프로젝트 오일러 문제 6

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.

합의 제곱 - 제곱의 합 을 구하는 문제.

long long int Problem006(unsigned int number)
{
	long long int sum = SumUpToScope(number);
	long long int Squares = 0;
	for (unsigned int i = 1; i <= number; i++)
	{
		Squares += i * i;
	}
	return sum * sum - Squares;
}

int SumUpToScope(int scope)
{
	return ((1 + scope) * scope) / 2;
}