:
运行过了
输出e=2.718282
不知是否满意
望采纳
#include
void
main(){
long
fun(int
n)
int
i
double
e=0
double
eps=1e-6//eps表示精度
此处指10的-6次方
for(i=01.0/fun(i)>epsi++)
{
e+=1.0/fun(i)
}
printf("e=%lf\n",e)
}
long
fun(int
n)//求n!的函数
{
if(n==0)
return
1
else
return
n*fun(n-1)
}
// 我先提供一种/* e = 1 + 1/1! + 1/2! + 1/3!+........1/n!+.... ... */#include <math.h>
#include <stdio.h>void main() {
double e = 1.0,delta
int factorial = 1,i = 1
do {
delta = 1.0/factorial
e = e + delta
i++
factorial = factorial*i
} while(fabs(delta) >1.0e-6)
printf("e = %lf\n",e)
}
#include "stdio.h"void main()
{
int k,j
long m
double e=0
for(k=0k++)
{
// k!
for(j=1,m=1j<=kj++)
{
m*=j
}
e+=1.0/m
if(1.0/m <0.000001)
break
}
printf("e=%lf",e)
printf("\npress any key to exit:\n")
getch()
}
运行结果:
e=2.718282
press any key to exit: