Skip to content
Programmingoneonone
Programmingoneonone

Learn everything about programming

  • Home
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone
Programmingoneonone

Learn everything about programming

HackerEarth Friendship and diseases problem solution

YASH PAL, 31 July 2024
In this 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

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';
}
}
coding problems solutions

Post navigation

Previous post
Next post

Pages

  • About US
  • Contact US
  • Privacy Policy

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2025 Programmingoneonone | WordPress Theme by SuperbThemes