HDU 1012（u Calculate e）

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于 2019-01-23 14:55:52 发布

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分类专栏： HDU刷题集文章标签： ACM OJ 大学生程序设计大赛

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本文介绍了一种通过数学公式计算自然对数底e的近似值的方法，并提供了使用C++编程实现的具体代码示例，展示了如何计算从n=0到n=9时e的逐步逼近过程。

											u Calculate e
		  Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                     Total Submission(s): 56198    Accepted Submission(s): 25854

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<stdio.h>
using namespace std;
int main()
{
    long long int i,j,t=1;
    double s=2.5;
    cout<<"n e"<<endl;
    cout<<"- -----------"<<endl;
    cout<<"0 1"<<endl<<"1 2"<<endl<<"2 2.5"<<endl;
    for(i=3;i<=9;i++)
    {
        cout<<i<<' ';
        for(j=1;j<=i;j++)
            t=t*j;
        s=s+1.0/t;
        t=1;
        printf("%.9lf\n",s);
    }
}