HDU 1012 u Calculate e

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

 

题意：按照格式输出n取0~9时，1.0/n!的值。

分析：可以先用数组将0~9的阶乘存放起来，前三列输出的e保留的小数位与后面的不同，所以可以单独输出。

AC源代码（C语言）：

#include<stdio.h>

int main()
{
    int i,N,fn[10]={1};
    double e=2.5;
    i=1;
    while(i<10)
        {
            fn[i]=i*fn[i-1];
            i++;
        }
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    i=3;
    while(i<10)
        {
            e+=1.0/fn[i];               /*如果用e+=1/fn[i]，就达不到累加的效果，因为，整形除以整形要去掉小数位，切记！！！！！*/
            printf("%d %.9lf\n",i,e);
            i++;
        }
    return 0;
}

2013-04-11