HDU-1012 u Calculate e

Time Limit : 2000/1000ms (Java/Other)
Memory Limit : 65536/32768K (Java/Other)
 

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

 
题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1012
 

分析

题意：公式已经给出，按照这个公署来算，输出前10个的结果就可以了，最多保留9位小数。
 
思路：水题，也没坑，注意前3个输出的结果就可以了。
 

代码

#include <iostream>
#include <iomanip>
#include <stdio.h>
using namespace std;
int main()
{
    cout<<"n e"<<endl<<"- -----------"<<endl;
    double sum=2.5;
    int n=2;
    for (int i=0;i<10;i++)
    {
        if (i==0)
        {
            cout<<"0 1"<<endl;
            continue;
        }
        if (i==1)
        {
            cout<<"1 2"<<endl;
            continue;
        }
        if (i==2)
        {
            cout<<"2 2.5"<<endl;
            continue;
        }
        n*=i;
        sum+=1.0/n;
        printf("%d %.9lf\n",i,sum);
    }
    return 0;
}