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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iomanip>
int MOD=1e9+7;
using namespace std;
int main(){
    int n;
    cin >> n;
    vector<double> vec(n);
    for(int i=0;i<n;i++){
        double x;
        cin >> x;
        vec[i]=x;
    }
    double dp[n+1][n+2];
    for(int i=0;i<2;i++){
        if(i==0){
            dp[1][0]=1-vec[0];
        }else{
            dp[1][1]=vec[0];
        }
    }
    for(int i=2;i<=n;i++){
        for(int j=0;j<i;j++){
            dp[i][j+1] += dp[i-1][j]*vec[i-1];
            dp[i][j] += dp[i-1][j]*(1.0-vec[i-1]);
        }
    }
    double res=0;
  //  cout<<dp[2][2]<<endl;
    for(int i=n/2+1;i<=n;i++){
        res += dp[n][i];
       // cout<<dp[n][i]<<endl;
    }
    cout<<setprecision(10)<<res<<endl;
}