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
}

conan_002:14, Dec 13Anupam SinghIt 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;

Please make sure the answer is not too short