zoukankan      html  css  js  c++  java
  • HDU 4336——Card Collector——————【概率dp】

    Card Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3407    Accepted Submission(s): 1665
    Special Judge


    Problem Description
    In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people in the famous novel Water Margin, you will win an amazing award. 

    As a smart boy, you notice that to win the award, you must buy much more snacks than it seems to be. To convince your friends not to waste money any more, you should find the expected number of snacks one should buy to collect a full suit of cards.
     
    Input
    The first line of each test case contains one integer N (1 <= N <= 20), indicating the number of different cards you need the collect. The second line contains N numbers p1, p2, ..., pN, (p1 + p2 + ... + pN <= 1), indicating the possibility of each card to appear in a bag of snacks. 

    Note there is at most one card in a bag of snacks. And it is possible that there is nothing in the bag.
     
    Output
    Output one number for each test case, indicating the expected number of bags to buy to collect all the N different cards.

    You will get accepted if the difference between your answer and the standard answer is no more that 10^-4.
     
    Sample Input
    1
    0.1
    2
    0.1 0.4
     
    Sample Output
    10.000
    10.500
     
    Source
     
     
    题目大意:要收集方便面中的人物卡片,n是要收集n种卡片,下面给n种卡片的出现概率,问你收集全n种卡片的期望值。
     
    解题思路:概率dp。
     
    #include<bits/stdc++.h>
    using namespace std;
    const int maxn=1<<21;
    double dp[maxn];
    double p[21];
    int main(){
        int n;
        while(scanf("%d",&n)!=EOF){
            for(int i=0;i<n;i++)
                scanf("%lf",&p[i]);
            dp[(1<<n)-1]=0;
            for(int s=(1<<n)-2;s>=0;s--){
                double sum=1.0,sump=0;
                for(int j=0;j<n;j++){
                    if(!((1<<j)&s)){
                        sum+=dp[s|(1<<j)]*p[j];
                        sump+=p[j];
                    }
                }
                dp[s]=sum/sump;
            }
            cout<<dp[0]<<"++++"<<endl;
            printf("%.5f
    ",dp[0]);
        }
        return 0;
    }
    

      

  • 相关阅读:
    系统重启
    Linux驱动程序开发
    Linux 下实现控制屏幕显示信息和光标的状态
    Linux C 语言 获取系统时间信息
    linux 获取系统屏幕分辨率
    Linux下得到显示屏参数的方法
    Linux如何关闭防火墙和查看防火墙的具体情况
    查看Linux下网卡状态或 是否连接(转)
    ArcGIS Engine中正确释放打开资源
    在ArcEngine下实现图层属性过滤的两种方法
  • 原文地址:https://www.cnblogs.com/chengsheng/p/4782845.html
Copyright © 2011-2022 走看看