#include <stdio.h>
int main()
{
double i,j;
double sum1=0;
double sum2=0;
double sum;
for(i=1;i<=99;i=i+2)
{
sum1=sum1+1/i;
}
for(j=2;j<=100;j=j+2)
{
sum2=sum2+1/j;
}
sum=sum1-sum2;
printf("%lf\n",sum);
return 0;
}计算1/1-1/2+1/3-1/4+1/5 …… + 1/99 - 1/100 的值
原文：http://10740329.blog.51cto.com/10730329/1700630