（Problem 6）Sum square difference

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.

 

题目大意：

前十个自然数的平方和是：1² + 2² + ... + 10² = 385

 

前十个自然数的和的平方是： (1 + 2 + ... + 10)² = 55² = 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