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 <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>

#include <boost/rational.hpp>
*/
#include <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
using ll = long long;
using ld = long double;
using pll = pair<ll,ll>;
#define rep2(i, s, n) for (ll i = (s); i < (ll)(n); i++)
#define rep3(i, n) for (ll i = 1; i < (ll)(n+1); i++)
#define rep4(i, s, n) for (ll i = (s); i < (ll)(n+1); i++)
#define repr(i,n) for (ll i = (n-1); i>=0;i--)
#define repr3(i,n) for(ll i = (n);i>0;i--)
#define repr4(i,s,n) for(ll i = (n);i>=(s);i--)
#define stlen(s) ll s.size()-1
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define cout(n) cout<<std::fixed<<std::setprecision(n)
using Graph = vector<vector<int>>;
using Graphw = vector<vector<pair<ll,ll>>>;
#define INF1  INT_MAX;
#define INF2  LLONG_MAX;
#define PI  3.14159265358979323846;
#define MOD 1000000007;
/*
namespace mp = boost::multiprecision;
// 任意長整数型
using Bint = mp::cpp_int;
// 仮数部長が32の浮動小数点数型
using Real32 = mp::number<mp::cpp_dec_float<32>>;
// 仮数部長が1024の浮動小数点数型
using Real1024 = mp::number<mp::cpp_dec_float<1024>>;
// 有理数型
using Rat = boost::rational<Bint>;
*/
const int mod = 1000000007;


struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}


int main(){
int N;
cin>>N;
ld p[N+1];
rep3(i,N)cin>>p[i];
ld dp[N+1][N+1];
rep(i,N+1)rep(j,N+1)dp[i][j]=0;
dp[0][0]=1;
rep(i,N+1)rep3(j,N){
    if(i>0){
        dp[i][j]+=p[j]*dp[i-1][j-1];
    }
    dp[i][j]+=(1-p[j])*dp[i][j-1];
}
ld ans=0;
rep4(i,N/2+1,N)ans+=dp[i][N];
cout(12)<<ans<<endl;
}