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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively. | instruction | 0 | 98,630 | 19 | 197,260 |
Tags: math, number theory
Correct Solution:
```
import math
x=int(input())
xr=math.ceil(math.sqrt(x))
LIST=[]
for i in range(1,xr+1):
if x%i==0:
LIST.append(i)
LIST.append(x//i)
LIST=set(LIST)
ANS=[]
for l in LIST:
ANS.append((x-x//l)*l//2+l)
ANS.sort()
for a in ANS:
print(a,end=" ")
``` | output | 1 | 98,630 | 19 | 197,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively. | instruction | 0 | 98,631 | 19 | 197,262 |
Tags: math, number theory
Correct Solution:
```
def prime(n):
l=[]
for i in range(1,int(n**(0.5))+1):
if n%i==0:
l.append(i)
l.append(n//i)
return l
n=int(input())
ans=[]
t=prime(n)
for i in t:
elm=n//i
ans.append(((n+2-i)*elm)//2)
ans=list(set(ans))
ans.sort()
print (*ans)
``` | output | 1 | 98,631 | 19 | 197,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively. | instruction | 0 | 98,632 | 19 | 197,264 |
Tags: math, number theory
Correct Solution:
```
from math import sqrt
n = int(input())
def c(x):
global n
cc = n//x
#print('calc:', n, x, cc + x*(cc-1)*(cc)//2)
return cc + x*(cc-1)*(cc)//2
ans = set()
for i in range(1, int(sqrt(n))+1):
if n%i==0:
ans.add(c(i))
ans.add(c(n//i))
ans = list(ans)
ans.sort()
print(' '.join(map(str, ans)))
``` | output | 1 | 98,632 | 19 | 197,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
import math
n = int(input())
st = set([])
for i in range(1, int(math.sqrt(n)) + 2):
if not (n % i):
k = i
st.add((k * (n // k - 1) * (n // k)) // 2 + n // k)
k = n // i
st.add((k * (n // k - 1) * (n // k)) // 2 + n // k)
print(*sorted(list(st)))
``` | instruction | 0 | 98,633 | 19 | 197,266 |
Yes | output | 1 | 98,633 | 19 | 197,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
n=int(input())
a=[]
i=1
while i*i<=n:
if n%i==0:
a.append(i)
a.append(n//i)
i+=1
a.sort(reverse=True)
temp=2*n+(n*n)
minus=n
l=[]
l.append(0)
for i in a:
ans=(temp//i-minus)//2
if l[-1]!=ans:
l.append(ans)
for i in l:
if i>0:
print(i,end=' ')
# print(i,end=' ')
``` | instruction | 0 | 98,634 | 19 | 197,268 |
Yes | output | 1 | 98,634 | 19 | 197,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
l = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
a = n//i
b = i
if a!=b :
s = (2+i*(a - 1))*a // 2
l.append(s)
b = (2 + a * (b-1)) * (b) // 2
print(b, end =' ')
for i in reversed(l):
print(i,end = ' ')
``` | instruction | 0 | 98,635 | 19 | 197,270 |
Yes | output | 1 | 98,635 | 19 | 197,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
import math
n=int(input())
l1=[]
l2=[]
for i in range(1,math.floor(math.sqrt(n))+1):
if n%i==0:
l=1+((n//i)-1)*i
p1=((n//i)*(1+l))//2
l1.append(p1)
l=1+((i-1)*(n//i))
p=int((i/2)*(1+l))
if p1!=p:
l2.append(p)
for j in range(len(l2)):
print(l2[j],end=" ")
for j in range(len(l1)):
print(l1[len(l1)-j-1],end=" ")
``` | instruction | 0 | 98,636 | 19 | 197,272 |
Yes | output | 1 | 98,636 | 19 | 197,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
import sys
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
n = int(input())
fac = factors(n)
sol = []
for f in fac:
if f == 1:
if n%2 == 0:
sq = (n+1)*(n//2)
sol.append(sq)
else:
sq = ((n+1)*(n//2)) + (n+1)//2
sol.append(sq)
elif f == n:
sol.append(1)
else:
end = n-f+1
if end%2 == 0:
sq = (end+1)*(end//2)
sol.append(sq)
else:
sq = ((end+1)*(end//2)) + (end+1)//2
sol.append(sq)
print(' '.join([str(i) for i in sorted(sol)]))
``` | instruction | 0 | 98,637 | 19 | 197,274 |
No | output | 1 | 98,637 | 19 | 197,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
def gbs(k,n):
c = n
while True:
if c%k == 0 and c%n == 0:
break
c += 1
return (c//k)
cnt = [0]*(n+1)
for k in range(1,n//2+2):
s = 0
c = 1
num = gbs(k,n)
if cnt[num] == 0:
for i in range(1,num):
if (c+k)%n == 0:
s += (c+k)%n+n
else:
s += (c+k)%n
c += k
s += 1
cnt[num] = 1
print(s,end=" ")
print(1,end= " ")
``` | instruction | 0 | 98,638 | 19 | 197,276 |
No | output | 1 | 98,638 | 19 | 197,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
a = set([])
i = 1
while i*i <= n:
if(n%i == 0):
x = i
y = n//i
# 1+x+1+2x+1+....+(y-1)x+1
# y + (y)(y-1)(x)/2
a.add(y+((y)*(y-1)*(x)/2))
x,y = y,x
a.add(y+((y)*(y-1)*(x)/2))
i += 1
a = sorted(list(a))
for i in a:
print(int(i),end=' ')
print()
``` | instruction | 0 | 98,639 | 19 | 197,278 |
No | output | 1 | 98,639 | 19 | 197,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
a = set([])
i = 1
while i*i <= n:
if(n%i == 0):
x = i
y = n//i
# 1+x+1+2x+1+....+(y-1)x+1
# y + (y)(y-1)(x)/2
a.add(y+(((y)*(y-1)*(x))/2))
x,y = y,x
a.add(y+(((y)*(y-1)*(x))/2))
i += 1
a = sorted(list(a))
for i in a:
print(int(i),end=' ')
print()
``` | instruction | 0 | 98,640 | 19 | 197,280 |
No | output | 1 | 98,640 | 19 | 197,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,119 | 19 | 198,238 |
Tags: brute force, implementation
Correct Solution:
```
n, m = map(int, input().split())
dignities = '23456789TJQKA'
colors = 'CDHS'
all_cards = []
for i in dignities:
for j in colors:
all_cards.append(i + j)
my_cards = [[] for x in range(n)]
joker1 = 0
joker2 = 0
for i in range(n):
my_cards[i] = input().split()
for j in range(m):
if my_cards[i][j][1] == '1':
joker1 = (i, j)
elif my_cards[i][j][1] == '2':
joker2 = (i, j)
else:
all_cards.remove(my_cards[i][j])
def intersect_segments(a, b, c, d):
if a > b:
a, b = b, a
if c > d:
c, d = d, c
return a <= d and b >= c
def intersect_squares(a, b):
return intersect_segments(a[0], a[2], b[0], b[2]) and intersect_segments(a[1], a[3], b[1], b[3])
def suitable(cards, x, y):
colors = set()
dignities = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
colors.add(cards[i][j][1])
dignities.add(cards[i][j][0])
return len(colors) == 1 or len(dignities) == 9
def ok(cards, n, m):
for a in range(n - 2):
for b in range(m - 2):
if suitable(cards, a, b):
for c in range(n - 2):
for d in range(m - 2):
if not intersect_squares((a, b, a + 2, b + 2), (c, d, c + 2, d + 2)) and suitable(cards, c, d):
return (a, b, c, d)
return 0
for i in all_cards:
for j in all_cards:
if i != j:
new_cards = my_cards
if joker1 != 0:
new_cards[joker1[0]][joker1[1]] = i
if joker2 != 0:
new_cards[joker2[0]][joker2[1]] = j
if ok(new_cards, n, m):
print('Solution exists.')
if joker1 == 0 and joker2 == 0:
print('There are no jokers.')
elif joker1 != 0 and joker2 != 0:
print('Replace J1 with %s and J2 with %s.' % (i, j))
elif joker1 != 0:
print('Replace J1 with %s.' % (i))
else:
print('Replace J2 with %s.' % (j))
tmp = ok(new_cards, n, m)
print('Put the first square to (%d, %d).' % (tmp[0] + 1, tmp[1] + 1))
print('Put the second square to (%d, %d).' % (tmp[2] + 1, tmp[3] + 1))
quit()
print('No solution.')
``` | output | 1 | 99,119 | 19 | 198,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,120 | 19 | 198,240 |
Tags: brute force, implementation
Correct Solution:
```
def get_cards():
n,m = map(int,input().split())
lis = [[]]*n
for i in range(n):
lis[i] = list(map(str,input().split()))
return (n,m,lis)
def find_remains(cards):
remains = []
ranks = ["2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K","A"]
suits = ["C", "D", "H","S"]
for r in ranks:
for s in suits:
remains.append(r+s)
remains += ["J1","J2"]
for i in range(cards[0]):
for j in range(cards[1]):
remains.remove(cards[2][i][j])
return remains
def condition_1(ranks):
r = set(ranks)
if len(r) < len(ranks):
return 1
else:
return 0
def condition_2(suits):
r = set(suits)
if len(r) == 1:
return 0
else:
return 1
def find_solution(cards):
lis = cards[2]
positions = []
result = 1
position = []
for i in range(cards[0]-2):
for j in range(cards[1]-2):
positions.append((i,j))
for p in positions:
positions_2 = positions[:]
if result == 0:
break
i = p[0]
j = p[1]
nine_1 = [lis[i][j],lis[i][j+1],lis[i][j+2],lis[i+1][j],lis[i+1][j+1],lis[i+1][j+2],lis[i+2][j],lis[i+2][j+1],lis[i+2][j+2]]
nine_1_p = []
for a in range(-2,3):
for b in range(-2,3):
nine_1_p.append((i+a,j+b))
for c in nine_1_p:
if c in positions_2:
positions_2.remove(c)
for q in positions_2:
if result == 0:
break
i = q[0]
j = q[1]
nine_2 = [lis[i][j],lis[i][j+1],lis[i][j+2],lis[i+1][j],lis[i+1][j+1],lis[i+1][j+2],lis[i+2][j],lis[i+2][j+1],lis[i+2][j+2]]
ranks_1,ranks_2 = [],[]
suits_1,suits_2 = [],[]
for card in nine_1:
ranks_1.append(card[0])
suits_1.append(card[1])
result_1_1 = condition_1(ranks_1)
result_1_2 = condition_2(suits_1)
for card in nine_2:
ranks_2.append(card[0])
suits_2.append(card[1])
result_2_1 = condition_1(ranks_2)
result_2_2 = condition_2(suits_2)
if result_1_1*result_1_2 == 0 and result_2_1*result_2_2 == 0:
result = 0
position = [p,q]
result = [result,position]
return result
def find_Jocker(cards):
n = cards[0]
m = cards[1]
J1,J2 = 0,0
for i in range(n):
for j in range(m):
if cards[2][i][j] == "J1":
J1 = (i,j)
if cards[2][i][j] == "J2":
J2 = (i,j)
positions = (J1,J2)
return positions
cards = get_cards()
if cards[0] < 3 or cards[1] < 3:
result = [1]
elif cards[0] < 6 and cards[1] < 6:
result = [1]
else:
remains = find_remains(cards)
jocker = find_Jocker(cards)
changed = [0,0]
if jocker[0] == 0 and jocker[1] == 0:
result = find_solution(cards)
elif jocker[0] != 0 and jocker[1] == 0:
remains.remove("J2")
for remain in remains:
cards[2][jocker[0][0]][jocker[0][1]] = remain
result = find_solution(cards)
if result[0] == 0:
changed = [remain,0]
break
elif jocker[1] != 0 and jocker[0] == 0:
remains.remove("J1")
for remain in remains:
cards[2][jocker[1][0]][jocker[1][1]] = remain
result = find_solution(cards)
if result[0] == 0:
changed = [0,remain]
break
else:
remains_2 = remains[:]
for remain_1 in remains:
cards[2][jocker[1][0]][jocker[1][1]] = remain_1
remains_2.remove(remain_1)
for remain_2 in remains_2:
cards[2][jocker[0][0]][jocker[0][1]] = remain_2
result = find_solution(cards)
if result[0] == 0:
changed[0] = remain_2
break
remains_2 = remains[:]
if result[0] == 0:
changed[1] = remain_1
break
if result[0] == 1:
print("No solution.")
if result[0] == 0:
print("Solution exists.")
if changed[0] != 0 and changed[1] != 0:
print("Replace J1 with "+changed[0]+" and J2 with "+changed[1]+".")
elif changed[0] != 0:
print("Replace J1 with "+changed[0]+".")
elif changed[1] != 0:
print("Replace J2 with "+changed[1]+".")
else:
print("There are no jokers.")
print("Put the first square to ("+str(result[1][0][0]+1)+", "+str(result[1][0][1]+1)+").")
print("Put the second square to ("+str(result[1][1][0]+1)+", "+str(result[1][1][1]+1)+").")
``` | output | 1 | 99,120 | 19 | 198,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,121 | 19 | 198,242 |
Tags: brute force, implementation
Correct Solution:
```
n, m = map(int, input().split())
dignities = ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']
colors = ['C', 'D', 'H', 'S']
all_cards = []
for i in dignities:
for j in colors:
all_cards.append(i + j)
my_cards = [[] for x in range(n)]
joker1 = 0
joker2 = 0
for i in range(n):
my_cards[i] = input().split()
for j in range(m):
if my_cards[i][j][1] == '1':
joker1 = (i, j)
elif my_cards[i][j][1] == '2':
joker2 = (i, j)
else:
all_cards.remove(my_cards[i][j])
def intersect_segments(a, b, c, d):
if a > b:
a, b = b, a
if c > d:
c, d = d, c
return a <= d and b >= c
def intersect_squares(a, b):
return intersect_segments(a[0], a[2], b[0], b[2]) and intersect_segments(a[1], a[3], b[1], b[3])
def suitable(cards, x, y):
colors = set()
dignities = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
colors.add(cards[i][j][1])
dignities.add(cards[i][j][0])
return len(colors) == 1 or len(dignities) == 9
def ok(cards, n, m):
for a in range(n - 2):
for b in range(m - 2):
if suitable(cards, a, b):
for c in range(n - 2):
for d in range(m - 2):
if not intersect_squares((a, b, a + 2, b + 2), (c, d, c + 2, d + 2)) and suitable(cards, c, d):
return (a, b, c, d)
return 0
for i in all_cards:
for j in all_cards:
if i != j:
new_cards = my_cards
if joker1 != 0:
ans1 = i[0] + i[1]
new_cards[joker1[0]][joker1[1]] = i
if joker2 != 0:
ans2 = j[0] + j[1]
new_cards[joker2[0]][joker2[1]] = j
if ok(new_cards, n, m):
print('Solution exists.')
if joker1 == 0 and joker2 == 0:
print('There are no jokers.')
elif joker1 != 0 and joker2 != 0:
print('Replace J1 with %s and J2 with %s.' % (ans1, ans2))
elif joker1 != 0:
print('Replace J1 with %s.' % (ans1))
else:
print('Replace J2 with %s.' % (ans2))
tmp = ok(new_cards, n, m)
print('Put the first square to (%d, %d).' % (tmp[0] + 1, tmp[1] + 1))
print('Put the second square to (%d, %d).' % (tmp[2] + 1, tmp[3] + 1))
quit()
print('No solution.')
``` | output | 1 | 99,121 | 19 | 198,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,122 | 19 | 198,244 |
Tags: brute force, implementation
Correct Solution:
```
n, m = map(int, input().split())
dignities = ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']
colors = ['C', 'D', 'H', 'S']
all_cards = []
for i in dignities:
for j in colors:
all_cards.append(i + j)
my_cards = [[] for x in range(n)]
joker1 = 0
joker2 = 0
for i in range(n):
my_cards[i] = input().split()
for j in range(m):
if my_cards[i][j][1] == '1':
joker1 = (i, j)
elif my_cards[i][j][1] == '2':
joker2 = (i, j)
else:
all_cards.remove(my_cards[i][j])
def intersect_segments(a, b, c, d):
if a > b:
a, b = b, a
if c > d:
c, d = d, c
return a <= d and b >= c
def intersect_squares(a, b):
return intersect_segments(a[0], a[2], b[0], b[2]) and intersect_segments(a[1], a[3], b[1], b[3])
def suitable(cards, x, y):
colors = set()
dignities = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
colors.add(cards[i][j][1])
dignities.add(cards[i][j][0])
return len(colors) == 1 or len(dignities) == 9
def ok(cards, n, m):
for a in range(n - 2):
for b in range(m - 2):
if suitable(cards, a, b):
for c in range(n - 2):
for d in range(m - 2):
if not intersect_squares((a, b, a + 2, b + 2), (c, d, c + 2, d + 2)) and suitable(cards, c, d):
return (a, b, c, d)
return 0
for i in all_cards:
for j in all_cards:
if i != j:
new_cards = my_cards
if joker1 != 0:
new_cards[joker1[0]][joker1[1]] = i
if joker2 != 0:
new_cards[joker2[0]][joker2[1]] = j
if ok(new_cards, n, m):
print('Solution exists.')
if joker1 == 0 and joker2 == 0:
print('There are no jokers.')
elif joker1 != 0 and joker2 != 0:
print('Replace J1 with %s and J2 with %s.' % (i, j))
elif joker1 != 0:
print('Replace J1 with %s.' % (i))
else:
print('Replace J2 with %s.' % (j))
tmp = ok(new_cards, n, m)
print('Put the first square to (%d, %d).' % (tmp[0] + 1, tmp[1] + 1))
print('Put the second square to (%d, %d).' % (tmp[2] + 1, tmp[3] + 1))
quit()
print('No solution.')
``` | output | 1 | 99,122 | 19 | 198,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,123 | 19 | 198,246 |
Tags: brute force, implementation
Correct Solution:
```
allCards = [value + suit for value in ["2","3","4","5","6","7","8","9","T","J","Q","K","A"] for suit in ["C","D","H","S"]]
n, m = map(lambda x: int(x), input().split(" "))
cards = []
for i in range(n):
cards.append(input().split(" "))
availableCards = allCards
for arr in cards:
for card in arr:
if card != "J1" and card != "J2":
availableCards.remove(card)
def joker2(i: int, j: int) -> [bool, bool, bool, set]:
usedValues = set()
didFoundJoker1 = False
didFoundJoker2 = False
for i1 in range(i, i + 3):
for j1 in range(j, j + 3):
if cards[i1][j1] == "J1":
didFoundJoker1 = True
elif cards[i1][j1] == "J2":
didFoundJoker2 = True
elif cards[i1][j1][0] not in usedValues:
usedValues.add(cards[i1][j1][0])
else:
return [False, False, False, set()]
res = set()
for card in availableCards:
if card[0] not in usedValues:
res.add(card)
isTrue = True
if didFoundJoker1 or didFoundJoker2:
isTrue = False
for card in res:
if isTrue: break
for i1 in range(i, i + 3):
if isTrue: break
for j1 in range(j, j + 3):
if cards[i1][j1] not in ["J1","J2"] and cards[i1][j1][0] != card[0]:
isTrue = True
break
return [isTrue, didFoundJoker1, didFoundJoker2, res]
ansi1, ansj1, ansi2, ansj2 = 0, 0, 0, 0
replaceJoker1 = None
replaceJoker2 = None
def joker1(i: int, j: int) -> bool:
global replaceJoker1
global replaceJoker2
global ansi1, ansj1, ansi2, ansj2
if not (i + 2 < n and j + 2 < m):
return False
didFound1, didFindJoker1, didFindJoker2, strings = joker2(i, j)
if not didFound1:
return False
for i1 in range(n):
for j1 in range(m):
if (i1 > i + 2 or j1 > j + 2) and i1 + 2 < n and j1 + 2 < m:
didFound2, didFindJoker21, didFindJoker22, strings2 = joker2(i1, j1)
if (didFindJoker1 or didFindJoker2) and (didFindJoker21 or didFindJoker22):
xor = (strings2 ^ strings)
x1 = xor & strings
x2 = xor & strings2
if x1 and x2:
if didFindJoker1 and didFindJoker22:
replaceJoker1 = x1.pop()
replaceJoker2 = x2.pop()
else:
replaceJoker1 = x2.pop()
replaceJoker2 = x1.pop()
ansi1, ansj1, ansi2, ansj2 = i, j, i1, j1
return True
elif didFindJoker21 and didFindJoker22:
if len(strings2) >= 2:
replaceJoker1 = strings2.pop()
replaceJoker2 = strings2.pop()
ansi1, ansj1, ansi2, ansj2 = i, j, i1, j1
return True
elif didFound2 and didFindJoker1 and didFindJoker2:
stringsx = list(strings)
ansi1, ansj1, ansi2, ansj2 = i, j, i1, j1
for i in range(len(stringsx)):
for j in range(i + 1, len(stringsx)):
if stringsx[i][0] != stringsx[j][0]:
replaceJoker1 = stringsx[i]
replaceJoker2 = stringsx[j]
return True
continue
elif didFound2 and didFindJoker21 and didFindJoker22:
stringsx = list(strings2)
ansi1, ansj1, ansi2, ansj2 = i, j, i1, j1
for i in range(len(stringsx)):
for j in range(i + 1, len(stringsx)):
if stringsx[i][0] != stringsx[j][0]:
replaceJoker1 = stringsx[i]
replaceJoker2 = stringsx[j]
return True
continue
elif didFound2:
if didFindJoker1:
if not strings:
continue
replaceJoker1 = strings.pop()
if didFindJoker2:
if not strings:
continue
replaceJoker2 = strings.pop()
if didFindJoker21:
if not strings2:
continue
replaceJoker1 = strings2.pop()
if didFindJoker22:
if not strings2:
continue
replaceJoker2 = strings2.pop()
ansi1, ansj1, ansi2, ansj2 = i, j, i1, j1
return True
return False
didFound = False
for i in range(n):
if didFound:
break
for j in range(m):
res = joker1(i, j)
if res:
print("Solution exists.")
j1f, j2f = False, False
for c in cards:
for card in c:
if card == "J1":
j1f = True
elif card == "J2":
j2f = True
if j1f and not replaceJoker1:
replaceJoker1 = availableCards.pop()
while availableCards and replaceJoker1 == replaceJoker2:
replaceJoker1 = availableCards.pop()
if j2f and not replaceJoker2:
replaceJoker2 = availableCards.pop()
if replaceJoker2 and replaceJoker1:
print(f"Replace J1 with {replaceJoker1} and J2 with {replaceJoker2}.")
elif replaceJoker1:
print(f"Replace J1 with {replaceJoker1}.")
elif replaceJoker2:
print(f"Replace J2 with {replaceJoker2}.")
else:
print("There are no jokers.")
print(f"Put the first square to ({ansi1+1}, {ansj1+1}).")
print(f"Put the second square to ({ansi2+1}, {ansj2+1}).")
didFound = True
break
if not didFound:
print("No solution.")
``` | output | 1 | 99,123 | 19 | 198,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,124 | 19 | 198,248 |
Tags: brute force, implementation
Correct Solution:
```
n, m = map(int, input().split())
num = '23456789TJQKA'
col = 'CDHS'
all = []
for i in num:
for j in col:
all.append(i + j)
my = [[] for x in range(n)]
joker1 = 0
joker2 = 0
for i in range(n):
my[i] = input().split()
for j in range(m):
if my[i][j][1] == '1':
joker1 = (i, j)
elif my[i][j][1] == '2':
joker2 = (i, j)
else:
all.remove(my[i][j])
def intersect_segments(a, b, c, d):
return a <= d and b >= c
def intersect_squares(a, b):
return intersect_segments(a[0], a[2], b[0], b[2]) and intersect_segments(a[1], a[3], b[1], b[3])
def suitable(cards, x, y):
col = set()
num = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
col.add(cards[i][j][1])
num.add(cards[i][j][0])
return len(col) == 1 or len(num) == 9
def ok(cards, n, m):
for a in range(n - 2):
for b in range(m - 2):
if suitable(cards, a, b):
for c in range(n - 2):
for d in range(m - 2):
if not intersect_squares((a, b, a + 2, b + 2), (c, d, c + 2, d + 2)) and suitable(cards, c, d):
return (a, b, c, d)
return 0
for i in all:
for j in all:
if i != j:
new = my
if joker1 != 0:
new[joker1[0]][joker1[1]] = i
if joker2 != 0:
new[joker2[0]][joker2[1]] = j
q = ok(new, n, m)
if q:
print('Solution exists.')
if joker1 == 0 and joker2 == 0:
print('There are no jokers.')
elif joker1 != 0 and joker2 != 0:
print('Replace J1 with %s and J2 with %s.' % (i, j))
elif joker1 != 0:
print('Replace J1 with %s.' % (i))
else:
print('Replace J2 with %s.' % (j))
print('Put the first square to (%d, %d).' % (q[0] + 1, q[1] + 1))
print('Put the second square to (%d, %d).' % (q[2] + 1, q[3] + 1))
quit()
print('No solution.')
``` | output | 1 | 99,124 | 19 | 198,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,125 | 19 | 198,250 |
Tags: brute force, implementation
Correct Solution:
```
ranks = '23456789TJQKA'
suits = 'CDHS'
n, m = [int(i) for i in input().split()]
b = [input().split() for _ in range(n)]
p = [r + s for r in ranks for s in suits]
j1, j2 = False, False
for r in b:
for c in r:
if c == 'J1':
j1 = True
elif c == 'J2':
j2 = True
else:
p.remove(c)
def valid(n, m):
r = set()
s = set()
for ni in range(n, n + 3):
for mi in range(m, m + 3):
c = b[ni][mi]
if c == 'J1':
c = j1v
if c == 'J2':
c = j2v
r.add(c[0])
s.add(c[1])
return len(r) == 9 or len(s) == 1
def solve():
global j1v, j2v, n0, m0, n1, m1
for j1v in p:
for j2v in p:
if j1v == j2v: continue
for n0 in range(n-2):
for m0 in range(m-2):
if not valid(n0, m0):
continue
for n1 in range(n-2):
for m1 in range(m-2):
if (n0 + 2 < n1 or n1 + 2 < n0 or
m0 + 2 < m1 or m1 + 2 < m0):
if valid(n1, m1):
return True
return False
if solve():
print('Solution exists.')
if j1 and j2:
print('Replace J1 with {} and J2 with {}.'.format(j1v, j2v))
elif j1:
print('Replace J1 with {}.'.format(j1v))
elif j2:
print('Replace J2 with {}.'.format(j2v))
else:
print('There are no jokers.')
print('Put the first square to ({}, {}).'.format(n0+1, m0+1))
print('Put the second square to ({}, {}).'.format(n1+1, m1+1))
else:
print('No solution.')
``` | output | 1 | 99,125 | 19 | 198,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats. | instruction | 0 | 99,126 | 19 | 198,252 |
Tags: brute force, implementation
Correct Solution:
```
n, m = map(int, input().split())
num = '23456789TJQKA'
col = 'CDHS'
all = []
for i in num:
for j in col:
all.append(i + j)
my = [[] for x in range(n)]
j1 = 0
j2 = 0
for i in range(n):
my[i] = input().split()
for j in range(m):
if my[i][j][1] == '1':
j1 = (i, j)
elif my[i][j][1] == '2':
j2 = (i, j)
else:
all.remove(my[i][j])
def i_seg(a, b, c, d):
return a <= d and b >= c
def i_sq(a, b):
return i_seg(a[0], a[2], b[0], b[2]) and i_seg(a[1], a[3], b[1], b[3])
def suit(cards, x, y):
col = set()
num = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
col.add(cards[i][j][1])
num.add(cards[i][j][0])
return len(col) == 1 or len(num) == 9
def ok(cards, n, m):
for a in range(n - 2):
for b in range(m - 2):
if suit(cards, a, b):
for c in range(n - 2):
for d in range(m - 2):
if not i_sq((a, b, a + 2, b + 2), (c, d, c + 2, d + 2)) and suit(cards, c, d):
return (a, b, c, d)
return 0
for i in all:
for j in all:
if i != j:
new = my
if j1 != 0:
new[j1[0]][j1[1]] = i
if j2 != 0:
new[j2[0]][j2[1]] = j
q = ok(new, n, m)
if q:
print('Solution exists.')
if j1 == 0 and j2 == 0:
print('There are no jokers.')
elif j1 != 0 and j2 != 0:
print('Replace J1 with %s and J2 with %s.' % (i, j))
elif j1 != 0:
print('Replace J1 with %s.' % (i))
else:
print('Replace J2 with %s.' % (j))
print('Put the first square to (%d, %d).' % (q[0] + 1, q[1] + 1))
print('Put the second square to (%d, %d).' % (q[2] + 1, q[3] + 1))
quit()
print('No solution.')
``` | output | 1 | 99,126 | 19 | 198,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
#!/usr/bin/env python3
import itertools
class Solitaire:
def __init__(self):
# Read inputs
(self.n,self.m) = map(int,input().split())
self.table = [input().split() for i in range(self.n)]
# Determine the unused cards and the positions of the jokers
ranks = ["2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K", "A"]
suits = ["C", "D", "H", "S"]
deck = set(("%s%s" % card) for card in itertools.product(ranks,suits))
self.J1_position = None
self.J2_position = None
for (r,c) in itertools.product(range(self.n), range(self.m)):
card = self.table[r][c]
deck.discard(card)
if (card == "J1"):
self.J1_position = (r,c)
elif (card == "J2"):
self.J2_position = (r,c)
self.unused_cards = list(deck)
self.unused_cards.sort() # Sort to make things deterministic (easier for testing)
def is_solved_square(self, r, c):
cards = [self.table[r+ro][c+co] for (ro, co) in itertools.product(range(3), range(3))]
suits = set(card[1] for card in cards)
ranks = set(card[0] for card in cards)
solved = (len(suits) == 1) or (len(ranks) == 9)
return solved
def is_solved(self):
all_positions = itertools.product(range(self.n-2), range(self.m-2))
all_solved_squares = [(r,c) for (r,c) in all_positions if self.is_solved_square(r,c)]
for ((r1, c1), (r2,c2)) in itertools.combinations(all_solved_squares, 2):
if (abs(r1-r2) >= 3) or (abs(c1-c2) >= 3):
return ((r1,c1),(r2,c2))
return None
def replace_card(self, position, card):
(r,c) = position
self.table[r][c] = card
def print_solution(self, joker_line, solution):
if (solution == None):
print("No solution.")
else:
((r1,c1),(r2,c2)) = solution
print("Solution exists.")
print(joker_line)
print("Put the first square to (%d, %d)." % (r1+1,c1+1))
print("Put the second square to (%d, %d)." % (r2+1,c2+1))
exit()
def solve(self):
if (self.J1_position != None) and (self.J2_position != None):
for (replacement1, replacement2) in itertools.permutations(self.unused_cards,2):
self.replace_card(self.J1_position, replacement1)
self.replace_card(self.J2_position, replacement2)
solution = self.is_solved()
if (solution):
self.print_solution("Replace J1 with %s and J2 with %s." % (replacement1, replacement2), solution)
elif (self.J1_position != None):
for replacement in self.unused_cards:
self.replace_card(self.J1_position, replacement)
solution = self.is_solved()
if (solution):
self.print_solution("Replace J1 with %s." % replacement, solution)
elif (self.J2_position != None):
for replacement in self.unused_cards:
self.replace_card(self.J2_position, replacement)
solution = self.is_solved()
if (solution):
self.print_solution("Replace J2 with %s." % replacement, solution)
else:
solution = self.is_solved()
self.print_solution("There are no jokers.", solution)
self.print_solution("", None)
Solitaire().solve()
``` | instruction | 0 | 99,127 | 19 | 198,254 |
Yes | output | 1 | 99,127 | 19 | 198,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
'''input
3 6
2H 3H 4H 5H 2S 3S
6H 7H 8H 9H 4S 5S
TH JH QH KH 6S 7S
'''
from sys import stdin
import math
from copy import deepcopy
def find_jokers(grid, n, m):
jokers = []
for i in range(n):
for j in range(m):
if grid[i][j] == 'J1' and len(jokers) > 0:
jokers.insert(0, [i, j])
elif (grid[i][j] == 'J1' or grid[i][j] == 'J2'):
jokers.append([i, j])
return jokers
def get_remain(grid, n, m):
total = set()
for typ in ['D', 'S', 'H', 'C']:
for rank in ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']:
total.add(rank + typ)
grid_set = set()
for i in range(n):
for j in range(m):
grid_set.add(grid[i][j])
r = total.difference(grid_set)
r = list(r)
return r
def replace(cgrid, x, y, item):
cgrid[x][y] = item
def first_condition(grid, x, y):
suit = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
suit.add(grid[i][j][1])
if len(suit) == 1:
return True
else:
return False
def second_condition(grid, x, y):
rank = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
rank.add(grid[i][j][0])
if len(rank) == 9:
return True
else:
return False
def check_mark(mark, x, y):
for i in range(x, x + 3):
for j in range(y, y + 3):
if mark[i][j] == True:
return False
else:
return True
def make_mark(mark, x, y):
for i in range(x, x + 3):
for j in range(y, y + 3):
mark[i][j] = True
def check(grid, n, m):
count = 0
mark = [[False for x in range(m)] for y in range(n)]
for i in range(n):
if i + 3 <= n:
for j in range(m):
if j + 3 <= m:
if check_mark(mark, i, j):
if first_condition(grid, i, j) or second_condition(grid, i, j):
count += 1
make_mark(mark, i, j)
#print(mark)
if count >= 2:
return True
else:
return False
def get_ans(grid, n, m):
ans = []
mark = [[False for x in range(m)] for y in range(n)]
for i in range(n):
if i + 3 <= n:
for j in range(m):
if j + 3 <= m:
if check_mark(mark, i, j):
if first_condition(grid, i, j) or second_condition(grid, i, j):
ans.append([i, j])
make_mark(mark, i, j)
return ans
# main starts
n, m = list(map(int, stdin.readline().split()))
grid = []
for _ in range(n):
grid.append(list(stdin.readline().split()))
jokers = find_jokers(grid, n, m)
remaining = get_remain(grid, n, m)
#print(remaining)
if len(jokers) == 2:
for i in range(len(remaining) - 1):
for j in range(i + 1, len(remaining)):
cgrid = deepcopy(grid)
fx, fy = jokers[0]
sx, sy = jokers[1]
replace(cgrid, fx, fy, remaining[i])
replace(cgrid, sx, sy, remaining[j])
if check(cgrid, n, m):
print('Solution exists.')
print('Replace J1 with ' + str(remaining[i])+ ' and J2 with '+ str(remaining[j])+ '.')
ans = get_ans(cgrid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
cgrid = deepcopy(grid)
replace(cgrid, sx, sy, remaining[i])
replace(cgrid, fx, fy, remaining[j])
if check(cgrid, n, m, ):
print('Solution exists.')
print('Replace J1 with ' + str(remaining[j])+ ' and J2 with '+ str(remaining[i])+ '.')
ans = get_ans(cgrid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
pass
elif len(jokers) == 1:
for i in range(len(remaining)):
cgrid = deepcopy(grid)
fx, fy = jokers[0]
replace(cgrid, fx, fy, remaining[i])
if check(cgrid, n, m):
print('Solution exists.')
print('Replace '+ str(grid[fx][fy]) +' with ' + str(remaining[i]) +'.')
ans = get_ans(cgrid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
continue
else:
if check(grid, n, m):
print('Solution exists.')
print("There are no jokers.")
ans = get_ans(grid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
pass
print("No solution.")
``` | instruction | 0 | 99,128 | 19 | 198,256 |
Yes | output | 1 | 99,128 | 19 | 198,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
suits = ['C', 'D', 'H', 'S', 'W']
ranks = ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']
def parse(c):
if c == 'J1':
return 52
if c == 'J2':
return 53
return 13 * suits.index(c[1]) + ranks.index(c[0])
def summarize(cards):
s = [0] * 4
r = [0] * 13
j1, j2 = False, False
for i in cards:
if i < 52:
s[i // 13] += 1
r[i % 13] += 1
elif i == 52:
j1 = True
else:
j2 = True
return s, r, j1, j2
n, m = map(int, input().split())
board = [
[parse(c) for c in input().split()]
for _ in range(n)
]
pack = set(range(54))
for r in board:
pack -= set(r)
ps, pr, pj1, pj2 = summarize(pack)
jc = [[0] * (m - 2) for _ in range(n - 2)]
ji = [[False] * (m - 2) for _ in range(n - 2)]
valid = [[False] * (m - 2) for _ in range(n - 2)]
subs = [[None] * (m - 2) for _ in range(n - 2)]
for ni in range(n - 2):
for mi in range(m - 2):
ss, sr, sj1, sj2 = summarize(
board[ni + nj][mi + mj] for nj in range(3) for mj in range(3))
if not sj1 and not sj2:
if max(ss) == 9 or max(sr) == 1:
valid[ni][mi] = True
elif sj1 and sj2:
jc[ni][mi] = 2
if max(ss) == 7:
ssi = ss.index(7)
for pi in pack:
if pi // 13 != ssi: continue
for pj in pack:
if pj == pi: continue
if pj // 13 != ssi: continue
subs[ni][mi] = [pi, pj]
valid[ni][mi] = True
if max(sr) == 1:
for pi in pack:
if sr[pi % 13] == 1: continue
for pj in pack:
if pj % 13 == pi % 13: continue
if sr[pj % 13] == 1: continue
subs[ni][mi] = [pi, pj]
valid[ni][mi] = True
else:
jc[ni][mi] = 1
ji[ni][mi] = 0 if sj1 else 1
pc = set()
if max(ss) == 8:
ssi = ss.index(8)
for p in pack:
if p // 13 == ssi:
pc.add(p)
if max(sr) == 1:
for p in pack:
if sr[p % 13] == 0:
pc.add(p)
if len(pc) > 0:
valid[ni][mi] = True
subs[ni][mi] = pc
def solve():
for ni0 in range(n - 2):
for mi0 in range(m - 2):
if not valid[ni0][mi0]: continue
for ni1 in range(n - 2):
for mi1 in range(m - 2):
if not valid[ni1][mi1]: continue
if (((ni1 >= ni0 and ni1 <= ni0 + 2)
or (ni0 >= ni1 and ni0 <= ni1 + 2))
and ((mi1 >= mi0 and mi1 <= mi0 + 2)
or (mi0 >= mi1 and mi0 <= mi1 + 2))):
continue
ja = [None, None]
if jc[ni0][mi0] == 0 and jc[ni1][mi1] == 0:
return True, (0, 0), ja, ((ni0, mi0), (ni1, mi1))
elif jc[ni0][mi0] == 1 and jc[ni1][mi1] == 1:
s0 = list(subs[ni0][mi0])
s1 = list(subs[ni1][mi1])
if len(s1) == 1 and s1[0] in s0:
s0.remove(s1[0])
if len(s0) == 0:
continue
ja[ji[ni0][mi0]] = s0[0]
if s0[0] in s1:
s1.remove(s0[0])
ja[ji[ni1][mi1]] = s1[0]
return True, (1, 1), ja, ((ni0, mi0), (ni1, mi1))
elif jc[ni0][mi0] == 2:
ja = subs[ni0][mi0]
elif jc[ni1][mi1] == 2:
ja = subs[ni1][mi1]
elif jc[ni0][mi0] == 1:
ja[ji[ni0][mi0]] = list(subs[ni0][mi0])[0]
else:
ja[ji[ni1][mi1]] = list(subs[ni1][mi1])[0]
return True, (jc[ni0][mi0], jc[ni1][mi1]), ja, ((ni0, mi0), (ni1, mi1))
return False, 0, [None, None], None
v, jc, j, c = solve()
if 52 in pack:
pack.remove(52)
if 53 in pack:
pack.remove(53)
if j[0] is not None:
pack.remove(j[0])
if j[1] is not None:
pack.remove(j[1])
if not pj1 and j[0] is None:
j[0] = pack.pop()
if not pj2 and j[1] is None:
j[1] = pack.pop()
for i in range(2):
if j[i] is not None:
j[i] = ranks[j[i] % 13] + suits[j[i] // 13]
if not v:
print('No solution.')
else:
print('Solution exists.')
if pj1 and pj2:
print('There are no jokers.')
elif not pj1 and not pj2:
print('Replace J1 with {} and J2 with {}.'.format(j[0], j[1]))
elif not pj1:
print('Replace J1 with {}.'.format(j[0]))
else:
print('Replace J2 with {}.'.format(j[1]))
print('Put the first square to ({}, {}).'.format(c[0][0]+1, c[0][1]+1))
print('Put the second square to ({}, {}).'.format(c[1][0]+1, c[1][1]+1))
``` | instruction | 0 | 99,129 | 19 | 198,258 |
Yes | output | 1 | 99,129 | 19 | 198,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
suits = ['C', 'D', 'H', 'S', 'W']
ranks = ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']
def parse(c):
if c == 'J1':
return 52
if c == 'J2':
return 53
return 13 * suits.index(c[1]) + ranks.index(c[0])
def summarize(cards):
s = [0] * 4
r = [0] * 13
j1, j2 = False, False
for i in cards:
if i < 52:
s[i // 13] += 1
r[i % 13] += 1
elif i == 52:
j1 = True
else:
j2 = True
return s, r, j1, j2
n, m = map(int, input().split())
board = [
[parse(c) for c in input().split()]
for _ in range(n)
]
pack = set(range(54))
for r in board:
pack -= set(r)
ps, pr, pj1, pj2 = summarize(pack)
jc = [[0] * (m - 2) for _ in range(n - 2)]
ji = [[False] * (m - 2) for _ in range(n - 2)]
valid = [[False] * (m - 2) for _ in range(n - 2)]
subs = [[None] * (m - 2) for _ in range(n - 2)]
for ni in range(n - 2):
for mi in range(m - 2):
ss, sr, sj1, sj2 = summarize(
board[ni + nj][mi + mj] for nj in range(3) for mj in range(3))
if not sj1 and not sj2:
if max(ss) == 9 or max(sr) == 1:
valid[ni][mi] = True
elif sj1 and sj2:
jc[ni][mi] = 2
if max(ss) == 7:
ssi = ss.index(7)
for pi in pack:
if pi // 13 != ssi: continue
for pj in pack:
if pj == pi: continue
if pj // 13 != ssi: continue
subs[ni][mi] = [pi, pj]
valid[ni][mi] = True
if max(sr) == 1:
for pi in pack:
if sr[pi % 13] == 1: continue
for pj in pack:
if pj % 13 == pi % 13: continue
subs[ni][mi] = [pi, pj]
valid[ni][mi] = True
else:
jc[ni][mi] = 1
ji[ni][mi] = 0 if sj1 else 1
pc = set()
if max(ss) == 8:
ssi = ss.index(8)
for p in pack:
if p // 13 == ssi:
pc.add(p)
if max(sr) == 1:
for p in pack:
if sr[p % 13] == 0:
pc.add(p)
if len(pc) > 0:
valid[ni][mi] = True
subs[ni][mi] = pc
def solve():
for ni0 in range(n - 2):
for mi0 in range(m - 2):
if not valid[ni0][mi0]: continue
for ni1 in range(n - 2):
for mi1 in range(m - 2):
if not valid[ni1][mi1]: continue
if (((ni1 >= ni0 and ni1 < ni0 + 2)
or (ni0 >= ni1 and ni0 < ni1 + 2))
and ((mi1 >= mi0 and mi1 < mi0 + 2)
or (mi0 >= mi1 and mi0 < mi1 + 2))):
continue
ja = [None, None]
if jc[ni0][mi0] == 0 and jc[ni1][mi1] == 0:
return True, (0, 0), ja, ((ni0, mi0), (ni1, mi1))
elif jc[ni0][mi0] == 1 and jc[ni1][mi1] == 1:
s0 = list(subs[ni0][mi0])
s1 = list(subs[ni1][mi1])
if len(s1) == 1 and s1[0] in s0:
s0.remove(s1[0])
if len(s0) == 0:
continue
ja[ji[ni0][mi0]] = s0[0]
if s0[0] in s1:
s1.remove(s0[0])
ja[ji[ni1][mi1]] = s1[0]
return True, (1, 1), ja, ((ni0, mi0), (ni1, mi1))
elif jc[ni0][mi0] == 2:
ja = subs[ni0][mi0]
elif jc[ni1][mi1] == 2:
ja = subs[ni1][mi1]
elif jc[ni0][mi0] == 1:
ja[ji[ni0][mi0]] = list(subs[ni0][mi0])[0]
else:
ja[ji[ni1][mi1]] = list(subs[ni1][mi1])[0]
return True, (jc[ni0][mi0], jc[ni1][mi1]), ja, ((ni0, mi0), (ni1, mi1))
return False, None, None, None
v, jc, j, c = solve()
if j is not None:
for i in range(2):
if j[i] is not None:
j[i] = ranks[j[i] % 13] + suits[j[i] // 13]
if not v:
print('No solution.')
else:
print('Solution exists.')
if pj1 and pj2:
print('There are no jokers.')
elif not pj1 and not pj2:
print('Replace J1 with {} and J2 with {}.'.format(j[0], j[1]))
elif not pj1:
print('Replace J1 with {}.'.format(j[0]))
else:
print('Replace J2 with {}.'.format(j[1]))
print('Put the first square to ({}, {}).'.format(c[0][0]+1, c[0][1]+1))
print('Put the second square to ({}, {}).'.format(c[1][0]+1, c[1][1]+1))
``` | instruction | 0 | 99,130 | 19 | 198,260 |
No | output | 1 | 99,130 | 19 | 198,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
def init_cards():
result = {}
for suit in ('C', 'D', 'H', 'S'):
for price in ('2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A'):
result[price + suit] = True
return result
def main(stdin):
cards = init_cards()
user_cards = []
n, m = next(stdin).split()
n, m = int(n), int(m)
for row in stdin:
cards_row = []
for card in row.split():
cards_row.append(card)
if card in cards:
del cards[card]
user_cards.append(cards_row)
founded = []
jockers = set()
row = 0
while row + 3 <= n:
col = 0
if len(founded) == 1:
if row < founded[0][0] + 2:
col = founded[0][1] + 2
while col + 3 <= m:
check_price = set()
for i in range(row, row + 3):
for j in range(col, col + 3):
if user_cards[i][j] in ('J1', 'J2'):
check_price.add(user_cards[i][j])
jockers.add(user_cards[i][j])
else:
check_price.add(user_cards[i][j][0])
if len(check_price) == 9:
founded.append((row + 1, col + 1))
col += 3
break
else:
col += 1
row += 1
jockers_founded = []
for i, jocker in enumerate(jockers):
item = cards.popitem()
jockers_founded.append((jocker, item[0]))
if len(founded) > 1:
print('Solution exists.')
if len(jockers_founded):
jockers_founded.sort(key=lambda x: x[0])
print('Replace ' + ' and '.join('%s with %s' % f for f in jockers_founded) + '.')
else:
print('There are no jokers.')
print('Put the first square to %s.' % repr(founded[0]))
print('Put the second square to %s.' % repr(founded[1]))
else:
print('No solution.')
if __name__ == '__main__':
import sys
main(sys.stdin)
``` | instruction | 0 | 99,131 | 19 | 198,262 |
No | output | 1 | 99,131 | 19 | 198,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
def init_cards():
result = {}
for suit in ('C', 'D', 'H', 'S'):
for price in ('2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A'):
result[price + suit] = True
return result
def main(stdin):
cards = init_cards()
user_cards = []
n, m = next(stdin).split()
n, m = int(n), int(m)
for row in stdin:
cards_row = []
for card in row.split():
cards_row.append(card)
if card in cards:
del cards[card]
user_cards.append(cards_row)
founded = []
jockers_founded = []
row = 0
while row + 3 <= n:
if len(founded) > 2:
break
col = 0
if len(founded) == 1:
if row < founded[0][0] + 2:
col = founded[0][1] + 2
while col + 3 <= m:
check_price = set()
jockers = []
for i in range(row, row + 3):
for j in range(col, col + 3):
if user_cards[i][j] in ('J1', 'J2'):
jockers.append(user_cards[i][j])
else:
check_price.add(user_cards[i][j][0])
for i, jocker in enumerate(jockers):
for card in cards:
if card[0] not in check_price:
check_price.add(card[0])
jockers[i] = (jocker, card)
del cards[card]
break
if len(check_price) == 9:
founded.append((row + 1, col + 1))
for jocker in jockers:
jockers_founded.append(jocker)
col += 3
break
else:
col += 1
row += 1
if len(founded) > 1:
print('Solution exists.')
if len(jockers_founded):
jockers_founded.sort(key=lambda x: x[0])
print('Replace ' + ' and '.join('%s with %s' % f for f in jockers_founded) + '.')
else:
print('There are no jokers.')
print('Put the first square to %s.' % repr(founded[0]))
print('Put the second square to %s.' % repr(founded[1]))
else:
print('No solution.')
if __name__ == '__main__':
import sys
main(sys.stdin)
``` | instruction | 0 | 99,132 | 19 | 198,264 |
No | output | 1 | 99,132 | 19 | 198,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pack of 54 cards (52 standard cards and 2 distinct jokers). That is all he has at the moment. Not to die from boredom, Vasya plays Solitaire with them.
Vasya lays out nm cards as a rectangle n × m. If there are jokers among them, then Vasya should change them with some of the rest of 54 - nm cards (which are not layed out) so that there were no jokers left. Vasya can pick the cards to replace the jokers arbitrarily. Remember, that each card presents in pack exactly once (i. e. in a single copy). Vasya tries to perform the replacements so that the solitaire was solved.
Vasya thinks that the solitaire is solved if after the jokers are replaced, there exist two non-overlapping squares 3 × 3, inside each of which all the cards either have the same suit, or pairwise different ranks.
Determine by the initial position whether the solitaire can be solved or not. If it can be solved, show the way in which it is possible.
Input
The first line contains integers n and m (3 ≤ n, m ≤ 17, n × m ≤ 52). Next n lines contain m words each. Each word consists of two letters. The jokers are defined as "J1" and "J2" correspondingly. For the rest of the cards, the first letter stands for the rank and the second one — for the suit. The possible ranks are: "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K" and "A". The possible suits are: "C", "D", "H" and "S". All the cards are different.
Output
If the Solitaire can be solved, print on the first line "Solution exists." without the quotes. On the second line print in what way the jokers can be replaced. Three variants are possible:
* "There are no jokers.", if there are no jokers in the input data.
* "Replace Jx with y.", if there is one joker. x is its number, and y is the card it should be replaced with.
* "Replace J1 with x and J2 with y.", if both jokers are present in the input data. x and y here represent distinct cards with which one should replace the first and the second jokers correspondingly.
On the third line print the coordinates of the upper left corner of the first square 3 × 3 in the format "Put the first square to (r, c).", where r and c are the row and the column correspondingly. In the same manner print on the fourth line the coordinates of the second square 3 × 3 in the format "Put the second square to (r, c).".
If there are several solutions to that problem, print any of them.
If there are no solutions, print of the single line "No solution." without the quotes.
See the samples to understand the output format better.
Examples
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H 5S TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
No solution.
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H J1 TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
Replace J1 with 2H.
Put the first square to (1, 1).
Put the second square to (2, 4).
Input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
Output
Solution exists.
There are no jokers.
Put the first square to (1, 1).
Put the second square to (2, 4).
Note
The pretests cover all the possible output formats.
Submitted Solution:
```
'''input
4 6
2S 3S 4S 7S 8S AS
5H 6H 7H QC TC AC
8H 9H TH 7C 8C 9C
2D 2C 3C 4C 5C 6C
'''
from sys import stdin
import math
from copy import deepcopy
def find_jokers(grid, n, m):
jokers = []
for i in range(n):
for j in range(m):
if grid[i][j] == 'J1' and len(jokers) > 0:
jokers.insert(0, [i, j])
elif (grid[i][j] == 'J1' or grid[i][j] == 'J2'):
jokers.append([i, j])
return jokers
def get_remain(grid, n, m):
total = set()
for typ in ['D', 'S', 'H', 'C']:
for rank in ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']:
total.add(rank + typ)
grid_set = set()
for i in range(n):
for j in range(m):
grid_set.add(grid[i][j])
r = total.difference(grid_set)
r = list(r)
return r
def replace(cgrid, x, y, item):
cgrid[x][y] = item
def first_condition(grid, x, y):
suit = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
suit.add(grid[i][j][1])
if len(suit) == 1:
return True
else:
return False
def second_condition(grid, x, y):
rank = set()
for i in range(x, x + 3):
for j in range(y, y + 3):
rank.add(grid[i][j][0])
if len(rank) == 9:
return True
else:
return False
def check_mark(mark, x, y):
for i in range(x, x + 3):
for j in range(y, y + 3):
if mark[i][j] == True:
return False
else:
return True
def make_mark(mark, x, y):
for i in range(x, x + 3):
for j in range(y, y + 3):
mark[i][j] == True
def check(grid, n, m):
count = 0
mark = [[False for x in range(m)] for y in range(n)]
for i in range(n):
if i + 3 <= n:
for j in range(m):
if j + 3 <= m:
if check_mark(mark, i, j):
if first_condition(grid, i, j) or second_condition(grid, i, j):
count += 1
make_mark(mark, i, j)
if count >= 2:
return True
else:
return False
def get_ans(grid, n, m):
ans = []
mark = [[False for x in range(m)] for y in range(n)]
for i in range(n):
if i + 3 <= n:
for j in range(m):
if j + 3 <= m:
if check_mark(mark, i, j):
if first_condition(grid, i, j) or second_condition(grid, i, j):
ans.append([i, j])
return ans
# main starts
n, m = list(map(int, stdin.readline().split()))
grid = []
for _ in range(n):
grid.append(list(stdin.readline().split()))
jokers = find_jokers(grid, n, m)
remaining = get_remain(grid, n, m)
#print(remaining)
if len(jokers) == 2:
for i in range(len(remaining) - 1):
for j in range(i + 1, len(remaining)):
cgrid = deepcopy(grid)
fx, fy = jokers[0]
sx, sy = jokers[1]
replace(cgrid, fx, fy, remaining[i])
replace(cgrid, sx, sy, remaining[j])
if check(cgrid, n, m):
print('Solution exists.')
print('Replace J1 with ' + str(remaining[i])+ ' and J2 with '+ str(remaining[j])+ '.')
ans = get_ans(grid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
cgrid = deepcopy(grid)
replace(cgrid, sx, sy, remaining[i])
replace(cgrid, fx, fy, remaining[j])
if check(cgrid, n, m, ):
print('Solution exists.')
print('Replace J1 with ' + str(remaining[i])+ ' and J2 with '+ str(remaining[j])+ '.')
ans = get_ans(grid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
pass
elif len(jokers) == 1:
for i in range(len(remaining)):
cgrid = deepcopy(grid)
fx, fy = jokers[0]
replace(cgrid, fx, fy, remaining[i])
if check(grid, n, m):
print('Solution exists.')
print('Replace J1 with ' + str(remaining[i]) +'.')
ans = get_ans(grid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
continue
else:
if check(grid, n, m):
print('Solution exists.')
print("There are no jokers.")
ans = get_ans(grid, n, m)
print('Put the first square to ' + '(' + str(ans[0][0] + 1) + ', ' + str(ans[0][1] + 1) + ').')
print('Put the second square to ' + '(' + str(ans[1][0] + 1) + ', ' + str(ans[1][1] + 1) + ').')
exit()
else:
pass
print("No solution.")
``` | instruction | 0 | 99,133 | 19 | 198,266 |
No | output | 1 | 99,133 | 19 | 198,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,239 | 19 | 198,478 |
Tags: math, probabilities
Correct Solution:
```
from math import gcd
v=list(map(int,input().split()))
a=max(v)
x=6-a+1
z=6
d=gcd(x,z)
x=x//d
z=z//d
print(str(x)+'/'+str(z))
``` | output | 1 | 99,239 | 19 | 198,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,240 | 19 | 198,480 |
Tags: math, probabilities
Correct Solution:
```
import math
a = max(map(int, input().split()))
num = 7 - a
den = 6
g = math.gcd(num, den)
print(str(num // g) + '/' + str(den // g))
``` | output | 1 | 99,240 | 19 | 198,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,241 | 19 | 198,482 |
Tags: math, probabilities
Correct Solution:
```
from fractions import Fraction
a_b = input()
ab = a_b.split()
ab = list(map(int,ab))
a,b = ab
m = max(a,b)
ans = Fraction(6-m+1,6)
if ans == 1:
print('1/1')
elif ans == 0:
print('0/1')
else:
print(ans)
``` | output | 1 | 99,241 | 19 | 198,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,242 | 19 | 198,484 |
Tags: math, probabilities
Correct Solution:
```
from fractions import Fraction
inp = list(map(int, input().rstrip().split()))
if max(inp) == 0:
print('0/1')
else:
prob = (6 - max(inp) + 1) / 6
prob = Fraction(prob).limit_denominator(6)
if prob == 1:
print('1/1')
elif prob <= 0:
print('0/1')
else:
print(prob)
``` | output | 1 | 99,242 | 19 | 198,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,243 | 19 | 198,486 |
Tags: math, probabilities
Correct Solution:
```
import fractions
while(1):
try:
y,w=map(int,input().split())
maxx=max(y,w)
prob=6-maxx+1
g=fractions.gcd(prob,6)
prob=prob//g
print(prob,end="/")
print(6//g)
except EOFError:
break
``` | output | 1 | 99,243 | 19 | 198,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,244 | 19 | 198,488 |
Tags: math, probabilities
Correct Solution:
```
a , b = map(int,input().split())
max1 = max(a,b)
k = 6-(max1-1)
if k==1:print("1"+"/"+"6")
elif k==2:print("1"+"/"+"3")
elif k==3:print("1"+"/"+"2")
elif k==4:print("2"+"/"+"3")
elif k==5:print("5"+"/"+"6")
elif k==6:print("1"+"/"+"1")
``` | output | 1 | 99,244 | 19 | 198,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,245 | 19 | 198,490 |
Tags: math, probabilities
Correct Solution:
```
from fractions import Fraction
x = list(map(lambda y: int(y), input().split(" ")))
y = 7 - max(x)
if y == 6:
print("1/1")
else:
print(Fraction(y, 6))
``` | output | 1 | 99,245 | 19 | 198,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | instruction | 0 | 99,246 | 19 | 198,492 |
Tags: math, probabilities
Correct Solution:
```
n,m=input().split()
n=int(n)
m=int(m)
if n>=m:
a=0
for i in range(1,7):
if n<=i:
a=a+1
else:
a=0
for i in range(1,7):
if m<=i:
a=a+1
if a==1:
print(str(1)+"/"+str(6))
if a==2:
print(str(1)+"/"+str(3))
if a==3:
print(str(1)+"/"+str(2))
if a==4:
print(str(2)+"/"+str(3))
if a==5:
print(str(5)+"/"+str(6))
if a==6:
print(str(1)+"/"+str(1))
``` | output | 1 | 99,246 | 19 | 198,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
x , y= map(int,input().split())
if x == 1 and y ==1:
print('1/1')
elif (6-max(x,y)+1) %2 ==0:
print(str(int((6-max(x,y)+1)/2))+'/'+str(int(6/2)))
elif (6-max(x,y)+1)%3 ==0:
print('1/2')
else:
print(str((6-max(x,y)+1))+'/'+'6')
``` | instruction | 0 | 99,247 | 19 | 198,494 |
Yes | output | 1 | 99,247 | 19 | 198,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
def sr(ch):
ch1=ch+' '
l=[]
p=''
for i in ch1:
if i!=' ':
p=p+i
else:
l.append(int(p))
p=''
return l
def pgcd(x,y):
if y==0:
return x
else:
r=x%y
return pgcd(y,r)
ch=str(input())
l=sr(ch)
x=l[0]
y=l[1]
z=max(x,y)
def prob(x,y):
ch='/'
return str(int(x/pgcd(x,y)))+ch+str(int(y/pgcd(x,y)))
print(prob(6-z+1,6))
``` | instruction | 0 | 99,248 | 19 | 198,496 |
Yes | output | 1 | 99,248 | 19 | 198,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
from array import array # noqa: F401
from math import gcd
x, y = map(int, input().split())
n = 6 - max(x, y) + 1
d = 6
g = gcd(n, d)
print(f'{n//g}/{d//g}')
``` | instruction | 0 | 99,249 | 19 | 198,498 |
Yes | output | 1 | 99,249 | 19 | 198,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
a, b = map(int, input().split(' '))
a = max(a, b)
if a == 1:
a = '1/1'
elif a == 2:
a = '5/6'
elif a == 3:
a = '2/3'
elif a == 4:
a = '1/2'
elif a == 5:
a = '1/3'
elif a == 6:
a = '1/6'
print(a)
``` | instruction | 0 | 99,250 | 19 | 198,500 |
Yes | output | 1 | 99,250 | 19 | 198,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
print((7 - max([int(x) for x in input().split()]))/6)
``` | instruction | 0 | 99,251 | 19 | 198,502 |
No | output | 1 | 99,251 | 19 | 198,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
from fractions import Fraction
y, w = map(int, input().split())
mx = max(y, w)
ans = 7 - mx
print(Fraction(ans/6))
``` | instruction | 0 | 99,252 | 19 | 198,504 |
No | output | 1 | 99,252 | 19 | 198,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
temp = [int(x) for x in input().split()]
y = temp[0]
w = temp[1]
d = 6-(max(y, w)-1)
if d == 0 or d == 6:
print(str(d%5) + '/1')
elif d%2 == 0:
print(str(d//2) + '/3')
elif d == 5:
print('5/6')
else:
print('1/2')
``` | instruction | 0 | 99,253 | 19 | 198,506 |
No | output | 1 | 99,253 | 19 | 198,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Examples
Input
4 2
Output
1/2
Note
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
Submitted Solution:
```
from fractions import Fraction
Y,W=map(int,input().split())
if Y>W:
print(Fraction((6-(Y-1)),6))
if Y<W:
print(Fraction((6-(W-1)),6))
if Y==W:
print(Fraction((6-(W-1)),6))
``` | instruction | 0 | 99,254 | 19 | 198,508 |
No | output | 1 | 99,254 | 19 | 198,509 |
Provide a correct Python 3 solution for this coding contest problem.
Consider the following game:
* The game is played using a row of N squares and many stones.
* First, a_i stones are put in Square i\ (1 \leq i \leq N).
* A player can perform the following operation as many time as desired: "Select an integer i such that Square i contains exactly i stones. Remove all the stones from Square i, and add one stone to each of the i-1 squares from Square 1 to Square i-1."
* The final score of the player is the total number of the stones remaining in the squares.
For a sequence a of length N, let f(a) be the minimum score that can be obtained when the game is played on a.
Find the sum of f(a) over all sequences a of length N where each element is between 0 and K (inclusive). Since it can be extremely large, find the answer modulo 1000000007 (= 10^9+7).
Constraints
* 1 \leq N \leq 100
* 1 \leq K \leq N
Input
Input is given from Standard Input in the following format:
N K
Output
Print the sum of f(a) modulo 1000000007 (= 10^9+7).
Examples
Input
2 2
Output
10
Input
20 17
Output
983853488 | instruction | 0 | 99,332 | 19 | 198,664 |
"Correct Solution:
```
"""
https://atcoder.jp/contests/cf17-final/tasks/cf17_final_g
1回の操作で捨てられる石は1個
いくつ捨てられるかを考えた方がいいだろう
大きい方から?→そんなことはない
小さいほうから?→そんなことはない
今i=aiを満たしている中で最小のものからやるべき？→これはそう
数え上げなので…？ → X回とれる盤面は何通りあるか(1)
または、あるindexより右をX回選択できるindexより右側の数列は何通りか(2) → 左は自由
(2)の方がやりやすそう(Nはでかいとする)
→ 右端は1通り (Nのみ)
→ 右から2番目は 1回選択が N-1,A / 2回選択が N-2,N
→ 右から3番目は 1回選択が N-2,(0回選択)
右から順番に見てく？
100*100→10000個しか取れない
毎回何の数字か全探索したとしても 100^3で間に合う
取れる個数は100log100
3300 * 100 * 100 = 3.3*10^7
間に合いそう
"""
N,K = map(int,input().split())
mod = 10**9+7
lis = [0] * 3300
lis[0] = 1
for i in range(N,0,-1):
nlis = [0] * 3300
for j in range(K+1):
for last in range(3300):
if i < j:
nlis[last] += lis[last]
nlis[last] %= mod
elif (last+j)//i + last < 3300:
nlis[last+(last+j)//i] += lis[last]
nlis[last+(last+j)//i] %= mod
lis = nlis
#print (lis[:20])
ans = K * (K+1) // 2 * pow(K+1,N-1,mod) * N
#print (ans)
for i in range(3300):
ans -= lis[i] * i
ans %= mod
print (ans)
"""
#test
ans = 0
for i in range(100,0,-1):
ans += (ans//i) + 1
print (ans)
"""
``` | output | 1 | 99,332 | 19 | 198,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from math import gcd
def solve_case():
n, k = [int(x) for x in input().split()]
a = list(sorted([int(x) for x in input().split()]))
g = 0
for i in range(1, n):
g = gcd(g, a[i] - a[i - 1])
for i in range(n):
if (k - a[i]) % g == 0:
print('YES')
return
print('NO')
def main():
for _ in range(int(input())):
solve_case()
main()
``` | instruction | 0 | 99,670 | 19 | 199,340 |
Yes | output | 1 | 99,670 | 19 | 199,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
import math, sys
from functools import reduce
input = sys.stdin.readline
for _ in range(int(input())):
_, k = map(int, input().split())
a = list(map(int, input().split()))
print('NO' if (k - a[0]) % reduce(math.gcd, [v - a[0] for v in a[1:]]) else 'YES')
``` | instruction | 0 | 99,671 | 19 | 199,342 |
Yes | output | 1 | 99,671 | 19 | 199,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
def gcd(x, y):
if y == 0:
return x
return gcd(y, x % y)
t = int(input())
for i in range(t):
n, k = map(int, input().split())
xlist = list(map(int, input().split()))
lm = 0
for i in xlist:
lm = gcd(lm, i-xlist[0])
flag = False
for x in xlist:
if abs(x - k) % lm == 0:
print("YES")
flag = True
break
if not flag:
print("NO")
``` | instruction | 0 | 99,672 | 19 | 199,344 |
Yes | output | 1 | 99,672 | 19 | 199,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
def gcd(x, y):
while y != 0:
(x, y) = (y, x % y)
return x
def main():
t=int(input())
allAns=[]
for _ in range(t):
n,k=readIntArr()
a=readIntArr()
cumugcd=0
for i in range(1,n):
cumugcd=gcd(cumugcd,abs(a[i]-a[0]))
ans='NO'
for x in a:
if abs(k-x)%cumugcd==0:
ans='YES'
break
allAns.append(ans)
multiLineArrayPrint(allAns)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main()
``` | instruction | 0 | 99,673 | 19 | 199,346 |
Yes | output | 1 | 99,673 | 19 | 199,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
#1800
import sys
input = sys.stdin.readline
getint = lambda: int(input())
getints = lambda: [int(a) for a in input().split()]
def calc_gcd(big, small):
if small == 0:
return big
return calc_gcd(small, big % small)
def solve():
n, k = getints()
x = getints()
a = min(x)
gcd = x[0] - a
if gcd == 0:
gcd = x[1] - 1
for xvalue in x:
if xvalue == a:
continue
comparer = gcd, xvalue - a
gcd = calc_gcd(max(comparer), min(comparer))
k -= a
if k % gcd == 0:
print("YES")
else:
print("NO")
if __name__ == '__main__':
t = getint()
for i in range(t):
solve()
``` | instruction | 0 | 99,674 | 19 | 199,348 |
No | output | 1 | 99,674 | 19 | 199,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
from math import gcd
def solve(n,k,x):
if k in x:
return 'YES'
y = x[1]-x[0]
for i in range(1,n):
y = gcd(y,x[i]-x[0])
if k%y:
return 'NO'
else:
return 'YES'
def main():
for _ in range(int(input())):
n,k = map(int,input().split())
x = list(map(int,input().split()))
print(solve(n,k,x))
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | instruction | 0 | 99,675 | 19 | 199,350 |
No | output | 1 | 99,675 | 19 | 199,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board.
Submitted Solution:
```
cases=int(input())
inp=[]
for i in range(cases*2):
inp.append([int(x) for x in input().split(" ")])
print(inp)
for i in range(0,len(inp),2):
k=inp[i][1]
board=inp[i+1]
true=True if k in board else False
even=0
odd=0
for i in board:
if i>=0 and i%2==0:
even+=1
elif i>=0:
odd=1
if true:
print("Yes")
elif even>=1 and odd>=1:
print("Yes")
else:
print("No")
``` | instruction | 0 | 99,676 | 19 | 199,352 |
No | output | 1 | 99,676 | 19 | 199,353 |
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