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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
double coins[3005];
int n;
double dp[3005][1600][3];
double func(int idx,int hc,int k){
    
    if(dp[idx][hc][k]>-10) return dp[idx][hc][k];
   
    double res=0;
    if(idx==n && hc<=(n/2)) res=0;
    else if(idx==n){
        if(k==1) res=coins[idx];
        else res=(1-coins[idx]);
    } 
    
    else{
        double t1,t2;
        
        if(k==0){
            t1=func(idx+1,hc,0)*(1-coins[idx]);
            t2=func(idx+1,hc+1,1)*(1-coins[idx]);
            }
        else{
             t1=func(idx+1,hc,0)*(coins[idx]);
            t2=func(idx+1,hc+1,1)*(coins[idx]);
        }
 
        res=t1+t2;
       
    }
   dp[idx][hc][k]=res;
  // cout << dp[idx][hc][k] << endl;
    return res;
}
int main(){
    for(int i=0;i<3005;i++) for(int j=0;j<1600;j++) for(int l=0;l<3;l++) dp[i][j][l]=-1000;
    cin >> n;
    for(int i=1;i<=n;i++) cin >> coins[i];
    printf("%0.10lf\n",func(0,0,0));
}