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 <iostream>
#include <cstring>
#include <map>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <list>
#include <numeric>
#include <stdio.h>
#include <string>
#include <cstdlib>
#include <math.h>
#include <stack>
#include <climits>
#include <bitset>
#include <utility>


using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
// const ll INF = 1LL <<60;
// const int INF = 10000;

//最大公約数
ll gcd(ll x, ll y) {
  ll tmp = 0;
  if (x < y){
    tmp=x;
    x=y;
    y=tmp;
  }
  while (y > 0) {
    ll r = x % y;
    x = y;
    y = r;
  }
  return x;
}

//最大公倍数
ll lcm(ll x,ll y){
  return x/gcd(x,y)*y;
}

const int MAX = 1e6+1;
const ll MOD = 1e9+7;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

vector<long long> nCr;

//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

//for(int i = ; i < ; i++){}

ll lmax(ll s,ll t){
  if(s>t){
    return s;
  }
  else{
    return t;
  }
}

ll lmin(ll s,ll t){
  if(s<t){
    return s;
  }
  else{
    return t;
  }
}

long long modpow(long long res,long long a, long long n, long long mod) {
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

//sieve(MAX);でベクトルIsPrimeに「MAXまでの数値の素数の真偽」を格納する。
vector<bool> IsPrime;
void sieve(size_t max){
    if(max+1 > IsPrime.size()){     // resizeで要素数が減らないように
        IsPrime.resize(max+1,true); // IsPrimeに必要な要素数を確保
    } 
    IsPrime[0] = false; // 0は素数ではない
    IsPrime[1] = false; // 1は素数ではない

    for(size_t i=2; i*i<=max; ++i) // 0からsqrt(max)まで調べる
        if(IsPrime[i]) // iが素数ならば
            for(size_t j=2; i*j<=max; ++j) // (max以下の)iの倍数は
                IsPrime[i*j] = false;      // 素数ではない
}

struct UnionFind {
    vector<int> par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};


// ここから開始
double dp[3000][3000];

int main(){
    int N;
    cin>>N;
    double p[N];
    for(int i = 0; i < N; i++) cin>>p[i];
    dp[0][0] = 1;
    for(int i = 1; i <= N; i++){
        for(int o = 0; o <= i; o++){
            int u = i - o ;//表o + 裏u = i回
            // cout<<o<<"+"<<u<<"="<<i<<" ";
            if(o-1>=0) dp[o][u] += dp[o-1][u]*p[i-1];
            if(u-1>=0) dp[o][u] += dp[o][u-1]*(1.0-p[i-1]);
            // cout<<dp[o][u]<<endl;
        }
    }
    double ans = 0;
    for(int u = 0; u <= N; u++){
        int o = N - u;
        if(o > u) {
            // cout<<o<<"+"<<u<<"="<<i<<" ";
            ans+=dp[o][u];
        }
    }
    cout<<fixed<<setprecision(10)<<ans<<endl;
    return 0;
}