（Problem 6）Sum square difference

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

前十个自然数的和的平方是：

(1 + 2 + ... + 10) ² = 55 ² = 3025

所以平方和与和的平方的差是3025  385 = 2640.

找出前一百个自然数的平方和与和平方的差。

#include <stdio.h>

#include <string.h>

#include <ctype.h>

#include <math.h>

  

#define N 100

  

int powplus(int n, int k)

{

    int s=1;

    while(k--)

    {

       s*=n;

    }

  return s;

}

  

int sum1(int n)

{

   return  powplus((n+1)*n/2,2);

} 

  

int sum2(int n)

{

   return (n*(n+1)*(2*n+1))/6;

}

  

void solve()

{

     printf("%d\n",sum1(N));

     printf("%d\n",sum2(N));

     printf("%d\n",sum1(N)-sum2(N));

} 

  

int main()

{

  solve();

  return 0;

}

Answer:

25164150