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>
using namespace std;
/*#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
/*template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
*/typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll,ll> pl;
typedef pair<int,int> pii;

#define LOCAL 0
#define dbg(x) cout << #x << " is " << x << "\n"
#define gll(x) scanf("%d",&x)
#define gll2(x,y) scanf("%d%d",&x,&y)
#define gll3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define gllarr(arr,n) f(i,n) gll(arr[i]);
#define sz(x) ((int)x.size())
#define s(x) sort(x.begin(),x.end())
#define all(v) v.begin(),v.end()
#define rs(v) { s(v) ; r(v) ; }
#define r(v) {reverse(all(v));}
#define pb push_back
#define f(i,n) for(int i=0;i<n;i++)
#define fr(i,n) for(int i=n-1;i>=0;i--)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repr(i,a,b) for(int i=a;i>=b;i--)

const ll mod = (ll)1e9 + 7;
const ll inf = (ll)1e16;
const ld eps = 1e-12;
const ll N = (int)1e5 + 5;
const ll LOGN = 19;
const ld PI = 3.14159265358979323846;
inline ll mul(ll a, ll b, ll m = mod) { return (ll)(a * b) % m;}
inline ll add(ll a, ll b, ll m = mod) { a += b; if(a >= m) a -= m; if(a < 0) a += m; return a;}
inline ll power(ll a, ll b, ll m = mod) { if(b == 0) return 1; if(b == 1) return (a % m); ll x = power(a, b / 2, m); x = mul(x, x, m); if(b % 2) x = mul(x, a, m); return x;}

int n;
ld p[3002];
ld dp[3002][3002];
bool vis[3002][3002];

ld rec(int pos, int heads){
    if(pos >= n){
        if(heads > 0){
            return (ld)0;
        }
        return (ld)1;
    }
    if(vis[pos][heads])
        return dp[pos][heads];
    ld A = 0, B = 0;
    if(heads > 0) A = rec(pos + 1, heads - 1) * p[pos];
    B = rec(pos + 1, heads) * (1 - p[pos]);
    vis[pos][heads] = true;
    return dp[pos][heads] = A + B;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    if (LOCAL) {
        freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\input.txt", "r", stdin);
        freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\output.txt", "w", stdout);
    }
    cin>>n;
    f(i, n) cin>>p[i];
    memset(vis, false, sizeof(vis));
    int to = (n + 1)/2;
    ld ans = 0;
    rep(i, to, n) ans += rec(0, i);
    cout<<setprecision(10)<<fixed<<ans;
    return 0;
}