題面：

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 7254	Accepted: 2118

Description

YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1-p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.

Input

The input contains many test cases ended with EOF.
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].

Output

For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.

Sample Input

1 0.5
2
2 0.5
2 4

Sample Output

0.5000000
0.2500000

題目大意：

一條路上有n個雷，走一步的可能性是p，走兩步的可能性是1-p，起點為1，求安全抵達n的概率。

題目分析：

矩陣相乘的題目，

如果k號位有雷，那麼安全通過這個雷只可能是在k-1號位選擇走兩步到k+1號位。因此，可以得到如下結論：在第i個雷的被處理掉的概率就是從a[i-1]+1號位到a[i]號位的概率。於是，可以用1減去就可以求出安全通過第i個雷的概率，最後乘起來即可，但是由於資料很大，所以需要用到矩陣快速冪，類似斐波那契數列，有ans[i]=p*ans[i-1]+(1-p)*ans[i-2]，構造矩陣為：

 

程式碼實現：

#include <iostream>
#include <stdio.h>
#include <algorithm>

using namespace std;

int n;
double p;
int a[20];
double b[5][5];
double s[5];
double tmp[5][5];
double temp[5];

double Count(int t)
{
    b[0][0]=p;
    b[0][1]=1-p;
    b[1][0]=1;
    b[1][1]=0;
    s[0]=1;
    s[1]=0;
    while(t!=0)
    {
        if(t%2==1)
        {
            for(int i=0;i<2;i++)
            {
                temp[i]=0;
                for(int j=0;j<2;j++)
                {
                    temp[i]+=b[i][j]*s[j];
                }
            }
            for(int i=0;i<2;i++)
            {
                s[i]=temp[i];
            }
        }
        t/=2;
        for(int i=0;i<2;i++)
        {
            for(int j=0;j<2;j++)
            {
                tmp[i][j]=0;
                for(int k=0;k<2;k++)
                {
                    tmp[i][j]+=b[i][k]*b[k][j];
                }
            }
        }
        for(int i=0;i<2;i++)
        {
            for(int j=0;j<2;j++)
            {
                b[i][j]=tmp[i][j];
            }
        }
    }
    return s[0];
}

int main()
{
    double ans=0;
    while(scanf("%d%lf",&n,&p)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        a[0]=0;
        sort(a,a+n+1);
        ans=1;
        for(int i=0;i<n;i++)
        {
            ans=ans*(1-Count(a[i+1]-a[i]-1));
        }
        printf("%.7lf\n",ans);
    }
    return 0;
}