Memory (Editorial)

The worst-case scenario is when at most two modules are working. To find this pair of working modules or ensure that it does not exist, one must pick all \frac{n(n-1)}{2} possible pairs of modules.

C++ implementation:

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;
    cout << n * (n-1) / 2;
}

Python implementation:

n = int(input())
print (n * (n-1) // 2)