概率dp硬币正反

传送门

 

Let $N$ be a positive odd number.

There are $\mathrm{NN}$ coins, numbered $\mathrm{1,2,\dots ,N}$ For each $i$ ($\mathrm{1\le i\le N}$), when Coin $\mathrm{ii}$ is tossed, it comes up heads with probability ${\mathrm{pipi}}^{}$ and tails with probability $\mathrm{1-pi}$.

Taro has tossed all the $\mathrm{NN}$ coins. Find the probability of having more heads than tails.

Constraints

 

• $\mathrm{NN}$ is an odd number.
• $\mathrm{1\le N\le 2999}$
• ${\mathrm{pipi}}^{}$ is a real number and has two decimal places.
• $\mathrm{0<pi<1}$

Input

 

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{p1p1}}^{}$ ${\mathrm{p2p2}}^{}$ $\dots \dots$ ${\mathrm{pNpN}}^{}$

Output

 

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than ${\mathrm{10-910-9}}^{}$.

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

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

Sample Input 2

 

1
0.50

Sample Output 2

 

0.5

Outputs such as 0.500, 0.500000001 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\n",ans);
}