C语言利用公式e=1+ 1/1! + 1/2! + 1/3! + ... + 1/n!,编程计算e的近似值,

#include "stdio.h"#include "stdlib.h" int main(){ double d,e,n,jc; int i,m; char ws[32]; printf("输入精确度："); scanf("%lf",&d); jc=1; e=1; n=1; do { jc=jc*n; e+=1/jc; n+=1; }while((1/jc)>d); if(e>1) pr