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<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
	ll N;
	cin>>N;
	double p[N];
	for(ll i=0;i<N;i++){
		cin>>p[i];
		//cout<<p[i]<<endl;
	}
	double dp[N+1][N+1];
	for(ll i=0;i<=N;i++){
		for(ll j=0;j<=N+1;j++){
			dp[i][j]=(double)0;
			//cout<<dp[i][j]<<" ";
		}
	}
	dp[0][0]=1.0;
	for(ll i=1;i<=N;i++){
		dp[0][i]=0.0;
	}
	for(ll i=1;i<=N;i++){
		dp[i][0]=dp[i-1][0]*(1-p[i-1]);
	}
	for(ll i=1;i<=N;i++){
		for(ll j=1;j<=N;j++){
			if(i==1 && j==1){
				dp[i][j]=p[i-1];
			}
			else if(j<=i){
				dp[i][j]=(double)dp[i-1][j]*(double)(1-p[i-1])+(double)dp[i-1][j-1]*(double)p[i-1];
			}
		}
	}
	double ans=0.0;
	for(ll i=N/2+1;i<=N;i++){
		ans+=dp[N][i];
	}
	cout<<setprecision(9)<<fixed<<ans<<endl;
}