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>

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define foru(i, n) for(int i = 1; i <= n; i++)
#define foru0(i, n) for(int i = 0; i < n; i++)
#define ford(i, n) for(int i = n; i >= 1; i--)
#define ford0(i, n) for(int i = n - 1; i >= 0; i--)
#define fors(i, a, b) for(int i = a; i <= b; i++)

using namespace std;

typedef long long ll;
typedef double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vii;
typedef vector<pll> v2l;
typedef pair<char, int> pci;
typedef vector<pci> vci;

const int N = 1e6 + 1;
const int NN = 1e3 + 1;
const ll MOD = 1e9 + 7;
const ll oo = 1e18 + 7;

void solve() {
    int n;
    cin >> n;
    vector<double> p(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> p[i];
    }
    vector<vector<double> > dp(n + 1, vector<double>(n + 1));
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= i; j++) {
            if (j < i) {
                dp[i][j] = dp[i-1][j] * (1 - p[i]) + dp[i-1][j-1] * p[i];
            } else {
                dp[i][j] = dp[i-1][j-1] * p[i];
            }
        }
    }
    double res = 0;
    for (int i = n / 2 + 1; i <= n; i++) {
        res += dp[n][i];
    }
    cout << res;
}

int main() {
    #ifdef DEBUG
        freopen("solve.inp", "r", stdin);
        freopen("solve.out", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    cout.precision(10);
    solve();
    cerr << "Time elapsed: " << 1.0 * clock() << " ms.\n";
}