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>
#define Mx 1000000000005 
#define ll long long  
#define pi 3.141592653
#define vi vector<ll>
#define Mo 1000000007
#define mp make_pair
#define M map<ll,ll>
#define fr(i,a,b) for(ll i=a;i<b;i++)
#define vii vector<pair<ll,ll> > 
#define fi first
#define se second
#define pb push_back
#define ld long double
#define fill(a,x) memset(a,x,sizeof(a))
using namespace std;
bool comp(const pair<ll,ll> &a, const pair<ll,ll> &b)
{
  return a.second<b.second;
}
/*vi b=v;
   sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  ll t; cin>>t; fr(f,0,t){ 
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
 */
int main()
{
  ll n; cin>>n;
  ld a[n],ans=0.0; 
  fr(i,0,n)
    cin>>a[i];

  vector<vector<ld>> dp(n+1,vector<ld>(n+1,0.0));
  dp[1][0]=1-a[0]; dp[1][1]=a[0];
  fr(i,2,n+1)
  {
    fr(j,0,n+1)
    {
      if(j==0) dp[i][j]=dp[i-1][0]*(1-a[i-1]);
      else if(i==j) dp[i][j]=dp[i-1][j-1]*a[i-1];

      else dp[i][j]=dp[i-1][j]*(1-a[i-1])+dp[i-1][j-1]*a[i-1];
    }
  }

  for(ll j=n;j>n/2;j--)
    ans+=dp[n][j];
  printf("%.9Lf",ans);
}