u Calculate e

FJNU.1093
PKU.1517

Description
A simple mathematical formula for e is
e=Σ_0<=i<=n1/i!
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Input
No input

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 Input
No input

Sample Output
n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
...

Source
Greater New York 2000

My Program

#include<iostream>
#define N 10
using namespace std;

int main()
...{
    int i,j,m,k=1;
    double e=1.0;
    long n;
    cout<<"n e"<<endl;
    cout<<"- -----------"<<endl;
    for(i=0;i<N;i++)
    ...{
        cout<<i<<" ";
        if(i!=0)
        ...{
            k*=i;
            e+=1.0/(k*1.0);
        }
        m=(int)e;
        if(m==e)
            printf("%d",m);
        else
        ...{
            n=(e-m)*1000000000;
            j=0;
            while(n%10==0)
            ...{
                n/=10;
                j++;
            }
            switch(j)
            ...{
            case 1:printf("%.8lf",e);break;
            case 2:printf("%.7lf",e);break;
            case 3:printf("%.6lf",e);break;
            case 4:printf("%.5lf",e);break;
            case 5:printf("%.4lf",e);break;
            case 6:printf("%.3lf",e);break;
            case 7:printf("%.2lf",e);break;
            case 8:printf("%.1lf",e);break;
            default:printf("%.9lf",e);
            }
        }
        cout<<endl;
    }
    return 0;
}

YOYO's Note: 
我不知道输出的格式能不能按变量调的？
所以只好用case……