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<bits/stdc++.h>
#define MAX 3000
#define ll unsigned long long
#define mod (1000000000+7)
#define Fast ios_base::sync_with_stdio(false); cin.tie(NULL);
//#define N 500001;

double dp[3005][3005];

using namespace std;

double solve(double a[],int n)
{
    int i,j;
    int x=(n+1)/2;
    for(i=0;i<=x;i++)
    dp[0][i]=0.0;
    for(i=0;i<=n;i++)
    dp[i][0]=1.0;
    dp[0][0]=1.0;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=x;j++)
        {
            dp[i][j]=a[i]*dp[i-1][j-1]+(1-a[i])*dp[i-1][j];
        }
    }
    /*for(i=0;i<=n;i++)
    {
        for(j=0;j<=x;j++)
        {
            cout<<dp[i][j]<<" ";
        }
        cout<<"\n";
    }*/
    return dp[n][x];
}
int main() {
    int n,i;
    cin>>n;
    double a[3005];
    for(i=1;i<=n;i++)
    cin>>a[i];
    memset(dp,-1,sizeof(dp));
    double k=solve(a,n);
    cout<<fixed<<setprecision(10)<<k<<"\n";
	return 0;
}