计算e=1+1/1!+1/2!+1/3!+1/4!+......+1/n!(精度为1e-6)
#include "stdio.h"
void main()
{
double sum=1.0;
int m=1,i,n;
while (sum>=1e-3)
{
for(i=n=1;i<=m;i++) n*=i;
sum+=1.0/n;
m++;
}
printf("e=%f\n",sum);
}