HDU 1012 u Calculate e(简单数学)

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15679    Accepted Submission(s): 6773

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

Source

Greater New York 2000

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解题报告：简单的数学题。因为这个题目没有输入，每次输出的结果一样，输出9的结果，又因前3组结果的保留形式，所以直接打印处即可，剩余的计算再按照规律。

代码如下：

#include <iostream>
using namespace std;
int main()
{
    double a[10], ans[10];
    int i;
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    a[2] = 0.5;
    ans[2]= 2.5;
    for (i = 3; i <= 9; ++i)
    {
        a[i] = a[i - 1]*(1.0/i);
        ans[i] = ans[i - 1] + a[i];
        printf("%d %.9lf\n", i, ans[i]);
    }
    return 0;
}
        

过完年的第一道题！嘿嘿！