C语言：求求e=1/1!+1/2!+1/3!+....+1/n!

#include<stdio.h>
// 求e=1/1!+1/2!+1/3!+....+1/n! 
main()
{
    double e=1,a,s=1;
    int i;
    for(i=1;i<=20;i++)
    {
        s*=i;
        a=1/s;
        e+=a;
    }
    printf("%f",e);
}