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)

#pragma GCC optimze("Ofast")
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define mod 1000000007
#define md 998244353
#define trace(...)
#define ll long long int
#define ld long double
#define fi first
#define se second
#define fr(i,a,b) for(ll i=a;i<=b;i++)
#define frr(i,b,a) for(ll i=b;i>=a;i--)
#define mem(a) memset(a, 0, sizeof(a))
#define vpi std::vector<pair<int,int>>
#define pii pair<int,int>
#define pll pair<ll,ll> 
#define pb push_back
#define sall(v) sort(v.begin(),v.end())
#define all(v) (v.begin(),v.end())
#define minf -10000000
#define vi vector<int>
#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int infi=0x3f3f3f3f;
typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
    uniform_int_distribution<int> uid(0,lim-1);
    return uid(rang);
}
ll power(ll x, ll y,ll p) { 
    ll res = 1;      // Initialize result 
    x = x % p;  // Update x if it is more than or equal to p 
    while (y > 0) {  
        if (y & 1) // If y is odd, multiply x with result
            res = (res*x) % p;  
        y = y>>1; // y = y/2 // y must be even now
        x = (x*x) % p;   
    } 
    return res; 
}
ll power(ll x, ll y) { 
    ll res = 1;      // Initialize result 
    while (y > 0) {    
        if (y & 1)// If y is odd, multiply x with result 
            res = (res*x) ;  
        y = y>>1; // y = y/2 // y must be even now 
        x = (x*x);   
    } 
    return res; 
}
 
ll max3(ll a,ll b,ll c){
    return max(a,max(b,c));
}
ll min3(ll a,ll b,ll c){
    return min(a,min(b,c));
}
bool isPrime(ll n) { // Corner cases 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
    if (n % 2 == 0 || n % 3 == 0) 
        return false; 
    for (ll i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
    return true; 
}
const int d4i[4]={-1, 0, 1, 0}, d4j[4]={0, 1, 0, -1};
const int d8i[8]={-1, -1, 0, 1, 1, 1, 0, -1}, d8j[8]={0, 1, 1, 1, 0, -1, -1, -1};
const int nax=1e5+10;

void solve(){
    int n;cin>>n;
    double dp[n+1];
    fr(i,0,n)dp[i]=0;   
    dp[0]=1;
    double p_heads;
    fr(i,0,n-1){
        cin>>p_heads;
        frr(j,i+1,0){
            dp[j]=((j==0?0:dp[j-1]*p_heads)+dp[j]*(1-p_heads));
        }
    }
    // fr(i,0,n)cout<<dp[i]<<" ";
    // cout<<endl;
    double ans=0;
    fr(i,n/2+1,n){
        ans+=dp[i];
    }
    printf("%.10lf\n",ans);
    return;
}

int main(){
    ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
    cout<<fixed<<setprecision(8);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    #endif
    int t;
    // cin>>t;
    t=1;
    while(t--){
        solve();
    }
    #ifndef ONLINE_JUDGE
    cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
    #endif
    return 0;
}