欧拉计划 第六题

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.

前十个自然数的平方和是，

1 2 + 2 2 + ... + 10 2 = 385

前十个自然数之和的平方是，

（1 + 2 + ... + 10）2 = 55 2 = 3025

因此，前十个自然数的平方和与总和的平方之差为3025-385 = 2640。

找出前100个自然数的平方和与总和的平方之间的差

思路：

这道题很简单啦

利用for循环将i从1遍历到一百，遍历的过程中计算出和： sum1，平方的和：sum2，sum1*sum1和sum2做差即可求出

#include <stdio.h>

int main(){
	int sum1 = 0, sum2 = 0;
	for(int i = 1; i <= 100; ++i){
		sum1 += i;
		sum2 += i * i;
	}	
	printf("%d\n", sum1 * sum1 - sum2);
	
	return 0;
}