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)

//
//  CPlibrary
//
//  Created by Tanmay Singh on 3/03/2020.
// INDIAN INSTITUTE OF TECHNOLOGY (BHU) , VARANASI
//  Copyright © 2020 Tanmay Singh. All rights reserved.
//#pragma GCC target ("avx2")
//#pragma GCC optimization ("O3")
//#pragma GCC optimization ("unroll-loops")
 
#include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <climits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <ratio>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
 
#define FAST(); ios_base::sync_with_stdio(false);cin.tie(NULL);
#define endl "\n"
#define mod 1000000007
typedef long long int  bigint ;
typedef std::vector< int > vi;
typedef std::vector< long long int > vll;
typedef std::vector< vi > vvi;

using namespace std ;

double coins( double probability[] , int N ){
    double dp[N+1][N+1] ;
    memset( dp, 0, sizeof(dp) ) ;
    dp[0][0] = 1 ;
    for( int i = 1 ; i <= N ; i++ )
        dp[i][0] = dp[i-1][0] * ( 1 - probability[i-1] ) ;
    for( int i = 1 ; i <= N ; i++ ){
        for( int j = 1 ; j <= N ; j++ ){
            dp[j][i] = probability[j-1] * dp[j-1][i-1] + ( 1 - probability[j-1] ) * dp[j-1][i] ;
        }
    }
    double ans = 0 ;
    for( int i = N / 2 + 1 ; i <= N ; i++ ){
        ans += dp[N][i] ;
    }
    return ans ;
}
    
int main(){
    FAST();
    int N ; cin >> N ;
    double arr[N] ;
    for( int i = 0 ; i < N ; i++ )
        cin >> arr[i] ;
    cout << coins(arr , N ) << endl ;
    return 0 ;
}