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;

typedef long double ll;
#define    mod      1000000007
#define    speed    ios_base::sync_with_stdio(false);
#define    up       cin.tie(NULL);
#define    pb       push_back
#define    mp       make_pair
#define    T        int t; cin >> t; while(t--)

ll dp[3005][3005] , arr[3005];
int  n, k;
ll solve(int i , int cnt){
    if(cnt > k ){
        return 0;
    }
    if(i > n){
        return 1.0;
    }
    ll &ans = dp[cnt][i];
    if(ans != -1){
        return ans;
    }
    ll temp1 = (1.0 - arr[i])*solve(i+1 , cnt + 1);
    ll temp2 = arr[i]*solve(i+1 , cnt);
    return (ans = temp1 + temp2);
}

int main(){
   speed up
   for(int i = 0; i <= 3000; i++){
       for(int j = 0; j <= 3000; j++){
           dp[i][j] = -1.0;
       }
   }
   cin >> n;
   for(int i = 1; i <= n; i++){
       cin >> arr[i];
   }
   k = n/2;
   if(n%2==0){
       k--;
   }
   cout << setprecision(12) << solve(1 , 0) << '\n';
}