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;

#define MOD (1000 * 1000 * 1000 + 7)
#define INF (1LL << 60)
#define pb push_back
#define po pop_back
#define fi first
#define se second

typedef long long ll;
typedef long double ld;

#define maxn 3005

ld A[maxn];
int N;

ld dp[maxn][maxn];

ld solution(int n, int h) {
  if (n >= N && h > (int)(N / 2)) {
    return 1;
  } else if(n >= N) {
    return 0;
  }

  if (dp[n][h] != -1.0)
    return dp[n][h];

  dp[n][h] = solution(n + 1, h + 1) * A[n] +
             solution(n + 1, h) * (1.0 - A[n]);

  return dp[n][h];
}

int main() {
  scanf("%d", &N);

  for (int i = 0; i < N; ++i)
    cin >> A[i];

  for (int i = 0; i < maxn; ++i) {
    for (int j = 0; j < maxn; ++j) {
      dp[i][j] = -1.0;
    }
  }

  cout << setprecision(10) << solution(0, 0) << endl;

  return 0;
}