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;
double dp[3001][3001];

double get(int n,  int must , int tail ,  vector <double> p )
{
	if(n==0)
      	return 1;
  
 	if(dp[n][tail]>-0.9)
 	    return dp[n][tail];
  
  if(tail<must)
  {
  	dp[n][tail]=p[n-1]*get(n-1 , must , tail,  p  ) + (1-p[n-1])*get(n-1 , must, tail+1 ,p );
    
  }
  else if(tail==must)
    dp[n][tail]=p[n-1]*get(n-1 , must ,tail ,p);
  
  return dp[n][tail];
}

int main()
{
  int n;
  cin>>n;
  vector<double> p(n);
  for(int i=0  ; i<n; i++)
    cin>>p[i];
    int must=n/2 ,  tail=0;
 
 
 memset(dp , -1 , sizeof(dp));
  
  cout<< get(n , must, 0, p);
  
  return 0;
 
}