Scout YYF I  POJ - 3744 

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

有n個地雷，分別是1-100000000之間

你在位置1，每次走一步的概率是p，走兩步的概率是1-p

問你每一個地雷都避過的概率是多少

每一次，如果該點是地雷就不能走，概率為0，那麼每兩個地雷中間有多少，就走多少正常的

如果遇見地雷了，就把那一步的概率變為0，記得到了最後要在走一步，因為要完美避開所有地雷，就是要走到最後一個地雷後一格

因為最多可能走1e8，不能線性推，但是dp[n]=dp[n-1]*p+dp[n-2]*(1-p)，可以用矩陣快速冪

特別注意，你人的起始位置是1 

and   POJ輸出請用f

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n,mine[15];
double p;

struct node
{
    double a[3][3];
};

node matrix(node x,node y)
{
    node r;
    memset(r.a,0,sizeof(r.a));
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            for(int k=0;k<2;k++)
            {
                r.a[i][j]+=x.a[i][k]*y.a[k][j];
            }
        }
    }
    return r;
}

double quickpow(int x,node r)
{
    node c;
    memset(c.a,0,sizeof(c.a));
    c.a[0][0]=p,c.a[0][1]=1,c.a[1][0]=1-p;
    while(x)
    {
        if(x%2) r=matrix(r,c);
        c=matrix(c,c);
        x/=2;
    }
    return r.a[0][0];
}

int main()
{
    while(~scanf("%d%lf",&n,&p))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&mine[i]);
        if(mine[1]==1)
        {
            printf("0.0000000\n");
            continue;
        }
        double q=0;
        mine[0]=1;
        node r;
        r.a[0][0]=1,r.a[0][1]=0;
        int t=0;
        for(int i=1;i<=n;i++)
        {
            if(i!=1)
            {
                r.a[0][0]=0,r.a[0][1]=q;
            }
            t=mine[i]-mine[i-1]-1;
            q=quickpow(t,r);
        }
        r.a[0][0]=0,r.a[0][1]=q;
        q=quickpow(1,r);
        printf("%.7f\n",q);
    }
    return 0;
}