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;

using ll = long long;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int ddx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll MOD = 1000000007;
const ll INF = 1000000000000000000L;
#ifdef __DEBUG

/**
 * For DEBUG
 * https://github.com/ta7uw/cpp-pyprint
 */
#include "cpp-pyprint/pyprint.h"

#endif

double dp[3000][3000];

void Main() {
    int N;
    cin >> N;
    vector<double> P(N);
    rep(i, N) cin >> P[i];

    dp[0][0] = 1;
    rep(i, N) {
        double p = P[i];
        rrep(j, N - 1, 0) {
            dp[i + 1][j + 1] = dp[i][j] * p + dp[i][j + 1] * (1 - p);
            dp[i + 1][j] = dp[i][j] * (1 - p);
        }
    }
    double ans = 0;
    rep(i, N + 1) {
        if (i > N - i) {
            ans += dp[N][i];
        }
    }
    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
    return 0;
}