# HackerEarth Number of triangles problem solution

In this HackerEarth Number of triangles problem solution you are given a polygon of N sides with vertices numbered from 1, 2, ..., N. Now, exactly 2 vertices of the polygons are colored black, and the remaining are colored white. You are required to find the number of triangles denoted by A such that:
1. The triangle is formed by joining only the white-colored vertices of the polygon.
2. The triangle shares at least one side with the polygon.

## HackerEarth Number of triangles problem solution.

`#include<bits/stdc++.h>using namespace std;int main() {        int t;    cin>>t;    while(t--) {                int n,b1,b2;        cin>>n>>b1>>b2;        long long one_side = 0;        long long two_side = 0;        int b1_one_side = n-4 + (2)*(n-4);        int b2_one_side = n-4 + (2)*(n-4);        int b1_b2_one_side = 0;        if((b1+1)%n==b2%n||(b2+1)%n==b1%n) {            b1_b2_one_side = n-4;        }        else if((b1+2)%n==b2%n||(b2+2)%n==b1%n) {            b1_b2_one_side = 2;        }        else b1_b2_one_side = 4;        one_side  = 1ll*n*(n-4) - b2_one_side - b1_one_side + b1_b2_one_side ;                int b1_two_side = 3;        int b2_two_side = 3;        int b1_b2_two_side = 0;        if((b1+1)%n==b2%n||(b2+1)%n==b1%n) {            b1_b2_two_side = 2;        }        else if((b1+2)%n==b2%n||(b2+2)%n==b1%n) {            b1_b2_two_side = 1;        }        two_side = n - b1_two_side - b2_two_side + b1_b2_two_side ;        cout<<one_side+two_side <<endl;    }    return 0;}`

### second solution

`#include <bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 1e3 + 4;int main(){    ios::sync_with_stdio(0), cin.tie(0);    int t;    cin >> t;    while(t--){        int n, b1, b2;        cin >> n >> b1 >> b2;        b1--, b2--;        ll ans = 0;        auto bad = [&](int i){            return i == b1 || i == b2;        };        for(int i = 0; i < n; i++){            if(!bad(i) && !bad((i + 1) % n))                ans += n - 4;            if(!bad(i) && !bad((i + 1) % n) && !bad((i + 2) % n))                ans--;        }        cout << ans << '\n';    }}`