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.

#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 ;
}

Friday, November 26, 2021