zoukankan      html  css  js  c++  java
  • HDU 5645 DZY Loves Balls 水题

    DZY Loves Balls

    题目连接:

    http://acm.hdu.edu.cn/showproblem.php?pid=5645

    Description

    DZY loves playing balls.

    He has n balls in a big box. On each ball there is an integer written.

    One day he decides to pick two balls from the box. First he randomly picks a ball from the box, and names it A. Next, without putting A back into the box, he randomly picks another ball from the box, and names it B.

    If the number written on A is strictly greater than the number on B, he will feel happy.

    Now you are given the numbers on each ball. Please calculate the probability that he feels happy.

    Input

    First line contains t denoting the number of testcases.

    t testcases follow. In each testcase, first line contains n, second line contains n space-separated positive integers ai, denoting the numbers on the balls.

    (1≤t≤300,2≤n≤300,1≤ai≤300)

    Output

    For each testcase, output a real number with 6 decimal places.

    Sample Input

    2
    3
    1 2 3
    3
    100 100 100

    Sample Output

    0.500000
    0.000000

    Hint

    题意

    有n个球,你首先拿一个球A,然后不放回,再拿一个球B

    问你球A大于球B的概率是多少

    题解:

    数据范围很小,直接暴力就好了

    nlogn的话,写个权值线段树/树状数组就好了

    代码

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 301;
    int a[maxn];
    void solve()
    {
        int n;scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        double ans = 0;
        for(int i=1;i<=n;i++)
        {
            int add = 0;
            for(int j=1;j<=n;j++)
                if(a[i]>a[j])add++;
            ans=ans+add;
        }
        printf("%.6f
    ",ans/((1.0*n*(n-1))));
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)solve();
    }
  • 相关阅读:
    linux下解压命令大全
    关于伸展树的详细解析(E文)
    数据结构与算法汇总
    Linux下的压缩解压缩命令详解
    Linux Netcat 命令——网络工具中的瑞士军刀
    gethostbyname
    Html 转化为 PDF
    返回一个表
    sqlserver创建函数
    取不同类别的第一条数据
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5310041.html
Copyright © 2011-2022 走看看