zoukankan      html  css  js  c++  java
  • lightoj-1050

    1050 - Marbles
    PDF (English) Statistics Forum
    Time Limit: 2 second(s) Memory Limit: 32 MB
    Your friend Jim has challenged you to a game. He has a bag containing red and blue marbles. There will be an odd number of marbles in the bag, and you go first. On your turn, you reach into the bag and remove a random marble from the bag; each marble may be selected with equal probability. After your turn is over, Jim will reach into the bag and remove a blue marble; if there is no blue marble for Jim to remove, then he wins. If the final marble removed from the bag is blue (by you or Jim), you will win. Otherwise, Jim wins.

    Given the number of red and blue marbles in the bag, determine the probability that you win the game.

    Input
    Input starts with an integer T (≤ 10000), denoting the number of test cases.

    Each case begins with two integers R and B denoting the number of red and blue marbles respectively. You can assume that 0 ≤ R, B ≤ 500 and R+B is odd.

    Output
    For each case of input you have to print the case number and your winning probability. Errors less than 10-6 will be ignored.

    Sample Input
    Output for Sample Input
    5
    1 2
    2 3
    2 5
    11 6
    4 11
    Case 1: 0.3333333333
    Case 2: 0.13333333
    Case 3: 0.2285714286
    Case 4: 0
    Case 5: 0.1218337218

    解题思路: 以dp[R][B]作为状态方程,通过观察发现,但R>=B时,dp[R][B]肯定是为0的

    当R<B时 我们可以发现dp[R][B] = R/(R+b)*dp[R-1][B-1] + B/(R+B)*dp[R][B-2];

    根据这两个状态预处理好dp方程就可以了;

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    
    double dp[600][600];
    
    void init(){
        for(int i=1;i<510;i++) dp[0][i] = 1;
        for(int i=1;i<510;i++){
            for(int j=i+1;j<510;j++){
                dp[i][j] = (double)i*1.0/(i+j)*dp[i-1][j-1] + (double)j*1.0/(i+j)*dp[i][j-2];
            }
        }
        return ;
        
    }
    
    int main(){
        init();
        int T,r,b;
        scanf("%d",&T);
        for(int t=1;t<=T;t++){
            
            scanf("%d%d",&r,&b);
            printf("Case %d: %.6lf
    ",t,dp[r][b]);
        }
        
    }
  • 相关阅读:
    python2.7下同步华为云照片的爬虫程序实现
    python 下字符串格式时间比较
    C# Socket通信 小案例
    win 10 安装 mysql解压版 步骤
    Android 连接 SQL Server (jtds方式)——下
    Android 连接 SQL Server (jtds方式)——上
    Android 项目建立步骤
    ubuntu 配置android开发环境
    ubuntu 安装eclipse
    ubuntu 配置Java jdk
  • 原文地址:https://www.cnblogs.com/yuanshixingdan/p/5565609.html
Copyright © 2011-2022 走看看