zoukankan      html  css  js  c++  java
  • 概率dp硬币正反

    传送门

    Let N be a positive odd number.

    There are NN coins, numbered 1,2,,N For each i (1iN), when Coin ii is tossed, it comes up heads with probability pipi and tails with probability 1pi.

    Taro has tossed all the NN coins. Find the probability of having more heads than tails.

    Constraints

     

    • NN is an odd number.
    • 1N2999
    • pipi is a real number and has two decimal places.
    • 0<pi<1
    Input

     

    Input is given from Standard Input in the following format:

    NN
    p1p1 p2p2  pNpN
    
    Output

     

    Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10910−9.

    Sample Input 1

     

    3
    0.30 0.60 0.80
    
    Sample Output 1

     

    0.612
    

    The probability of each case where we have more heads than tails is as follows:

    • The probability of having (Coin1,Coin2,Coin3)=(Head,Head,Head)(Coin1,Coin2,Coin3)=(Head,Head,Head) is 0.3×0.6×0.8=0.1440.3×0.6×0.8=0.144;
    • The probability of having (Coin1,Coin2,Coin3)=(Tail,Head,Head)(Coin1,Coin2,Coin3)=(Tail,Head,Head) is 0.7×0.6×0.8=0.3360.7×0.6×0.8=0.336;
    • The probability of having (Coin1,Coin2,Coin3)=(Head,Tail,Head)(Coin1,Coin2,Coin3)=(Head,Tail,Head) is 0.3×0.4×0.8=0.0960.3×0.4×0.8=0.096;
    • The probability of having (Coin1,Coin2,Coin3)=(Head,Head,Tail)(Coin1,Coin2,Coin3)=(Head,Head,Tail) is 0.3×0.6×0.2=0.0360.3×0.6×0.2=0.036.

    Thus, the probability of having more heads than tails is 0.144+0.336+0.096+0.036=0.6120.144+0.336+0.096+0.036=0.612.

    Sample Input 2

     

    1
    0.50
    
    Sample Output 2

     

    0.5
    

    Outputs such as 0.5000.500000001 and 0.499999999 are also considered correct.

    Sample Input 3

     

    5
    0.42 0.01 0.42 0.99 0.42
    
    Sample Output 3

     

    0.3821815872

    题意:给N个硬币,每一个硬币扔向空中落地是正面朝上的概率是p[i] ,让求扔了N个硬币,正面的数量大于背面数量的概率


    这个题就是的dp[i][j]代表的是前
    i个硬币时有j个是正面的概率,
    所以转移方程就是:
    j!=0时:dp[i][j]=1.0*dp[i-1][j-1]*p[i]+1.0*dp[i-1][j]*(1.0-p[i]);
    j==0时:dp[i][j]=dp[i-1][j]*(1-p[i])

    #include<iostream>
    #include<algorithm>
    #include<cstring> 
    using namespace std;
    typedef long long ll;
    const int maxn=3e3+100;
    double p[maxn];
    double dp[maxn][maxn];//前i个硬币,有j个正面朝上的概率 
    int n;
    int main(){
        cin>>n;
        for(int i=1;i<=n;i++){
            cin>>p[i];
        } 
        dp[0][0]=1;
        for(int i=1;i<=n;i++){
            dp[i][0]=dp[i-1][0]*(1.0-p[i]);
        } 
        dp[1][1]=p[1]; 
        for(int i=2;i<=n;i++){
            for(int j=1;j<=i;j++){
                dp[i][j]=1.0*dp[i-1][j-1]*p[i]+1.0*dp[i-1][j]*(1.0-p[i]);
            }
        }
        double ans=0;
        for(int i=(n/2+1);i<=n;i++){
            ans+=dp[n][i];
        }
        printf("%.9lf
    ",ans);
    }
    
    
    
     
    
    
  • 相关阅读:
    UVALive 7352 Dance Recital
    [ An Ac a Day ^_^ ] UVALive 7270 Osu! Master
    vim配置文件
    数据结构 链表
    [ An Ac a Day ^_^ ] hrbust 2291 Help C5 分形
    [ An Ac a Day ^_^ ] hdu 2553 N皇后问题 搜索
    [ An Ac a Day ^_^ ] HihoCoder 1249 Xiongnu's Land 线性扫描
    hdu 5874 Friends and Enemies icpc大连站网络赛 1007 数学
    hdu 5876 Sparse Graph icpc大连站网络赛 1009 补图最短路
    6.Z字变换 direction
  • 原文地址:https://www.cnblogs.com/lipu123/p/14506450.html
Copyright © 2011-2022 走看看