C++编程：根据公式e=1+1/1!+1/2!+1/3!+......+1/n!，求e的近似值，精确到最后一项小于10-5为止。咋写？

Question

如题所述

Answer 1

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;

}

Answer 2

#include <stdio.h>

int main()

{

    int i,t;

    double sum;

    for(i=0,t=1,sum=0;1.0/t>1e-6;t*=++i)

        sum+=1.0/t;

    printf("%lf\n",sum);

    return 0;

}