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;

#define int long long
#define pii pair<int,int>
#define pqi priority_queue<int>
#define pb push_back()
#define INF 1000000000000000000

#define pi acos(-1)

const int modu= (1e9)+7;

void testCase(int ti){
    cout<<"Case "<<ti<<": ";
}

int bigmod(int n,int p){
    if(p==0) return 1;
    int x=bigmod(n,p/2)%modu;
    x=(x*x)%modu;
    if(p%2) return (x*n)%modu;
    return x;
}

int modInverse(int a){
    return bigmod(a,modu-2)%modu;
}

void debug(string pr,vector<int> var,bool out=true){
    #ifdef OFFLINE
        ofstream fl;
        fl.open("debug.txt");
        fl<<pr;
        for(int i=0;i<var.size();i++) fl<<" "<<var[i];
        fl<<endl;
        fl.close();
    if(out){
        cout<<pr;
        for(int i=0;i<var.size();i++) cout<<" "<<var[i];
        cout<<endl;
    }
    #endif
}

void faster(){
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
}

void defInit(){
    //faster();
    #ifdef OFFLINE
      freopen("in.txt", "r", stdin);
      freopen("out.txt", "w", stdout);
    #endif
}

int add(int x,int y){
    int z =(x%modu + y%modu)%modu;
    while(z<0) z=(z+modu)%modu;
    return z;
}

int mul(int x,int y){
    int z = (x%modu * y%modu)%modu;
    while(z<0) z=(z+modu)%modu;
    return z;
}
// starts here
#define float long double
const int MAXN = 3000;
int n;
float ara[MAXN];
vector<vector<float>> dp(MAXN,vector<float>(MAXN,-1));

float calc(int idx,int rem){
    float &now=dp[idx][rem];
    if(now!=-1) return now;
    if(idx==n) return now=1;
    float ans1=0,ans2=0;
    if(rem) ans1=(1-ara[idx])*calc(idx+1,rem-1);
    ans2=ara[idx]*calc(idx+1,rem);
    return now=ans1+ans2;
}

int32_t main(){
    defInit();
    //Now I will start writing codes and start preparation to get errors
    cin>>n;
    for(int i=0;i<n;i++) cin>>ara[i];
    int rem=(n-1)/2;
    cout<<fixed<<setprecision(10);
    cout<<calc(0,rem);
    return 0;
}