c语言计算1到1000之间能被13或20整除的所有自然数的倒数之和

Question

如题所述

Answer 1

#include<stdio.h>
void main()
{
double sum=0;
int n;
for(n=1;n<=1000;n++)
if(n%13==0 || n%20==0)
sum+=1.0/n;
printf("%lf\n",sum);
}

Answer 2

include<stdio.h>

int main()
{
int i;
float sum=0.;

for(i=13;i<=1000;i++)
if((i%13==0)||(i%20==0))
sum+=1.0/i;
printf("%.4f\n",sum);
return 0;
}