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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n-1; i >= 0; i--)
#define ALL(x) x.begin(),x.end()

using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vc = vector<char>;

const int mod = 1e9+7;
const ll INF = 1e18;
const int MAX = 1e6;

// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

struct edge {int to, cost;}; // 辺
// vector<edge> graph[MAX]; // 隣接リスト
// bool visit[MAX]; // 訪問状況

int n;
double p[3000];
double dp[3000][3000];

int main() {
// input
    cin >> n;
    rep(i, n) {
        cin >> p[i];
    }
// solve
    dp[0][0] = 1.0;

    // DP ループ
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j <= i; ++j) {
            dp[i+1][j+1] += dp[i][j] * p[i];
            dp[i+1][j] += dp[i][j] * (1.0 - p[i]);
        }
    }
// output
    double ans = 0;
    for (int i = n; i > n/2; i--) {
        ans += dp[n][i];
    }
    cout << ans << endl;
}