根据公式e=1+1/1!+1/2!+..1/n 求e的近似值，用python来求。

根据公式e=1+1\/1!+1\/2!+..1\/n 求e的近似值,用python来求。。请大神帮 ...
sum += 1\/tmp print("e的近似值（精度为10**-6）为%.6f"%sum)输出结果是2.718282

根据公式e=1+1\/1!+1\/2!+..1\/n 求e的近似值,用python来求。
1、阶乘随着n的增大计算难度也增大，所以不能无限进行下去 2、代码如下（计算前10位）import math def e(accuracy): return sum(1.0\/math.factorial(i) for i in range(accuracy)) if __name__ == "__main__": print(e(10))

TC编程:用公式e=1+1\/1!+1\/2!+1\/3!+……+1\/n!,求e的近似值,直到1\/n...
int a=1,i;double b;for(i=1;i<=n;i++)a*=i;b=(double)1\/a;return b;} void main(){ double s=0;int i;for(i=1;Jie(i)>=1e-15;i++)s+=Jie(i);printf("%lf",s);} OK了

(填空)按照公式 e=1+1\/1!+1\/2!+1\/3!+…+1\/n! 求e的近似值,至n=10...
include <stdio.h>void main(){ float e=1.00; int n=1,i; for(i=1;i<=10;i++) { n=n*i; e=e+1\/(float)n; } printf("e=%f\\n",e); }你这个题目中的头文件应该是stdio.h写错了。题解：n用来存放阶乘，i用来自增，e用来求和；这里容易出错的地方就是 n是整型...

e=1+1\/1!+1\/2!+1\/3!+...1\/n! 编写程序用这个公式计算e的近似值这里是n...
include<stdio.h> int main(){ int n ,i,t=1; float e=1; scanf("%d", &n); for(i=1;i<=n;i++){ t*=i; e+=1.0\/t; } printf("%f", e); return 0; } 麻烦采纳，谢谢!

...1+1\/(1!)+1\/(2!)+1\/(3!)+…+1\/(n!)计算e的近似值
e=1+1\/1!+1\/2!+1\/3!+...C代码：include<stdio.h> void main(){ double e=1;double jc=1;\/\/求阶乘，并存入jc中 int i=1;while(1\/jc>=1e-6){ e=e+1\/jc;i++;jc=jc*i;} printf("e=%f\\n",e);} 》其他参考答案》》：http:\/\/zhidao.baidu.com\/question\/56549128.html?

...根据公式e=1+1\/1!+1\/2!+1\/3!+…,求e的近似值,精度要求为10^(-_百度...
include<stdio.h> int jc(int x){int i,s=1;for(i=1;i<=x;i++)s*=i;return s;} void main(){int i;double n=0;for(i=1;1.\/jc(i)>1e-6;i++)n+=1.\/jc(i);printf("%lf\\n",n+1);}

vb编程:用近似公式e=1\/1!+1\/2!+...+1\/n!计算自然对数的底e的近似值...
n = 2 Do While n <= 100 e = e + 1# \/ Factorial(n) n = n + 1 Loop Print eEnd SubPrivate Function Factorial(ByVal n As Integer) As Double Dim f As Double Dim k As Integer f = 1 For k = 1 To n f = f * k Next Factoria...

...下面公式,求自然对数e的近似值 e=1+1\/1!+1\/2!+1\/3!+...
\/\/C语言中，求e=1\/1!+1\/2!+…+1\/n!精确到10ˉ8#include <stdio.h>int main(void){ long n = 0, ns = 1; double x = 0.0f, y=0.0f, e = 1.0f; for(;;) { n++; \/*计算n*\/ ns *= n; \/*计算n!*\/ x = ns; y = 1.0f \/ x; \/*计算1\/n!*\/ ...

...项的值小于10^(-4)是结束运算.用公式e=1+1\/1!+1\/2!+1\/n
1,e应当定义为“！”;"%"是整型。2,e的初值应设置为 “e＝1"3,将 “ While s < 10000”改为 “ While s >10000 "