#include<stdio.h>
#include<math.h>
int main()
{
double i=1.0,sum=0.0;
for(i=1.0;i<=100.0;i++)
{
sum=sum+(1/i)*pow(-1,i+1);
}
printf("%f\n",sum);
return 0;
}本文出自 “Vs吕小布” 博客,请务必保留此出处http://survive.blog.51cto.com/10728490/1700658
【C语言】求1-1/2+1/3-1/4+1/5....+1/99-1/100
原文地址:http://survive.blog.51cto.com/10728490/1700658