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: Accepted)

#include<bits/stdc++.h>
using namespace std;
 
#define int long long 
#define ll long long 
#define pb push_back
#define mp make_pair
#define pii pair<ll,ll>
#define ff first
#define ss second
#define vi vector<ll>
#define mod 1000000007
#define MOD 1000000007
#define full(a) a.begin(),a.end()
#define inf INT_MAX
#define minf INT_MIN
#define dd double
#define fori(x,n) for(ll i=x;i<n;i++)
#define modd 998244353
#define umap unordered_map<ll,ll>
double dp[3000][3000];
double p[3000];
ll n;

double count(ll i,ll j){
  if(i==n){
    if(j>n-j)return 1;
    return 0;
  }
  if(dp[i][j]!=-1)return dp[i][j];
  return dp[i][j]=p[i]*count(i+1,j+1)+(1-p[i])*count(i+1,j);
}


int32_t main() 
{ 
  
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    
    cin>>n;
    fori(0,3000)
    for(int j=0;j<3000;j++)dp[i][j]=-1;

    fori(0,n)cin>>p[i];

    cout<<setprecision(12)<<count(0,0);
   

    
   


return 0;
}