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### How to approach this question? Link- https://codeforces.com/problemset/problem/1036/A

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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.

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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
}```

Anupam_Singh last updated on April 7, 2019, 6:34 p.m.

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