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 ll long long
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

int main() {
    fastio
    
	ll n,i,j,x,t,k,w;
	cin>>n;
	
	long double a[n+1];
	
	for(j=1;j<=n;j++)
	cin>>a[j];
	
	vector<vector<long double> > dp(n+1,vector<long double> (n+1,0));
	long double ans=0;
	dp[0][0]=1;
	
	for(i=1;i<=n;i++){
	    for(j=0;j<=n;j++){
	        if(j==0) dp[i][j]=dp[i-1][j]*(1-a[i]);
	        else if(i==j) dp[i][j]=dp[i-1][j-1]*a[i];
	        else if(j<i) dp[i][j]=dp[i-1][j-1]*a[i]+dp[i-1][j]*(1-a[i]);
	    }
	}
	
	for(j=(n+1)/2;j<=n;j++)
	ans+=dp[n][j];
	
	cout<<ans<<endl;
	
	return 0;
}