Project Euler Problem 6: Sum square difference

Sum square difference

Problem 6

The sum of the squares of the first ten natural numbers is,

1² + 2² + ... + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

C++:

#include <iostream>

using namespace std;

int main()
{
    int n, sum_sq, sum;

    while(cin >> n) {
        sum_sq = 0;
        sum = 0;

        for(int i=1; i<=n; i++) {
            sum_sq += i * i;
            sum += i;
        }

        cout << sum * sum - sum_sq << endl;
    }

    return 0;
}

C++:

#include <iostream>

using namespace std;

int main()
{
    int n, sum_sq, sum;

    while(cin >> n) {
        sum_sq = (2 * n + 1) * (n + 1) * n / 6;
        sum = n * (n + 1) / 2;

        cout << sum * sum - sum_sq << endl;
    }

    return 0;
}