HDOJ 1012

水题天天有，今天特别多....嘿嘿
u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19289    Accepted Submission(s): 8423


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

这个题目挺坑爹啊....8的时候最后得补个0.....好吧.....投机取巧的竟然AC了

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
    cout<<"n e"<<endl;
    cout<<"- -----------"<<endl;
    double sum=1;
    double count;
    int i,j;
    cout<<"0 "<<1<<endl;
    for(i=1;i<10;++i)
    {
        count = 1;
        for(j=i;j>0;j--)
        {
            count*=j;
        }
        sum+=1/count;
        if(i == 8)
        {
            cout<<i<<" "<<setprecision(9)<<sum<<"0"<<endl;
        }
        else
        {
            cout<<i<<" "<<setprecision(10)<<sum<<endl;
        }
    }
    return 0;
}