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>
typedef long long LL;
using namespace std;

const int maxn = 3005;
int n;
double pro[maxn];
double d;

int32_t main()
{
    // faster_input
    ios:: sync_with_stdio(0), cin.tie(0), cout.tie(0);

    //input
    cin >> n;
    for(int i = 1;i<=n;i++){
        cin >> pro[i];
    }

    // finding the answer


    // array_dp[i][j] --> i coins and j heads
    double dp[maxn][maxn];
    dp[0][0] = 1;
    dp[1][0] = 1 - pro[1];
    for(int i = 0;i<=n;i++){
        for(int j = 0;j<=n;j++){
            if(i == 0 && j == 0){
                continue;
            }
            dp[i][j] = (dp[i-1][j]*(1-pro[i])+dp[i-1][j-1]*pro[i]);
        }
    }




    // output
    double ans = 0;
    d = n;
    d /= 2;
    for(int i = 0;i<=n;i++){
        if(i > d){
            ans += dp[n][i];
        }
    }
    cout << setprecision(11) << fixed << endl;
    cout << ans << endl;

}