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;

void solve()
{
    int n;
    cin >> n;

    vector<long double> v(n, 0);
    for(int i = 0; i < n; ++i)
        cin >> v[i];

    int base = n;
    vector<vector<long double> > dp(n, vector<long double>(2 * base + 1));
    dp[0][base + 1] = v[0];
    dp[0][base - 1] = 1.0 - v[0];

    for(int i = 1; i < n; ++i)
    {
        for(int j = -i; j <= i; ++j)
        {
            dp[i][j + 1 + base] += dp[i - 1][j + base] * v[i];
            dp[i][j - 1 + base] += dp[i - 1][j + base] * (1.0 - v[i]);
        }
    }

    long double answer = 0;
    for(int i = 1; i <= n; ++i)
        answer += dp[n - 1][i + base];

    cout.precision(10);
    cout << fixed << answer << '\n';
}

main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //srand (time(0));

    solve();
}