Problem:
Score : 100 points

Problem Statement

Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

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

Constraints

N is an odd number.

1 \leq N \leq 2999

p_i is a real number and has two decimal places.

0 < p_i < 1

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 \ldots p_N

Output

Print the probability of having more heads than tails.
The output is considered correct when the absolute error is not greater than 10^{-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 (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

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

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

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<unordered_map>
#include<vector>
#include<set>
#include<list>
#include<iomanip>


#define prDouble(x) cout<<fixed<<setprecision(10)<<x //to print decimal numbers

using namespace std;
typedef long long ll;
int modulo = 1e9 + 7;

double dp[3001][3001];

double heads_prob(double prob[], int n, int x){
    if(x == 0){
        return 1;
    }
    if(n == 0){
        return 0;
    }
    if(dp[n][x]> -0.9){
        return dp[n][x];
    }
    dp[n][x] = (prob[n]*heads_prob(prob,n-1,x-1) + (1-prob[n])*heads_prob(prob,n-1,x));
    return dp[n][x];
}

int main(){
    int n;
    cin>>n;
    double prob[n+1];
    for(int i = 1; i<=n; i++){
        cin>>prob[i];
    }
    int x = (n+1)/2;
    for(int i = 0; i<=n; i++){
        for(int j = 0; j<=x; j++){
            dp[i][j] = -1;
        }
    }
    prDouble(heads_prob(prob,n,x));

    // double dp[n+1][x+1];
    // dp[0][0] = 1.0;
    // for(int i = 1; i<=x; i++){
    //     dp[0][x] = 0;
    // }
    // for(int i = 1; i<=n; i++){
    //     dp[i][0] = 1;
    // }
    // for(int i = 1; i<=x; i++){
    //     dp[i][i] = prob[i]*dp[i-1][i-1];
    // }
    // for(int i = 1; i<=n; i++){
    //     for(int j = 1; j<=x; j++){
    //         dp[i][j] = dp[i-1][j-1]*prob[i] + (1-prob[i])*dp[i-1][i];
    //     }
    // }
    // cout<<dp[n][x]<<endl;
}