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  • 概率dp硬币正反

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    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);
    }
    
    
    
     
    
    
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  • 原文地址:https://www.cnblogs.com/lipu123/p/14506450.html
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