 0 | [
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input":... | 1,616,274,381 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 93 | 819,200 | import re
def solve(s):
return re.sub(r'ogo(?:go)*','***',s)
def main() :
n = input()
s = input()
# arr = list(map(int, input().split(' ')))
# arr = []
# for _ in range(4):
# i = input()
# arr.append(i)
print(solve(s))
main()
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | ```python
import re
def solve(s):
return re.sub(r'ogo(?:go)*','***',s)
def main() :
n = input()
s = input()
# arr = list(map(int, input().split(' ')))
# arr = []
# for _ in range(4):
# i = input()
# arr.append(i)
print(solve(s))
main()
``` | 3 | |
339 | A | Helpful Maths | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings",
"strings"
] | null | null | Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum. | The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long. | Print the new sum that Xenia can count. | [
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] | [
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] | none | 500 | [
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": ... | 1,697,553,499 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 21 | 92 | 0 | n = input()
a = b = c = 0
for i in range(0, len(n), 2):
if n[i] == '1':
a += 1
elif n[i] == '2':
b += 1
else:
c += 1
d = "1+" * a + "2+" * b + "3+" * c
print(d[:-1])
| Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none | ```python
n = input()
a = b = c = 0
for i in range(0, len(n), 2):
if n[i] == '1':
a += 1
elif n[i] == '2':
b += 1
else:
c += 1
d = "1+" * a + "2+" * b + "3+" * c
print(d[:-1])
``` | 3 | |
991 | B | Getting an A | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.
The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$.
This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo.
Help Vasya — calculate the minimum amount of lab works Vasya has to redo. | The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$).
The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works. | Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$. | [
"3\n4 4 4\n",
"4\n5 4 5 5\n",
"4\n5 3 3 5\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first sample, it is enough to redo two lab works to make two $4$s into $5$s.
In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$.
In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$. | 1,000 | [
{
"input": "3\n4 4 4",
"output": "2"
},
{
"input": "4\n5 4 5 5",
"output": "0"
},
{
"input": "4\n5 3 3 5",
"output": "1"
},
{
"input": "1\n5",
"output": "0"
},
{
"input": "4\n3 2 5 4",
"output": "2"
},
{
"input": "5\n5 4 3 2 5",
"output": "2"
},
... | 1,628,446,478 | 2,147,483,647 | Python 3 | OK | TESTS | 61 | 77 | 6,963,200 | from sys import *
def main():
n = int(stdin.readline())
target = 4.5*n
currentTotal = 0
minimumRetakes = 0
listOfScores = [int(x) for x in stdin.readline().split()]
for x in listOfScores:
currentTotal+=x
listOfScores.sort()
while(currentTotal < target):
currentTotal+=(5-listOfScores[minimumRetakes])
minimumRetakes += 1
print(minimumRetakes)
if __name__ == '__main__':
main() | Title: Getting an A
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.
The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$.
This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo.
Help Vasya — calculate the minimum amount of lab works Vasya has to redo.
Input Specification:
The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$).
The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works.
Output Specification:
Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$.
Demo Input:
['3\n4 4 4\n', '4\n5 4 5 5\n', '4\n5 3 3 5\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first sample, it is enough to redo two lab works to make two $4$s into $5$s.
In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$.
In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$. | ```python
from sys import *
def main():
n = int(stdin.readline())
target = 4.5*n
currentTotal = 0
minimumRetakes = 0
listOfScores = [int(x) for x in stdin.readline().split()]
for x in listOfScores:
currentTotal+=x
listOfScores.sort()
while(currentTotal < target):
currentTotal+=(5-listOfScores[minimumRetakes])
minimumRetakes += 1
print(minimumRetakes)
if __name__ == '__main__':
main()
``` | 3 | |
30 | A | Accounting | PROGRAMMING | 1,400 | [
"brute force",
"math"
] | A. Accounting | 2 | 256 | A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.
The total income *A* of his kingdom during 0-th year is known, as well as the total income *B* during *n*-th year (these numbers can be negative — it means that there was a loss in the correspondent year).
King wants to show financial stability. To do this, he needs to find common coefficient *X* — the coefficient of income growth during one year. This coefficient should satisfy the equation:
Surely, the king is not going to do this job by himself, and demands you to find such number *X*.
It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient *X* must be integers. The number *X* may be zero or negative. | The input contains three integers *A*, *B*, *n* (|*A*|,<=|*B*|<=≤<=1000, 1<=≤<=*n*<=≤<=10). | Output the required integer coefficient *X*, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them. | [
"2 18 2\n",
"-1 8 3\n",
"0 0 10\n",
"1 16 5\n"
] | [
"3",
"-2",
"5",
"No solution"
] | none | 500 | [
{
"input": "2 18 2",
"output": "3"
},
{
"input": "-1 8 3",
"output": "-2"
},
{
"input": "0 0 10",
"output": "5"
},
{
"input": "1 16 5",
"output": "No solution"
},
{
"input": "0 1 2",
"output": "No solution"
},
{
"input": "3 0 4",
"output": "0"
},... | 1,680,580,362 | 2,147,483,647 | Python 3 | OK | TESTS | 68 | 92 | 0 | import sys
q,w,num = map(int,input().split())
for p in range(-1000,1001):
if q*(p**num)== w:
print(p)
sys.exit()
print("No solution") | Title: Accounting
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.
The total income *A* of his kingdom during 0-th year is known, as well as the total income *B* during *n*-th year (these numbers can be negative — it means that there was a loss in the correspondent year).
King wants to show financial stability. To do this, he needs to find common coefficient *X* — the coefficient of income growth during one year. This coefficient should satisfy the equation:
Surely, the king is not going to do this job by himself, and demands you to find such number *X*.
It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient *X* must be integers. The number *X* may be zero or negative.
Input Specification:
The input contains three integers *A*, *B*, *n* (|*A*|,<=|*B*|<=≤<=1000, 1<=≤<=*n*<=≤<=10).
Output Specification:
Output the required integer coefficient *X*, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them.
Demo Input:
['2 18 2\n', '-1 8 3\n', '0 0 10\n', '1 16 5\n']
Demo Output:
['3', '-2', '5', 'No solution']
Note:
none | ```python
import sys
q,w,num = map(int,input().split())
for p in range(-1000,1001):
if q*(p**num)== w:
print(p)
sys.exit()
print("No solution")
``` | 3.977 |
285 | A | Slightly Decreasing Permutations | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
The decreasing coefficient of permutation *p*1,<=*p*2,<=...,<=*p**n* is the number of such *i* (1<=≤<=*i*<=<<=*n*), that *p**i*<=><=*p**i*<=+<=1.
You have numbers *n* and *k*. Your task is to print the permutation of length *n* with decreasing coefficient *k*. | The single line contains two space-separated integers: *n*,<=*k* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*k*<=<<=*n*) — the permutation length and the decreasing coefficient. | In a single line print *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* — the permutation of length *n* with decreasing coefficient *k*.
If there are several permutations that meet this condition, print any of them. It is guaranteed that the permutation with the sought parameters exists. | [
"5 2\n",
"3 0\n",
"3 2\n"
] | [
"1 5 2 4 3\n",
"1 2 3\n",
"3 2 1\n"
] | none | 500 | [
{
"input": "5 2",
"output": "1 5 2 4 3"
},
{
"input": "3 0",
"output": "1 2 3"
},
{
"input": "3 2",
"output": "3 2 1"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "1 2"
},
{
"input": "2 1",
"output": "2 1"
},
{
"input":... | 1,683,263,419 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 28 | 186 | 9,728,000 | a, b = map(int, input().split())
m = []
j = 0
for i in range(a, 1, -1):
if j != b:
m.append(i)
j += 1
if j == b:
break
print(*m, end = ' ')
v = []
for i in range(1, a-b+1):
v.append(i)
print(*v)
| Title: Slightly Decreasing Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
The decreasing coefficient of permutation *p*1,<=*p*2,<=...,<=*p**n* is the number of such *i* (1<=≤<=*i*<=<<=*n*), that *p**i*<=><=*p**i*<=+<=1.
You have numbers *n* and *k*. Your task is to print the permutation of length *n* with decreasing coefficient *k*.
Input Specification:
The single line contains two space-separated integers: *n*,<=*k* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*k*<=<<=*n*) — the permutation length and the decreasing coefficient.
Output Specification:
In a single line print *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* — the permutation of length *n* with decreasing coefficient *k*.
If there are several permutations that meet this condition, print any of them. It is guaranteed that the permutation with the sought parameters exists.
Demo Input:
['5 2\n', '3 0\n', '3 2\n']
Demo Output:
['1 5 2 4 3\n', '1 2 3\n', '3 2 1\n']
Note:
none | ```python
a, b = map(int, input().split())
m = []
j = 0
for i in range(a, 1, -1):
if j != b:
m.append(i)
j += 1
if j == b:
break
print(*m, end = ' ')
v = []
for i in range(1, a-b+1):
v.append(i)
print(*v)
``` | 3 | |
653 | A | Bear and Three Balls | PROGRAMMING | 900 | [
"brute force",
"implementation",
"sortings"
] | null | null | Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above. | The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball. | Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes). | [
"4\n18 55 16 17\n",
"6\n40 41 43 44 44 44\n",
"8\n5 972 3 4 1 4 970 971\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971. | 500 | [
{
"input": "4\n18 55 16 17",
"output": "YES"
},
{
"input": "6\n40 41 43 44 44 44",
"output": "NO"
},
{
"input": "8\n5 972 3 4 1 4 970 971",
"output": "YES"
},
{
"input": "3\n959 747 656",
"output": "NO"
},
{
"input": "4\n1 2 2 3",
"output": "YES"
},
{
... | 1,458,694,381 | 2,147,483,647 | Python 3 | OK | TESTS | 84 | 62 | 5,120,000 | #http://codeforces.com/contest/653/problem/A
import sys
len_input = input()
seq_input = input().split()
#seq_input = '18 55 16 17'.split()
# seq_input = '40 41 43 44 44 44'.split()
# seq_input = '5 972 3 4 1 4 970 971'.split()
# seq_input = ''.split()
# seq_input = '998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543'.split()
seq = sorted([int(x) for x in seq_input])
seq_red = sorted([int(x) for x in list(set(seq))])
# print(seq)
# print(seq_red)
if len(seq_red) < 3:
print('NO') #less than two distinct options
sys.exit()
seq_sub = [seq_red[i] - seq_red[i+2] for i in range(len(seq_red)-2)]
seq_truth = [1 if abs(x) < 3 else 0 for x in seq_sub]
seq_sum = sum(seq_truth)
if seq_sum == 0:
print('NO') #all values are too far apart
elif seq_sum > 0:
print('YES') | Title: Bear and Three Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input Specification:
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Output Specification:
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Demo Input:
['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971. | ```python
#http://codeforces.com/contest/653/problem/A
import sys
len_input = input()
seq_input = input().split()
#seq_input = '18 55 16 17'.split()
# seq_input = '40 41 43 44 44 44'.split()
# seq_input = '5 972 3 4 1 4 970 971'.split()
# seq_input = ''.split()
# seq_input = '998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543'.split()
seq = sorted([int(x) for x in seq_input])
seq_red = sorted([int(x) for x in list(set(seq))])
# print(seq)
# print(seq_red)
if len(seq_red) < 3:
print('NO') #less than two distinct options
sys.exit()
seq_sub = [seq_red[i] - seq_red[i+2] for i in range(len(seq_red)-2)]
seq_truth = [1 if abs(x) < 3 else 0 for x in seq_sub]
seq_sum = sum(seq_truth)
if seq_sum == 0:
print('NO') #all values are too far apart
elif seq_sum > 0:
print('YES')
``` | 3 | |
863 | A | Quasi-palindrome | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number. | The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes. | Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes). | [
"131\n",
"320\n",
"2010200\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 0 | [
{
"input": "131",
"output": "YES"
},
{
"input": "320",
"output": "NO"
},
{
"input": "2010200",
"output": "YES"
},
{
"input": "1",
"output": "YES"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "999999999",
"output": "YES"
},
{
"i... | 1,579,274,871 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 109 | 0 | s=input()
s=s[::-1]
s=str(int(s))
if s==s[::-1]:
print('YES')
else:
print('NO') | Title: Quasi-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
Input Specification:
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
Output Specification:
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
Demo Input:
['131\n', '320\n', '2010200\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
s=input()
s=s[::-1]
s=str(int(s))
if s==s[::-1]:
print('YES')
else:
print('NO')
``` | 3 | |
471 | A | MUH and Sticks | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:
- Four sticks represent the animal's legs, these sticks should have the same length. - Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.
Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it. | The single line contains six space-separated integers *l**i* (1<=≤<=*l**i*<=≤<=9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks. | If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes). | [
"4 2 5 4 4 4\n",
"4 4 5 4 4 5\n",
"1 2 3 4 5 6\n"
] | [
"Bear",
"Elephant",
"Alien"
] | If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue. | 500 | [
{
"input": "4 2 5 4 4 4",
"output": "Bear"
},
{
"input": "4 4 5 4 4 5",
"output": "Elephant"
},
{
"input": "1 2 3 4 5 6",
"output": "Alien"
},
{
"input": "5 5 5 5 5 5",
"output": "Elephant"
},
{
"input": "1 1 1 2 3 5",
"output": "Alien"
},
{
"input": "... | 1,412,091,275 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 77 | 0 |
sticks = input().split()
four = False
two = False
for stick in sticks:
if sticks.count(stick) >= 4:
four = True
if sticks.count(stick) == 2 or sticks.count(stick) == 6:
two = True
if four:
if two:
print("Elephant")
else:
print("Bear")
else:
print("Alien") | Title: MUH and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:
- Four sticks represent the animal's legs, these sticks should have the same length. - Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.
Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it.
Input Specification:
The single line contains six space-separated integers *l**i* (1<=≤<=*l**i*<=≤<=9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks.
Output Specification:
If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).
Demo Input:
['4 2 5 4 4 4\n', '4 4 5 4 4 5\n', '1 2 3 4 5 6\n']
Demo Output:
['Bear', 'Elephant', 'Alien']
Note:
If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue. | ```python
sticks = input().split()
four = False
two = False
for stick in sticks:
if sticks.count(stick) >= 4:
four = True
if sticks.count(stick) == 2 or sticks.count(stick) == 6:
two = True
if four:
if two:
print("Elephant")
else:
print("Bear")
else:
print("Alien")
``` | 3 | |
467 | A | George and Accommodation | PROGRAMMING | 800 | [
"implementation"
] | null | null | George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity. | Print a single integer — the number of rooms where George and Alex can move in. | [
"3\n1 1\n2 2\n3 3\n",
"3\n1 10\n0 10\n10 10\n"
] | [
"0\n",
"2\n"
] | none | 500 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "0"
},
{
"input": "3\n1 10\n0 10\n10 10",
"output": "2"
},
{
"input": "2\n36 67\n61 69",
"output": "2"
},
{
"input": "3\n21 71\n10 88\n43 62",
"output": "3"
},
{
"input": "3\n1 2\n2 3\n3 4",
"output": "0"
},
{
"... | 1,694,139,163 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 27 | 62 | 0 | c = 0
for room in range(int(input())):
p, q = map(int, input().split())
if p + 2 <= q:
c += 1
else:
c += 0
print(c)
| Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none | ```python
c = 0
for room in range(int(input())):
p, q = map(int, input().split())
if p + 2 <= q:
c += 1
else:
c += 0
print(c)
``` | 3 | |
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"
},... | 1,492,358,199 | 1,299 | Python 3 | OK | TESTS | 35 | 62 | 5,529,600 | x = [i for i in input()]
y = [i for i in input()]
z = []
possible = True
for i in range(len(x)):
if x[i] == y[i]:
z.append('z')
elif x[i] > y[i]:
z.append(y[i])
else:
possible = False
break
if possible:
print("".join(z))
else:
print(-1)
| 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 = [i for i in input()]
y = [i for i in input()]
z = []
possible = True
for i in range(len(x)):
if x[i] == y[i]:
z.append('z')
elif x[i] > y[i]:
z.append(y[i])
else:
possible = False
break
if possible:
print("".join(z))
else:
print(-1)
``` | 3 | |
743 | E | Vladik and cards | PROGRAMMING | 2,200 | [
"binary search",
"bitmasks",
"brute force",
"dp"
] | null | null | Vladik was bored on his way home and decided to play the following game. He took *n* cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:
- the number of occurrences of each number from 1 to 8 in the subsequence doesn't differ by more then 1 from the number of occurrences of any other number. Formally, if there are *c**k* cards with number *k* on them in the subsequence, than for all pairs of integers the condition |*c**i*<=-<=*c**j*|<=≤<=1 must hold. - if there is at least one card with number *x* on it in the subsequence, then all cards with number *x* in this subsequence must form a continuous segment in it (but not necessarily a continuous segment in the original sequence). For example, the subsequence [1,<=1,<=2,<=2] satisfies this condition while the subsequence [1,<=2,<=2,<=1] doesn't. Note that [1,<=1,<=2,<=2] doesn't satisfy the first condition.
Please help Vladik to find the length of the longest subsequence that satisfies both conditions. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards in Vladik's sequence.
The second line contains the sequence of *n* positive integers not exceeding 8 — the description of Vladik's sequence. | Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions. | [
"3\n1 1 1\n",
"8\n8 7 6 5 4 3 2 1\n",
"24\n1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8\n"
] | [
"1",
"8",
"17"
] | In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition. | 2,500 | [
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "8\n8 7 6 5 4 3 2 1",
"output": "8"
},
{
"input": "24\n1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8",
"output": "17"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "2\n5 4",
"output": "2"
},
{
"i... | 1,675,524,759 | 4,359 | PyPy 3 | OK | TESTS | 41 | 233 | 24,780,800 | import bisect
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v):
return (m + 1) * u + v
n = int(input())
a = list(map(int, input().split()))
m = 8
x = [[] for _ in range(m)]
for i in range(n):
x[a[i] - 1].append(i)
s = 0
for y in x:
s += min(len(y), 1)
if s < m:
ans = s
print(ans)
exit()
pow2 = [1]
for _ in range(m):
pow2.append(2 * pow2[-1])
pm = pow2[m]
inf = pow(10, 9) + 1
ans = 8
ok = 1
for c in range(1, n // 8 + 3):
dp = [inf] * ((m + 1) * pm)
dp[0] = -1
for i in range(pm):
for j in range(m):
u = f(i, j)
if dp[u] == inf:
break
dpu = dp[u]
for k in range(m):
if i & pow2[k]:
continue
l = i ^ pow2[k]
xk = x[k]
z = bisect.bisect_left(xk, dpu) + c
for y in range(2):
if y + z - 1 < len(xk):
v = f(l, j + y)
dp[v] = min(dp[v], xk[y + z - 1])
for i in range(1, m + 1):
if dp[f(pm - 1, i)] == inf:
ok = 0
break
ans += 1
if not ok:
break
print(ans) | Title: Vladik and cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vladik was bored on his way home and decided to play the following game. He took *n* cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:
- the number of occurrences of each number from 1 to 8 in the subsequence doesn't differ by more then 1 from the number of occurrences of any other number. Formally, if there are *c**k* cards with number *k* on them in the subsequence, than for all pairs of integers the condition |*c**i*<=-<=*c**j*|<=≤<=1 must hold. - if there is at least one card with number *x* on it in the subsequence, then all cards with number *x* in this subsequence must form a continuous segment in it (but not necessarily a continuous segment in the original sequence). For example, the subsequence [1,<=1,<=2,<=2] satisfies this condition while the subsequence [1,<=2,<=2,<=1] doesn't. Note that [1,<=1,<=2,<=2] doesn't satisfy the first condition.
Please help Vladik to find the length of the longest subsequence that satisfies both conditions.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards in Vladik's sequence.
The second line contains the sequence of *n* positive integers not exceeding 8 — the description of Vladik's sequence.
Output Specification:
Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions.
Demo Input:
['3\n1 1 1\n', '8\n8 7 6 5 4 3 2 1\n', '24\n1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8\n']
Demo Output:
['1', '8', '17']
Note:
In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition. | ```python
import bisect
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v):
return (m + 1) * u + v
n = int(input())
a = list(map(int, input().split()))
m = 8
x = [[] for _ in range(m)]
for i in range(n):
x[a[i] - 1].append(i)
s = 0
for y in x:
s += min(len(y), 1)
if s < m:
ans = s
print(ans)
exit()
pow2 = [1]
for _ in range(m):
pow2.append(2 * pow2[-1])
pm = pow2[m]
inf = pow(10, 9) + 1
ans = 8
ok = 1
for c in range(1, n // 8 + 3):
dp = [inf] * ((m + 1) * pm)
dp[0] = -1
for i in range(pm):
for j in range(m):
u = f(i, j)
if dp[u] == inf:
break
dpu = dp[u]
for k in range(m):
if i & pow2[k]:
continue
l = i ^ pow2[k]
xk = x[k]
z = bisect.bisect_left(xk, dpu) + c
for y in range(2):
if y + z - 1 < len(xk):
v = f(l, j + y)
dp[v] = min(dp[v], xk[y + z - 1])
for i in range(1, m + 1):
if dp[f(pm - 1, i)] == inf:
ok = 0
break
ans += 1
if not ok:
break
print(ans)
``` | 3 | |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,613,824,362 | 2,147,483,647 | Python 3 | OK | TESTS | 88 | 62 | 307,200 | n=int(input())
dp=[[0 for i in range(3)]for j in range(n+1)]
ar=tuple(map(str,input().split()))
for i in range(1,n+1):
x=ar[i-1]
dp[i][0]=min(dp[i-1])+1
if(x=='1' or x=='3'):
dp[i][1]=min(dp[i-1][0],dp[i-1][2])
else:
dp[i][1]=dp[i-1][1]+1
if(x=='2' or x=='3'):
dp[i][2]=min(dp[i-1][0],dp[i-1][1])
else:
dp[i][2]=dp[i-1][2]+1
print(min(dp[-1]))
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
n=int(input())
dp=[[0 for i in range(3)]for j in range(n+1)]
ar=tuple(map(str,input().split()))
for i in range(1,n+1):
x=ar[i-1]
dp[i][0]=min(dp[i-1])+1
if(x=='1' or x=='3'):
dp[i][1]=min(dp[i-1][0],dp[i-1][2])
else:
dp[i][1]=dp[i-1][1]+1
if(x=='2' or x=='3'):
dp[i][2]=min(dp[i-1][0],dp[i-1][1])
else:
dp[i][2]=dp[i-1][2]+1
print(min(dp[-1]))
``` | 3 | |
402 | C | Searching for Graph | PROGRAMMING | 1,500 | [
"brute force",
"constructive algorithms",
"graphs"
] | null | null | Let's call an undirected graph of *n* vertices *p*-interesting, if the following conditions fulfill:
- the graph contains exactly 2*n*<=+<=*p* edges; - the graph doesn't contain self-loops and multiple edges; - for any integer *k* (1<=≤<=*k*<=≤<=*n*), any subgraph consisting of *k* vertices contains at most 2*k*<=+<=*p* edges.
A subgraph of a graph is some set of the graph vertices and some set of the graph edges. At that, the set of edges must meet the condition: both ends of each edge from the set must belong to the chosen set of vertices.
Your task is to find a *p*-interesting graph consisting of *n* vertices. | The first line contains a single integer *t* (1<=≤<=*t*<=≤<=5) — the number of tests in the input. Next *t* lines each contains two space-separated integers: *n*, *p* (5<=≤<=*n*<=≤<=24; *p*<=≥<=0; ) — the number of vertices in the graph and the interest value for the appropriate test.
It is guaranteed that the required graph exists. | For each of the *t* tests print 2*n*<=+<=*p* lines containing the description of the edges of a *p*-interesting graph: the *i*-th line must contain two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — two vertices, connected by an edge in the resulting graph. Consider the graph vertices numbered with integers from 1 to *n*.
Print the answers to the tests in the order the tests occur in the input. If there are multiple solutions, you can print any of them. | [
"1\n6 0\n"
] | [
"1 2\n1 3\n1 4\n1 5\n1 6\n2 3\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6\n"
] | none | 1,500 | [
{
"input": "1\n6 0",
"output": "1 2\n1 3\n1 4\n1 5\n1 6\n2 3\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6"
},
{
"input": "1\n5 0",
"output": "1 2\n1 3\n1 4\n1 5\n2 3\n2 4\n2 5\n3 4\n3 5\n4 5"
},
{
"input": "5\n6 0\n5 0\n7 0\n8 0\n9 0",
"output": "1 2\n1 3\n1 4\n1 5\n1 6\n2 3\n2 4\n2 5\n2 6\n3 4\n3 ... | 1,636,386,845 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 21 | 77 | 33,587,200 | def construct(n, p):
for i in range (2*n + p):
pos = i % n
# 1, ...
cycle = (i // n) + 1
neighbor = (pos + cycle) % n
# convert to 1-index
print(pos+1, neighbor+1)
for _ in range(int(input())):
n, p = [int(x) for x in input().split()]
construct(n,p)
| Title: Searching for Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call an undirected graph of *n* vertices *p*-interesting, if the following conditions fulfill:
- the graph contains exactly 2*n*<=+<=*p* edges; - the graph doesn't contain self-loops and multiple edges; - for any integer *k* (1<=≤<=*k*<=≤<=*n*), any subgraph consisting of *k* vertices contains at most 2*k*<=+<=*p* edges.
A subgraph of a graph is some set of the graph vertices and some set of the graph edges. At that, the set of edges must meet the condition: both ends of each edge from the set must belong to the chosen set of vertices.
Your task is to find a *p*-interesting graph consisting of *n* vertices.
Input Specification:
The first line contains a single integer *t* (1<=≤<=*t*<=≤<=5) — the number of tests in the input. Next *t* lines each contains two space-separated integers: *n*, *p* (5<=≤<=*n*<=≤<=24; *p*<=≥<=0; ) — the number of vertices in the graph and the interest value for the appropriate test.
It is guaranteed that the required graph exists.
Output Specification:
For each of the *t* tests print 2*n*<=+<=*p* lines containing the description of the edges of a *p*-interesting graph: the *i*-th line must contain two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — two vertices, connected by an edge in the resulting graph. Consider the graph vertices numbered with integers from 1 to *n*.
Print the answers to the tests in the order the tests occur in the input. If there are multiple solutions, you can print any of them.
Demo Input:
['1\n6 0\n']
Demo Output:
['1 2\n1 3\n1 4\n1 5\n1 6\n2 3\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6\n']
Note:
none | ```python
def construct(n, p):
for i in range (2*n + p):
pos = i % n
# 1, ...
cycle = (i // n) + 1
neighbor = (pos + cycle) % n
# convert to 1-index
print(pos+1, neighbor+1)
for _ in range(int(input())):
n, p = [int(x) for x in input().split()]
construct(n,p)
``` | 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": "tymbzjyqhymeda... | 1,602,611,129 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 109 | 0 | s = input()
word = "hello"
ans = 0
i = 0
for char in s:
if char == word[i]:
i += 1
if i == len(word):
ans = 1
break
if ans == 1:
print("YES")
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = input()
word = "hello"
ans = 0
i = 0
for char in s:
if char == word[i]:
i += 1
if i == len(word):
ans = 1
break
if ans == 1:
print("YES")
else:
print("NO")
``` | 3.9455 |
182 | E | Wooden Fence | PROGRAMMING | 1,800 | [
"dp"
] | null | null | Vasya has recently bought some land and decided to surround it with a wooden fence.
He went to a company called "Wooden board" that produces wooden boards for fences. Vasya read in the catalog of products that the company has at its disposal *n* different types of wood. The company uses the *i*-th type of wood to produce a board of this type that is a rectangular *a**i* by *b**i* block.
Vasya decided to order boards in this company and build a fence from them. It turned out that the storehouse of the company is so large that Vasya can order arbitrary number of boards of every type. Note that Vasya is allowed to turn the boards as he builds the fence. However, Vasya cannot turn square boards.
Vasya is required to construct a fence of length *l*, however, an arbitrary fence won't do. Vasya wants his fence to look beautiful. We'll say that a fence is beautiful if and only if the following two conditions are fulfilled:
- there are no two successive boards of the same type - the first board of the fence has an arbitrary length, and the length of each subsequent board equals the width of the previous one
In other words, the fence is considered beautiful, if the type of the *i*-th board in the fence is different from the *i*<=-<=1-th board's type; besides, the *i*-th board's length is equal to the *i*<=-<=1-th board's width (for all *i*, starting from 2).
Now Vasya wonders, how many variants of arranging a fence for his land exist. Your task is to count the number of different beautiful fences of length *l*.
Two fences will be considered the same if the corresponding sequences of fence boards types and rotations are the same, otherwise the fences are different. Since the sought number can be large enough, you need to calculate the answer modulo 1000000007 (109<=+<=7). | The first line contains two integers *n* and *l* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*l*<=≤<=3000) — the number of different board types and the fence length, correspondingly. Next *n* lines contain descriptions of board types: the *i*-th line contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100) — the sizes of the board of the *i*-th type. All numbers on the lines are separated by spaces. | Print a single integer — the sought number of variants modulo 1000000007 (109<=+<=7). | [
"2 3\n1 2\n2 3\n",
"1 2\n2 2\n",
"6 6\n2 1\n3 2\n2 5\n3 3\n5 1\n2 1\n"
] | [
"2\n",
"1\n",
"20\n"
] | In the first sample there are exactly two variants of arranging a beautiful fence of length 3:
- As the first fence board use the board of the first type of length 1 and width 2. As the second board use board of the second type of length 2 and width 3. - Use one board of the second type after you turn it. That makes its length equal 3, and width — 2. | 1,500 | [
{
"input": "2 3\n1 2\n2 3",
"output": "2"
},
{
"input": "1 2\n2 2",
"output": "1"
},
{
"input": "6 6\n2 1\n3 2\n2 5\n3 3\n5 1\n2 1",
"output": "20"
},
{
"input": "4 3\n1 2\n1 1\n3 1\n2 2",
"output": "4"
},
{
"input": "4 6\n1 1\n1 2\n3 1\n5 10",
"output": "0"
... | 1,680,289,086 | 2,147,483,647 | PyPy 3 | OK | TESTS | 50 | 498 | 4,710,400 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v):
return 2 * n * u + v
n, l = map(int, input().split())
mod = pow(10, 9) + 7
a, b = [], []
for _ in range(n):
a0, b0 = map(int, input().split())
a.append(a0)
b.append(b0)
if a0 == b0:
a0, b0 = l + 1, l + 1
a.append(b0)
b.append(a0)
G = [[] for _ in range(2 * n)]
for i in range(2 * n):
for j in range(2 * n):
if i // 2 == j // 2:
continue
if b[i] == a[j]:
G[i].append(j)
m = 2 * n * (l + 1)
dp = [0] * m
for i in range(2 * n):
if a[i] <= l:
dp[f(a[i], i)] = 1
for i in range(1, l):
for j in range(2 * n):
u = f(i, j)
if not dp[u]:
continue
for k in G[j]:
if i + a[k] <= l:
dp[f(i + a[k], k)] += dp[u]
dp[f(i + a[k], k)] %= mod
ans = 0
for i in range(2 * n):
ans += dp[f(l, i)]
ans %= mod
print(ans) | Title: Wooden Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has recently bought some land and decided to surround it with a wooden fence.
He went to a company called "Wooden board" that produces wooden boards for fences. Vasya read in the catalog of products that the company has at its disposal *n* different types of wood. The company uses the *i*-th type of wood to produce a board of this type that is a rectangular *a**i* by *b**i* block.
Vasya decided to order boards in this company and build a fence from them. It turned out that the storehouse of the company is so large that Vasya can order arbitrary number of boards of every type. Note that Vasya is allowed to turn the boards as he builds the fence. However, Vasya cannot turn square boards.
Vasya is required to construct a fence of length *l*, however, an arbitrary fence won't do. Vasya wants his fence to look beautiful. We'll say that a fence is beautiful if and only if the following two conditions are fulfilled:
- there are no two successive boards of the same type - the first board of the fence has an arbitrary length, and the length of each subsequent board equals the width of the previous one
In other words, the fence is considered beautiful, if the type of the *i*-th board in the fence is different from the *i*<=-<=1-th board's type; besides, the *i*-th board's length is equal to the *i*<=-<=1-th board's width (for all *i*, starting from 2).
Now Vasya wonders, how many variants of arranging a fence for his land exist. Your task is to count the number of different beautiful fences of length *l*.
Two fences will be considered the same if the corresponding sequences of fence boards types and rotations are the same, otherwise the fences are different. Since the sought number can be large enough, you need to calculate the answer modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two integers *n* and *l* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*l*<=≤<=3000) — the number of different board types and the fence length, correspondingly. Next *n* lines contain descriptions of board types: the *i*-th line contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100) — the sizes of the board of the *i*-th type. All numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the sought number of variants modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\n1 2\n2 3\n', '1 2\n2 2\n', '6 6\n2 1\n3 2\n2 5\n3 3\n5 1\n2 1\n']
Demo Output:
['2\n', '1\n', '20\n']
Note:
In the first sample there are exactly two variants of arranging a beautiful fence of length 3:
- As the first fence board use the board of the first type of length 1 and width 2. As the second board use board of the second type of length 2 and width 3. - Use one board of the second type after you turn it. That makes its length equal 3, and width — 2. | ```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v):
return 2 * n * u + v
n, l = map(int, input().split())
mod = pow(10, 9) + 7
a, b = [], []
for _ in range(n):
a0, b0 = map(int, input().split())
a.append(a0)
b.append(b0)
if a0 == b0:
a0, b0 = l + 1, l + 1
a.append(b0)
b.append(a0)
G = [[] for _ in range(2 * n)]
for i in range(2 * n):
for j in range(2 * n):
if i // 2 == j // 2:
continue
if b[i] == a[j]:
G[i].append(j)
m = 2 * n * (l + 1)
dp = [0] * m
for i in range(2 * n):
if a[i] <= l:
dp[f(a[i], i)] = 1
for i in range(1, l):
for j in range(2 * n):
u = f(i, j)
if not dp[u]:
continue
for k in G[j]:
if i + a[k] <= l:
dp[f(i + a[k], k)] += dp[u]
dp[f(i + a[k], k)] %= mod
ans = 0
for i in range(2 * n):
ans += dp[f(l, i)]
ans %= mod
print(ans)
``` | 3 | |
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 4... | 1,511,860,530 | 2,147,483,647 | Python 3 | OK | TESTS | 93 | 420 | 6,348,800 | n=int(input())
a=[int (i) for i in input().split()]
min=361
for i in range(n):
for j in range(i, n):
if -min<sum(a[0:i])+sum(a[j:])-sum(a[i:j])<min:
min=sum(a[0:i])+sum(a[j:])-sum(a[i:j])
if min<0:
min*=-1
print(min) | Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
n=int(input())
a=[int (i) for i in input().split()]
min=361
for i in range(n):
for j in range(i, n):
if -min<sum(a[0:i])+sum(a[j:])-sum(a[i:j])<min:
min=sum(a[0:i])+sum(a[j:])-sum(a[i:j])
if min<0:
min*=-1
print(min)
``` | 3 | |
278 | A | Circle Line | PROGRAMMING | 800 | [
"implementation"
] | null | null | The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces. | Print a single number — the length of the shortest path between stations number *s* and *t*. | [
"4\n2 3 4 9\n1 3\n",
"4\n5 8 2 100\n4 1\n",
"3\n1 1 1\n3 1\n",
"3\n31 41 59\n1 1\n"
] | [
"5\n",
"15\n",
"1\n",
"0\n"
] | In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0. | 500 | [
{
"input": "4\n2 3 4 9\n1 3",
"output": "5"
},
{
"input": "4\n5 8 2 100\n4 1",
"output": "15"
},
{
"input": "3\n1 1 1\n3 1",
"output": "1"
},
{
"input": "3\n31 41 59\n1 1",
"output": "0"
},
{
"input": "5\n16 13 10 30 15\n4 2",
"output": "23"
},
{
"inpu... | 1,547,180,761 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 248 | 0 | nb_elems = [int(x) for x in input().split()]
elems = [int(x) for x in input().split()]
src, tgt = [int(x) for x in input().split()]
mins = min(src, tgt)
maxs = max(src, tgt)
sums = sum(elems)
clockwise = sum(elems[mins-1:maxs-1])
aclocwise = sums - clockwise
print(min(clockwise, aclocwise))
| Title: Circle Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single number — the length of the shortest path between stations number *s* and *t*.
Demo Input:
['4\n2 3 4 9\n1 3\n', '4\n5 8 2 100\n4 1\n', '3\n1 1 1\n3 1\n', '3\n31 41 59\n1 1\n']
Demo Output:
['5\n', '15\n', '1\n', '0\n']
Note:
In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0. | ```python
nb_elems = [int(x) for x in input().split()]
elems = [int(x) for x in input().split()]
src, tgt = [int(x) for x in input().split()]
mins = min(src, tgt)
maxs = max(src, tgt)
sums = sum(elems)
clockwise = sum(elems[mins-1:maxs-1])
aclocwise = sums - clockwise
print(min(clockwise, aclocwise))
``` | 3 | |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
... | 1,698,000,279 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 92 | 0 | n=int(input())
res=0
for i in range(n):
messi=input()
a,b,c=messi.split()
a=int(a)
b=int(b)
c=int(c)
s=a+b+c
if s>=2:
res=res+1
print(res)
| Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
n=int(input())
res=0
for i in range(n):
messi=input()
a,b,c=messi.split()
a=int(a)
b=int(b)
c=int(c)
s=a+b+c
if s>=2:
res=res+1
print(res)
``` | 3 | |
855 | B | Marvolo Gaunt's Ring | PROGRAMMING | 1,500 | [
"brute force",
"data structures",
"dp"
] | null | null | Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative. | First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). | Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. | [
"5 1 2 3\n1 2 3 4 5\n",
"5 1 2 -3\n-1 -2 -3 -4 -5\n"
] | [
"30\n",
"12\n"
] | In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | 1,000 | [
{
"input": "5 1 2 3\n1 2 3 4 5",
"output": "30"
},
{
"input": "5 1 2 -3\n-1 -2 -3 -4 -5",
"output": "12"
},
{
"input": "5 886327859 82309257 -68295239\n-731225382 354766539 -48222231 -474691998 360965777",
"output": "376059240645059046"
},
{
"input": "4 -96405765 -495906217 6... | 1,675,766,309 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 82 | 77 | 14,233,600 | def main():
n, p, q, r = map(int, input().split())
a = list(map(int, input().split()))
ans = -float('inf')
mx = -float('inf')
mx1 = -float('inf')
for i in range(n):
mx = max(mx, a[i] * p)
mx1 = max(mx1, mx + a[i] * q)
ans = max(ans, mx1 + a[i] * r)
print(ans)
if __name__ == '__main__':
main() | Title: Marvolo Gaunt's Ring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.
Input Specification:
First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.
Demo Input:
['5 1 2 3\n1 2 3 4 5\n', '5 1 2 -3\n-1 -2 -3 -4 -5\n']
Demo Output:
['30\n', '12\n']
Note:
In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | ```python
def main():
n, p, q, r = map(int, input().split())
a = list(map(int, input().split()))
ans = -float('inf')
mx = -float('inf')
mx1 = -float('inf')
for i in range(n):
mx = max(mx, a[i] * p)
mx1 = max(mx1, mx + a[i] * q)
ans = max(ans, mx1 + a[i] * r)
print(ans)
if __name__ == '__main__':
main()
``` | 3 | |
711 | A | Bus to Udayland | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details. | If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them. | [
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] | [
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] | Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | 500 | [
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
... | 1,635,760,692 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 62 | 4,505,600 | n = int(input())
a = []
checker = True
for i in range(n):
a.append(input().split('|'))
if a[i][0] == 'OO' and checker:
checker = False
a[i][0] = "++"
elif a[i][1] == "OO" and checker:
checker = False
a[i][1] = "++"
if checker:
print("NO")
else:
print("YES")
for i in range(n):
print(f"{a[i][0]}|{a[i][1]}")
| Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | ```python
n = int(input())
a = []
checker = True
for i in range(n):
a.append(input().split('|'))
if a[i][0] == 'OO' and checker:
checker = False
a[i][0] = "++"
elif a[i][1] == "OO" and checker:
checker = False
a[i][1] = "++"
if checker:
print("NO")
else:
print("YES")
for i in range(n):
print(f"{a[i][0]}|{a[i][1]}")
``` | 3 | |
785 | A | Anton and Polyhedrons | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. | Output one number — the total number of faces in all the polyhedrons in Anton's collection. | [
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] | [
"42\n",
"28\n"
] | In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | 500 | [
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosa... | 1,674,567,841 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 295 | 10,444,800 | lists=[]
for _ in range(int(input())):
lists.append(input())
s=0
for i in lists:
if i=='Tetrahedron':
s+=4
elif i=='Cube':
s+=6
elif i=='Octahedron':
s+=8
elif i=='Dodecahedron':
s+=12
elif i=='Icosahedron':
s+=20
print(s) | Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | ```python
lists=[]
for _ in range(int(input())):
lists.append(input())
s=0
for i in lists:
if i=='Tetrahedron':
s+=4
elif i=='Cube':
s+=6
elif i=='Octahedron':
s+=8
elif i=='Dodecahedron':
s+=12
elif i=='Icosahedron':
s+=20
print(s)
``` | 3 | |
699 | A | Launch of Collider | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point. | The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order. | In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen. | [
"4\nRLRL\n2 4 6 10\n",
"3\nLLR\n40 50 60\n"
] | [
"1\n",
"-1\n"
] | In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point. | 500 | [
{
"input": "4\nRLRL\n2 4 6 10",
"output": "1"
},
{
"input": "3\nLLR\n40 50 60",
"output": "-1"
},
{
"input": "4\nRLLR\n46 230 264 470",
"output": "92"
},
{
"input": "6\nLLRLLL\n446 492 650 844 930 970",
"output": "97"
},
{
"input": "8\nRRLLLLLL\n338 478 512 574 59... | 1,643,022,038 | 2,147,483,647 | Python 3 | OK | TESTS | 85 | 202 | 16,384,000 | n = int(input())
s = input()
arr = list(map(int, input().split()))
ans = []
for i in range(n):
if(i == n-1):
break
if((s[i] == "R" and s[i+1] == "L")):
ans.append(arr[i+1]-arr[i])
for i in range(len(ans)):
ans[i] = int(ans[i]/2)
if not ans:
print(-1)
else:
print(min(ans)) | Title: Launch of Collider
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.
Output Specification:
In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen.
Demo Input:
['4\nRLRL\n2 4 6 10\n', '3\nLLR\n40 50 60\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point. | ```python
n = int(input())
s = input()
arr = list(map(int, input().split()))
ans = []
for i in range(n):
if(i == n-1):
break
if((s[i] == "R" and s[i+1] == "L")):
ans.append(arr[i+1]-arr[i])
for i in range(len(ans)):
ans[i] = int(ans[i]/2)
if not ans:
print(-1)
else:
print(min(ans))
``` | 3 | |
729 | A | Interview with Oleg | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters. | Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. | [
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] | [
"a***b\n",
"***gmg***\n",
"*********\n"
] | The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | 500 | [
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input":... | 1,679,298,530 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 48 | 93 | 2,764,800 | import sys, io, os
import math
import bisect
import heapq
import string
from collections import defaultdict,Counter,deque
input = sys.stdin.readline
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
def WRITE(out):
return print('\n'.join(map(str, out)))
def WS(out):
return print(' '.join(map(str, out)))
def WNS(out):
return print(''.join(map(str, out)))
'''
n%4 == 0 and all_cnts <= n//4
'''
def solve():
n = II()
s = I().strip()
ans = []
l = 0
while l < n:
# print(l, s[l:])
if l+2 < n and s[l:l+3] == 'ogo':
l += 3
for i in range(l, n-1, 2):
if s[i:i+2] == 'go':
l = i+2
else:
break
ans.append('***')
else:
ans.append(s[l])
l += 1
WNS(ans)
solve() | Title: Interview with Oleg
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | ```python
import sys, io, os
import math
import bisect
import heapq
import string
from collections import defaultdict,Counter,deque
input = sys.stdin.readline
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
def WRITE(out):
return print('\n'.join(map(str, out)))
def WS(out):
return print(' '.join(map(str, out)))
def WNS(out):
return print(''.join(map(str, out)))
'''
n%4 == 0 and all_cnts <= n//4
'''
def solve():
n = II()
s = I().strip()
ans = []
l = 0
while l < n:
# print(l, s[l:])
if l+2 < n and s[l:l+3] == 'ogo':
l += 3
for i in range(l, n-1, 2):
if s[i:i+2] == 'go':
l = i+2
else:
break
ans.append('***')
else:
ans.append(s[l])
l += 1
WNS(ans)
solve()
``` | 3 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,586,515,707 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 216 | 0 | a,b = map(int,input().split())
v = max(a,b)
v = (6-v)+1
if(v==1):
print("1/6")
elif(v==2):
print("1/3")
elif(v==3):
print("1/2")
elif(v==4):
print("2/3")
elif(v==5):
print("5/6")
else:
print("1/1")
| Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
a,b = map(int,input().split())
v = max(a,b)
v = (6-v)+1
if(v==1):
print("1/6")
elif(v==2):
print("1/3")
elif(v==3):
print("1/2")
elif(v==4):
print("2/3")
elif(v==5):
print("5/6")
else:
print("1/1")
``` | 3.892 |
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"
},... | 1,492,358,665 | 1,765 | Python 3 | OK | TESTS | 35 | 62 | 5,529,600 | x=input()
y=input()
z=[]
boom=0
for i in range(len(x)):
if x[i]<y[i]:
boom+=1
elif x[i]==y[i]:
z.append("z")
else:
z.append(y[i])
if boom==0:
print("".join(z))
else:
print(-1) | 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=input()
y=input()
z=[]
boom=0
for i in range(len(x)):
if x[i]<y[i]:
boom+=1
elif x[i]==y[i]:
z.append("z")
else:
z.append(y[i])
if boom==0:
print("".join(z))
else:
print(-1)
``` | 3 | |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "... | 1,601,034,669 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 218 | 0 | n, m = [int(x) for x in input().split(' ')]
a = [int(x) for x in input().split(' ') if int(x) < 0]
a.sort()
print(-sum(a[:min(len(a), m)]))
| Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n, m = [int(x) for x in input().split(' ')]
a = [int(x) for x in input().split(' ') if int(x) < 0]
a.sort()
print(-sum(a[:min(len(a), m)]))
``` | 3.9455 |
991 | A | If at first you don't succeed... | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time? | The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$). | If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$. | [
"10 10 5 20\n",
"2 2 0 4\n",
"2 2 2 1\n"
] | [
"5",
"-1",
"-1"
] | The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | 500 | [
{
"input": "10 10 5 20",
"output": "5"
},
{
"input": "2 2 0 4",
"output": "-1"
},
{
"input": "2 2 2 1",
"output": "-1"
},
{
"input": "98 98 97 100",
"output": "1"
},
{
"input": "1 5 2 10",
"output": "-1"
},
{
"input": "5 1 2 10",
"output": "-1"
}... | 1,642,578,243 | 2,147,483,647 | PyPy 3 | OK | TESTS | 76 | 77 | 0 | a, b, c, n = map(int, input().split())
tmp = n - (a + b - c)
if tmp < 1 or a < c or b < c:
print(-1)
else:
print(tmp) | Title: If at first you don't succeed...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?
Input Specification:
The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).
Output Specification:
If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.
Demo Input:
['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n']
Demo Output:
['5', '-1', '-1']
Note:
The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | ```python
a, b, c, n = map(int, input().split())
tmp = n - (a + b - c)
if tmp < 1 or a < c or b < c:
print(-1)
else:
print(tmp)
``` | 3 | |
331 | C1 | The Great Julya Calendar | PROGRAMMING | 1,100 | [
"dp"
] | null | null | Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero. | The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3). | Print a single integer — the minimum number of subtractions that turns the magic number to a zero. | [
"24\n"
] | [
"5"
] | In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | 20 | [
{
"input": "24",
"output": "5"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "31",
"output": "6"
},
... | 1,586,089,525 | 2,147,483,647 | PyPy 3 | OK | TESTS1 | 24 | 310 | 1,536,000 | n=int(input())
l=list(str(n))
n=int(n)
c=0
while n!=0:
m=max(list(str(n)))
n-=int(m)
c+=1
print(c) | Title: The Great Julya Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
Input Specification:
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
Output Specification:
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
Demo Input:
['24\n']
Demo Output:
['5']
Note:
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | ```python
n=int(input())
l=list(str(n))
n=int(n)
c=0
while n!=0:
m=max(list(str(n)))
n-=int(m)
c+=1
print(c)
``` | 3 | |
16 | B | Burglar and Matches | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | B. Burglar and Matches | 0 | 64 | A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer. | Output the only number — answer to the problem. | [
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] | [
"62\n",
"7\n"
] | none | 0 | [
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"ou... | 1,623,587,774 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 124 | 0 | n,m=map(int,input().split())
l=[]
for i in range(m):
a,b=map(int,input().split())
l.append([b,a])
l.sort(reverse=True)
i=c=0
while n>0 and i<len(l):
if l[i][1]<n:
n-=l[i][1]
c+=l[i][0]*l[i][1]
i+=1
else:
# print(n)
d=n
n-=d
c+=l[i][0]*d
print(c) | Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none | ```python
n,m=map(int,input().split())
l=[]
for i in range(m):
a,b=map(int,input().split())
l.append([b,a])
l.sort(reverse=True)
i=c=0
while n>0 and i<len(l):
if l[i][1]<n:
n-=l[i][1]
c+=l[i][0]*l[i][1]
i+=1
else:
# print(n)
d=n
n-=d
c+=l[i][0]*d
print(c)
``` | 3 |
802 | M | April Fools' Problem (easy) | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | The marmots have prepared a very easy problem for this year's HC2 – this one. It involves numbers *n*, *k* and a sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. They also came up with a beautiful and riveting story for the problem statement. It explains what the input means, what the program should output, and it also reads like a good criminal.
However I, Heidi, will have none of that. As my joke for today, I am removing the story from the statement and replacing it with these two unhelpful paragraphs. Now solve the problem, fools! | The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2200). The second line contains *n* space-separated integers *a*1,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104). | Output one number. | [
"8 5\n1 1 1 1 1 1 1 1\n",
"10 3\n16 8 2 4 512 256 32 128 64 1\n",
"5 1\n20 10 50 30 46\n",
"6 6\n6 6 6 6 6 6\n",
"1 1\n100\n"
] | [
"5",
"7",
"10",
"36",
"100"
] | none | 0 | [
{
"input": "8 5\n1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "10 3\n16 8 2 4 512 256 32 128 64 1",
"output": "7"
},
{
"input": "5 1\n20 10 50 30 46",
"output": "10"
},
{
"input": "6 6\n6 6 6 6 6 6",
"output": "36"
},
{
"input": "1 1\n100",
"output": "100"
... | 1,660,065,581 | 2,147,483,647 | Python 3 | OK | TESTS | 14 | 31 | 0 | a,b=map(int,input().split())
t=[int(i) for i in input().split()]
count=0
t.sort()
for i in range(b):count+=t[i]
print(count) | Title: April Fools' Problem (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The marmots have prepared a very easy problem for this year's HC2 – this one. It involves numbers *n*, *k* and a sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. They also came up with a beautiful and riveting story for the problem statement. It explains what the input means, what the program should output, and it also reads like a good criminal.
However I, Heidi, will have none of that. As my joke for today, I am removing the story from the statement and replacing it with these two unhelpful paragraphs. Now solve the problem, fools!
Input Specification:
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2200). The second line contains *n* space-separated integers *a*1,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104).
Output Specification:
Output one number.
Demo Input:
['8 5\n1 1 1 1 1 1 1 1\n', '10 3\n16 8 2 4 512 256 32 128 64 1\n', '5 1\n20 10 50 30 46\n', '6 6\n6 6 6 6 6 6\n', '1 1\n100\n']
Demo Output:
['5', '7', '10', '36', '100']
Note:
none | ```python
a,b=map(int,input().split())
t=[int(i) for i in input().split()]
count=0
t.sort()
for i in range(b):count+=t[i]
print(count)
``` | 3 | |
987 | B | High School: Become Human | PROGRAMMING | 1,100 | [
"math"
] | null | null | Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him. | On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$). | If $x^y < y^x$, then print '<' (without quotes). If $x^y > y^x$, then print '>' (without quotes). If $x^y = y^x$, then print '=' (without quotes). | [
"5 8\n",
"10 3\n",
"6 6\n"
] | [
">\n",
"<\n",
"=\n"
] | In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '>'.
In the second example $10^3 = 1000 < 3^{10} = 59049$.
In the third example $6^6 = 46656 = 6^6$. | 1,000 | [
{
"input": "5 8",
"output": ">"
},
{
"input": "10 3",
"output": "<"
},
{
"input": "6 6",
"output": "="
},
{
"input": "14 1",
"output": ">"
},
{
"input": "2 4",
"output": "="
},
{
"input": "987654321 123456987",
"output": "<"
},
{
"input": "... | 1,628,583,448 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 62 | 6,963,200 | x, y = map(int, input().split())
if ((x == y) or ((x == 2) and (y == 4)) or ((x == 4) and (y == 2))) : print('=')
else :
if (((x == 2) and (y == 3)) or (x == 1)) : print('<')
else :
if (((x == 3) and (y == 2)) or (y == 1)) : print('>')
else :
if (x < y) : print('>')
else :
if (x > y) : print('<') | Title: High School: Become Human
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him.
Input Specification:
On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$).
Output Specification:
If $x^y < y^x$, then print '<' (without quotes). If $x^y > y^x$, then print '>' (without quotes). If $x^y = y^x$, then print '=' (without quotes).
Demo Input:
['5 8\n', '10 3\n', '6 6\n']
Demo Output:
['>\n', '<\n', '=\n']
Note:
In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '>'.
In the second example $10^3 = 1000 < 3^{10} = 59049$.
In the third example $6^6 = 46656 = 6^6$. | ```python
x, y = map(int, input().split())
if ((x == y) or ((x == 2) and (y == 4)) or ((x == 4) and (y == 2))) : print('=')
else :
if (((x == 2) and (y == 3)) or (x == 1)) : print('<')
else :
if (((x == 3) and (y == 2)) or (y == 1)) : print('>')
else :
if (x < y) : print('>')
else :
if (x > y) : print('<')
``` | 3 | |
413 | A | Data Recovery | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Not so long ago company R2 bought company R1 and consequently, all its developments in the field of multicore processors. Now the R2 laboratory is testing one of the R1 processors.
The testing goes in *n* steps, at each step the processor gets some instructions, and then its temperature is measured. The head engineer in R2 is keeping a report record on the work of the processor: he writes down the minimum and the maximum measured temperature in his notebook. His assistant had to write down all temperatures into his notebook, but (for unknown reasons) he recorded only *m*.
The next day, the engineer's assistant filed in a report with all the *m* temperatures. However, the chief engineer doubts that the assistant wrote down everything correctly (naturally, the chief engineer doesn't doubt his notes). So he asked you to help him. Given numbers *n*, *m*, *min*, *max* and the list of *m* temperatures determine whether you can upgrade the set of *m* temperatures to the set of *n* temperatures (that is add *n*<=-<=*m* temperatures), so that the minimum temperature was *min* and the maximum one was *max*. | The first line contains four integers *n*,<=*m*,<=*min*,<=*max* (1<=≤<=*m*<=<<=*n*<=≤<=100; 1<=≤<=*min*<=<<=*max*<=≤<=100). The second line contains *m* space-separated integers *t**i* (1<=≤<=*t**i*<=≤<=100) — the temperatures reported by the assistant.
Note, that the reported temperatures, and the temperatures you want to add can contain equal temperatures. | If the data is consistent, print 'Correct' (without the quotes). Otherwise, print 'Incorrect' (without the quotes). | [
"2 1 1 2\n1\n",
"3 1 1 3\n2\n",
"2 1 1 3\n2\n"
] | [
"Correct\n",
"Correct\n",
"Incorrect\n"
] | In the first test sample one of the possible initial configurations of temperatures is [1, 2].
In the second test sample one of the possible initial configurations of temperatures is [2, 1, 3].
In the third test sample it is impossible to add one temperature to obtain the minimum equal to 1 and the maximum equal to 3. | 500 | [
{
"input": "2 1 1 2\n1",
"output": "Correct"
},
{
"input": "3 1 1 3\n2",
"output": "Correct"
},
{
"input": "2 1 1 3\n2",
"output": "Incorrect"
},
{
"input": "3 1 1 5\n3",
"output": "Correct"
},
{
"input": "3 2 1 5\n1 5",
"output": "Correct"
},
{
"input... | 1,597,905,534 | 2,147,483,647 | Python 3 | OK | TESTS | 68 | 124 | 0 | a,b,c,d=map(int,input().split())
z=list(map(int,input().split()))
k=z.copy()
if c in k:k.remove(c)
if d in k:k.remove(d)
if (max(z)>d or min(z)<c or len(k)+2>a ):print("Incorrect")
else:print("Correct") | Title: Data Recovery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Not so long ago company R2 bought company R1 and consequently, all its developments in the field of multicore processors. Now the R2 laboratory is testing one of the R1 processors.
The testing goes in *n* steps, at each step the processor gets some instructions, and then its temperature is measured. The head engineer in R2 is keeping a report record on the work of the processor: he writes down the minimum and the maximum measured temperature in his notebook. His assistant had to write down all temperatures into his notebook, but (for unknown reasons) he recorded only *m*.
The next day, the engineer's assistant filed in a report with all the *m* temperatures. However, the chief engineer doubts that the assistant wrote down everything correctly (naturally, the chief engineer doesn't doubt his notes). So he asked you to help him. Given numbers *n*, *m*, *min*, *max* and the list of *m* temperatures determine whether you can upgrade the set of *m* temperatures to the set of *n* temperatures (that is add *n*<=-<=*m* temperatures), so that the minimum temperature was *min* and the maximum one was *max*.
Input Specification:
The first line contains four integers *n*,<=*m*,<=*min*,<=*max* (1<=≤<=*m*<=<<=*n*<=≤<=100; 1<=≤<=*min*<=<<=*max*<=≤<=100). The second line contains *m* space-separated integers *t**i* (1<=≤<=*t**i*<=≤<=100) — the temperatures reported by the assistant.
Note, that the reported temperatures, and the temperatures you want to add can contain equal temperatures.
Output Specification:
If the data is consistent, print 'Correct' (without the quotes). Otherwise, print 'Incorrect' (without the quotes).
Demo Input:
['2 1 1 2\n1\n', '3 1 1 3\n2\n', '2 1 1 3\n2\n']
Demo Output:
['Correct\n', 'Correct\n', 'Incorrect\n']
Note:
In the first test sample one of the possible initial configurations of temperatures is [1, 2].
In the second test sample one of the possible initial configurations of temperatures is [2, 1, 3].
In the third test sample it is impossible to add one temperature to obtain the minimum equal to 1 and the maximum equal to 3. | ```python
a,b,c,d=map(int,input().split())
z=list(map(int,input().split()))
k=z.copy()
if c in k:k.remove(c)
if d in k:k.remove(d)
if (max(z)>d or min(z)<c or len(k)+2>a ):print("Incorrect")
else:print("Correct")
``` | 3 | |
467 | A | George and Accommodation | PROGRAMMING | 800 | [
"implementation"
] | null | null | George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity. | Print a single integer — the number of rooms where George and Alex can move in. | [
"3\n1 1\n2 2\n3 3\n",
"3\n1 10\n0 10\n10 10\n"
] | [
"0\n",
"2\n"
] | none | 500 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "0"
},
{
"input": "3\n1 10\n0 10\n10 10",
"output": "2"
},
{
"input": "2\n36 67\n61 69",
"output": "2"
},
{
"input": "3\n21 71\n10 88\n43 62",
"output": "3"
},
{
"input": "3\n1 2\n2 3\n3 4",
"output": "0"
},
{
"... | 1,691,077,472 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | def count_rooms_with_free_space(n, rooms):
count = 0
for i in range(n):
pi, qi = rooms[i]
if qi - pi >= 2:
count += 1
return count
# Read input
n = int(input())
rooms = [tuple(map(int, input().split())) for _ in range(n)]
# Count the number of rooms with free space for George and Alex
result = count_rooms_with_free_space(n, rooms)
# Print the output
print(result)
| Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none | ```python
def count_rooms_with_free_space(n, rooms):
count = 0
for i in range(n):
pi, qi = rooms[i]
if qi - pi >= 2:
count += 1
return count
# Read input
n = int(input())
rooms = [tuple(map(int, input().split())) for _ in range(n)]
# Count the number of rooms with free space for George and Alex
result = count_rooms_with_free_space(n, rooms)
# Print the output
print(result)
``` | 3 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "... | 1,675,782,705 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | s = input().split("WUB")
output = ""
for k in s:
if k != '':
output += k + " "
print(output) | Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
s = input().split("WUB")
output = ""
for k in s:
if k != '':
output += k + " "
print(output)
``` | 3 | |
750 | A | New Year and Hurry | PROGRAMMING | 800 | [
"binary search",
"brute force",
"implementation",
"math"
] | null | null | Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party? | The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house. | Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier. | [
"3 222\n",
"4 190\n",
"7 1\n"
] | [
"2\n",
"4\n",
"7\n"
] | In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems. | 500 | [
{
"input": "3 222",
"output": "2"
},
{
"input": "4 190",
"output": "4"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "10 135",
"output": "6"
},
{
"input": "10 136",
"output": "5"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 240",
... | 1,698,582,765 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 46 | 0 | def solve(n, k):
l = 240 - k
for i in range(1, n+1):
if l >= 5*i:
l -= 5*i
else:
return i-1
return n
n, k = list(map(int, input().split()))
print(solve(n, k)) | Title: New Year and Hurry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
Input Specification:
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
Output Specification:
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
Demo Input:
['3 222\n', '4 190\n', '7 1\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems. | ```python
def solve(n, k):
l = 240 - k
for i in range(1, n+1):
if l >= 5*i:
l -= 5*i
else:
return i-1
return n
n, k = list(map(int, input().split()))
print(solve(n, k))
``` | 3 | |
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
... | 1,593,255,407 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 109 | 6,963,200 | def Remove(s):
a = []
for i in range(len(s)):
if len(s[i]) == 0:
a.append(i)
a.reverse()
for i in a:
del s[i]
return s
def Print(s):
print(len(s))
for i in s:
print(len(i),end=' ')
n = int(input())
s = list(map(str,input().split('W')))
s = Remove(s)
Print(s) | Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
def Remove(s):
a = []
for i in range(len(s)):
if len(s[i]) == 0:
a.append(i)
a.reverse()
for i in a:
del s[i]
return s
def Print(s):
print(len(s))
for i in s:
print(len(i),end=' ')
n = int(input())
s = list(map(str,input().split('W')))
s = Remove(s)
Print(s)
``` | 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": "chn... | 1,683,655,208 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 122 | 0 | y=input("")
z=y.upper()
count=0
for i in range(len(z)):
if y[i]==z[i]:
count=count+1
if count>len(y)/2:
y=y.upper()
else:
y=y.lower()
print(y) | 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
y=input("")
z=y.upper()
count=0
for i in range(len(z)):
if y[i]==z[i]:
count=count+1
if count>len(y)/2:
y=y.upper()
else:
y=y.lower()
print(y)
``` | 3.9695 |
439 | A | Devu, the Singer and Churu, the Joker | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event. | The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100). | If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event. | [
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] | [
"5\n",
"-1\n"
] | Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | 500 | [
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 ... | 1,578,119,411 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 109 | 0 | n,d=map(int,input().split())
s=0
j=0
t=list(map(int,input().split()))
for i in range(n):
s=s+t[i]
s=s+(n-1)*10
if(s<=d):
j=2*(n-1)+(d-s)//5
print(j)
else:
print(-1) | Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | ```python
n,d=map(int,input().split())
s=0
j=0
t=list(map(int,input().split()))
for i in range(n):
s=s+t[i]
s=s+(n-1)*10
if(s<=d):
j=2*(n-1)+(d-s)//5
print(j)
else:
print(-1)
``` | 3 | |
690 | C2 | Brain Network (medium) | PROGRAMMING | 1,500 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Further research on zombie thought processes yielded interesting results. As we know from the previous problem, the nervous system of a zombie consists of *n* brains and *m* brain connectors joining some pairs of brains together. It was observed that the intellectual abilities of a zombie depend mainly on the topology of its nervous system. More precisely, we define the distance between two brains *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*) as the minimum number of brain connectors used when transmitting a thought between these two brains. The brain latency of a zombie is defined to be the maximum distance between any two of its brains. Researchers conjecture that the brain latency is the crucial parameter which determines how smart a given zombie is. Help them test this conjecture by writing a program to compute brain latencies of nervous systems.
In this problem you may assume that any nervous system given in the input is valid, i.e., it satisfies conditions (1) and (2) from the easy version. | The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100000) denoting the number of brains (which are conveniently numbered from 1 to *n*) and the number of brain connectors in the nervous system, respectively. In the next *m* lines, descriptions of brain connectors follow. Every connector is given as a pair of brains *a* *b* it connects (1<=≤<=*a*,<=*b*<=≤<=*n* and *a*<=≠<=*b*). | Print one number – the brain latency. | [
"4 3\n1 2\n1 3\n1 4\n",
"5 4\n1 2\n2 3\n3 4\n3 5\n"
] | [
"2",
"3"
] | none | 0 | [
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "3 2\n2 1\n3 2",
"output": "2"
},
{
"input": "10 9\n5 1\n1 2\n9 3\n10 5\n6 3\n8 5\n2 7\n2 3\n9 4",
"output": "6"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "2"
},
{
"input": "5 4\n1 2\n2 3\n3 4\n3 5",
"outp... | 1,695,531,292 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 405 | 19,865,600 | def solution(idx):
visited = [False] * n # visited
max_distance = (0, idx) # (distance, index)
stack = [max_distance] # stack
while stack:
distance, idx = stack.pop() # distance, index
visited[idx] = True
if distance > max_distance[0]:
max_distance = (distance, idx)
stack += [(distance + 1, j) for j in links[idx] if not visited[j]]
return max_distance
if __name__ == '__main__':
n, m = map(int, input().split())
links = [[] for i in range(n)]
for j in range(m):
a, b = map(int, input().split())
links[a - 1].append(b - 1)
links[b - 1].append(a - 1)
print(solution(solution(0)[1])[0])
| Title: Brain Network (medium)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Further research on zombie thought processes yielded interesting results. As we know from the previous problem, the nervous system of a zombie consists of *n* brains and *m* brain connectors joining some pairs of brains together. It was observed that the intellectual abilities of a zombie depend mainly on the topology of its nervous system. More precisely, we define the distance between two brains *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*) as the minimum number of brain connectors used when transmitting a thought between these two brains. The brain latency of a zombie is defined to be the maximum distance between any two of its brains. Researchers conjecture that the brain latency is the crucial parameter which determines how smart a given zombie is. Help them test this conjecture by writing a program to compute brain latencies of nervous systems.
In this problem you may assume that any nervous system given in the input is valid, i.e., it satisfies conditions (1) and (2) from the easy version.
Input Specification:
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100000) denoting the number of brains (which are conveniently numbered from 1 to *n*) and the number of brain connectors in the nervous system, respectively. In the next *m* lines, descriptions of brain connectors follow. Every connector is given as a pair of brains *a* *b* it connects (1<=≤<=*a*,<=*b*<=≤<=*n* and *a*<=≠<=*b*).
Output Specification:
Print one number – the brain latency.
Demo Input:
['4 3\n1 2\n1 3\n1 4\n', '5 4\n1 2\n2 3\n3 4\n3 5\n']
Demo Output:
['2', '3']
Note:
none | ```python
def solution(idx):
visited = [False] * n # visited
max_distance = (0, idx) # (distance, index)
stack = [max_distance] # stack
while stack:
distance, idx = stack.pop() # distance, index
visited[idx] = True
if distance > max_distance[0]:
max_distance = (distance, idx)
stack += [(distance + 1, j) for j in links[idx] if not visited[j]]
return max_distance
if __name__ == '__main__':
n, m = map(int, input().split())
links = [[] for i in range(n)]
for j in range(m):
a, b = map(int, input().split())
links[a - 1].append(b - 1)
links[b - 1].append(a - 1)
print(solution(solution(0)[1])[0])
``` | 3 | |
401 | A | Vanya and Cards | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*. | The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value. | Print a single number — the answer to the problem. | [
"3 2\n-1 1 2\n",
"2 3\n-2 -2\n"
] | [
"1\n",
"2\n"
] | In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value. | 500 | [
{
"input": "3 2\n-1 1 2",
"output": "1"
},
{
"input": "2 3\n-2 -2",
"output": "2"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "2 2\n-1 -1",
"output": "1"
},
{
"input": "15 5\n-2 -1 2 -4 -3 4 -4 -2 -2 2 -2 -1 1 -4 -2",
"output": "4"
},
{
"... | 1,618,350,535 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 62 | 0 | n,x=map(int,input().split())
l=list(map(int,input().split()))
import math as m
print(m.ceil(abs(sum(l))/x)) | Title: Vanya and Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*.
Input Specification:
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['3 2\n-1 1 2\n', '2 3\n-2 -2\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value. | ```python
n,x=map(int,input().split())
l=list(map(int,input().split()))
import math as m
print(m.ceil(abs(sum(l))/x))
``` | 3 | |
78 | B | Easter Eggs | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | B. Easter Eggs | 2 | 256 | The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible. | The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100). | Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them. | [
"8\n",
"13\n"
] | [
"ROYGRBIV\n",
"ROYGBIVGBIVYG\n"
] | The way the eggs will be painted in the first sample is shown on the picture: | 1,000 | [
{
"input": "8",
"output": "ROYGBIVG"
},
{
"input": "13",
"output": "ROYGBIVOYGBIV"
},
{
"input": "7",
"output": "ROYGBIV"
},
{
"input": "10",
"output": "ROYGBIVYGB"
},
{
"input": "14",
"output": "ROYGBIVROYGBIV"
},
{
"input": "50",
"output": "ROYGB... | 1,533,133,929 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 248 | 0 | n=int(input())
arr=['R','O','Y','G']
print("BIV",end='')
for i in range(n-3):
print(arr[i%4],end='') | Title: Easter Eggs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible.
Input Specification:
The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100).
Output Specification:
Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them.
Demo Input:
['8\n', '13\n']
Demo Output:
['ROYGRBIV\n', 'ROYGBIVGBIVYG\n']
Note:
The way the eggs will be painted in the first sample is shown on the picture: | ```python
n=int(input())
arr=['R','O','Y','G']
print("BIV",end='')
for i in range(n-3):
print(arr[i%4],end='')
``` | 3.938 |
181 | A | Series of Crimes | PROGRAMMING | 800 | [
"brute force",
"geometry",
"implementation"
] | null | null | The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements. | Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right. | [
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] | [
"1 1\n",
"2 3\n"
] | none | 500 | [
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n....... | 1,585,811,201 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 218 | 307,200 | r,c = map(int,input().split())
ind_c =[]
ind = None
for i in range(r):
s = input()
if s.count('*') == 1:
ind = [i,s.index('*')]
elif s.count('*')>1:
for j in range(c):
if s[j] == '*':
ind_c.append(j)
if ind[1] == ind_c[1]:
print(ind[0]+1,ind_c[0]+1)
else:
print(ind[0]+1,ind_c[1]+1)
| Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none | ```python
r,c = map(int,input().split())
ind_c =[]
ind = None
for i in range(r):
s = input()
if s.count('*') == 1:
ind = [i,s.index('*')]
elif s.count('*')>1:
for j in range(c):
if s[j] == '*':
ind_c.append(j)
if ind[1] == ind_c[1]:
print(ind[0]+1,ind_c[0]+1)
else:
print(ind[0]+1,ind_c[1]+1)
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — 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": "chn... | 1,597,768,871 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 310 | 0 | s=input()
up_count=0
low_count=0
for i in s:
if(i.islower()):
low_count+=1
else:
up_count+=1
if(low_count>=up_count):
print(s.lower())
else:
print(s.upper())
| Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
s=input()
up_count=0
low_count=0
for i in s:
if(i.islower()):
low_count+=1
else:
up_count+=1
if(low_count>=up_count):
print(s.lower())
else:
print(s.upper())
``` | 3.9225 |
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": "tymbzjyqhymeda... | 1,619,244,895 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 78 | 0 | k=input().lower()
a='hello';j=0
for i in range(len(k)):
if k[i]==a[j]:
j+=1
if j==5:
break
print(['NO','YES'][j>4])
| 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
k=input().lower()
a='hello';j=0
for i in range(len(k)):
if k[i]==a[j]:
j+=1
if j==5:
break
print(['NO','YES'][j>4])
``` | 3.961 |
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": "... | 1,615,248,249 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 109 | 2,764,800 | from operator import xor
from typing import Callable, List, TypeVar, cast
F = TypeVar("F", bound=Callable[[], None])
def my_decorator(func: F) -> F:
def wrapper():
for _ in range(int(input("Times: "))):
func()
return cast(F, wrapper)
def main() -> None:
VALUES: List[str] = [input() for _ in range(2)]
print(f"{xor(*(int(value,2)for value in VALUES)):b}".zfill(len(VALUES[0])))
if __name__ == "__main__":
main()
| 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
from operator import xor
from typing import Callable, List, TypeVar, cast
F = TypeVar("F", bound=Callable[[], None])
def my_decorator(func: F) -> F:
def wrapper():
for _ in range(int(input("Times: "))):
func()
return cast(F, wrapper)
def main() -> None:
VALUES: List[str] = [input() for _ in range(2)]
print(f"{xor(*(int(value,2)for value in VALUES)):b}".zfill(len(VALUES[0])))
if __name__ == "__main__":
main()
``` | 3.9676 |
548 | A | Mike and Fax | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"strings"
] | null | null | While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string *s*.
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly *k* messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string *s* is a concatenation of *k* palindromes of the same length. | The first line of input contains string *s* containing lowercase English letters (1<=≤<=|*s*|<=≤<=1000).
The second line contains integer *k* (1<=≤<=*k*<=≤<=1000). | Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise. | [
"saba\n2\n",
"saddastavvat\n2\n"
] | [
"NO\n",
"YES\n"
] | Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | 500 | [
{
"input": "saba\n2",
"output": "NO"
},
{
"input": "saddastavvat\n2",
"output": "YES"
},
{
"input": "aaaaaaaaaa\n3",
"output": "NO"
},
{
"input": "aaaaaa\n3",
"output": "YES"
},
{
"input": "abaacca\n2",
"output": "NO"
},
{
"input": "a\n1",
"output"... | 1,657,257,121 | 2,147,483,647 | Python 3 | OK | TESTS | 56 | 46 | 0 | string = input()
partskitnekarnehai = int(input())
if len(string)%partskitnekarnehai!=0:
print('NO')
exit()
nayalength = len(string) // partskitnekarnehai
# o is length of part
for j in range(partskitnekarnehai):
seedha_part = string[ j*nayalength : (j + 1)*nayalength]
# print(seedha_part)
ulta_part=reversed(seedha_part)
# print("".join(ulta_part))
if seedha_part != "".join(ulta_part):
print("NO")
exit()
print("YES")
'''
https://codeforces.com/problemset/problem/548/A
1. Aaj exit() seekha, kaafi useful, i always thought agar aise kuch
karna hai to i'll have to define a function, fir use return to get
out of the code.
4. Abe yaar mai bhi kya gadha hu, like agar NO hai answer,
to exit karna chahiye, i exited on YES wala case
3. Are bhai yaaar wo sabse upar wala block mai bhool hi gaya tha,
gahdam ek aur wrong submission
2. Also line 7 me agar -1 kia to it only check first part of string
''' | Title: Mike and Fax
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string *s*.
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly *k* messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string *s* is a concatenation of *k* palindromes of the same length.
Input Specification:
The first line of input contains string *s* containing lowercase English letters (1<=≤<=|*s*|<=≤<=1000).
The second line contains integer *k* (1<=≤<=*k*<=≤<=1000).
Output Specification:
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Demo Input:
['saba\n2\n', 'saddastavvat\n2\n']
Demo Output:
['NO\n', 'YES\n']
Note:
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | ```python
string = input()
partskitnekarnehai = int(input())
if len(string)%partskitnekarnehai!=0:
print('NO')
exit()
nayalength = len(string) // partskitnekarnehai
# o is length of part
for j in range(partskitnekarnehai):
seedha_part = string[ j*nayalength : (j + 1)*nayalength]
# print(seedha_part)
ulta_part=reversed(seedha_part)
# print("".join(ulta_part))
if seedha_part != "".join(ulta_part):
print("NO")
exit()
print("YES")
'''
https://codeforces.com/problemset/problem/548/A
1. Aaj exit() seekha, kaafi useful, i always thought agar aise kuch
karna hai to i'll have to define a function, fir use return to get
out of the code.
4. Abe yaar mai bhi kya gadha hu, like agar NO hai answer,
to exit karna chahiye, i exited on YES wala case
3. Are bhai yaaar wo sabse upar wala block mai bhool hi gaya tha,
gahdam ek aur wrong submission
2. Also line 7 me agar -1 kia to it only check first part of string
'''
``` | 3 | |
411 | A | Password Check | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q. | The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_". | If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes). | [
"abacaba\n",
"X12345\n",
"CONTEST_is_STARTED!!11\n"
] | [
"Too weak\n",
"Too weak\n",
"Correct\n"
] | none | 0 | [
{
"input": "abacaba",
"output": "Too weak"
},
{
"input": "X12345",
"output": "Too weak"
},
{
"input": "CONTEST_is_STARTED!!11",
"output": "Correct"
},
{
"input": "1zA__",
"output": "Correct"
},
{
"input": "1zA_",
"output": "Too weak"
},
{
"input": "zA_... | 1,618,994,030 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 93 | 819,200 | import string
def selection_sort(l):
for i in range(len(l)):
min_val = l[i]
min_val_index = i
for j in range(i, len(l)):
if l[j] < min_val:
min_val = l[j]
min_val_index = j
l[i], l[min_val_index] = l[min_val_index], l[i]
return l
def merge_sort(l):
def merge(l1, l2):
output_list = []
i, j = 0, 0
while i < len(l1) and j < len(l2):
if l1[i] < l2[j]:
output_list.append(l1[i])
i += 1
else:
output_list.append(l2[j])
j += 1
if i == len(l1):
output_list += l2[j:]
else:
output_list += l1[i:]
return output_list
def sort(li):
if len(li) < 2:
return li
mid = len(li) // 2
left = sort(li[:mid])
right = sort(li[mid:])
return merge(left, right)
return sort(l)
def main_function():
is_len_greater_than_5 = False
does_contain_digit = False
does_contain_upper = False
does_contain_lower = False
password = input()
if len(password) >= 5:
is_len_greater_than_5 = True
for i in password:
if i.isdigit():
does_contain_digit = True
elif i.isupper():
does_contain_upper = True
elif i.islower():
does_contain_lower = True
if is_len_greater_than_5 and does_contain_lower and does_contain_upper and does_contain_digit:
return "Correct"
return "Too weak"
print(main_function()) | Title: Password Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input Specification:
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output Specification:
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Demo Input:
['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n']
Demo Output:
['Too weak\n', 'Too weak\n', 'Correct\n']
Note:
none | ```python
import string
def selection_sort(l):
for i in range(len(l)):
min_val = l[i]
min_val_index = i
for j in range(i, len(l)):
if l[j] < min_val:
min_val = l[j]
min_val_index = j
l[i], l[min_val_index] = l[min_val_index], l[i]
return l
def merge_sort(l):
def merge(l1, l2):
output_list = []
i, j = 0, 0
while i < len(l1) and j < len(l2):
if l1[i] < l2[j]:
output_list.append(l1[i])
i += 1
else:
output_list.append(l2[j])
j += 1
if i == len(l1):
output_list += l2[j:]
else:
output_list += l1[i:]
return output_list
def sort(li):
if len(li) < 2:
return li
mid = len(li) // 2
left = sort(li[:mid])
right = sort(li[mid:])
return merge(left, right)
return sort(l)
def main_function():
is_len_greater_than_5 = False
does_contain_digit = False
does_contain_upper = False
does_contain_lower = False
password = input()
if len(password) >= 5:
is_len_greater_than_5 = True
for i in password:
if i.isdigit():
does_contain_digit = True
elif i.isupper():
does_contain_upper = True
elif i.islower():
does_contain_lower = True
if is_len_greater_than_5 and does_contain_lower and does_contain_upper and does_contain_digit:
return "Correct"
return "Too weak"
print(main_function())
``` | 3 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,603,101,859 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 218 | 0 | m = max(list(map(int,input().split())))
if m==1: print('1/1')
elif m==2: print('5/6')
elif m==3: print('2/3')
elif m==4: print('1/2')
elif m==5: print('1/3')
else: print('1/6') | Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
m = max(list(map(int,input().split())))
if m==1: print('1/1')
elif m==2: print('5/6')
elif m==3: print('2/3')
elif m==4: print('1/2')
elif m==5: print('1/3')
else: print('1/6')
``` | 3.891 |
701 | A | Cards | PROGRAMMING | 800 | [
"greedy",
"implementation"
] | null | null | There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible. | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card. | Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them. | [
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] | [
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] | In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable. | 500 | [
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ... | 1,570,361,214 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 109 | 0 | n=int(input())
l=list(map(int,input().split()))
a=[]
for q in range(n):
a.append([l[q],q])
a.sort()
for q in range(n//2):
print(a[q][1]+1,a[n-q-1][1]+1) | Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable. | ```python
n=int(input())
l=list(map(int,input().split()))
a=[]
for q in range(n):
a.append([l[q],q])
a.sort()
for q in range(n//2):
print(a[q][1]+1,a[n-q-1][1]+1)
``` | 3 | |
937 | A | Olympiad | PROGRAMMING | 800 | [
"implementation",
"sortings"
] | null | null | The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score. | Print a single integer — the desired number of ways. | [
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] | [
"3\n",
"1\n",
"1\n"
] | There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | 500 | [
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
... | 1,671,085,105 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 46 | 0 | n=int(input())
a=list(map(int,input().split()))
sa=set(a)
arr=list(sa)
count=0
for i in sa:
if i!=0:
count+=1
print(count)
| Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | ```python
n=int(input())
a=list(map(int,input().split()))
sa=set(a)
arr=list(sa)
count=0
for i in sa:
if i!=0:
count+=1
print(count)
``` | 3 | |
990 | B | Micro-World | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope. | The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have. | Print the only integer — minimal possible number of bacteria can remain. | [
"7 1\n101 53 42 102 101 55 54\n",
"6 5\n20 15 10 15 20 25\n",
"7 1000000\n1 1 1 1 1 1 1\n"
] | [
"3\n",
"1\n",
"7\n"
] | The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria. | 0 | [
{
"input": "7 1\n101 53 42 102 101 55 54",
"output": "3"
},
{
"input": "6 5\n20 15 10 15 20 25",
"output": "1"
},
{
"input": "7 1000000\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 4\n8",
"output": "1"
},
{
"inp... | 1,553,015,310 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 358 | 14,643,200 | n, K = map(int, input().split())
a = sorted(list(map(int, input().split())))
num = 0
for i in a:
while a[num] < i:
if i <= K+a[num]:
n-=1
num += 1
print(n)
| Title: Micro-World
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
Input Specification:
The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have.
Output Specification:
Print the only integer — minimal possible number of bacteria can remain.
Demo Input:
['7 1\n101 53 42 102 101 55 54\n', '6 5\n20 15 10 15 20 25\n', '7 1000000\n1 1 1 1 1 1 1\n']
Demo Output:
['3\n', '1\n', '7\n']
Note:
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria. | ```python
n, K = map(int, input().split())
a = sorted(list(map(int, input().split())))
num = 0
for i in a:
while a[num] < i:
if i <= K+a[num]:
n-=1
num += 1
print(n)
``` | 3 | |
268 | A | Games | PROGRAMMING | 800 | [
"brute force"
] | null | null | Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively. | In a single line print the number of games where the host team is going to play in the guest uniform. | [
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] | [
"1\n",
"5\n",
"0\n"
] | In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first). | 500 | [
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n... | 1,697,790,922 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | teams=int(input())
matches=[]
for i in range(teams):
z=list(map(int,input().split(' ')))
matches.append(z)
out=0
for i in matches:
for m in matches:
if i[0]==m[1]:
out+=1
print(out)
| Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first). | ```python
teams=int(input())
matches=[]
for i in range(teams):
z=list(map(int,input().split(' ')))
matches.append(z)
out=0
for i in matches:
for m in matches:
if i[0]==m[1]:
out+=1
print(out)
``` | 3 | |
415 | A | Mashmokh and Lights | PROGRAMMING | 900 | [
"implementation"
] | null | null | Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to *n*. There are *n* buttons in Mashmokh's room indexed from 1 to *n* as well. If Mashmokh pushes button with index *i*, then each light with index not less than *i* that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed *m* distinct buttons *b*1,<=*b*2,<=...,<=*b**m* (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button *b**i* is actually *b**i*, not *i*.
Please, help Mashmokh, print these indices. | The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of the factory lights and the pushed buttons respectively. The next line contains *m* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*).
It is guaranteed that all lights will be turned off after pushing all buttons. | Output *n* space-separated integers where the *i*-th number is index of the button that turns the *i*-th light off. | [
"5 4\n4 3 1 2\n",
"5 5\n5 4 3 2 1\n"
] | [
"1 1 3 4 4 \n",
"1 2 3 4 5 \n"
] | In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off. | 500 | [
{
"input": "5 4\n4 3 1 2",
"output": "1 1 3 4 4 "
},
{
"input": "5 5\n5 4 3 2 1",
"output": "1 2 3 4 5 "
},
{
"input": "16 11\n8 5 12 10 14 2 6 3 15 9 1",
"output": "1 2 2 2 5 5 5 8 8 8 8 8 8 8 8 8 "
},
{
"input": "79 22\n76 32 48 28 33 44 58 59 1 51 77 13 15 64 49 72 74 21 6... | 1,556,639,925 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 109 | 0 | def main():
[n, m] = [int(_) for _ in input().split()]
buttons = [int(_) for _ in input().split()]
# index = 0
# while index < len(buttons) - 1:
# button = buttons[index]
# buttons = buttons[:(index + 1)] + [buttons[i] for i in range(index + 1, len(buttons)) if buttons[i] < button]
# index += 1
lights = [0] * (n + 1)
for button in buttons:
i = button
while i <= n and lights[i] == 0:
lights[i] = button
i += 1
result = ' '.join([str(lights[i]) for i in range(1, n + 1)])
print(result)
if __name__ == '__main__':
main()
| Title: Mashmokh and Lights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to *n*. There are *n* buttons in Mashmokh's room indexed from 1 to *n* as well. If Mashmokh pushes button with index *i*, then each light with index not less than *i* that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed *m* distinct buttons *b*1,<=*b*2,<=...,<=*b**m* (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button *b**i* is actually *b**i*, not *i*.
Please, help Mashmokh, print these indices.
Input Specification:
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of the factory lights and the pushed buttons respectively. The next line contains *m* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*).
It is guaranteed that all lights will be turned off after pushing all buttons.
Output Specification:
Output *n* space-separated integers where the *i*-th number is index of the button that turns the *i*-th light off.
Demo Input:
['5 4\n4 3 1 2\n', '5 5\n5 4 3 2 1\n']
Demo Output:
['1 1 3 4 4 \n', '1 2 3 4 5 \n']
Note:
In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off. | ```python
def main():
[n, m] = [int(_) for _ in input().split()]
buttons = [int(_) for _ in input().split()]
# index = 0
# while index < len(buttons) - 1:
# button = buttons[index]
# buttons = buttons[:(index + 1)] + [buttons[i] for i in range(index + 1, len(buttons)) if buttons[i] < button]
# index += 1
lights = [0] * (n + 1)
for button in buttons:
i = button
while i <= n and lights[i] == 0:
lights[i] = button
i += 1
result = ' '.join([str(lights[i]) for i in range(1, n + 1)])
print(result)
if __name__ == '__main__':
main()
``` | 3 | |
897 | A | Scarborough Fair | PROGRAMMING | 800 | [
"implementation"
] | null | null | Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space. | Output string *s* after performing *m* operations described above. | [
"3 1\nioi\n1 1 i n\n",
"5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n"
] | [
"noi",
"gaaak"
] | For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | 500 | [
{
"input": "3 1\nioi\n1 1 i n",
"output": "noi"
},
{
"input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g",
"output": "gaaak"
},
{
"input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n... | 1,619,169,874 | 2,147,483,647 | PyPy 3 | OK | TESTS | 47 | 108 | 2,560,000 | n, m = map(int, input().rstrip().split())
s=list(input())
for i in range(m):
l, r, c1,c2 = map(str,input().rstrip().split())
for j in range(int(l)-1,int(r)):
if s[j]==c1:
s[j]=c2
print("".join(s)) | Title: Scarborough Fair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
Output Specification:
Output string *s* after performing *m* operations described above.
Demo Input:
['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n']
Demo Output:
['noi', 'gaaak']
Note:
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | ```python
n, m = map(int, input().rstrip().split())
s=list(input())
for i in range(m):
l, r, c1,c2 = map(str,input().rstrip().split())
for j in range(int(l)-1,int(r)):
if s[j]==c1:
s[j]=c2
print("".join(s))
``` | 3 | |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,666,420,095 | 2,147,483,647 | Python 3 | OK | TESTS | 88 | 62 | 716,800 | from calendar import calendar
def List():
return list(map(int,input().split()))
def Lst():
return list(input())
def Int():
return int(input())
def two():
return map(int,input().split())
numOfVactions = Int()
calendar = List()
outPut = 0
prevState = 0
for currState in calendar:
if currState == 3:
prevState = currState - prevState
elif currState == 0 or currState == prevState:
outPut +=1
prevState = 0
else:
prevState = currState
print(outPut)
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
from calendar import calendar
def List():
return list(map(int,input().split()))
def Lst():
return list(input())
def Int():
return int(input())
def two():
return map(int,input().split())
numOfVactions = Int()
calendar = List()
outPut = 0
prevState = 0
for currState in calendar:
if currState == 3:
prevState = currState - prevState
elif currState == 0 or currState == prevState:
outPut +=1
prevState = 0
else:
prevState = currState
print(outPut)
``` | 3 | |
265 | A | Colorful Stones (Simplified Edition) | PROGRAMMING | 800 | [
"implementation"
] | null | null | There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. | The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. | Print the final 1-based position of Liss in a single line. | [
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] | [
"2\n",
"3\n",
"15\n"
] | none | 500 | [
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
... | 1,647,890,424 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 186 | 6,041,600 | #A. Colorful Stones
from tkinter import S
s = input()
t = input()
i = 0
j = 0
counter = 1
while j < len(t):
if s[i] != t[j]:
j += 1
elif s[i] == t[j]:
i += 1
j += 1
counter += 1
print(counter) | Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none | ```python
#A. Colorful Stones
from tkinter import S
s = input()
t = input()
i = 0
j = 0
counter = 1
while j < len(t):
if s[i] != t[j]:
j += 1
elif s[i] == t[j]:
i += 1
j += 1
counter += 1
print(counter)
``` | 3 | |
616 | C | The Labyrinth | PROGRAMMING | 1,600 | [
"dfs and similar"
] | null | null | You are given a rectangular field of *n*<=×<=*m* cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with '.', impassable cells are marked with '*'. Let's call two empty cells adjacent if they share a side.
Let's call a connected component any non-extendible set of cells such that any two of them are connected by the path of adjacent cells. It is a typical well-known definition of a connected component.
For each impassable cell (*x*,<=*y*) imagine that it is an empty cell (all other cells remain unchanged) and find the size (the number of cells) of the connected component which contains (*x*,<=*y*). You should do it for each impassable cell independently.
The answer should be printed as a matrix with *n* rows and *m* columns. The *j*-th symbol of the *i*-th row should be "." if the cell is empty at the start. Otherwise the *j*-th symbol of the *i*-th row should contain the only digit —- the answer modulo 10. The matrix should be printed without any spaces.
To make your output faster it is recommended to build the output as an array of *n* strings having length *m* and print it as a sequence of lines. It will be much faster than writing character-by-character.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. | The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the field.
Each of the next *n* lines contains *m* symbols: "." for empty cells, "*" for impassable cells. | Print the answer as a matrix as described above. See the examples to precise the format of the output. | [
"3 3\n*.*\n.*.\n*.*\n",
"4 5\n**..*\n..***\n.*.*.\n*.*.*\n"
] | [
"3.3\n.5.\n3.3\n",
"46..3\n..732\n.6.4.\n5.4.3\n"
] | In first example, if we imagine that the central cell is empty then it will be included to component of size 5 (cross). If any of the corner cell will be empty then it will be included to component of size 3 (corner). | 0 | [
{
"input": "3 3\n*.*\n.*.\n*.*",
"output": "3.3\n.5.\n3.3"
},
{
"input": "4 5\n**..*\n..***\n.*.*.\n*.*.*",
"output": "46..3\n..732\n.6.4.\n5.4.3"
},
{
"input": "1 1\n*",
"output": "1"
},
{
"input": "1 1\n.",
"output": "."
},
{
"input": "1 10\n**********",
"ou... | 1,594,530,330 | 2,147,483,647 | PyPy 3 | OK | TESTS | 19 | 810 | 70,348,800 | # n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for i in range(m):
# for _ in range(int(input())):
from collections import Counter
#from fractions import Fraction
#n=int(input())
#arr=list(map(int,input().split()))
#ls = [list(map(int, input().split())) for i in range(n)]
from math import log2
#for _ in range(int(input())):
#n, m = map(int, input().split())
# for _ in range(int(input())):
from math import gcd
#n=int(input())
# for i in range(m):
# for i in range(int(input())):
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=sys.stdin.readline()
# n=int(n)
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=int(inaput())
#for _ in range(int(input())):
#arr=list(map(int,input().split()))
from collections import deque
dx=[-1,0,0,1]
dy=[0,-1,1,0]
def bfs(x,y):
global total
total+=1
q=deque([(x,y)])
v[x][y]=True
h[x][y]=comp
#q.append()
while q:
x,y=q.pop()
for i in range(4):
nx=x+dx[i]
ny=y+dy[i]
#print("nx,y",nx, ny)
if (nx>=0 and nx<n) and (ny>=0 and ny<m) and (v[nx][ny]==False) and (g[nx][ny]=="."):
q.appendleft((nx,ny))
total+=1
v[nx][ny]=True
h[nx][ny]=comp
#global g,h,r,comp,total
n, m = map(int, input().split())
h=[[-1 for i in range(m)] for j in range(n)]
g=[]
v=[[False for i in range(m)]for j in range(n)]
for i in range(n):
g.append(list(input()))
component=[]
for i in range(n):
for j in range(m):
if v[i][j]==False and g[i][j]==".":############
comp=len(component)
#global total
total=0
bfs(i,j)
component.append(total)
#print(component)
for x in range(n):
for y in range(m):
if g[x][y] == "*":
ans = 0
s = set()
for k in range(4):
nx = x + dx[k]
ny = y + dy[k]
if nx >= 0 and nx < n and ny >= 0 and ny < m and g[nx][ny] == ".":
s.add(h[nx][ny])
for itm in s:
ans += component[itm]
ans += 1
ans %= 10
g[x][y] = str(ans)
for i in range(n):
print("".join(g[i]))
| Title: The Labyrinth
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular field of *n*<=×<=*m* cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with '.', impassable cells are marked with '*'. Let's call two empty cells adjacent if they share a side.
Let's call a connected component any non-extendible set of cells such that any two of them are connected by the path of adjacent cells. It is a typical well-known definition of a connected component.
For each impassable cell (*x*,<=*y*) imagine that it is an empty cell (all other cells remain unchanged) and find the size (the number of cells) of the connected component which contains (*x*,<=*y*). You should do it for each impassable cell independently.
The answer should be printed as a matrix with *n* rows and *m* columns. The *j*-th symbol of the *i*-th row should be "." if the cell is empty at the start. Otherwise the *j*-th symbol of the *i*-th row should contain the only digit —- the answer modulo 10. The matrix should be printed without any spaces.
To make your output faster it is recommended to build the output as an array of *n* strings having length *m* and print it as a sequence of lines. It will be much faster than writing character-by-character.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the field.
Each of the next *n* lines contains *m* symbols: "." for empty cells, "*" for impassable cells.
Output Specification:
Print the answer as a matrix as described above. See the examples to precise the format of the output.
Demo Input:
['3 3\n*.*\n.*.\n*.*\n', '4 5\n**..*\n..***\n.*.*.\n*.*.*\n']
Demo Output:
['3.3\n.5.\n3.3\n', '46..3\n..732\n.6.4.\n5.4.3\n']
Note:
In first example, if we imagine that the central cell is empty then it will be included to component of size 5 (cross). If any of the corner cell will be empty then it will be included to component of size 3 (corner). | ```python
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for i in range(m):
# for _ in range(int(input())):
from collections import Counter
#from fractions import Fraction
#n=int(input())
#arr=list(map(int,input().split()))
#ls = [list(map(int, input().split())) for i in range(n)]
from math import log2
#for _ in range(int(input())):
#n, m = map(int, input().split())
# for _ in range(int(input())):
from math import gcd
#n=int(input())
# for i in range(m):
# for i in range(int(input())):
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=sys.stdin.readline()
# n=int(n)
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=int(inaput())
#for _ in range(int(input())):
#arr=list(map(int,input().split()))
from collections import deque
dx=[-1,0,0,1]
dy=[0,-1,1,0]
def bfs(x,y):
global total
total+=1
q=deque([(x,y)])
v[x][y]=True
h[x][y]=comp
#q.append()
while q:
x,y=q.pop()
for i in range(4):
nx=x+dx[i]
ny=y+dy[i]
#print("nx,y",nx, ny)
if (nx>=0 and nx<n) and (ny>=0 and ny<m) and (v[nx][ny]==False) and (g[nx][ny]=="."):
q.appendleft((nx,ny))
total+=1
v[nx][ny]=True
h[nx][ny]=comp
#global g,h,r,comp,total
n, m = map(int, input().split())
h=[[-1 for i in range(m)] for j in range(n)]
g=[]
v=[[False for i in range(m)]for j in range(n)]
for i in range(n):
g.append(list(input()))
component=[]
for i in range(n):
for j in range(m):
if v[i][j]==False and g[i][j]==".":############
comp=len(component)
#global total
total=0
bfs(i,j)
component.append(total)
#print(component)
for x in range(n):
for y in range(m):
if g[x][y] == "*":
ans = 0
s = set()
for k in range(4):
nx = x + dx[k]
ny = y + dy[k]
if nx >= 0 and nx < n and ny >= 0 and ny < m and g[nx][ny] == ".":
s.add(h[nx][ny])
for itm in s:
ans += component[itm]
ans += 1
ans %= 10
g[x][y] = str(ans)
for i in range(n):
print("".join(g[i]))
``` | 3 | |
553 | B | Kyoya and Permutation | PROGRAMMING | 1,900 | [
"binary search",
"combinatorics",
"constructive algorithms",
"greedy",
"implementation",
"math"
] | null | null | Let's define the permutation of length *n* as an array *p*<==<=[*p*1,<=*p*2,<=...,<=*p**n*] consisting of *n* distinct integers from range from 1 to *n*. We say that this permutation maps value 1 into the value *p*1, value 2 into the value *p*2 and so on.
Kyota Ootori has just learned about cyclic representation of a permutation. A cycle is a sequence of numbers such that each element of this sequence is being mapped into the next element of this sequence (and the last element of the cycle is being mapped into the first element of the cycle). The cyclic representation is a representation of *p* as a collection of cycles forming *p*. For example, permutation *p*<==<=[4,<=1,<=6,<=2,<=5,<=3] has a cyclic representation that looks like (142)(36)(5) because 1 is replaced by 4, 4 is replaced by 2, 2 is replaced by 1, 3 and 6 are swapped, and 5 remains in place.
Permutation may have several cyclic representations, so Kyoya defines the standard cyclic representation of a permutation as follows. First, reorder the elements within each cycle so the largest element is first. Then, reorder all of the cycles so they are sorted by their first element. For our example above, the standard cyclic representation of [4,<=1,<=6,<=2,<=5,<=3] is (421)(5)(63).
Now, Kyoya notices that if we drop the parenthesis in the standard cyclic representation, we get another permutation! For instance, [4,<=1,<=6,<=2,<=5,<=3] will become [4,<=2,<=1,<=5,<=6,<=3].
Kyoya notices that some permutations don't change after applying operation described above at all. He wrote all permutations of length *n* that do not change in a list in lexicographic order. Unfortunately, his friend Tamaki Suoh lost this list. Kyoya wishes to reproduce the list and he needs your help. Given the integers *n* and *k*, print the permutation that was *k*-th on Kyoya's list. | The first line will contain two integers *n*, *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=*min*{1018,<=*l*} where *l* is the length of the Kyoya's list). | Print *n* space-separated integers, representing the permutation that is the answer for the question. | [
"4 3\n",
"10 1\n"
] | [
"1 3 2 4\n",
"1 2 3 4 5 6 7 8 9 10\n"
] | The standard cycle representation is (1)(32)(4), which after removing parenthesis gives us the original permutation. The first permutation on the list would be [1, 2, 3, 4], while the second permutation would be [1, 2, 4, 3]. | 500 | [
{
"input": "4 3",
"output": "1 3 2 4"
},
{
"input": "10 1",
"output": "1 2 3 4 5 6 7 8 9 10"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "50 1",
"output": "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 ... | 1,435,166,677 | 3,277 | Python 3 | OK | TESTS | 65 | 62 | 0 | #!/usr/bin/python3
arr = [1] * 51
for i in range(2, 51):
arr[i] = arr[i - 1] + arr[i - 2]
ans = []
def generate(i, n, to):
if i == n:
assert to == 1
print(" ".join(map(str, ans)))
return
if i + 1 == n:
ans.append(n)
generate(i + 1, n, to)
return
if arr[n - i - 1] < to:
ans.append(i + 2)
ans.append(i + 1)
generate(i + 2, n, to - arr[n - i - 1])
else:
ans.append(i + 1)
generate(i + 1, n, to)
n, k = map(int, input().split())
generate(0, n, k)
| Title: Kyoya and Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's define the permutation of length *n* as an array *p*<==<=[*p*1,<=*p*2,<=...,<=*p**n*] consisting of *n* distinct integers from range from 1 to *n*. We say that this permutation maps value 1 into the value *p*1, value 2 into the value *p*2 and so on.
Kyota Ootori has just learned about cyclic representation of a permutation. A cycle is a sequence of numbers such that each element of this sequence is being mapped into the next element of this sequence (and the last element of the cycle is being mapped into the first element of the cycle). The cyclic representation is a representation of *p* as a collection of cycles forming *p*. For example, permutation *p*<==<=[4,<=1,<=6,<=2,<=5,<=3] has a cyclic representation that looks like (142)(36)(5) because 1 is replaced by 4, 4 is replaced by 2, 2 is replaced by 1, 3 and 6 are swapped, and 5 remains in place.
Permutation may have several cyclic representations, so Kyoya defines the standard cyclic representation of a permutation as follows. First, reorder the elements within each cycle so the largest element is first. Then, reorder all of the cycles so they are sorted by their first element. For our example above, the standard cyclic representation of [4,<=1,<=6,<=2,<=5,<=3] is (421)(5)(63).
Now, Kyoya notices that if we drop the parenthesis in the standard cyclic representation, we get another permutation! For instance, [4,<=1,<=6,<=2,<=5,<=3] will become [4,<=2,<=1,<=5,<=6,<=3].
Kyoya notices that some permutations don't change after applying operation described above at all. He wrote all permutations of length *n* that do not change in a list in lexicographic order. Unfortunately, his friend Tamaki Suoh lost this list. Kyoya wishes to reproduce the list and he needs your help. Given the integers *n* and *k*, print the permutation that was *k*-th on Kyoya's list.
Input Specification:
The first line will contain two integers *n*, *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=*min*{1018,<=*l*} where *l* is the length of the Kyoya's list).
Output Specification:
Print *n* space-separated integers, representing the permutation that is the answer for the question.
Demo Input:
['4 3\n', '10 1\n']
Demo Output:
['1 3 2 4\n', '1 2 3 4 5 6 7 8 9 10\n']
Note:
The standard cycle representation is (1)(32)(4), which after removing parenthesis gives us the original permutation. The first permutation on the list would be [1, 2, 3, 4], while the second permutation would be [1, 2, 4, 3]. | ```python
#!/usr/bin/python3
arr = [1] * 51
for i in range(2, 51):
arr[i] = arr[i - 1] + arr[i - 2]
ans = []
def generate(i, n, to):
if i == n:
assert to == 1
print(" ".join(map(str, ans)))
return
if i + 1 == n:
ans.append(n)
generate(i + 1, n, to)
return
if arr[n - i - 1] < to:
ans.append(i + 2)
ans.append(i + 1)
generate(i + 2, n, to - arr[n - i - 1])
else:
ans.append(i + 1)
generate(i + 1, n, to)
n, k = map(int, input().split())
generate(0, n, k)
``` | 3 | |
361 | A | Levko and Table | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them. | The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000). | Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them. | [
"2 4\n",
"4 7\n"
] | [
"1 3\n3 1\n",
"2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n"
] | In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.
In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements. | 500 | [
{
"input": "2 4",
"output": "4 0 \n0 4 "
},
{
"input": "4 7",
"output": "7 0 0 0 \n0 7 0 0 \n0 0 7 0 \n0 0 0 7 "
},
{
"input": "1 8",
"output": "8 "
},
{
"input": "9 3",
"output": "3 0 0 0 0 0 0 0 0 \n0 3 0 0 0 0 0 0 0 \n0 0 3 0 0 0 0 0 0 \n0 0 0 3 0 0 0 0 0 \n0 0 0 0 3 0... | 1,614,523,427 | 2,147,483,647 | Python 3 | OK | TESTS | 22 | 77 | 307,200 | n, k = map(int, input().split());
a = list(range(n));
for i in range (0, n):
a[i] = list(range(n));
for j in range (0, n):
if j == i:
a[i][j] = k;
else: a[i][j] = 0;
print(*a[i], sep=" ") | Title: Levko and Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them.
Input Specification:
The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000).
Output Specification:
Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them.
Demo Input:
['2 4\n', '4 7\n']
Demo Output:
['1 3\n3 1\n', '2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n']
Note:
In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.
In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements. | ```python
n, k = map(int, input().split());
a = list(range(n));
for i in range (0, n):
a[i] = list(range(n));
for j in range (0, n):
if j == i:
a[i][j] = k;
else: a[i][j] = 0;
print(*a[i], sep=" ")
``` | 3 | |
582 | A | GCD Table | PROGRAMMING | 1,700 | [
"constructive algorithms",
"greedy",
"number theory"
] | null | null | The GCD table *G* of size *n*<=×<=*n* for an array of positive integers *a* of length *n* is defined by formula
Let us remind you that the greatest common divisor (GCD) of two positive integers *x* and *y* is the greatest integer that is divisor of both *x* and *y*, it is denoted as . For example, for array *a*<==<={4,<=3,<=6,<=2} of length 4 the GCD table will look as follows:
Given all the numbers of the GCD table *G*, restore array *a*. | The first line contains number *n* (1<=≤<=*n*<=≤<=500) — the length of array *a*. The second line contains *n*2 space-separated numbers — the elements of the GCD table of *G* for array *a*.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array *a*. | In the single line print *n* positive integers — the elements of array *a*. If there are multiple possible solutions, you are allowed to print any of them. | [
"4\n2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2\n",
"1\n42\n",
"2\n1 1 1 1\n"
] | [
"4 3 6 2",
"42 ",
"1 1 "
] | none | 750 | [
{
"input": "4\n2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2",
"output": "2 3 4 6 "
},
{
"input": "1\n42",
"output": "42 "
},
{
"input": "2\n1 1 1 1",
"output": "1 1 "
},
{
"input": "2\n54748096 1 641009859 1",
"output": "54748096 641009859 "
},
{
"input": "3\n1 7 923264237 374... | 1,592,034,328 | 2,147,483,647 | PyPy 3 | OK | TESTS | 54 | 451 | 21,811,200 | import math as ma
import sys
input=sys.stdin.readline
def fu(b):
for i in b:
if b[i]!=0:
return i
return -1
def gcd(a,b):
if a%b==0:
return b
else:
return gcd(b,a%b)
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
b={}
for i in range(n*n):
if a[i] in b.keys():
b[a[i]]+=1
else:
b[a[i]]=1
c=[]
for i in b:
c.append(i)
b[i]-=1
break
while 1>0:
if len(c)<n:
a=fu(b)
if a==-1:
break
else:
b[a]-=1
for i in range(len(c)):
b[gcd(a,c[i])]-=2
c.append(a)
else:
break
print(*c) | Title: GCD Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The GCD table *G* of size *n*<=×<=*n* for an array of positive integers *a* of length *n* is defined by formula
Let us remind you that the greatest common divisor (GCD) of two positive integers *x* and *y* is the greatest integer that is divisor of both *x* and *y*, it is denoted as . For example, for array *a*<==<={4,<=3,<=6,<=2} of length 4 the GCD table will look as follows:
Given all the numbers of the GCD table *G*, restore array *a*.
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=500) — the length of array *a*. The second line contains *n*2 space-separated numbers — the elements of the GCD table of *G* for array *a*.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array *a*.
Output Specification:
In the single line print *n* positive integers — the elements of array *a*. If there are multiple possible solutions, you are allowed to print any of them.
Demo Input:
['4\n2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2\n', '1\n42\n', '2\n1 1 1 1\n']
Demo Output:
['4 3 6 2', '42 ', '1 1 ']
Note:
none | ```python
import math as ma
import sys
input=sys.stdin.readline
def fu(b):
for i in b:
if b[i]!=0:
return i
return -1
def gcd(a,b):
if a%b==0:
return b
else:
return gcd(b,a%b)
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
b={}
for i in range(n*n):
if a[i] in b.keys():
b[a[i]]+=1
else:
b[a[i]]=1
c=[]
for i in b:
c.append(i)
b[i]-=1
break
while 1>0:
if len(c)<n:
a=fu(b)
if a==-1:
break
else:
b[a]-=1
for i in range(len(c)):
b[gcd(a,c[i])]-=2
c.append(a)
else:
break
print(*c)
``` | 3 | |
102 | B | Sum of Digits | PROGRAMMING | 1,000 | [
"implementation"
] | B. Sum of Digits | 2 | 265 | Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit? | The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes. | Print the number of times a number can be replaced by the sum of its digits until it only contains one digit. | [
"0\n",
"10\n",
"991\n"
] | [
"0\n",
"1\n",
"3\n"
] | In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit. | 1,000 | [
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output... | 1,647,802,245 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 124 | 0 | s=input()
c=0
while(len(s)>1):
sum=0
for i in s:
sum+=int(i)
s=str(sum)
c+=1
print(c) | Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit. | ```python
s=input()
c=0
while(len(s)>1):
sum=0
for i in s:
sum+=int(i)
s=str(sum)
c+=1
print(c)
``` | 3.969 |
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": "... | 1,553,772,872 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 124 | 0 | m = input()
oo = []
mm = list(m)
n = input()
nn = list(n)
for i in range(0,len(m)):
oo.append(int(mm[i]) + int(nn[i]))
for i in range(0,len(oo)):
oo[i] = str(oo[i])
str1 = ''.join(oo)
finfin = str1.replace('2','0')
print(finfin) | 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 = input()
oo = []
mm = list(m)
n = input()
nn = list(n)
for i in range(0,len(m)):
oo.append(int(mm[i]) + int(nn[i]))
for i in range(0,len(oo)):
oo[i] = str(oo[i])
str1 = ''.join(oo)
finfin = str1.replace('2','0')
print(finfin)
``` | 3.969 |
764 | A | Taymyr is calling you | PROGRAMMING | 800 | [
"brute force",
"implementation",
"math"
] | null | null | Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. | The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). | Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. | [
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] | [
"10\n",
"2\n",
"1\n"
] | Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | 500 | [
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"i... | 1,486,136,023 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 61 | 5,939,200 | n, m, z = (int(i) for i in input().split())
calls = set(i for i in range(0, z + 1, n))
art = set(i for i in range(0, z + 1, m))
print(len(calls & art) - 1) | Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | ```python
n, m, z = (int(i) for i in input().split())
calls = set(i for i in range(0, z + 1, n))
art = set(i for i in range(0, z + 1, m))
print(len(calls & art) - 1)
``` | 3 | |
78 | A | Haiku | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Haiku | 2 | 256 | Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not. | The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification. | Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes). | [
"on codeforces \nbeta round is running\n a rustling of keys \n",
"how many gallons\nof edo s rain did you drink\n cuckoo\n"
] | [
"YES",
"NO"
] | none | 500 | [
{
"input": "on codeforces \nbeta round is running\n a rustling of keys ",
"output": "YES"
},
{
"input": "how many gallons\nof edo s rain did you drink\n cuckoo",
"output": "NO"
},
{
"input": " hatsu shigure\n saru mo komino wo\nhoshige nari",
... | 1,572,794,228 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 218 | 0 | a=input()
b=input()
c=input()
p=['a','e','i','o','u']
d=0
e=0
f=0
for i in a:
if i in p:
d=d+1
for j in b:
if j in p:
e=e+1
for k in c:
if k in p:
f=f+1
if d==5 and e==7 and f==5:
print('YES')
else:
print('NO') | Title: Haiku
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input Specification:
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output Specification:
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Demo Input:
['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n']
Demo Output:
['YES', 'NO']
Note:
none | ```python
a=input()
b=input()
c=input()
p=['a','e','i','o','u']
d=0
e=0
f=0
for i in a:
if i in p:
d=d+1
for j in b:
if j in p:
e=e+1
for k in c:
if k in p:
f=f+1
if d==5 and e==7 and f==5:
print('YES')
else:
print('NO')
``` | 3.9455 |
965 | A | Paper Airplanes | PROGRAMMING | 800 | [
"math"
] | null | null | To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy? | The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively. | Print a single integer — the minimum number of packs they should buy. | [
"5 3 2 3\n",
"5 3 100 1\n"
] | [
"4\n",
"5\n"
] | In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets. | 500 | [
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51... | 1,552,765,663 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 109 | 0 | import math
k,n,s,p=map(int,input().split())
print(math.ceil(k*math.ceil(n/s)/p))
| Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets. | ```python
import math
k,n,s,p=map(int,input().split())
print(math.ceil(k*math.ceil(n/s)/p))
``` | 3 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"outp... | 1,696,332,928 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 92 | 0 | line = input()
ans = "NO"
for i in range(len(line) - 6):
if line[i: i + 7].count('0') == 7 or line[i: i + 7].count('1') == 7:
ans = "YES"
break
print(ans)
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
line = input()
ans = "NO"
for i in range(len(line) - 6):
if line[i: i + 7].count('0') == 7 or line[i: i + 7].count('1') == 7:
ans = "YES"
break
print(ans)
``` | 3.977 |
560 | A | Currency System in Geraldion | PROGRAMMING | 1,000 | [
"implementation",
"sortings"
] | null | null | A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum? | The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes. | Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1. | [
"5\n1 2 3 4 5\n"
] | [
"-1\n"
] | none | 500 | [
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"o... | 1,450,812,026 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 46 | 0 | n = int(input())
num = map(int, input().rstrip().split())
if 1 in num:
print(-1)
else:
print(1) | Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none | ```python
n = int(input())
num = map(int, input().rstrip().split())
if 1 in num:
print(-1)
else:
print(1)
``` | 3 | |
462 | A | Appleman and Easy Task | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | null | null | Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces. | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | [
"3\nxxo\nxox\noxx\n",
"4\nxxxo\nxoxo\noxox\nxxxx\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "3\nxxo\nxox\noxx",
"output": "YES"
},
{
"input": "4\nxxxo\nxoxo\noxox\nxxxx",
"output": "NO"
},
{
"input": "1\no",
"output": "YES"
},
{
"input": "2\nox\nxo",
"output": "YES"
},
{
"input": "2\nxx\nxo",
"output": "NO"
},
{
"input": "3\nooo\no... | 1,648,535,153 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | num = int(input())
li = []
for x in range(num):
l = input()
for i in l:
if i == "x":
li.append(0)
else:
li.append(1)
li = [0 for x in range(num)] + li + [0 for x in range(num)]
lis = []
for x in range(num,len(li)-num):
if x%num==0:
lis.append(li[x+1]+ li[x-num]+ li[x+num])
elif (x+1)%num==0:
lis.append(li[x-1]+ li[x-num]+ li[x+num])
else:
lis.append(li[x-1]+ li[x+1]+ li[x-num]+ li[x+num])
yes = True
for i in lis:
if i%2 == 0:
yes = True
else:
yes = False
break
if yes: print("YES")
else: print("NO") | Title: Appleman and Easy Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.
Output Specification:
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
Demo Input:
['3\nxxo\nxox\noxx\n', '4\nxxxo\nxoxo\noxox\nxxxx\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
num = int(input())
li = []
for x in range(num):
l = input()
for i in l:
if i == "x":
li.append(0)
else:
li.append(1)
li = [0 for x in range(num)] + li + [0 for x in range(num)]
lis = []
for x in range(num,len(li)-num):
if x%num==0:
lis.append(li[x+1]+ li[x-num]+ li[x+num])
elif (x+1)%num==0:
lis.append(li[x-1]+ li[x-num]+ li[x+num])
else:
lis.append(li[x-1]+ li[x+1]+ li[x-num]+ li[x+num])
yes = True
for i in lis:
if i%2 == 0:
yes = True
else:
yes = False
break
if yes: print("YES")
else: print("NO")
``` | 3 | |
855 | B | Marvolo Gaunt's Ring | PROGRAMMING | 1,500 | [
"brute force",
"data structures",
"dp"
] | null | null | Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative. | First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). | Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. | [
"5 1 2 3\n1 2 3 4 5\n",
"5 1 2 -3\n-1 -2 -3 -4 -5\n"
] | [
"30\n",
"12\n"
] | In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | 1,000 | [
{
"input": "5 1 2 3\n1 2 3 4 5",
"output": "30"
},
{
"input": "5 1 2 -3\n-1 -2 -3 -4 -5",
"output": "12"
},
{
"input": "5 886327859 82309257 -68295239\n-731225382 354766539 -48222231 -474691998 360965777",
"output": "376059240645059046"
},
{
"input": "4 -96405765 -495906217 6... | 1,691,231,512 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 82 | 108 | 17,920,000 | n,p,q,r=map(int,input().split())
arr=list(map(int,input().strip().split()))
pi,qj,rk= [],[],[]
for i in range(n):
x=arr[i]
pi.append(x*p)
qj.append(x*q)
rk.append(x*r)
for i in range(1,n):
pi[i]=max(pi[i-1],pi[i])
rk[n-i-1]=max(rk[n-i],rk[n-i-1])
s=pi[0]+qj[0]+rk[0]
for i in range(1,n):s=max(s,pi[i]+qj[i]+rk[i])
print(s) | Title: Marvolo Gaunt's Ring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.
Input Specification:
First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.
Demo Input:
['5 1 2 3\n1 2 3 4 5\n', '5 1 2 -3\n-1 -2 -3 -4 -5\n']
Demo Output:
['30\n', '12\n']
Note:
In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | ```python
n,p,q,r=map(int,input().split())
arr=list(map(int,input().strip().split()))
pi,qj,rk= [],[],[]
for i in range(n):
x=arr[i]
pi.append(x*p)
qj.append(x*q)
rk.append(x*r)
for i in range(1,n):
pi[i]=max(pi[i-1],pi[i])
rk[n-i-1]=max(rk[n-i],rk[n-i-1])
s=pi[0]+qj[0]+rk[0]
for i in range(1,n):s=max(s,pi[i]+qj[i]+rk[i])
print(s)
``` | 3 | |
186 | A | Comparing Strings | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not. | The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length. | Print "YES", if the dwarves belong to the same race. Otherwise, print "NO". | [
"ab\nba\n",
"aa\nab\n"
] | [
"YES\n",
"NO\n"
] | - First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b". | 500 | [
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "aa\nab",
"output": "NO"
},
{
"input": "a\nza",
"output": "NO"
},
{
"input": "vvea\nvvae",
"output": "YES"
},
{
"input": "rtfabanpc\natfabrnpc",
"output": "YES"
},
{
"input": "mt\ntm",
"output": "Y... | 1,615,054,742 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 154 | 716,800 | a = input()
b = input()
d = 0
f = []
if not len(a) == len(b):
print('NO')
else:
for i in range(len(a)):
if not a[i] == b[i]:
d += 1
if d == 3:
break
f.append([a[i], b[i]])
if d == 2:
print('YES' if f[0][0] == f[1][1] and f[0][1] == f[1][0] else 'NO')
else:
print('NO')
| Title: Comparing Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
Input Specification:
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
Output Specification:
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
Demo Input:
['ab\nba\n', 'aa\nab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b". | ```python
a = input()
b = input()
d = 0
f = []
if not len(a) == len(b):
print('NO')
else:
for i in range(len(a)):
if not a[i] == b[i]:
d += 1
if d == 3:
break
f.append([a[i], b[i]])
if d == 2:
print('YES' if f[0][0] == f[1][1] and f[0][1] == f[1][0] else 'NO')
else:
print('NO')
``` | 3 | |
358 | B | Dima and Text Messages | PROGRAMMING | 1,500 | [
"brute force",
"strings"
] | null | null | Seryozha has a very changeable character. This time he refused to leave the room to Dima and his girlfriend (her hame is Inna, by the way). However, the two lovebirds can always find a way to communicate. Today they are writing text messages to each other.
Dima and Inna are using a secret code in their text messages. When Dima wants to send Inna some sentence, he writes out all words, inserting a heart before each word and after the last word. A heart is a sequence of two characters: the "less" characters (<) and the digit three (3). After applying the code, a test message looks like that: <3*word*1<3*word*2<3 ... *word**n*<3.
Encoding doesn't end here. Then Dima inserts a random number of small English characters, digits, signs "more" and "less" into any places of the message.
Inna knows Dima perfectly well, so she knows what phrase Dima is going to send her beforehand. Inna has just got a text message. Help her find out if Dima encoded the message correctly. In other words, find out if a text message could have been received by encoding in the manner that is described above. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of words in Dima's message. Next *n* lines contain non-empty words, one word per line. The words only consist of small English letters. The total length of all words doesn't exceed 105.
The last line contains non-empty text message that Inna has got. The number of characters in the text message doesn't exceed 105. A text message can contain only small English letters, digits and signs more and less. | In a single line, print "yes" (without the quotes), if Dima decoded the text message correctly, and "no" (without the quotes) otherwise. | [
"3\ni\nlove\nyou\n<3i<3love<23you<3\n",
"7\ni\nam\nnot\nmain\nin\nthe\nfamily\n<3i<>3am<3the<3<main<3in<3the<3><3family<3\n"
] | [
"yes\n",
"no\n"
] | Please note that Dima got a good old kick in the pants for the second sample from the statement. | 1,000 | [
{
"input": "3\ni\nlove\nyou\n<3i<3love<23you<3",
"output": "yes"
},
{
"input": "7\ni\nam\nnot\nmain\nin\nthe\nfamily\n<3i<>3am<3the<3<main<3in<3the<3><3family<3",
"output": "no"
},
{
"input": "3\ni\nlove\nyou\n<3i<3lo<3ve<3y<<<<<<<ou3<3",
"output": "yes"
},
{
"input": "4\na\n... | 1,653,941,980 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 702 | 10,035,200 | s = []
for i in range(int(input())):
s.append(input())
s = "<3" + '<3'.join(s) + "<3"
cur = 0
t = input()
for i in t:
if i == s[cur]: cur += 1
if cur == len(s): break
print('yes' if cur == len(s) else 'no') | Title: Dima and Text Messages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Seryozha has a very changeable character. This time he refused to leave the room to Dima and his girlfriend (her hame is Inna, by the way). However, the two lovebirds can always find a way to communicate. Today they are writing text messages to each other.
Dima and Inna are using a secret code in their text messages. When Dima wants to send Inna some sentence, he writes out all words, inserting a heart before each word and after the last word. A heart is a sequence of two characters: the "less" characters (<) and the digit three (3). After applying the code, a test message looks like that: <3*word*1<3*word*2<3 ... *word**n*<3.
Encoding doesn't end here. Then Dima inserts a random number of small English characters, digits, signs "more" and "less" into any places of the message.
Inna knows Dima perfectly well, so she knows what phrase Dima is going to send her beforehand. Inna has just got a text message. Help her find out if Dima encoded the message correctly. In other words, find out if a text message could have been received by encoding in the manner that is described above.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of words in Dima's message. Next *n* lines contain non-empty words, one word per line. The words only consist of small English letters. The total length of all words doesn't exceed 105.
The last line contains non-empty text message that Inna has got. The number of characters in the text message doesn't exceed 105. A text message can contain only small English letters, digits and signs more and less.
Output Specification:
In a single line, print "yes" (without the quotes), if Dima decoded the text message correctly, and "no" (without the quotes) otherwise.
Demo Input:
['3\ni\nlove\nyou\n<3i<3love<23you<3\n', '7\ni\nam\nnot\nmain\nin\nthe\nfamily\n<3i<>3am<3the<3<main<3in<3the<3><3family<3\n']
Demo Output:
['yes\n', 'no\n']
Note:
Please note that Dima got a good old kick in the pants for the second sample from the statement. | ```python
s = []
for i in range(int(input())):
s.append(input())
s = "<3" + '<3'.join(s) + "<3"
cur = 0
t = input()
for i in t:
if i == s[cur]: cur += 1
if cur == len(s): break
print('yes' if cur == len(s) else 'no')
``` | 3 | |
250 | A | Paper Work | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as *n* days. Right now his task is to make a series of reports about the company's performance for the last *n* days. We know that the main information in a day report is value *a**i*, the company's profit on the *i*-th day. If *a**i* is negative, then the company suffered losses on the *i*-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the *n* days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (*a**i*<=<<=0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence *a**i*, will print the minimum number of folders. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100), *n* is the number of days. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=100), where *a**i* means the company profit on the *i*-th day. It is possible that the company has no days with the negative *a**i*. | Print an integer *k* — the required minimum number of folders. In the second line print a sequence of integers *b*1, *b*2, ..., *b**k*, where *b**j* is the number of day reports in the *j*-th folder.
If there are multiple ways to sort the reports into *k* days, print any of them. | [
"11\n1 2 3 -4 -5 -6 5 -5 -6 -7 6\n",
"5\n0 -1 100 -1 0\n"
] | [
"3\n5 3 3 ",
"1\n5 "
] | Here goes a way to sort the reports from the first sample into three folders:
In the second sample you can put all five reports in one folder. | 500 | [
{
"input": "11\n1 2 3 -4 -5 -6 5 -5 -6 -7 6",
"output": "3\n5 3 3 "
},
{
"input": "5\n0 -1 100 -1 0",
"output": "1\n5 "
},
{
"input": "1\n0",
"output": "1\n1 "
},
{
"input": "1\n-1",
"output": "1\n1 "
},
{
"input": "2\n0 0",
"output": "1\n2 "
},
{
"inp... | 1,667,083,853 | 2,147,483,647 | Python 3 | OK | TESTS | 69 | 92 | 0 | days = int(input())
sequence = input().split()
folder = 1
bad_rep = 0
rep = 0
ans = ""
for profit in sequence:
rep += 1
if int(profit) < 0:
bad_rep += 1
if bad_rep == 3:
folder += 1
ans = ans + str(rep - 1) + " "
bad_rep = 1
rep = 1
ans += str(rep)
print(folder)
print(ans)
| Title: Paper Work
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as *n* days. Right now his task is to make a series of reports about the company's performance for the last *n* days. We know that the main information in a day report is value *a**i*, the company's profit on the *i*-th day. If *a**i* is negative, then the company suffered losses on the *i*-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the *n* days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (*a**i*<=<<=0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence *a**i*, will print the minimum number of folders.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), *n* is the number of days. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=100), where *a**i* means the company profit on the *i*-th day. It is possible that the company has no days with the negative *a**i*.
Output Specification:
Print an integer *k* — the required minimum number of folders. In the second line print a sequence of integers *b*1, *b*2, ..., *b**k*, where *b**j* is the number of day reports in the *j*-th folder.
If there are multiple ways to sort the reports into *k* days, print any of them.
Demo Input:
['11\n1 2 3 -4 -5 -6 5 -5 -6 -7 6\n', '5\n0 -1 100 -1 0\n']
Demo Output:
['3\n5 3 3 ', '1\n5 ']
Note:
Here goes a way to sort the reports from the first sample into three folders:
In the second sample you can put all five reports in one folder. | ```python
days = int(input())
sequence = input().split()
folder = 1
bad_rep = 0
rep = 0
ans = ""
for profit in sequence:
rep += 1
if int(profit) < 0:
bad_rep += 1
if bad_rep == 3:
folder += 1
ans = ans + str(rep - 1) + " "
bad_rep = 1
rep = 1
ans += str(rep)
print(folder)
print(ans)
``` | 3 | |
394 | A | Counting Sticks | PROGRAMMING | 0 | [
"brute force",
"implementation"
] | null | null | When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us? | The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100. | If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples. | [
"||+|=|||||\n",
"|||||+||=||\n",
"|+|=||||||\n",
"||||+||=||||||\n"
] | [
"|||+|=||||\n",
"Impossible\n",
"Impossible\n",
"||||+||=||||||\n"
] | In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks. | 500 | [
{
"input": "||+|=|||||",
"output": "|||+|=||||"
},
{
"input": "|||||+||=||",
"output": "Impossible"
},
{
"input": "|+|=||||||",
"output": "Impossible"
},
{
"input": "||||+||=||||||",
"output": "||||+||=||||||"
},
{
"input": "||||||||||||+|||||||||||=||||||||||||||... | 1,524,645,638 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 93 | 7,372,800 | n = input()
lth = len(n)
a = n.find('+')
b = n.find('=') - a - 1
c = len(n) - n.find('=') - 1
if a + b == c:
print(n)
elif a + b == c - 2:
print('|',n[:lth-1],sep="")
elif a + b == c + 2:
if a >= 2:
print(n[1:],'|',sep="")
else:
print(n[:a+1],n[a+2:],'|',sep="")
else:
print('Impossible')
| Title: Counting Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
Input Specification:
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
Output Specification:
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
Demo Input:
['||+|=|||||\n', '|||||+||=||\n', '|+|=||||||\n', '||||+||=||||||\n']
Demo Output:
['|||+|=||||\n', 'Impossible\n', 'Impossible\n', '||||+||=||||||\n']
Note:
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks. | ```python
n = input()
lth = len(n)
a = n.find('+')
b = n.find('=') - a - 1
c = len(n) - n.find('=') - 1
if a + b == c:
print(n)
elif a + b == c - 2:
print('|',n[:lth-1],sep="")
elif a + b == c + 2:
if a >= 2:
print(n[1:],'|',sep="")
else:
print(n[:a+1],n[a+2:],'|',sep="")
else:
print('Impossible')
``` | 3 | |
912 | A | Tricky Alchemy | PROGRAMMING | 800 | [
"implementation"
] | null | null | During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls. | The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained. | Print a single integer — the minimum number of crystals that Grisha should acquire in addition. | [
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] | [
"2\n",
"1\n",
"2147483648\n"
] | In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output"... | 1,681,511,351 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 77 | 0 | a, b = map(int, input().split())
x, y, z = map(int, input().split())
need_yellow = x * 2 + y
need_blue = y + z * 3
yellow_diff = max(0, need_yellow - a)
blue_diff = max(0, need_blue - b)
total_diff = yellow_diff + blue_diff
print(total_diff)
| Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | ```python
a, b = map(int, input().split())
x, y, z = map(int, input().split())
need_yellow = x * 2 + y
need_blue = y + z * 3
yellow_diff = max(0, need_yellow - a)
blue_diff = max(0, need_blue - b)
total_diff = yellow_diff + blue_diff
print(total_diff)
``` | 3 | |
965 | C | Greedy Arkady | PROGRAMMING | 2,000 | [
"math"
] | null | null | $k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$. | The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies. | Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$. | [
"20 4 5 2\n",
"30 9 4 1\n"
] | [
"8\n",
"4\n"
] | In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time. | 1,500 | [
{
"input": "20 4 5 2",
"output": "8"
},
{
"input": "30 9 4 1",
"output": "4"
},
{
"input": "2 2 1 1",
"output": "1"
},
{
"input": "42 20 5 29",
"output": "5"
},
{
"input": "1000000000000000000 135 1000000000000000 1000",
"output": "8325624421831635"
},
{
... | 1,525,286,181 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 109 | 7,065,600 | n, k, m, d = map(int, input().split())
ans = int(0)
for i in range(1, d + 1):
l, r, mid = 1, m + 1, 0
while l + 1 < r:
mid = (l + r) // 2
if (n - mid) // (k * mid) < i - 1:
r = mid
else:
l = mid
if (n - l) // (k * l) < i - 1:
break
ans = max(ans, l * i)
print(ans)
| Title: Greedy Arkady
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
Input Specification:
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
Output Specification:
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
Demo Input:
['20 4 5 2\n', '30 9 4 1\n']
Demo Output:
['8\n', '4\n']
Note:
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time. | ```python
n, k, m, d = map(int, input().split())
ans = int(0)
for i in range(1, d + 1):
l, r, mid = 1, m + 1, 0
while l + 1 < r:
mid = (l + r) // 2
if (n - mid) // (k * mid) < i - 1:
r = mid
else:
l = mid
if (n - l) // (k * l) < i - 1:
break
ans = max(ans, l * i)
print(ans)
``` | 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": "... | 1,602,569,375 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 108 | 0 | num1=input()
num2=input()
list=[]
for i in range(len(num1)):
if num1[i]==num2[i]:
list.append(0)
else:
list.append(1)
for e in list:
print(e,end="") | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (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
num1=input()
num2=input()
list=[]
for i in range(len(num1)):
if num1[i]==num2[i]:
list.append(0)
else:
list.append(1)
for e in list:
print(e,end="")
``` | 3.973 |
146 | A | Lucky Ticket | PROGRAMMING | 800 | [
"implementation"
] | 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 loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky. | The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros. | On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes). | [
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] | [
"NO\n",
"NO\n",
"YES\n"
] | In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | 500 | [
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
... | 1,631,361,611 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 124 | 6,963,200 | n = int(input())
num = int(input())
lis = [i for i in str(num)]
lis1 = lis[:int(n/2)]
lis2 = lis[int(n/2):]
lis11 = [int(x) for x in lis1]
lis22 = [int(x) for x in lis2]
if( lis.count('7') + lis.count('4')) == n and sum(lis11) == sum(lis22):
print('YES')
else:
print('NO')
| Title: Lucky Ticket
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 loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | ```python
n = int(input())
num = int(input())
lis = [i for i in str(num)]
lis1 = lis[:int(n/2)]
lis2 = lis[int(n/2):]
lis11 = [int(x) for x in lis1]
lis22 = [int(x) for x in lis2]
if( lis.count('7') + lis.count('4')) == n and sum(lis11) == sum(lis22):
print('YES')
else:
print('NO')
``` | 3 | |
6 | C | Alice, Bob and Chocolate | PROGRAMMING | 1,200 | [
"greedy",
"two pointers"
] | C. Alice, Bob and Chocolate | 2 | 64 | Alice and Bob like games. And now they are ready to start a new game. They have placed *n* chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.
How many bars each of the players will consume? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the amount of bars on the table. The second line contains a sequence *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000), where *t**i* is the time (in seconds) needed to consume the *i*-th bar (in the order from left to right). | Print two numbers *a* and *b*, where *a* is the amount of bars consumed by Alice, and *b* is the amount of bars consumed by Bob. | [
"5\n2 9 8 2 7\n"
] | [
"2 3\n"
] | none | 0 | [
{
"input": "5\n2 9 8 2 7",
"output": "2 3"
},
{
"input": "1\n1",
"output": "1 0"
},
{
"input": "2\n1 1",
"output": "1 1"
},
{
"input": "3\n1 1 1",
"output": "2 1"
},
{
"input": "4\n1 1 1 1",
"output": "2 2"
},
{
"input": "1\n6",
"output": "1 0"
}... | 1,682,004,027 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 248 | 14,131,200 | import sys
from math import *
from collections import *
from heapq import *
from bisect import *
input = sys.stdin.readline
n = int(input())
arr = list(map(int, input().split()))
i = 0
j = n - 1
a, al = 0, 0
b, bo = 0, 0
while i <= j:
if al <= bo:
al += arr[i]
a += 1
i += 1
else:
bo += arr[j]
b += 1
j -= 1
print(a, b)
| Title: Alice, Bob and Chocolate
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Alice and Bob like games. And now they are ready to start a new game. They have placed *n* chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.
How many bars each of the players will consume?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the amount of bars on the table. The second line contains a sequence *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000), where *t**i* is the time (in seconds) needed to consume the *i*-th bar (in the order from left to right).
Output Specification:
Print two numbers *a* and *b*, where *a* is the amount of bars consumed by Alice, and *b* is the amount of bars consumed by Bob.
Demo Input:
['5\n2 9 8 2 7\n']
Demo Output:
['2 3\n']
Note:
none | ```python
import sys
from math import *
from collections import *
from heapq import *
from bisect import *
input = sys.stdin.readline
n = int(input())
arr = list(map(int, input().split()))
i = 0
j = n - 1
a, al = 0, 0
b, bo = 0, 0
while i <= j:
if al <= bo:
al += arr[i]
a += 1
i += 1
else:
bo += arr[j]
b += 1
j -= 1
print(a, b)
``` | 3.832714 |
501 | B | Misha and Changing Handles | PROGRAMMING | 1,100 | [
"data structures",
"dsu",
"strings"
] | null | null | Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. | The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. | In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description. | [
"5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n"
] | [
"3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n"
] | none | 500 | [
{
"input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov",
"output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123"
},
{
"input": "1\nMisha Vasya",
"output": "1\nMisha Vasya"
},
{
"input": "10\na b\nb c\nc d\nd... | 1,599,593,132 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 109 | 307,200 | n=int(input())
a=[]
for i in range(n):
b=input().split()
if b[0] in a:
ind=a.index(b[0])
a.insert(ind,b[1])
a.pop(ind+1)
else:
a.append(b[0])
a.append(b[1])
print(len(a)//2)
for i in range(len(a)//2):
print(*a[:2])
a=a[2:]
| Title: Misha and Changing Handles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input Specification:
The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
Output Specification:
In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Demo Input:
['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n']
Demo Output:
['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n']
Note:
none | ```python
n=int(input())
a=[]
for i in range(n):
b=input().split()
if b[0] in a:
ind=a.index(b[0])
a.insert(ind,b[1])
a.pop(ind+1)
else:
a.append(b[0])
a.append(b[1])
print(len(a)//2)
for i in range(len(a)//2):
print(*a[:2])
a=a[2:]
``` | 3 | |
656 | D | Rosetta Problem | PROGRAMMING | 2,100 | [
"*special"
] | null | null | The input contains a single integer *a* (0<=≤<=*a*<=≤<=1<=000<=000). | Output a single integer. | [
"129\n"
] | [
"1\n"
] | none | 0 | [
{
"input": "129",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "6",
"output": "0"
},
{
"input": "16",
"output": "0"
},
{
"input": "299593",
"output": "7"
},
{
"input": "398929",
"output":... | 1,679,112,009 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 25 | 61 | 0 | n = int(input())
c = 0
while n:
c += n%8 == 1
n//=8
print(c) | Title: Rosetta Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input contains a single integer *a* (0<=≤<=*a*<=≤<=1<=000<=000).
Output Specification:
Output a single integer.
Demo Input:
['129\n']
Demo Output:
['1\n']
Note:
none | ```python
n = int(input())
c = 0
while n:
c += n%8 == 1
n//=8
print(c)
``` | 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": "chn... | 1,595,847,899 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 436 | 22,220,800 | import string
word = input()
count_lower = 0
count_upper = 0
for i in word:
if "a" <= i <= "z":
count_lower +=1
else:
count_upper +=1
if count_lower >= count_upper:
print(word.lower())
else:
print(word.upper()) | 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
import string
word = input()
count_lower = 0
count_upper = 0
for i in word:
if "a" <= i <= "z":
count_lower +=1
else:
count_upper +=1
if count_lower >= count_upper:
print(word.lower())
else:
print(word.upper())
``` | 3.849611 |
509 | B | Painting Pebbles | PROGRAMMING | 1,300 | [
"constructive algorithms",
"greedy",
"implementation"
] | null | null | There are *n* piles of pebbles on the table, the *i*-th pile contains *a**i* pebbles. Your task is to paint each pebble using one of the *k* given colors so that for each color *c* and any two piles *i* and *j* the difference between the number of pebbles of color *c* in pile *i* and number of pebbles of color *c* in pile *j* is at most one.
In other words, let's say that *b**i*,<=*c* is the number of pebbles of color *c* in the *i*-th pile. Then for any 1<=≤<=*c*<=≤<=*k*, 1<=≤<=*i*,<=*j*<=≤<=*n* the following condition must be satisfied |*b**i*,<=*c*<=-<=*b**j*,<=*c*|<=≤<=1. It isn't necessary to use all *k* colors: if color *c* hasn't been used in pile *i*, then *b**i*,<=*c* is considered to be zero. | The first line of the input contains positive integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100), separated by a space — the number of piles and the number of colors respectively.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) denoting number of pebbles in each of the piles. | If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then *n* lines should follow, the *i*-th of them should contain *a**i* space-separated integers. *j*-th (1<=≤<=*j*<=≤<=*a**i*) of these integers should be equal to the color of the *j*-th pebble in the *i*-th pile. If there are several possible answers, you may output any of them. | [
"4 4\n1 2 3 4\n",
"5 2\n3 2 4 1 3\n",
"5 4\n3 2 4 3 5\n"
] | [
"YES\n1\n1 4\n1 2 4\n1 2 3 4\n",
"NO\n",
"YES\n1 2 3\n1 3\n1 2 3 4\n1 3 4\n1 1 2 3 4\n"
] | none | 0 | [
{
"input": "4 4\n1 2 3 4",
"output": "YES\n1 \n1 1 \n1 1 2 \n1 1 2 3 "
},
{
"input": "5 2\n3 2 4 1 3",
"output": "NO"
},
{
"input": "5 4\n3 2 4 3 5",
"output": "YES\n1 1 1 \n1 1 \n1 1 1 2 \n1 1 1 \n1 1 1 2 3 "
},
{
"input": "4 3\n5 6 7 8",
"output": "YES\n1 1 1 1 1 \n1 1 ... | 1,596,118,938 | 2,147,483,647 | PyPy 3 | OK | TESTS | 24 | 171 | 23,040,000 | keeper=input()
n, m=keeper.split(' ')
keeper2=input()
tp=keeper2.split(' ')
map_object=map(int, tp)
arr=list(map_object)
if (max(arr))-(min(arr)) > int(m): print("NO")
else:
print("YES")
for x in arr:
for y in range(0,int(x)): print(int(y)%int(m)+1, end=" ")
print() | Title: Painting Pebbles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* piles of pebbles on the table, the *i*-th pile contains *a**i* pebbles. Your task is to paint each pebble using one of the *k* given colors so that for each color *c* and any two piles *i* and *j* the difference between the number of pebbles of color *c* in pile *i* and number of pebbles of color *c* in pile *j* is at most one.
In other words, let's say that *b**i*,<=*c* is the number of pebbles of color *c* in the *i*-th pile. Then for any 1<=≤<=*c*<=≤<=*k*, 1<=≤<=*i*,<=*j*<=≤<=*n* the following condition must be satisfied |*b**i*,<=*c*<=-<=*b**j*,<=*c*|<=≤<=1. It isn't necessary to use all *k* colors: if color *c* hasn't been used in pile *i*, then *b**i*,<=*c* is considered to be zero.
Input Specification:
The first line of the input contains positive integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100), separated by a space — the number of piles and the number of colors respectively.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) denoting number of pebbles in each of the piles.
Output Specification:
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then *n* lines should follow, the *i*-th of them should contain *a**i* space-separated integers. *j*-th (1<=≤<=*j*<=≤<=*a**i*) of these integers should be equal to the color of the *j*-th pebble in the *i*-th pile. If there are several possible answers, you may output any of them.
Demo Input:
['4 4\n1 2 3 4\n', '5 2\n3 2 4 1 3\n', '5 4\n3 2 4 3 5\n']
Demo Output:
['YES\n1\n1 4\n1 2 4\n1 2 3 4\n', 'NO\n', 'YES\n1 2 3\n1 3\n1 2 3 4\n1 3 4\n1 1 2 3 4\n']
Note:
none | ```python
keeper=input()
n, m=keeper.split(' ')
keeper2=input()
tp=keeper2.split(' ')
map_object=map(int, tp)
arr=list(map_object)
if (max(arr))-(min(arr)) > int(m): print("NO")
else:
print("YES")
for x in arr:
for y in range(0,int(x)): print(int(y)%int(m)+1, end=" ")
print()
``` | 3 | |
151 | A | Soft Drinking | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make? | The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space. | Print a single integer — the number of toasts each friend can make. | [
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] | [
"2\n",
"3\n",
"0\n"
] | A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | 500 | [
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
... | 1,670,478,528 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | n,k,l,c,d,p,nl,dp = map(int,input().strip().split())
toasts = (k*l) // nl
limes = c * d
salt = p // dp
print(min(toasts,min(limes,salt)) // n) | Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | ```python
n,k,l,c,d,p,nl,dp = map(int,input().strip().split())
toasts = (k*l) // nl
limes = c * d
salt = p // dp
print(min(toasts,min(limes,salt)) // n)
``` | 3 | |
462 | B | Appleman and Card Game | PROGRAMMING | 1,300 | [
"greedy"
] | null | null | Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get? | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman. | Print a single integer – the answer to the problem. | [
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] | [
"82\n",
"4\n"
] | In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | 1,000 | [
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output":... | 1,587,234,750 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 155 | 614,400 | firstLine = input();
n, k = firstLine.split();
n = int(n);
k = int(k);
s = input();
result = 0;
count = [0] * 26
for i in range(0, len(s)):
index = ord(s[i]) - 65;
count[index] = count[index] + 1;
count.sort(reverse=True)
for j in range(0, len(count)):
if(count[j] <= k):
result = result + (count[j]*count[j]);
k = k - count[j];
if(k==0): break
else:
result = result + (k*k);
break
print(result)
| Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | ```python
firstLine = input();
n, k = firstLine.split();
n = int(n);
k = int(k);
s = input();
result = 0;
count = [0] * 26
for i in range(0, len(s)):
index = ord(s[i]) - 65;
count[index] = count[index] + 1;
count.sort(reverse=True)
for j in range(0, len(count)):
if(count[j] <= k):
result = result + (count[j]*count[j]);
k = k - count[j];
if(k==0): break
else:
result = result + (k*k);
break
print(result)
``` | 3 | |
181 | A | Series of Crimes | PROGRAMMING | 800 | [
"brute force",
"geometry",
"implementation"
] | null | null | The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements. | Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right. | [
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] | [
"1 1\n",
"2 3\n"
] | none | 500 | [
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n....... | 1,620,291,270 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 218 | 21,811,200 | N, M = [int(X) for X in input().split()]
A = []
for i in range (N):
B = input()
A.append([i for i in B])
pts = []
for i in range (N):
for j in range (M):
if A[i][j] == '*':
pts.append((i,j))
# print (pts)
X = []
Y = []
for i in pts:
X.append(i[0])
Y.append(i[1])
# print (X)
# print (Y)
x = None
y = None
for i in X:
if X.count(i) == 1:
x = i
break
for i in Y:
if Y.count(i) == 1:
y = i
break
print (x+1,y+1) | Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none | ```python
N, M = [int(X) for X in input().split()]
A = []
for i in range (N):
B = input()
A.append([i for i in B])
pts = []
for i in range (N):
for j in range (M):
if A[i][j] == '*':
pts.append((i,j))
# print (pts)
X = []
Y = []
for i in pts:
X.append(i[0])
Y.append(i[1])
# print (X)
# print (Y)
x = None
y = None
for i in X:
if X.count(i) == 1:
x = i
break
for i in Y:
if Y.count(i) == 1:
y = i
break
print (x+1,y+1)
``` | 3 |
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