Forest Gathering - CCDSAP (Binary_Search)

Chef is the head of commercial logging industry that recently bought a farm containing N trees. 
You are given initial height of the i-th tree by Hi and the rate of growth of height as Ri meters per month. 
For simplicity, you can assume that all the trees are perfect cylinders of equal radius. 
This allows us to consider only the height of trees when we talk about the amount of wood.

In Chef's country, laws don't allow one to cut a tree partially, so one has to cut the tree completely for gathering wood. 
Also, laws prohibit cutting trees of heights (strictly) lower than L meters.

Today Chef received an order of W meters (of height) of wood. 
Chef wants to deliver this order as soon as possible. 
Find out how minimum number of months he should wait after which he will able to fulfill the order. 
You can assume that Chef's company's sawing machines are very efficient and take negligible amount of time to cut the trees.

//  MULTIPLICATION OF LONG DOUBLE IS FAST AND IT REMOVES OVERFLOW ERROR FOR VALUES OF MULTIPLICATION OF VALUES 
INT ORDER OF 1e18

#include <bits/stdc++.h>
typedef long long  ll;
using namespace std;
long double h[100004],r[100004];
ll w,l;
int n;
int check(ll m)
{   long double sum=0;
    long double t=m;
    for(int i=0;i<n;i++)
    {
        if(t>=(l-h[i])/r[i])
        sum+=(h[i]+r[i]*m);
        if(sum>=w)return 1;
    }
    return (sum>=w);
}
int main() {
     ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> n >> w >> l;
    for(int i=0;i<n;i++)
    {
        cin >> h[i] >> r[i];
    }

    ll low=0;
    ll high=1e18,ans;
    while(low<=high)
    {
        ll mid=low+(high-low)/2;
        if(check(mid))
        {
            ans=mid;
            high=mid-1;
        }
        else
        {
            low=mid+1;
        }
    }
    cout<<ans<<'\n';
}