[Project Euler]加入欧拉 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.

 

 

这个问题比较好解决有2个公式

一，平方和公式

平方和公式n(n+1)(2n+1)/6

　　即1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6 (注：n^2=n的平方)

二， 求和公式

1+2+..+n = (1+n)n/2

 

所以上面的就好做了

欧拉项目第六题#include <stdio.h>

#include <stdlib.h>



int main(int argc, char *argv[])

{

    int n = 100;

    unsigned long int sum1, sum2;

    

    sum1 = n * n * (n + 1) * (n + 1) / 4;

    

    sum2 = n * (n + 1) * (2*n + 1) /6;

    

    printf("%lu", sum1 - sum2);

  

  system("PAUSE");	

  return 0;

}