Which test case am i missing?

I was solving a question. But I am missing something. There's some test case for which my solution is wrong. I need to find that test case.
For the other subtask, it is saying that execution took too long.

Chef will have access to Discourse if his knowledge and power become exactly equal to N and M respectively. Initially, he has power
1 and knowledge 1.

Chef can perform actions of the following types to improve his skills:

Chef can only install ShareChat at most once. The remaining actions may be performed any number of times and the actions may be
performed in any order.

Help Chef find out whether it is possible to move from Discuss to Discourse.

Input

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.

The first and only line of each test case contains four space-separated integers N, M, X and Y.

Output

For each test case, print a single line containing the string "Chefirnemo" if it is possible to reach the required knowledge and
power or "Pofik" if it is impossible

you can also visit this link to read the question
Link to question https://www.codechef.com/SEPT18B/problems/CHEFADV
My solution-

#include<stdio.h>
int main()

int t;
scanf("%d",&t);
while(t--)

long int n,m,x,y,power,know;
scanf("%ld%ld%ld%ld",&n,&m,&x,&y);
power=1;
know=1;
if(power+1==m && know+1==n)
printf("Chefirnemon");
else
while(power+y<m)
power=power+y;
if(power+y==m)
power=power+y;
while(know+x<n)
know=know+x;
if(know+x==n)
know=know+x;
if(power==m && know==n)
printf("Chefirnemon");
else if(power+1==m && know+1==n)
printf("Chefirnemon");
else
printf("Pofikn");


return 0;



%

C++

C

3 Answers
3

Your code fails if either x==1 and (m-1)%y==1 or vice versa (i.e. y==1 and (n-1)%x==1). Then your iteration will increase power to n and know to m-1 when installing ShareChat cannot help. However, had you increased power only to n-1, installing ShareChat would have obtained a match.

x==1

(m-1)%y==1

y==1

(n-1)%x==1

power

n

know

m-1

power

n-1

A correct solution should treat the installation of ShareChat more careful. For example

bool chefirnemo(int n, int m, int x, int y)

// 0 check that input is okay
assert(1<=n && 1<=m && 1<=x && 1<=y);
// 1 change to the equivalent problem where power
// and knowledge are 0 initially
--n; --m;
// 2 try w/o installation of ShareChat:
if (n>=0 && (n%x)==0 && m>=0 && (m%y)==0)
return true;
// 3 install ShareChat
--n; --m;
// 4 try again
return (n>=0 && (n%x)==0 && m>=0 && (m%y)==0);



Your rules don't guarantee that the inputs are positive. If X or Y is zero, your loops never terminate. (Note, this is a problem with your paraphrase, the challenge rules do assure positive values)

Test case: 2 2 0 0

Test case: 2 4 0 2

Test case: 4 4 1 0

How many times you apply (+X,0) and (0,+Y) should depend on whether the (+1,+1) action is used... but you have already locked in your decision.

Test case: 5 6 1 2

although Walter does point out the possible hints.
the solution does not seem to be working.

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