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>
#include <stdlib.h>

int main() {
    int N;
    scanf("%d", &N);
    float *p = malloc((N + 1) * sizeof(float));
    float **DP = malloc((N + 1) * sizeof(float*));
    for (int i = 1; i <= N; i++) {
        scanf("%f", &p[i]);
        DP[i] = malloc((N + 1) * sizeof(float));
    }

    // initialize
    DP[1][0] = 1 - p[1];
    DP[1][1] = p[1];
    for (int i = 2; i <= N; i++) {
        DP[1][i] = 0;
    }

    for (int i = 3; i <= N; i += 2) {
        DP[i][0] = DP[i - 2][0] * (1 - p[i -1]) * (1 - p[i]); // どっちも裏
        DP[i][1] = DP[i - 2][0] * (p[i -1] * (1 - p[i]) + (1 - p[i - 1]) * p[i]) // どっちか表
                + DP[i - 2][1] * (1 - p[i -1]) * (1 - p[i]); // どっちも裏
        for (int j = 2; j <= i; j++) {
            DP[i][j] = (DP[i-2][j - 2] * p[i-1] * p[i])// 2枚表が少ない状態から二枚とも表
                      + (DP[i-2][j - 1] * (p[i-1] * (1 - p[i]) + (1 - p[i-1]) * p[i])) // 1枚少ない状態からどっちかが表
                      + (DP[i-2][j] * (1 - p[i-1]) * (1 - p[i])); // 足りてる状態からどっちも裏
        }
    }

    float ans = 0;
    for (int i = (N - 1)/2 + 1; i <= N; i++) {
        ans += DP[N][i];
    }
    printf("%f\n", ans);

    return 0;
}