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  • HDU 1012 u Calculate e(简单阶乘计算)

    传送门:

    http://acm.hdu.edu.cn/showproblem.php?pid=1012

    u Calculate e

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 52607    Accepted Submission(s): 24106


    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
     分析:
    没有什么好说的
    水题
    code:
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    int f(int x)
    {
        if(x==0)
            return 1;
        if(x==1)
            return 1;
        LL sum=1;
        for(int i=1;i<=x;i++)
        {
            sum*=i;
        }
        return sum;
    }
    int main()
    {
        printf("n e
    ");
        printf("- -----------
    ");
        printf("0 1
    1 2
    2 2.5
    ");
        for(int i=3;i<=9;i++)
        {
            double result=0;
            for(int j=0;j<=i;j++)
            {
                result+=(1.0/(f(j)*1.0));
            }
            printf("%d %0.9lf
    ",i,result);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yinbiao/p/9313265.html
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