hdu-oj 1012 u Calculate e

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30311    Accepted Submission(s): 13530

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
题目大意：计算e的精确值
解题思路：定义double类型
附代码：
#include<stdio.H>
int main()
{
    int n,i;
    printf("n e\n");
    printf("- -----------\n");
    for(n=0;n<=9;n++)
    {
        double s=2.5;
        if(n==0)  printf("%d 1\n",n);
        else if(n==1)
            printf("%d 2\n",n);
        else if(n==2)
            printf("%d 2.5\n",n);
        else
        {
            int k=2;
            for(i=3;i<=n;i++)
            {
                k*=i;
                s+=1.0/k;
            }
            printf("%d %.9lf\n",n,s);
        }
    }
    return 0;
}