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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
using namespace std;
 
#define mod 1000000007
#define FOR(x,to) for(int x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define long long long
inline int rei(){int x;cin>>x;return x;}
inline long rel(){long x;cin>>x;return x;}
inline string res(){string x;cin>>x;return x;}
//------------------------------------------------------- 
double DP[3000*3000];
void Calc(){
	int N = rei();
	DP[0] = 1;
	for(int i=0;i<N;i++){
		double prob;
		cin >> prob;
		for(int j=0;j<3000;j++){
			DP[(i+1)*3000+j] += DP[i*3000+j] * (1-prob);
			if(j != 2999){
				DP[(i+1)*3000+j+1] += DP[i*3000+j] * prob;
			}
		}
	}
	double sum = 0;
	for(int i=N/2+1;i<=N;i++){
		sum += DP[N*3000+i];
	}
	cout << sum << endl;
}
int main(int argc,char** argv){
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0); Calc(); return 0;
}