u Calculate e

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.       

              

Input

There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.       

              

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

    
    

     
      1 2
3 2 3 1
0 
    
    

              

Sample Output

    
    

     
      n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333 
    
    
    
    

     
      
    
    
    
    

     
     这个题就是一水题。很简单的，但要考虑精度问题。。
    
    
    
    

     
      
    
    
    
    

     
     
#include<stdio.h>
int main()
{
    double a[10],c[10];
    int b[10];
    int i,j;
    a[0]=1;
    b[0]=1;
    for(i=1; i<10; i++)
    {
        b[i]=b[i-1]*i;
        c[i]=b[i]*1.0;
    }
    for(i=1; i<10; i++)
    {
        a[i]=a[i-1]+1.0/c[i];
    }
    printf("n e\n");
    printf("- -----------\n");
    for(i=0; i<10; i++)
    {
        if(i==0) printf("0 1\n");
        if(i==1) printf("1 2\n");
        if(i==2) printf("2 2.5\n");
        if(i>2)
            printf("%d %.9lf\n",i,a[i]);
    }
    return 0;
}