In this HackerEarth Code Apocalypse 3.0 ..Coming SOON ![Easy-Medium] problem solution Cypher guys are ready for the grand event Code Apocalypse!!.
Questions are prepared, Promotions just begun ( from this question, LOL! ). But there is one problem they are facing. What should they do if there is a tie on the final day.
To solve this dilemma, Ayush came up with a tie breaker problem (although this can't be disclosed), give a try to solve it :

You have an army of N unsullied soldiers, you can upgrade a soldier to be stronger by feeding him X magical apples (found beyond the wall!). Initially, you have M magical apples.
You can buy extra magical apples ( from the Iron Bank! :P ), by selling one of your soldiers for Y apples.
Calculate what is maximum number of upgraded soldiers you can have for the long night.


HackerEarth Code Apocalypse 3.0 ..Coming SOON ![Easy-Medium] problem solution


HackerEarth Code Apocalypse 3.0 ..Coming SOON ![Easy-Medium] problem solution.

#include<bits/stdc++.h>
#define ll long long int

using namespace std ;

int main()

{

ios_base::sync_with_stdio(false) ;

cin.tie(NULL) ;

cout.tie(NULL) ;

ll t ; cin >> t ;

while(t--)

{

ll n , m , x ,y ; cin >> n >> m >> x >> y ;

ll l = 0 , r = n , mid , ans ;


while( l <= r )

{

mid = l + (r-l)/2 ;




ll temp = m+mid*y ;

ll g = temp/x ;

if( g >= n-mid )

{

ans = mid ;

r = mid-1 ;



}

else{

l = mid+1 ;

}



}


cout << n-ans <<"\n" ;

}


return 0 ;

}