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: Accepted)

#include <iostream>
#include <fstream>
#include <stdio.h>
#include <math.h>
#include <time.h>
#include <string>
#include <tuple>
#include <vector>
#include <map>
#include <unordered_map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <cstdlib>
#include <algorithm>
#include <random>
#include <cassert>
#include <iomanip>
using namespace std;
#define LL long long
#define MP(a, b) make_pair(a, b)
#define MMP(a, b, c) make_pair(make_pair(a, b), c)
#define MAX 1000000000
#undef INT_MIN
#undef INT_MAX
#define INT_MIN -2147483648
#define INT_MAX 2147483647
#define LL_MIN -9223372036854775808
#define LL_MAX 9223372036854775807
#define PI 3.14159265359

int N;
double p[3000];

double dp[3000][6001];
double solve(int now, int diff){
	if(now == N-1 && diff == 1) return p[now];
	if(now == N-1 && diff == -1) return 1-p[now];
	if(now == N-1) return 0;
	if(dp[now][diff+3000] != -1) return dp[now][diff+3000];
	return dp[now][diff+3000] = solve(now+1,diff-1)*p[now] + solve(now+1,diff+1)*((double)1-p[now]);
}

int main(){
	ios::sync_with_stdio(false);
	
	cin >> N;
	for(int i=0; i<N; i++){
		cin >> p[i];
		for(int j=0; j<=6001; j++) dp[i][j] = -1;
	}

	double ans = solve(0, 1);
	for(int i=3; i<=N; i+=2) ans += solve(0, i);
	cout << std::setprecision(14) << ans << endl;

	return 0;
}