Can any one solve this problem (a Game) in c or c++ or any other language?

Question

A person has two piles of stones with him, one has n1 stones and the other has n2 stones. Fired up by boredom, he invented a game with the two piles.

1. Before the start of the game person chooses an integer m.
2. In the j-th move: He chooses a number xj such that 1 ≤ xj ≤ m, and removes xj stones from both the piles (this is only possible when both the piles have ≥ xj stones).
3. The number chosen must be unique over all the moves in the game. That is, for all k < j, xj ≠ xk.
4. The game stops when person is unable to make any more moves.
   Note: Person wants to make the moves in such a way that the sum of the number of stones remaining in the two piles is minimized.

Please help person find this.

Input
The first line of input contains an integer T denoting the number of test cases.
Each test case consists of 1 line with three integers — n1, n2 and m — separated by single spaces.

Output
For each test case, output a single line containing the minimum sum of the number of stones of two piles.

Answer 1

#include<stdio.h>
int main()
 {
   int t;
   long int m;
   long long int n1,n2;
   scanf("%d",&t);
   if(t<0)
     return 1;
   while(t--)
    {

      scanf("%lld%lld%ld",&n1,&n2,&m);
       if(n1<0 || n2<0 || m<0)
         return 1;
      label:

        if(m>n1 || m>n2)
        {
          if(n2>n1)
           printf("%lld\n",n1+n2-(2*n1));
          else
           printf("%lld\n",n1+n2-(2*n2));
          continue;
        }

          if(m==0 || (n1==0 || n2==0))
            {
              printf("%lld\n",(n1+n2));
            }  
           else
            {
             n1=n1-m;
             n2=n2-m;
             m--;
            goto label;
            }
    }  
   return 0;
 }