求C语言编程,e=1+1\/1!+1\/2!+1\/3!+...+1\/n!,精确度1e-6
int i=1;double e=1.0,d=1.0;while(d>=1e-6){ d=d\/i;e+=d;i++;} cout<<"e的值为:"<<e<<endl;return 0;}
用C语言编写程序计算:e=1+1\/1!+2\/2!+1\/3!+···
include "stdio.h"int main(int argv,char *argc[]){int n;double s,t;for(t=s=n=1;t>=1.0E-6;s+=t\/=n++);printf("e≈%f\\n",s);return 0; }运行结果:
c语言编程题,求e的近似值,e=1\/1!+1\/2!+1\/3!+...+1\/n!,累加项小于1
正确的公式为:e=1+1\/1!+1\/2!+1\/3!+...+1\/n!代码实现如下:include<stdio.h> int fun(int n){ if(n == 1)return 1;return n*fun(n-1);} int main(){ double sum =1.0 ;int i = 1;while((1.0\/fun(i))>=1e-8){ sum +=(1.0\/fun(i));i++;} printf("%.8...
C语言利用公式e=1+1\/1!+1\/2!+```+1\/n!计算e的值最后一项小于10的负6次...
\/\/e=1+1\/1!+1\/2!+...+1\/n!int main(){ int i = 1;double e=1.0, f = 1.0;while(f>1e-6) { f = f\/i;e += f;i++;} printf("%f",e);return 0;} \/\/答案:2.718282
C语言 编写程序求e的值 e=1+1\/1!+1\/2!+1\/3!+...
;getchar();return 0;} 或:public class Test{ public static void main(String args[]){ double sum=1,n=1;double e=1;for(n=1;n<=10;n++){ sum=sum*n;e+=1\/sum;} System.out.println("e="+e);} } 得到的值为:2.7182818011463845...随着n值越专大得到的e值越精属确。
c语言求e=1+1\/1!+1\/2!+1\/3!+1\/4!+...
double e = 1.0;\/\/声明变量e并初始化 for (i = 2; i <= n; i++)\/\/声明for循环 { e += 1.0 \/ jiecheng(i);\/\/e加上当前计算结果(即1\/1!,1\/2!等)} printf("%f", e);\/\/输出e return 0;\/\/结束程序 } 大部分代码我都做了注释,希望你能理解这个程序。希望能帮到您。
C语言编程: 根据公式e=1+1\/1!+1\/2!+1\/3!+…,求e的近似值,精度要求为...
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);}
C语言编写,求e的值.e≈1+1\/1!+1\/2!+1\/3!+……+1\/n!
int main(int argc, char* argv[]){ double e=1,t=1;int t1;for (t1=2;t1<=N;++t1){ t*=(double)1\/t1;e+=t;} printf("%lf",e);return 0;} \/\/--- 2.\/\/--- include <stdio.h> const double eps=1
C++编程:根据公式e=1+1\/1!+1\/2!+1\/3!+...+1\/n!,求e的近似值,精确到最后...
代码文本:\/\/#include "stdafx.h"\/\/vc++ 6.0? Maybe should add this line.include <iostream> using namespace std;int main(int argc,char *argv[]){ double e,t;int n;for(t=e=n=1;t>=1.0E-5;e+=t\/=n++);cout << "e≈" << e << endl;return 0;} ...
c 语言编程题:求e e=1+1\/1!+1\/2!+1\/3!+...
N取得越大,精度越高!\/***以下为代码***\/ int main(){ float e = 1;int i, j, den, N = 100000;for(i=1,i<N,i++){ den = 1;for(j=i,j>0,j--)den = den * j;e = e + 1\/den;} printf("%6.2f",&e);return 0;} ...