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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <cmath>
#include <iomanip>

typedef long long ll;
const ll MOD = 1e9 + 7;
const long long INF = 1LL<<60;

using namespace std;
double dp[3010][3010];

int main(){
    ll n;
    cin >> n;
    double p[n];
    for(int i = 0; i < n; i++){
        cin >> p[i];
    }

    for(int i = 0; i < 3010; i++){
        for(int j = 0; j < 3010; j++){
            dp[i][j] = 0;
        }
    }
    dp[0][0] = 1;
    double one = 1.000000000;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            dp[i+1][j] += dp[i][j]*(one-p[i]);
            dp[i+1][j+1] += dp[i][j]*p[i];
        }
    }

    double ans = 0.000000000;
    for(int i = n/2+1; i <= n; i++){
        ans += dp[n][i];
    }
    cout << fixed;
    cout << setprecision(10) << ans << endl;

    return 0;
}