
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(h1)<k<=nh => h1<(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 }