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 <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <iomanip>
#include <cassert>
#include <random>
#include <tuple>
#include <cstdint>

#define rep(i,n) for (int i = 0; i < (n); ++i)

using namespace std;
typedef long long ll;

const int INF = 1 << 30;
const int MOD = 1e9 + 7;


int main() {
    int n; cin >> n;
    vector<double> p(n); rep(i, n) cin >> p[i];
    vector<vector<double>> dp(n+1, vector<double>(n+1, 0));
    dp[0][0] = 1;
    rep(i, n) rep(j, n+1) {
        if (j+1 <= n) dp[i+1][j+1] += dp[i][j] * p[i];
        dp[i+1][j] += dp[i][j] * (1-p[i]);
    }

    double ans = 0;
    for (int i = n/2 + 1; i <= n; i++) {
        ans += dp[n][i];
    }
    cout << setprecision(10) << ans << endl;

    return 0;
}