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:

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=;
while(k--)
{
s*=n;
}
return s;
} int sum1(int n)
{
return powplus((n+)*n/,);
} int sum2(int n)
{
return (n*(n+)*(*n+))/;
} void solve()
{
printf("%d\n",sum1(N));
printf("%d\n",sum2(N));
printf("%d\n",sum1(N)-sum2(N));
} int main()
{
solve();
return ;
}
Answer:
25164150

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