message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One of the oddest traditions of the town of Gameston may be that even the town mayor of the next term is chosen according to the result of a game. When the expiration of the term of the mayor approaches, at least three candidates, including the mayor of the time, play a game of pebbles, and the winner will be the next mayor.
The rule of the game of pebbles is as follows. In what follows, n is the number of participating candidates.
Requisites A round table, a bowl, and plenty of pebbles. Start of the Game A number of pebbles are put into the bowl; the number is decided by the Administration Commission using some secret stochastic process. All the candidates, numbered from 0 to n-1 sit around the round table, in a counterclockwise order. Initially, the bowl is handed to the serving mayor at the time, who is numbered 0. Game Steps When a candidate is handed the bowl and if any pebbles are in it, one pebble is taken out of the bowl and is kept, together with those already at hand, if any. If no pebbles are left in the bowl, the candidate puts all the kept pebbles, if any, into the bowl. Then, in either case, the bowl is handed to the next candidate to the right. This step is repeated until the winner is decided. End of the Game When a candidate takes the last pebble in the bowl, and no other candidates keep any pebbles, the game ends and that candidate with all the pebbles is the winner.
A math teacher of Gameston High, through his analysis, concluded that this game will always end within a finite number of steps, although the number of required steps can be very large.
Input
The input is a sequence of datasets. Each dataset is a line containing two integers n and p separated by a single space. The integer n is the number of the candidates including the current mayor, and the integer p is the total number of the pebbles initially put in the bowl. You may assume 3 β€ n β€ 50 and 2 β€ p β€ 50.
With the settings given in the input datasets, the game will end within 1000000 (one million) steps.
The end of the input is indicated by a line containing two zeros separated by a single space.
Output
The output should be composed of lines corresponding to input datasets in the same order, each line of which containing the candidate number of the winner. No other characters should appear in the output.
Sample Input
3 2
3 3
3 50
10 29
31 32
50 2
50 50
0 0
Output for the Sample Input
1
0
1
5
30
1
13
Example
Input
3 2
3 3
3 50
10 29
31 32
50 2
50 50
0 0
Output
1
0
1
5
30
1
13
Submitted Solution:
```
while 1:
n,p=map(int,input().split())
if n+p==0:
break
x=[0]*n
i=0
while 1:
if p==0:
p+=x[i]
x[i]=0
else:
x[i]+=1
p-=1
if x[i]==n:
print (i)
break
i+=1
if i==n:
i=0
``` | instruction | 0 | 31,739 | 19 | 63,478 |
No | output | 1 | 31,739 | 19 | 63,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One of the oddest traditions of the town of Gameston may be that even the town mayor of the next term is chosen according to the result of a game. When the expiration of the term of the mayor approaches, at least three candidates, including the mayor of the time, play a game of pebbles, and the winner will be the next mayor.
The rule of the game of pebbles is as follows. In what follows, n is the number of participating candidates.
Requisites A round table, a bowl, and plenty of pebbles. Start of the Game A number of pebbles are put into the bowl; the number is decided by the Administration Commission using some secret stochastic process. All the candidates, numbered from 0 to n-1 sit around the round table, in a counterclockwise order. Initially, the bowl is handed to the serving mayor at the time, who is numbered 0. Game Steps When a candidate is handed the bowl and if any pebbles are in it, one pebble is taken out of the bowl and is kept, together with those already at hand, if any. If no pebbles are left in the bowl, the candidate puts all the kept pebbles, if any, into the bowl. Then, in either case, the bowl is handed to the next candidate to the right. This step is repeated until the winner is decided. End of the Game When a candidate takes the last pebble in the bowl, and no other candidates keep any pebbles, the game ends and that candidate with all the pebbles is the winner.
A math teacher of Gameston High, through his analysis, concluded that this game will always end within a finite number of steps, although the number of required steps can be very large.
Input
The input is a sequence of datasets. Each dataset is a line containing two integers n and p separated by a single space. The integer n is the number of the candidates including the current mayor, and the integer p is the total number of the pebbles initially put in the bowl. You may assume 3 β€ n β€ 50 and 2 β€ p β€ 50.
With the settings given in the input datasets, the game will end within 1000000 (one million) steps.
The end of the input is indicated by a line containing two zeros separated by a single space.
Output
The output should be composed of lines corresponding to input datasets in the same order, each line of which containing the candidate number of the winner. No other characters should appear in the output.
Sample Input
3 2
3 3
3 50
10 29
31 32
50 2
50 50
0 0
Output for the Sample Input
1
0
1
5
30
1
13
Example
Input
3 2
3 3
3 50
10 29
31 32
50 2
50 50
0 0
Output
1
0
1
5
30
1
13
Submitted Solution:
```
#!/usr/bin/python
# -*- coding: utf-8 -*-
# having_amount[i] represents the amount of pebbles the
# candidate numbered i is having in the "turn"
winner_number = []
while True:
mayors, pebbles_in_bowl = map(int, input().split())
# action of a candidate
if mayors == 0 and pebbles_in_bowl == 0:
break
else:
turn = 0
having_amount = [0 for i in range(mayors)]
while True:
if pebbles_in_bowl == 0:
pebbles_in_bowl += having_amount[turn]
having_amount[turn] = 0
elif pebbles_in_bowl == 1:
# if the candidate in the turn wins
if having_amount.count(0) == len(having_amount) - 1:
print(turn)
break
else:
having_amount[turn] += 1
pebbles_in_bowl -= 1
else:
having_amount[turn] += 1
pebbles_in_bowl -= 1
# go to the next turn
if turn == mayors - 1:
turn = 0
else:
turn += 1
``` | instruction | 0 | 31,740 | 19 | 63,480 |
No | output | 1 | 31,740 | 19 | 63,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One of the oddest traditions of the town of Gameston may be that even the town mayor of the next term is chosen according to the result of a game. When the expiration of the term of the mayor approaches, at least three candidates, including the mayor of the time, play a game of pebbles, and the winner will be the next mayor.
The rule of the game of pebbles is as follows. In what follows, n is the number of participating candidates.
Requisites A round table, a bowl, and plenty of pebbles. Start of the Game A number of pebbles are put into the bowl; the number is decided by the Administration Commission using some secret stochastic process. All the candidates, numbered from 0 to n-1 sit around the round table, in a counterclockwise order. Initially, the bowl is handed to the serving mayor at the time, who is numbered 0. Game Steps When a candidate is handed the bowl and if any pebbles are in it, one pebble is taken out of the bowl and is kept, together with those already at hand, if any. If no pebbles are left in the bowl, the candidate puts all the kept pebbles, if any, into the bowl. Then, in either case, the bowl is handed to the next candidate to the right. This step is repeated until the winner is decided. End of the Game When a candidate takes the last pebble in the bowl, and no other candidates keep any pebbles, the game ends and that candidate with all the pebbles is the winner.
A math teacher of Gameston High, through his analysis, concluded that this game will always end within a finite number of steps, although the number of required steps can be very large.
Input
The input is a sequence of datasets. Each dataset is a line containing two integers n and p separated by a single space. The integer n is the number of the candidates including the current mayor, and the integer p is the total number of the pebbles initially put in the bowl. You may assume 3 β€ n β€ 50 and 2 β€ p β€ 50.
With the settings given in the input datasets, the game will end within 1000000 (one million) steps.
The end of the input is indicated by a line containing two zeros separated by a single space.
Output
The output should be composed of lines corresponding to input datasets in the same order, each line of which containing the candidate number of the winner. No other characters should appear in the output.
Sample Input
3 2
3 3
3 50
10 29
31 32
50 2
50 50
0 0
Output for the Sample Input
1
0
1
5
30
1
13
Example
Input
3 2
3 3
3 50
10 29
31 32
50 2
50 50
0 0
Output
1
0
1
5
30
1
13
Submitted Solution:
```
while 1:
n,p = map(int,input().split())
if n==0:break
man = [i for i in [0]*n]
ball,a = p,0
while p not in man:
if ball >0:ball,man[a%n] =ball-1, man[a%n]+1
elif ball == 0:ball,man[a%n] = man[a%n],ball
a+=1
print(man.index(max(man)))
``` | instruction | 0 | 31,741 | 19 | 63,482 |
No | output | 1 | 31,741 | 19 | 63,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx | instruction | 0 | 32,027 | 19 | 64,054 |
Tags: constructive algorithms, implementation
Correct Solution:
```
a,b=[int(i) for i in input().split()]
if(a==0):
print(-b**2)
print('x'*b)
elif(b==0):
print(a**2)
print('o'*a)
elif(b==1):
print(a**2-1)
print('x'+'o'*a)
else:
ans=-float('inf')
gr_no=None
for i in range(2,min(a+2,b+1)):
v1=(a+2-i)**2 + i-2
quo=b//i
rem=b%i
v2=rem*((quo+1)**2) + (i-rem)*((quo**2))
if(v1-v2>ans):
gr_no=i
ans=v1-v2
quo=b//gr_no
rem=b%gr_no
if(rem>0):
s='x'*(quo+1)+'o'*(a+2-gr_no)
rem-=1
else:
s='x'*(quo)+'o'*(a+2-gr_no)
gr_no-=1
s1='x'*(quo+1)+'o'
s2='x'*quo + 'o'
for i in range(rem):
s+=s1
for i in range(gr_no-rem-1):
s+=s2
s+='x'*(quo)
print(ans)
print(s)
``` | output | 1 | 32,027 | 19 | 64,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx | instruction | 0 | 32,028 | 19 | 64,056 |
Tags: constructive algorithms, implementation
Correct Solution:
```
a,b=map(int,input().split())
def sqr(x):
return x*x
def work( num, flag=0 ):
ans=sqr(a-num+1)+num-1
could = min(b, num+1)
cc=b//could
res=b%could
ans-=res * sqr(cc+1) + (could-res)*sqr(cc)
if flag:
print(ans)
list=''
res2=could-res
if could==num+1:
list+='x'*cc
res2-=1
ta=a
list+='o'*(a-num+1)
ta-=a-num+1
while ta>0:
u=cc+int(res>0)
if res>0:
res-=1
else:
res2-=1
list+='x'*u
list+='o'
ta-=1
if res>0 or res2>0:
list+='x'*(cc+int(res>0))
print(str(list))
return ans
if a==0:
print(-sqr(b))
print('x'*b)
elif b==0:
print(sqr(a))
print('o'*a)
else:
now=1
for i in range(1,a+1):
if i-1<=b and work(i)>work(now):
now=i
work(now,1)
``` | output | 1 | 32,028 | 19 | 64,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx | instruction | 0 | 32,029 | 19 | 64,058 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import math
import sys
inp = input().split()
a = int(inp[0])
b = int(inp[1])
if a == 0:
print(str(-b**2))
out = "x" * b
print(out)
elif b == 0:
print(str(a**2))
out = "o" * a
print(out)
else:
ind = -1
buk = -1
quo = -1
rem = -1
sco = -sys.maxsize - 1
for i in range(a):
buk_i = i + 2
quo_i = int(b / buk_i)
rem_i = b % buk_i
sco_i = (a - i)**2 + i - rem_i * (quo_i + 1)**2 - (buk_i - rem_i) * quo_i**2
if sco_i > sco:
buk = buk_i
quo = quo_i
rem = rem_i
sco = sco_i
ind = i
print(sco)
out = "x" * quo + "o" * (a - ind - 1)
for i in range(rem):
out += "o" + "x" * (quo + 1)
for i in range(buk - rem - 1):
out += "o" + "x" * quo
print(out)
``` | output | 1 | 32,029 | 19 | 64,059 |
Provide tags and a correct Python 3 solution for this coding contest problem.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx | instruction | 0 | 32,030 | 19 | 64,060 |
Tags: constructive algorithms, implementation
Correct Solution:
```
# Made By Mostafa_Khaled
bot = True
a,b=map(int,input().split())
def sqr(x):
return x*x
def work( num, flag=0 ):
ans=sqr(a-num+1)+num-1
could = min(b, num+1)
cc=b//could
res=b%could
ans-=res * sqr(cc+1) + (could-res)*sqr(cc)
if flag:
print(ans)
list=''
res2=could-res
if could==num+1:
list+='x'*cc
res2-=1
ta=a
list+='o'*(a-num+1)
ta-=a-num+1
while ta>0:
u=cc+int(res>0)
if res>0:
res-=1
else:
res2-=1
list+='x'*u
list+='o'
ta-=1
if res>0 or res2>0:
list+='x'*(cc+int(res>0))
print(str(list))
return ans
if a==0:
print(-sqr(b))
print('x'*b)
elif b==0:
print(sqr(a))
print('o'*a)
else:
now=1
for i in range(1,a+1):
if i-1<=b and work(i)>work(now):
now=i
work(now,1)
# Made By Mostafa_Khaled
``` | output | 1 | 32,030 | 19 | 64,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx | instruction | 0 | 32,031 | 19 | 64,062 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def main():
a, b = map(int, input().split(' '))
answer = -b ** 2
a_cluster = a
if a:
for i in range(a):
answer_candicate = (a - i) ** 2 + i
quotient = b // (i + 2)
remainder = b % (i + 2)
answer_candicate -= ((i + 2) - remainder) * quotient ** 2 \
+ remainder * (quotient + 1) ** 2
if answer_candicate > answer:
answer = answer_candicate
a_cluster = a - i
print(answer)
if a:
quotient = b // (a - a_cluster + 2)
remainder = b % (a - a_cluster + 2)
print('x' * quotient, end='')
print('o' * a_cluster, end='')
print('x' * (quotient + (1 if remainder else 0)), end='')
remainder = max(remainder - 1, 0)
for i in range(a - a_cluster):
print('o', end='')
print('x' * (quotient + (1 if remainder else 0)), end='')
remainder = max(remainder - 1, 0)
print()
else:
print('x' * b)
if __name__ == '__main__':
main()
``` | output | 1 | 32,031 | 19 | 64,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx
Submitted Solution:
```
a,b=[int(i) for i in input().split()]
if(a==0):
print(-b**2)
print('x'*b)
elif(b==0):
print(a**2)
print('o'*a)
elif(b==1):
print(a**2-1)
print('x'+'o'*a)
else:
ans=-float('inf')
gr_no=None
for i in range(2,min(a+2,b+1)):
v1=(a+2-i)**2 + i-1
quo=b//i
rem=b%i
v2=rem*((quo+1)**2) + (i-rem)*((quo**2))
if(v1-v2>ans):
gr_no=i
ans=v1-v2
quo=b//gr_no
rem=b%gr_no
if(rem>0):
s='x'*(quo+1)+'o'*(a+2-gr_no)
rem-=1
else:
s='x'*(quo)+'o'*(a+2-gr_no)
gr_no-=1
s1='x'*(quo+1)+'o'
s2='x'*quo + 'o'
for i in range(rem):
s+=s1
for i in range(gr_no-rem-1):
s+=s2
s+='x'*(quo)
print(ans)
print(s)
``` | instruction | 0 | 32,032 | 19 | 64,064 |
No | output | 1 | 32,032 | 19 | 64,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx
Submitted Solution:
```
a,b=[int(i) for i in input().split()]
if(a==0):
print(-b**2)
print('x'*b)
elif(b==0):
print(a**2)
print('o'*a)
else:
ans=-float('inf')
gr_no=None
for i in range(2,min(a+2,b+1)):
v1=(a+2-i)**2 + i-1
quo=b//i
rem=b%i
v2=rem*((quo+1)**2) + (i-rem)*((quo**2))
if(v1-v2>ans):
gr_no=i
ans=v1-v2
quo=b//gr_no
rem=b%gr_no
if(rem>0):
s='x'*(quo+1)+'o'*(a+2-gr_no)
rem-=1
else:
s='x'*(quo)+'o'*(a+2-gr_no)
gr_no-=1
s1='x'*(quo+1)+'o'
s3='x'*quo + 'o'
for i in range(rem):
s+=s1
for i in range(gr_no-rem-1):
s+=s3
s+='x'*(quo)
print(s)
``` | instruction | 0 | 32,033 | 19 | 64,066 |
No | output | 1 | 32,033 | 19 | 64,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx
Submitted Solution:
```
a,b=[int(i) for i in input().split()]
if(a==0):
print(-b**2)
print('x'*b)
elif(b==0):
print(a**2)
print('o'*a)
else:
ans=-float('inf')
gr_no=None
for i in range(2,min(a+2,b+1)):
v1=(a+2-i)**2 + i-1
quo=b//i
rem=b%i
v2=rem*((quo+1)**2) + (i-rem)*((quo**2))
if(v1-v2>ans):
gr_no=i
ans=v1-v2
quo=b//gr_no
rem=b%gr_no
s='x'*(quo+1)+'o'*(a+2-gr_no)
s1='x'*(quo+1)+'o'
s2='x'*quo + 'o'
for i in range(rem-1):
s+=s1
for i in range(gr_no-rem-1):
s+=s2
s+='x'*(quo)
print(s)
``` | instruction | 0 | 32,034 | 19 | 64,068 |
No | output | 1 | 32,034 | 19 | 64,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
User ainta loves to play with cards. He has a cards containing letter "o" and b cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0.
2. For each block of contiguous "o"s with length x the score increases by x2.
3. For each block of contiguous "x"s with length y the score decreases by y2.
For example, if a = 6, b = 3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22 - 12 + 32 - 22 + 12 = 9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input
The first line contains two space-separated integers a and b (0 β€ a, b β€ 105; a + b β₯ 1) β the number of "o" cards and the number of "x" cards.
Output
In the first line print a single integer v β the maximum score that ainta can obtain.
In the second line print a + b characters describing the deck. If the k-th card of the deck contains "o", the k-th character must be "o". If the k-th card of the deck contains "x", the k-th character must be "x". The number of "o" characters must be equal to a, and the number of "x " characters must be equal to b. If there are many ways to maximize v, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
2 3
Output
-1
xoxox
Input
4 0
Output
16
oooo
Input
0 4
Output
-16
xxxx
Submitted Solution:
```
import math
import sys
inp = input().split()
a = int(inp[0])
b = int(inp[1])
if a == 0:
print(str(-b**2))
out = "x" * b
print(out)
elif b == 0:
print(str(a**2))
out = "o" * a
print(out)
elif a == b:
print(str(0))
out = "xo" * a
print(out)
else:
ind = -1
buk = -1
quo = -1
rem = -1
sco = -sys.maxsize - 1
for i in range(a):
buk_i = i + 2
quo_i = int(b / buk_i)
rem_i = b % buk_i
sco_i = (a - i)**2 + i - rem_i * (quo_i + 1)**2 - (buk_i - rem_i) * quo_i**2
if sco_i > sco:
buk = buk_i
quo = quo_i
rem = rem_i
sco = sco_i
ind = i
print(sco)
out = "x" * quo + "o" * (a - ind - 1)
for i in range(rem):
out += "o" + "x" * (quo + 1)
for i in range(buk - rem - 1):
out += "o" + "x" * quo
print(out)
``` | instruction | 0 | 32,035 | 19 | 64,070 |
No | output | 1 | 32,035 | 19 | 64,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,039 | 19 | 64,078 |
Tags: greedy, implementation
Correct Solution:
```
# I Wanna Be the Guy
a=int(input())
d=[]
b=list(map(int,input().split()))
w=list(map(int,input().split()))
b.pop(0)
w.pop(0)
w.extend(b)
for c in w:
if c==0:
w.remove(0)
d=set(w)
r=len(d)
if r==a:
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | output | 1 | 32,039 | 19 | 64,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,040 | 19 | 64,080 |
Tags: greedy, implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 30 16:17:49 2020
@author: CCLAB
"""
n=int(input())
a=input().split()
b=input().split()
p=int(a[0])
q=int(b[0])
l=[]
for i in range(p):
l+=[int(a[i+1])]
for i in range(q):
l+=[int(b[i+1])]
for i in range(1,n+1):
if l.count(i)==0:
print('Oh, my keyboard!')
break
else:
print('I become the guy.')
``` | output | 1 | 32,040 | 19 | 64,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,041 | 19 | 64,082 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
x = input().split()
y = input().split()
x = [int(x) for x in x[1:]]
y = [int(y) for y in y[1:]]
a = [False]*n
for i in x:
a[i-1] = True
for i in y:
a[i-1] = True
if False in a:
print('Oh, my keyboard!')
else:
print('I become the guy.')
``` | output | 1 | 32,041 | 19 | 64,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,042 | 19 | 64,084 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = set(input().split()[1:])
b = set(input().split()[1:])
if len(a.union(b)) == n:
print('I become the guy.')
else:
print('Oh, my keyboard!')
``` | output | 1 | 32,042 | 19 | 64,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,043 | 19 | 64,086 |
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
l=[]
for i in range(1,len(x)):
if x[i] not in l and x[i]!=0:
l.append(x[i])
for i in range(1,len(y)):
if y[i] not in l and y[i]!=0:
l.append(y[i])
if(len(l)==n):
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | output | 1 | 32,043 | 19 | 64,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,044 | 19 | 64,088 |
Tags: greedy, implementation
Correct Solution:
```
import sys
import math
n=int(input())
#p1n=int(input().split())
p1=list(map(int,input().split()))
#p2n=int(input().split())
p1n=p1[:1]
p2=list(map(int,input().split()))
p2n=p2[:1]
p1=set(p1[1:])
p2=set(p2[1:])
pf=p1.union(p2)
x=set(range(1,n+1))
if(x == pf):
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | output | 1 | 32,044 | 19 | 64,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,045 | 19 | 64,090 |
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
lis=set(input().split()[1:]+input().split()[1:])
if (len(lis))>=n:
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | output | 1 | 32,045 | 19 | 64,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4. | instruction | 0 | 32,046 | 19 | 64,092 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()[1:]))
b = list(map(int, input().split()[1:]))
a.extend(b)
print('I become the guy.' if len(set(a))==n else 'Oh, my keyboard!')
``` | output | 1 | 32,046 | 19 | 64,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
n = int(input())
X = set([int(i) for i in input().split()][1:])
Y = set([int(i) for i in input().split()][1:])
if len(X.union(Y)) == n:
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | instruction | 0 | 32,047 | 19 | 64,094 |
Yes | output | 1 | 32,047 | 19 | 64,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
n = int(input())
s = set()
l = list(map(int, input().split()))
p = l[0]
for i in range(1, p+1):
s.add(l[i])
l = list(map(int, input().split()))
q = l[0]
for i in range(1, q+1):
s.add(l[i])
if len(s)==n:
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | instruction | 0 | 32,048 | 19 | 64,096 |
Yes | output | 1 | 32,048 | 19 | 64,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
n=int(input())
a=set(input().split()[1:])
b=set(input().split()[1:])
if len(set.union(a,b)) == n:
print('I become the guy.')
else:
print('Oh, my keyboard!')
``` | instruction | 0 | 32,049 | 19 | 64,098 |
Yes | output | 1 | 32,049 | 19 | 64,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
a=int(input())
b=list(map(int,input().split(' ')))
c=list(map(int,input().split(' ')))
n=[]
for i in range(0,a):
n.append(0)
for i in range(1,len(b)):
if n[b[i]-1]==0:
n[b[i]-1]=1
for i in range(1,len(c)):
if n[c[i]-1]==0:
n[c[i]-1]=1
if 0 in n:
print("Oh, my keyboard!")
else:
print("I become the guy.")
``` | instruction | 0 | 32,050 | 19 | 64,100 |
Yes | output | 1 | 32,050 | 19 | 64,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
a=int(input())
b=list(set(input().split()+input().split()))
print('I become the guy.' if len(b)==a else 'Oh, my keyboard!')
``` | instruction | 0 | 32,051 | 19 | 64,102 |
No | output | 1 | 32,051 | 19 | 64,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
x=int(input())
arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
a=set(arr1)
b=set(arr2)
a.update(b)
if len(a)==x:
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | instruction | 0 | 32,052 | 19 | 64,104 |
No | output | 1 | 32,052 | 19 | 64,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
a=int(input())
b=list(map(int,input().split()))
c=list(map(int,input().split()))
for i in range ( 1 , a+1 ):
if (i not in b) and (i not in c):
print("Oh, my keyboard!")
exit()
print("I become the guy.")
``` | instruction | 0 | 32,053 | 19 | 64,106 |
No | output | 1 | 32,053 | 19 | 64,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input
The first line contains a single integer n (1 β€ n β€ 100).
The next line contains an integer p (0 β€ p β€ n) at first, then follows p distinct integers a1, a2, ..., ap (1 β€ ai β€ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.
Output
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
Submitted Solution:
```
n = int(input())
p = list(map(int,input().split()))
q = list(map(int,input().split()))
l = p + q
l.sort()
s1 = set(l)
l = list(s1)
count = 0
for i in range(0,len(l)):
print(i)
if (l[i] == i + 1 or l[i] == i):
count = count + 1
if (l[len(l) - 1] == n):
count = count + 1
if (count == len(l) + 1):
print("I become the guy.")
else:
print("Oh, my keyboard!")
``` | instruction | 0 | 32,054 | 19 | 64,108 |
No | output | 1 | 32,054 | 19 | 64,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,562 | 19 | 65,124 |
Tags: constructive algorithms, math
Correct Solution:
```
n, k = map(int, input().split())
if 2 * k <= (n + 1):
print(3 * n + k - 1)
else:
print(4 * n - k)
``` | output | 1 | 32,562 | 19 | 65,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,563 | 19 | 65,126 |
Tags: constructive algorithms, math
Correct Solution:
```
import sys
import math,bisect
sys.setrecursionlimit(10 ** 5)
from collections import defaultdict
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,OrderedDict
def I(): return int(sys.stdin.readline())
def neo(): return map(int, sys.stdin.readline().split())
def Neo(): return list(map(int, sys.stdin.readline().split()))
n,k = neo()
Ans = 3*n
if k != 1 and k != n:
Ans += min(k-1,n-k)
print(Ans)
``` | output | 1 | 32,563 | 19 | 65,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,564 | 19 | 65,128 |
Tags: constructive algorithms, math
Correct Solution:
```
t = input().split(" ")
n = int(t[0])
k = int(t[1])
m = min(k-1, n-k)
print(3 * n + m)
``` | output | 1 | 32,564 | 19 | 65,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,565 | 19 | 65,130 |
Tags: constructive algorithms, math
Correct Solution:
```
n, m = map(int, input().split())
s = 3
if m - 1 <= n - m:
s += (m - 1)*2 + 2*(m - 1)
s += (n - m)*2 + n - m
else:
s += (n - m)*2 + 2*(n - m)
s += (m - 1) * 2 + m - 1
print(s)
``` | output | 1 | 32,565 | 19 | 65,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,566 | 19 | 65,132 |
Tags: constructive algorithms, math
Correct Solution:
```
n, k = [int(x) for x in input().split()]
summ = 0
if (n - k) < k:
summ += (n-k+1)*3-1
summ += (n-k)+2
summ += (k-1)*3-1
else:
summ += (k+1)*3-1
summ += k+2
summ += (n-k-1)*3-2
print(summ)
``` | output | 1 | 32,566 | 19 | 65,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,567 | 19 | 65,134 |
Tags: constructive algorithms, math
Correct Solution:
```
data = input().split()
n = int(data[0])
k = int(data[1])
print(str(min(n-k, k - 1) + 3*n))
``` | output | 1 | 32,567 | 19 | 65,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,568 | 19 | 65,136 |
Tags: constructive algorithms, math
Correct Solution:
```
'''input
5 1
'''
import sys
from collections import defaultdict as dd
from itertools import permutations as pp
from itertools import combinations as cc
from collections import Counter as ccd
from random import randint as rd
from bisect import bisect_left as bl
import heapq
mod=10**9+7
def ri(flag=0):
if flag==0:
return [int(i) for i in sys.stdin.readline().split()]
else:
return int(sys.stdin.readline())
n,k=ri()
ok=2*n+1
print(ok+min(n-1+k-1,n-k+n-1))
``` | output | 1 | 32,568 | 19 | 65,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win. | instruction | 0 | 32,569 | 19 | 65,138 |
Tags: constructive algorithms, math
Correct Solution:
```
line = input().split(' ')
n = int(line[0])
possition = int(line[1])
if possition < (n / 2) + 1:
p1 = (3 * (possition - 1))
p2 = possition - 1
p3 = (3 * (n - possition + 1)) - 1
print(p1 + p2 + p3 + 1)
else:
p1 = (3 * (n - possition))
p2 = (n - possition)
p3 = (3 * possition) - 1
print(p1 + p2 + p3 + 1)
``` | output | 1 | 32,569 | 19 | 65,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
def ans(n, k):
if(n == 2):
return(6)
return(min(k - 1, n - k) + 6 + 3 *(n - 2))
if __name__ == '__main__':
nk = list(map(int, input().strip().split()))
print(ans(nk[0], nk[1]))
``` | instruction | 0 | 32,570 | 19 | 65,140 |
Yes | output | 1 | 32,570 | 19 | 65,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
A,B=input().split()
if(int(B)==int(A) or int(B)==1):
print(int(A)*3)
else:
X=int(A)-int(B)+1
Y=int(B)
if(X>=Y):
print(int(A)*2+(int(B)-1)*2+(int(A)-int(B)+1))
else:
print(int(A)*2+(X-1)*2+int(B))
``` | instruction | 0 | 32,571 | 19 | 65,142 |
Yes | output | 1 | 32,571 | 19 | 65,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
n, k = map(int, input().split())
count=0
if k==1 or k==n:
count = 5 + 3*(n-2) + 1
else:
if k>n//2:
k = n-k+1
count = 5 + 3*(k-2)+1 + k + 2 + 3*(n-k-1)
print(count)
``` | instruction | 0 | 32,572 | 19 | 65,144 |
Yes | output | 1 | 32,572 | 19 | 65,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
n,k=map(int,input().split())
ans=n*3
u=min(n-k,k-1)
ans=ans+u
print(ans)
``` | instruction | 0 | 32,573 | 19 | 65,146 |
Yes | output | 1 | 32,573 | 19 | 65,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
# problem/1136/B
n,k = [int(n) for n in input().split(" ")]
# stones = [1 for i in range(n)]
# # if closer to left, go left first
# # print(k < n/2)
# if k < (n-1)/2:
# moves = 0
# # if k == 0 then we can't go left
# if k != 0:
# # go left
# for i in range(k,-1,-1):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[i+1] += st
# moves += st
# # pick up coin
# moves += 1
# if i != 0:
# # move to the left
# moves += 1
# # we're now at index 0
# # move k moves back to where we started
# moves += k + 1
# k_ = k + 1
# else:
# k_ = k
# # go right
# for i in range(k_,n):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[(i-1)%n] += st
# moves += st
# # pick up coin
# moves += 1
# if i != n-1:
# # move to the right
# moves += 1
# # if closer to right, go right first
# else:
# moves = 0
# # if k == n then we can only go left
# if k != n-1:
# # go right
# for i in range(k,n):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[i-1] += st
# moves += st
# # pick up coin
# moves += 1
# if i != n-1:
# # move to the right
# moves += 1
# # we're now at index n
# # move k moves back to where we started - 1
# moves += (n-k) + 1
# k_ = k - 1
# else:
# k_ = k
# # go left
# for i in range(k_,-1,-1):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[(i+1)%n] += st
# moves += st
# # pick up coin
# moves += 1
# if i != 0:
# # move to the left
# moves += 1
# print(moves)
# print(k < (n-1)/2)
if k == 1:
# pickup n+1 stones, move n-1 times and pickup n gold coins
print(3*n)
elif k == n:
# pickup n+1 stones, move n-1 times and pickup n gold coins
print(3*n)
# go left
elif k < (n-1)/2:
steps = 0
# pickup k stones, move k-1 times and pickup k gold coins
steps += 3*k - 1
# move k steps
steps += k
# pickup (n-k)+1 stones, move (n-k)-1 times and pickup (n-k) gold coins
steps += (n-k)*3
print(steps)
# go right
else:
steps = 0
# pickup (n-k) stones, move (n-k)-1 times and pickup (n-k) gold coins
steps += (n-k)*3 - 1
# move (n-k) steps
steps += (n-k)
# pickup k+1 stones, move k-1 times and pickup k gold coins
steps += 3*k
print(steps)
``` | instruction | 0 | 32,574 | 19 | 65,148 |
No | output | 1 | 32,574 | 19 | 65,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
# problem/1136/B
n,k = [int(n) for n in input().split(" ")]
# stones = [1 for i in range(n)]
# # if closer to left, go left first
# # print(k < n/2)
# if k < (n-1)/2:
# moves = 0
# # if k == 0 then we can't go left
# if k != 0:
# # go left
# for i in range(k,-1,-1):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[i+1] += st
# moves += st
# # pick up coin
# moves += 1
# if i != 0:
# # move to the left
# moves += 1
# # we're now at index 0
# # move k moves back to where we started
# moves += k + 1
# k_ = k + 1
# else:
# k_ = k
# # go right
# for i in range(k_,n):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[(i-1)%n] += st
# moves += st
# # pick up coin
# moves += 1
# if i != n-1:
# # move to the right
# moves += 1
# # if closer to right, go right first
# else:
# moves = 0
# # if k == n then we can only go left
# if k != n-1:
# # go right
# for i in range(k,n):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[i-1] += st
# moves += st
# # pick up coin
# moves += 1
# if i != n-1:
# # move to the right
# moves += 1
# # we're now at index n
# # move k moves back to where we started - 1
# moves += (n-k) + 1
# k_ = k - 1
# else:
# k_ = k
# # go left
# for i in range(k_,-1,-1):
# if stones[i] > 0:
# st = stones[i]
# stones[i] -= st
# stones[(i+1)%n] += st
# moves += st
# # pick up coin
# moves += 1
# if i != 0:
# # move to the left
# moves += 1
# print(moves)
# print(k < (n-1)/2)
if k == 1:
# pickup n+1 stones, move n-1 times and pickup n gold coins
print(3*n)
elif k == n:
# pickup n+1 stones, move n-1 times and pickup n gold coins
print(3*n)
# go left
elif k < (n-1)/2:
steps = 0
# pickup k stones, move k-1 times and pickup k gold coins
steps += 3*k - 1
# move k steps
steps += k
# pickup (n-k)+1 stones, move (n-k)-1 times and pickup (n-k) gold coins
steps += (n-k)*3
print(steps)
# go right
else:
steps = 0
# pickup (n-k+1) stones, move (n-k+1)-1 times and pickup (n-k+1) gold coins
steps += (n-k+1)*3 - 1
# move (n-k+1) steps
steps += (n-k+1)
# pickup (k-1)+1 stones, move (k-1)-1 times and pickup (k-1) gold coins
steps += 3*(k-1)
print(steps)
``` | instruction | 0 | 32,575 | 19 | 65,150 |
No | output | 1 | 32,575 | 19 | 65,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
n, k = map(int, input().split())
if k == 1 or k == n:
print(6 + 3 * (n - 2))
else:
if k * 2 > n:
k = n - k
b = 5 + 3 * (n - 2) + k
print(b)
``` | instruction | 0 | 32,576 | 19 | 65,152 |
No | output | 1 | 32,576 | 19 | 65,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finished her homework, Nastya decided to play computer games. Passing levels one by one, Nastya eventually faced a problem. Her mission is to leave a room, where a lot of monsters live, as quickly as possible.
There are n manholes in the room which are situated on one line, but, unfortunately, all the manholes are closed, and there is one stone on every manhole. There is exactly one coin under every manhole, and to win the game Nastya should pick all the coins. Initially Nastya stands near the k-th manhole from the left. She is thinking what to do.
In one turn, Nastya can do one of the following:
* if there is at least one stone on the manhole Nastya stands near, throw exactly one stone from it onto any other manhole (yes, Nastya is strong).
* go to a neighboring manhole;
* if there are no stones on the manhole Nastya stays near, she can open it and pick the coin from it. After it she must close the manhole immediately (it doesn't require additional moves).
<image> The figure shows the intermediate state of the game. At the current position Nastya can throw the stone to any other manhole or move left or right to the neighboring manholes. If she were near the leftmost manhole, she could open it (since there are no stones on it).
Nastya can leave the room when she picks all the coins. Monsters are everywhere, so you need to compute the minimum number of moves Nastya has to make to pick all the coins.
Note one time more that Nastya can open a manhole only when there are no stones onto it.
Input
The first and only line contains two integers n and k, separated by space (2 β€ n β€ 5000, 1 β€ k β€ n) β the number of manholes and the index of manhole from the left, near which Nastya stays initially. Initially there is exactly one stone near each of the n manholes.
Output
Print a single integer β minimum number of moves which lead Nastya to pick all the coins.
Examples
Input
2 2
Output
6
Input
4 2
Output
13
Input
5 1
Output
15
Note
Let's consider the example where n = 2, k = 2. Nastya should play as follows:
* At first she throws the stone from the second manhole to the first. Now there are two stones on the first manhole.
* Then she opens the second manhole and pick the coin from it.
* Then she goes to the first manhole, throws two stones by two moves to the second manhole and then opens the manhole and picks the coin from it.
So, 6 moves are required to win.
Submitted Solution:
```
# your code goes here
n,k=map(int,input().split())
if k==1:
print(2+4+(n-2)*3)
elif n==2:
print(6)
else:
if k<=n//2:
print(2+4+(n-k)+2+(n-3)*3)
# print(2+4+(k+2)+(n-k+1)*3)
else:
print(2+4+(n-k+1)+2+(n-3)*3)
``` | instruction | 0 | 32,577 | 19 | 65,154 |
No | output | 1 | 32,577 | 19 | 65,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
n people sitting in a circle are trying to shuffle a deck of cards. The players are numbered from 1 to n, so that players i and i+1 are neighbours (as well as players 1 and n). Each of them has exactly k cards, where k is even. The left neighbour of a player i is player i - 1, and their right neighbour is player i + 1 (except for players 1 and n, who are respective neighbours of each other).
Each turn the following happens: if a player has x cards, they give β x / 2 β to their neighbour on the left and β x / 2 β cards to their neighbour on the right. This happens for all players simultaneously.
However, one player p is the impostor and they just give all their cards to their neighbour on the right. You know the number of players n and the number of cards k each player has initially, but p is unknown to you. Your task is to determine the value of p, by asking questions like "how many cards does player q have?" for an index q of your choice. After each question all players will make exactly one move and give their cards to their neighbours. You need to find the impostor by asking no more than 1000 questions.
Input
The first line contains two integers n and k (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even) β the number of players and the number of cards.
Interaction
You can ask questions by printing "? q". The answer to this question is the number of cards player q has now (1 β€ q β€ n). The shuffling process starts immediately after your first question, so the answer to the first one is always equal to k.
Once you have identified the impostor, you can output the answer by printing "! p", where p is the player who is the impostor (1 β€ p β€ n). Then you have to terminate your program.
You have to find the impostor by asking no more than 1000 questions.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
To make a hack, use the following test format.
The only line of input should contain three integers n, k and p (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even, 1 β€ p β€ n) β the number of people, the number of cards each person has initially, and the position of the impostor.
Example
Input
4 2
2
1
2
3
2
Output
? 1
? 1
? 2
? 3
? 4
! 2
Note
In the example the cards are transferred in the following way:
* 2 2 2 2 β player 1 has 2 cards.
* 1 2 3 2 β player 1 has 1 card.
After this turn the number of cards remains unchanged for each player.
Submitted Solution:
```
import sys
input=sys.stdin.readline
def main():
n,t=map(int,input().split())
m=n//200+1
x=1
y=-1
z=-1
while 1:
print("?",x)
sys.stdout.flush()
p=int(input())
if p>t:
y=x
elif p<t:
z=x
x+=m
x=(x-1)%n+1
if y!=-1 and z!=-1:
break
if y<z:
y+=n
while 1:
mm=(y+z)//2
tt=(mm-1)%n+1
print("?",tt)
sys.stdiout.flush()
p=int(input())
if p==t:
print("!",tt)
return
elif p>t:
y=mm
else:
z=mm
``` | instruction | 0 | 32,735 | 19 | 65,470 |
No | output | 1 | 32,735 | 19 | 65,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
n people sitting in a circle are trying to shuffle a deck of cards. The players are numbered from 1 to n, so that players i and i+1 are neighbours (as well as players 1 and n). Each of them has exactly k cards, where k is even. The left neighbour of a player i is player i - 1, and their right neighbour is player i + 1 (except for players 1 and n, who are respective neighbours of each other).
Each turn the following happens: if a player has x cards, they give β x / 2 β to their neighbour on the left and β x / 2 β cards to their neighbour on the right. This happens for all players simultaneously.
However, one player p is the impostor and they just give all their cards to their neighbour on the right. You know the number of players n and the number of cards k each player has initially, but p is unknown to you. Your task is to determine the value of p, by asking questions like "how many cards does player q have?" for an index q of your choice. After each question all players will make exactly one move and give their cards to their neighbours. You need to find the impostor by asking no more than 1000 questions.
Input
The first line contains two integers n and k (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even) β the number of players and the number of cards.
Interaction
You can ask questions by printing "? q". The answer to this question is the number of cards player q has now (1 β€ q β€ n). The shuffling process starts immediately after your first question, so the answer to the first one is always equal to k.
Once you have identified the impostor, you can output the answer by printing "! p", where p is the player who is the impostor (1 β€ p β€ n). Then you have to terminate your program.
You have to find the impostor by asking no more than 1000 questions.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
To make a hack, use the following test format.
The only line of input should contain three integers n, k and p (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even, 1 β€ p β€ n) β the number of people, the number of cards each person has initially, and the position of the impostor.
Example
Input
4 2
2
1
2
3
2
Output
? 1
? 1
? 2
? 3
? 4
! 2
Note
In the example the cards are transferred in the following way:
* 2 2 2 2 β player 1 has 2 cards.
* 1 2 3 2 β player 1 has 1 card.
After this turn the number of cards remains unchanged for each player.
Submitted Solution:
```
import sys, os
# SCRIPT STARTS HERE
class sim:
def __init__(self,n,k,p=0):
self.n=n
self.k=k
self.p=p
self.list=[k]*min(n,2001)
self.horizon=0
def shuffle(self):
self.horizon=min(self.horizon+1,self.n//2)
list_copy=self.list[:]
for i in range(-self.horizon,self.horizon+1):
if i==-1:
list_copy[-1]=-(-self.list[-2]//2)
elif i==1:
list_copy[1]=self.list[0]+self.list[2]//2
else:
list_copy[i]=-(-self.list[i-1]//2)+self.list[i+1]//2
self.list=list_copy
def get_i(self,index):
index-=self.p
if abs(index)>self.horizon:
return self.k
else:
return self.list[index]
def query(self,index):
self.shuffle()
return self.get_i(index)
if os.environ['USERNAME']=='kissz':
n,k,p=4,2,1 # 0-indexing!
simulator=sim(n,k,p)
inp=simulator.query
def debug(*args):
print(*args,file=sys.stderr)
else:
n,k=map(int,input().split())
def inp(index):
print('?',index+1)
sys.stdout.flush()
return int(input())
def debug(*args):
pass
possible=[1]*n
ktest=inp(0)
debug(ktest,k)
S=sim(n,k)
for r in range(1,10):
S.shuffle()
debug(r, S.list)
i=0
while not possible[i]: i+=1
x=inp(i)
debug(i,x)
if x==k:
for j in range(i-r,i): possible[j]=0
for j in range(i+1,i+r+1): possible[j]=0
else:
new_possible=[0]*n
for j in range(-r,r+1):
if x==S.list[j]: new_possible[i-j]=1
possible=new_possible
debug(possible)
if sum(possible)==1:
print('!',possible.index(1)+1) # 1-indexing
break
``` | instruction | 0 | 32,736 | 19 | 65,472 |
No | output | 1 | 32,736 | 19 | 65,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
n people sitting in a circle are trying to shuffle a deck of cards. The players are numbered from 1 to n, so that players i and i+1 are neighbours (as well as players 1 and n). Each of them has exactly k cards, where k is even. The left neighbour of a player i is player i - 1, and their right neighbour is player i + 1 (except for players 1 and n, who are respective neighbours of each other).
Each turn the following happens: if a player has x cards, they give β x / 2 β to their neighbour on the left and β x / 2 β cards to their neighbour on the right. This happens for all players simultaneously.
However, one player p is the impostor and they just give all their cards to their neighbour on the right. You know the number of players n and the number of cards k each player has initially, but p is unknown to you. Your task is to determine the value of p, by asking questions like "how many cards does player q have?" for an index q of your choice. After each question all players will make exactly one move and give their cards to their neighbours. You need to find the impostor by asking no more than 1000 questions.
Input
The first line contains two integers n and k (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even) β the number of players and the number of cards.
Interaction
You can ask questions by printing "? q". The answer to this question is the number of cards player q has now (1 β€ q β€ n). The shuffling process starts immediately after your first question, so the answer to the first one is always equal to k.
Once you have identified the impostor, you can output the answer by printing "! p", where p is the player who is the impostor (1 β€ p β€ n). Then you have to terminate your program.
You have to find the impostor by asking no more than 1000 questions.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
To make a hack, use the following test format.
The only line of input should contain three integers n, k and p (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even, 1 β€ p β€ n) β the number of people, the number of cards each person has initially, and the position of the impostor.
Example
Input
4 2
2
1
2
3
2
Output
? 1
? 1
? 2
? 3
? 4
! 2
Note
In the example the cards are transferred in the following way:
* 2 2 2 2 β player 1 has 2 cards.
* 1 2 3 2 β player 1 has 1 card.
After this turn the number of cards remains unchanged for each player.
Submitted Solution:
```
import sys
n,k=map(int,input().split())
curr=0
for i in range(500):
print(' '.join(['?',str(curr+1)]),flush=True)
m=int(input())
if m<k:
flag=-1
break
if m>k:
flag=1
break
curr=(curr+i+1)%n
if flag==-1:
print(' '.join(['?',str(curr+1)]),flush=True)
m=int(input())
if m==k:
print(' '.join(['!',str(curr+1)]),flush=True)
sys.exit()
curr=(curr+1)%n
if flag==1:
print(' '.join(['?',str(curr+1)]),flush=True)
m=int(input())
if m==k:
print(' '.join(['!',str(curr+1)]),flush=True)
sys.exit()
curr=(curr-1)%n
``` | instruction | 0 | 32,737 | 19 | 65,474 |
No | output | 1 | 32,737 | 19 | 65,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
n people sitting in a circle are trying to shuffle a deck of cards. The players are numbered from 1 to n, so that players i and i+1 are neighbours (as well as players 1 and n). Each of them has exactly k cards, where k is even. The left neighbour of a player i is player i - 1, and their right neighbour is player i + 1 (except for players 1 and n, who are respective neighbours of each other).
Each turn the following happens: if a player has x cards, they give β x / 2 β to their neighbour on the left and β x / 2 β cards to their neighbour on the right. This happens for all players simultaneously.
However, one player p is the impostor and they just give all their cards to their neighbour on the right. You know the number of players n and the number of cards k each player has initially, but p is unknown to you. Your task is to determine the value of p, by asking questions like "how many cards does player q have?" for an index q of your choice. After each question all players will make exactly one move and give their cards to their neighbours. You need to find the impostor by asking no more than 1000 questions.
Input
The first line contains two integers n and k (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even) β the number of players and the number of cards.
Interaction
You can ask questions by printing "? q". The answer to this question is the number of cards player q has now (1 β€ q β€ n). The shuffling process starts immediately after your first question, so the answer to the first one is always equal to k.
Once you have identified the impostor, you can output the answer by printing "! p", where p is the player who is the impostor (1 β€ p β€ n). Then you have to terminate your program.
You have to find the impostor by asking no more than 1000 questions.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
To make a hack, use the following test format.
The only line of input should contain three integers n, k and p (4 β€ n β€ 10^5, 2 β€ k β€ 10^9, k is even, 1 β€ p β€ n) β the number of people, the number of cards each person has initially, and the position of the impostor.
Example
Input
4 2
2
1
2
3
2
Output
? 1
? 1
? 2
? 3
? 4
! 2
Note
In the example the cards are transferred in the following way:
* 2 2 2 2 β player 1 has 2 cards.
* 1 2 3 2 β player 1 has 1 card.
After this turn the number of cards remains unchanged for each player.
Submitted Solution:
```
"""
rm /tmp/fifo && mkfifo /tmp/fifo && (python3 1471_e.py < /tmp/fifo)1>&2 | python3 1471_e_interactor.py > /tmp/fifo1>&2
"""
import sys
n, k = map(int, input().split())
def do():
for i in range(502):
print("? 1")
sys.stdout.flush()
res = int(input())
if res == k: continue
if res < k: # is left
res = i
res %= n
print("!", res + 1)
sys.stdout.flush()
return
if res > k:
res = -i
res %= n
print("!", res + 1)
sys.stdout.flush()
return
for i in range(400):
cur = 1 + 500*i
print("?", cur)
sys.stdout.flush()
res = int(input())
if res == k: continue # cannot judge
if res < k: # more right
l = cur
r = cur + 510
break
if res > k: # more left
l = cur - 501
r = cur
break
# start Nibutan
while l <= r:
mid = (l + r) // 2
print("{0} {1}=> mid={2}".format(l, r, mid), file=sys.stderr)
print("?", mid % n + 1)
sys.stdout.flush()
res = int(input())
if res == k:
print("!", mid % n + 1)
sys.stdout.flush()
return
if res < k: # more right
l = mid + 1
continue
if res > k: # more left
r = mid - 1
continue
print("?", l)
sys.stdout.flush()
res = int(input())
if res == k:
print("!", l)
sys.stdout.flush()
return
print("?", r)
res = int(input())
if res == k:
print("!", r)
sys.stdout.flush()
return
print("cannot!!!")
sys.stdout.flush()
do()
``` | instruction | 0 | 32,738 | 19 | 65,476 |
No | output | 1 | 32,738 | 19 | 65,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya are tossing a coin. Their friend Valera is appointed as a judge. The game is very simple. First Vasya tosses a coin x times, then Petya tosses a coin y times. If the tossing player gets head, he scores one point. If he gets tail, nobody gets any points. The winner is the player with most points by the end of the game. If boys have the same number of points, the game finishes with a draw.
At some point, Valera lost his count, and so he can not say exactly what the score is at the end of the game. But there are things he remembers for sure. He remembers that the entire game Vasya got heads at least a times, and Petya got heads at least b times. Moreover, he knows that the winner of the game was Vasya. Valera wants to use this information to know every possible outcome of the game, which do not contradict his memories.
Input
The single line contains four integers x, y, a, b (1 β€ a β€ x β€ 100, 1 β€ b β€ y β€ 100). The numbers on the line are separated by a space.
Output
In the first line print integer n β the number of possible outcomes of the game. Then on n lines print the outcomes. On the i-th line print a space-separated pair of integers ci, di β the number of heads Vasya and Petya got in the i-th outcome of the game, correspondingly. Print pairs of integers (ci, di) in the strictly increasing order.
Let us remind you that the pair of numbers (p1, q1) is less than the pair of numbers (p2, q2), if p1 < p2, or p1 = p2 and also q1 < q2.
Examples
Input
3 2 1 1
Output
3
2 1
3 1
3 2
Input
2 4 2 2
Output
0 | instruction | 0 | 32,793 | 19 | 65,586 |
Tags: brute force, implementation
Correct Solution:
```
l=list(map(int,input().split()))
x=l[0]
y=l[1]
a=l[2]
b=l[3]
c=0
k=[]
for i in range(a,x+1):
for j in range(b,y+1):
if i>j:
c+=1
s=str(i)+" "+str(j)
k.append(s)
print(c)
for i in range(c):
print(k[i])
``` | output | 1 | 32,793 | 19 | 65,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya are tossing a coin. Their friend Valera is appointed as a judge. The game is very simple. First Vasya tosses a coin x times, then Petya tosses a coin y times. If the tossing player gets head, he scores one point. If he gets tail, nobody gets any points. The winner is the player with most points by the end of the game. If boys have the same number of points, the game finishes with a draw.
At some point, Valera lost his count, and so he can not say exactly what the score is at the end of the game. But there are things he remembers for sure. He remembers that the entire game Vasya got heads at least a times, and Petya got heads at least b times. Moreover, he knows that the winner of the game was Vasya. Valera wants to use this information to know every possible outcome of the game, which do not contradict his memories.
Input
The single line contains four integers x, y, a, b (1 β€ a β€ x β€ 100, 1 β€ b β€ y β€ 100). The numbers on the line are separated by a space.
Output
In the first line print integer n β the number of possible outcomes of the game. Then on n lines print the outcomes. On the i-th line print a space-separated pair of integers ci, di β the number of heads Vasya and Petya got in the i-th outcome of the game, correspondingly. Print pairs of integers (ci, di) in the strictly increasing order.
Let us remind you that the pair of numbers (p1, q1) is less than the pair of numbers (p2, q2), if p1 < p2, or p1 = p2 and also q1 < q2.
Examples
Input
3 2 1 1
Output
3
2 1
3 1
3 2
Input
2 4 2 2
Output
0 | instruction | 0 | 32,794 | 19 | 65,588 |
Tags: brute force, implementation
Correct Solution:
```
def read_int():
return int(input())
def read_ints():
return list(map(int, input().split()))
def print_nums(nums):
print(" ".join(map(str, nums)))
x, y, a, b = read_ints()
res = []
for A in range(max(a, b+1), x+1):
for B in range(b, min(A-1, y) + 1):
res.append((A, B))
print(len(res))
for A, B in res:
print("%d %d" % (A, B))
``` | output | 1 | 32,794 | 19 | 65,589 |
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