1012.u Calculate e

原题描述：

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

没有输入，直接输出，前面的小数位数不一样，直接单独输出，后面的迭代就好了。

#include <stdio.h>
int main ( )
{
    int i ;
    double t = 1 , sum = 1 ;

    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    for ( i = 1 ; i < 10 ; i ++ )
    {
        t = 1.0/ i * t ;
        sum += t ;
        if ( i < 3 )
            continue ;
        else
            printf("%d %.9lf\n", i , sum);

    }
    return 0;
}