Die Hard 3 Hackerrank Solution

   The primary idea is to shift the water in the larger jug to smaller jug, hence the difference of  the capacities of the jugs will be used as a constraint to find out variable c volume. we will also terminate the process of finding c if the difference between both the jugs is zero.

     There are 4 conditions 
      1)    if  diff >= min then diff=diff-min_capacity
        2)    if  diff < min then diff=max_capacity - ( min_capacity - diff)
        3)    if  diff == c then return "YES"
          4)     if   diff  < 0 then return "NO"

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <iostream>
#include <algorithm>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
using namespace std;
int jug(int a,int b,int c)
{
       int mx=max(a,b);
        int mn=min(a,b);
         if(a==c || b==c)
             return 1;
        int diff=abs(b-a);
        while(diff > 0)
        {
            if( diff==c )
              return 1;
           
           
            if(diff < mn)
            {
                diff=mx-(mn-diff);
               
            } else if(diff >= mn)
                {
                    diff=diff-mn;
                }
        }
             
                
   
    return 0;
}

int main() {
     int a,b,c,mx,mn,diff,t;
      cin >> t;
      while(t--)
       {  
       cin >>a>>b>>c;
          if(jug(a,b,c)==1)
              printf("YES\n");
          else printf("NO\n");
     
      }
   
    return 0;
}