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 <iostream>
#include<string.h>
#define ll long long 
#include<vector>
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int mod=1e9+7;
int main(){
	IOS;
	int n;
	cin>>n;
	vector<double> p(n);
	for(int i=0;i<n;i++)
	cin>>p[i];
	// dp[i][j] prob of having j heads when i coins have been tossed
	// dp[i][j]=p(h)*dp[i-1][j-1]+p(t)*dp[i-1][j]
	// dp[0][0]=1;
	double dp[3000][3000];
	memset(dp,0, sizeof(dp));
	dp[0][0]=1-p[0];
	dp[0][1]=p[0];
	for(int i=1;i<n;i++){
		for(int j=0;j<=i;j++){
			dp[i][j]=p[i]*dp[i-1][j-1]+(1-p[i])*dp[i-1][j];
		}
	}
double ans=0;
	for(int j=n/2;j<n;j++)
	ans+=dp[n-1][j];
	cout<<ans;
	
	
}