Author : Jahirul Islam Hridoy

• Understanding the Distribution:

  • Hridoy gives away his tickets first from his collection of A tickets.

  • Kamona gives away her tickets only after Hridoy has exhausted his.

A total of exactly X tickets are given away.

• Key Observations:

  • If X ≤ A, then Hridoy alone distributes the tickets.Hridoy’s remaining tickets: A − X

  • Kamona's remaining tickets: B (since he doesn't distribute any)If X > A, then Hridoy distributes all his A tickets, and Kamona gives away the remaining (X − A) tickets.

  • Hridoy’s remaining tickets: 0

  • Kamona’s remaining tickets: B − (X − A)

• Formulating the Answer:

  • Hridoy’s tickets left: max(A − X, 0)

  • Kamona’s tickets left: • If X ≤ A, then it remains B. • Otherwise, it is B − (X − A).

• Example: Suppose A = 5, B = 3, and X = 7.

  • Since 7 > 5, Hridoy gives away all 5 tickets, and Kamona gives away 7 − 5 = 2 tickets.

  • Remaining tickets: • Hridoy: 0 • Kamona: 3 − 2 = 1

Code (C++) :

// Author : Kamonasish Roy (Bullet)
// Time : 2025-03-27 03:53:47

#include<bits/stdc++.h>
using namespace std;
const long long M=5e5,MOD=1e9+7;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    cin>>t;
    while(t--){
    	int hridoy,kamona,x;
    	cin>>hridoy>>kamona>>x;
    	cout<<hridoy-min(x,hridoy)<<" "<<kamona-max(0,min(kamona,x-hridoy))<<"\n";
    	
    	
    	
    	}
    	
    	   
    return 0;
 
}