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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
double sum = 0;
int n;

double dp[30001][3001];

int main(){
  int i,j;
  double p[3000];
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%lf",&p[i]);
  dp[0][0] = 1-p[0];
  dp[0][1] = p[0];

  for(i=0;i<n-1;i++){
    for(j=0;j<=i+1;j++){
      dp[i+1][j] += dp[i][j]*(1-p[i+1]);
      dp[i+1][j+1] += dp[i][j]*(p[i+1]);
    }
  }
  /*
  for(i=0;i<n+1;i++){
    for(j=0;j<n+1;j++){
      printf("%f ", dp[i][j]);
    }
    printf("\n");
  }*/
  double res=0;
  for(i=(n/2)+1;i<n+1;i++){
    res += dp[n-1][i];
  }

  printf("%.8f\n", res);

  return 0;


}