hdu 1012 u Calculate e

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45773    Accepted Submission(s): 21027

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

题目解析：

这个题的意思就是让你输出n从0到9公式求出的值，是一个非常简单的题，但是要注意输出。

代码：

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
    int i,j,jiecheng[10];
    double a[10];
    cout<<"n e"<<endl<<"- -----------"<<endl;
    for(i=0; i<=9; i++)
    {
        for(j=0; j<i; j++)
        {
            jiecheng[0]=1;
            jiecheng[i]=i*jiecheng[i-1];
        }
        a[0]=1;
        a[i]=a[i-1]+1.0/jiecheng[i];
        if(i==0) cout<<"0"<<" "<<"1"<<endl;
        if(i==1) cout<<"1"<<" "<<"2"<<endl;
        if(i==2) cout<<"2"<<" "<<"2.5"<<endl;
        if(i>=3)
        {
            cout<<i<<" ";
            printf("%.9lf\n",a[i]);
        }
    }
    return 0;
}