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Asked by: Shreyashkar_lal on April 7, 2019, 6:34 p.m. Last updated on April 7, 2019, 6:34 p.m.
It is given there are 2n+1 points so there will be n odd points and maximum n triangles . In worst situation if the height is h for all n triangles, max area = n*(2*h)/2=nh (base length is 2) . Given k should satisfy condition n(h-1)<k<=nh => h-1<(k/n)<=h . h is basically ceiling value of k/n.
Here is the code,
#include<bits/stdc++.h>
using namespace std;
int main()
{
ll n,k;
cin>>n>>k;
ll x=k/n;
if(n*x==k)
cout<<x;//equality condition k=nh
else
cout<<(x+1);//ceiling value
}