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  • 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

    Recommend

    JGShining

    Statistic | Submit | Discuss | Note

    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! 

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  • 原文地址:https://www.cnblogs.com/lidaojian/p/2343013.html
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