求e的近似值

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
    int n;
    double e=1.0,term=1.0;
    long fac=1;
    for(n=1;fabs(term)>=1e-5;n++){
        fac=fac*n;
        term=1.0/fac;
        e = e+term;
    }
    printf("e=%f,count=%d",e,n);
    return 0;
}