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: Runtime Error)



#include <bits/stdc++.h> 
#define ll long long
#define ull unsigned long long
#define endl '\n'
#define pb push_back
#define mp make_pair
#define sp(x) fixed<<setprecision(x)
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;

vector<vector<double>> dp(3001,vector<double>(3001,-1));
//double dp[3001][3001];
double fun(vector<double> &p, int i, int atleast_heads)
{
    //Base case
    if(atleast_heads == 0)
    return 1;

    if(i == 0)
    return 0;

    //Lookup
    if(dp[i][atleast_heads] > -0.9)
    return dp[i][atleast_heads];

    //Rec case
    double op1 = p[i]*fun(p,i-1,atleast_heads-1);
    double op2 = (1-p[i])*fun(p,i-1,atleast_heads);
    return dp[i][atleast_heads] = op1 + op2;
}

int main()
{
    fast_io;
    
    int n;
    cin>>n;
    vector<double> p(n);

    for(int i=1;i<=n;i++)
    cin>>p[i];

    cout<<sp(10)<<fun(p,n,(n+1)/2)<<endl;

    return 0;
}