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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
typedef long long int ll;

int main(){

    fastio ;

    ll n;
    vector<double> p(3005);
    vector<vector<double>> dp(3005 , vector<double>(3005));

    cin >> n;
    for(ll i = 1; i <= n; i++) cin >> p[i];

    // OPTIMAL SUB-PROBLEM :

    // BASE CASE :

    dp[1][1] = p[1];
    dp[0][0] = 1;
    for(ll i = 1; i <= n; i++) dp[i][0] = (1-p[i])*dp[i-1][0];
    for(ll i = 2; i <= n; i++) dp[1][i] = 0;

    // RECURSIVE RELATION :


    for(ll i = 2; i <= n; i++){
        for(ll j = 1; j <= n; j++){
            dp[i][j] = dp[i-1][j-1]*p[i] + dp[i-1][j]*(1-p[i]);
        }
    }

    // OPTIMAL SOLUION :

    double total  = 0;

    for(ll i = floor(n/2)+1 ; i <= n; i++) total += dp[n][i];
    cout << fixed << setprecision(9);

    cout << total;
}