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**HackerEarth Friendship and diseases problem solution**You want to arrange k people at a table. The table has n rows and m columns. Some of the cells are free and some of them are blocked.The score of the arrangement is based on two factors:

Friendship is always important, even with respect to disease. The ith person has a friendship value fi. A connected component of people with the sum of friendship value equal to s adds s*s to score. A connected component is a set of people adjacent to each other. Two people are adjacent if they share a side.

You need to take care of the health of people. The ith person has a danger value di. Two adjacent people i and j reduce the score by di x dj.

Your task is to print an arrangement with as high a score as possible. Note that you can drop a person and do not give him a seat at the table.

## HackerEarth Friendship and diseases problem solution.

`#include <bits/stdc++.h>`

using namespace std;

typedef long long ll;

const int MAX_N = 5e5 + 14, B = 25;

int a[MAX_N];

int main() {

ios::sync_with_stdio(0), cin.tie(0);

int t;

cin >> t;

while (t--) {

int n;

cin >> n;

for (int i = 0; i < n; ++i)

cin >> a[i];

ll ans = 0;

for (int i = 0; i < B; ++i) {

int l = 0;

while (l < n && (a[l] >> i & 1))

++l;

int r = n;

while (r > l && (a[r - 1] >> i & 1))

--r;

ans += (l == n ? n * ll(n + 1) / 2 - 1 : (l + 1) * ll(n - r + 1) - 1) * (1 << i);

}

cout << ans << 'n';

}

}