GDUFE ACM-1013

题目：http://acm.gdufe.edu.cn/Problem/read/id/1013

 

u Calculate e

Time Limit: 2000/1000ms (Java/Others)

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.

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
...

思路：按照题目写就好了呀==（我绝不承认我又偷懒了）

难度：简单

代码：

#include<stdio.h>
int main()
{
    int i,b;
    double n,a;
    printf("n e\n- -----------\n");
    printf("0 1\n1 2\n2 2.5\n");
    n=2.5;b=2;
    for(i=3;i<10;i++)
    {
        b=b*i;
        a=1.0/b;
        n=n+a;
        printf("%d %.9lf\n",i,n);
    }
}