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;
}