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 <cmath>
#include <cstring> 
#include <vector>
#include <algorithm>
#include <bits/stdc++.h>

using namespace std;
#define mod 1000000007

vector<double> prob;
int coins;

double dp[3001][3001];
double findMaxAns(int coin,int x){
    if(x==0){
        return 1;
    }
    if(coin<0){
        return 0;
    }
    if(dp[coin][x]>=0 && dp[coin][x]<=1){
        return dp[coin][x];
    }
    
    dp[coin][x] = prob[coin]*findMaxAns(coin-1,x-1) + (1 - prob[coin])*findMaxAns(coin-1,x);
    return dp[coin][x];
}

void solve(){
    memset(dp,-1,sizeof(dp));
    
    cin>>coins;
    for(int i=0;i<coins;i++){
        double a;
        cin>>a;
        prob.push_back(a);
    }
    
    cout<<findMaxAns(coins-1,(coins+1)/2);
}

int main()
{
    
    solve();
    
    return 0;
}