hdu1012u Calculate e

最新推荐文章于 2019-03-18 20:48:45 发布

原创最新推荐文章于 2019-03-18 20:48:45 发布 · 267 阅读

0 ·

CC 4.0 BY-SA版权

HDU 专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种通过数学公式计算自然常数e的近似值的方法，并提供了C++代码实现，该程序能够输出n从0到9时e的近似值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43717 Accepted Submission(s): 20030

Problem Description
A simple mathematical formula for e is

where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output
n e

0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

#include <iostream>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdio>
#include <set>
using namespace std;

int main()
{
    int i;
    double m,sum;
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n1 2\n2 2.5\n");
    m=2;
    sum=2.5;
    for(i=3;i<=9;i++)
    {
        m*=i;
        sum+=1/m;
        printf("%d %.9lf\n",i,sum);
    }
    return 0;
}