Scout YYF I

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9552		Accepted: 2793

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.

The input contains many test cases ended with EOF.

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

 0.5

 0.5
 

0.5000000

0.2500000

# include <iostream>

# include <cstdio>

# include <cstring>

# include <algorithm>

using namespace std;

int n;

int dis[];

struct fi{

    double data[][];

    fi(){

        for(int i = ;i < ;i++){

            for(int j = ;j < ;j++){

                data[i][j] = ;

            }

        }

    }

}A,T,O;

inline fi operator *(fi A,fi B){

    fi t;

    for(int i = ;i < ;i++){

        for(int j = ;j < ;j++){

            for(int k = ;k < ;k++){

                t.data[i][j] += A.data[i][k] * B.data[k][j];

            }

        }

    }

    return t;

}

fi cmd(fi C,int k){

    fi D = O;

    while(k){

        if(k & )D = D * C;

        k >>= ;

        C = C * C;

    }

    return D;

}

double p;

int main(){

    for(int i = ;i < ;i++)O.data[i][i] = 1.0;

    while(~scanf("%d %lf",&n,&p)){

    T.data[][] =  (1.0 - p);

    T.data[][] = p;

    T.data[][] = 1.0;

    T.data[][] = 0.0;

        for(int i = ;i < ;i++){

            for(int j = ;j < ;j++){

                A.data[i][j] = 0.0;

            }

        }

        A.data[][] = 1.0;

    for(int i = ;i <= n;i++){

        scanf("%d",&dis[i]);

    }

    sort(dis + ,dis + n + );

    for(int i = ;i <= n;i++){

        A = A * cmd(T,dis[i] - dis[i - ] - );

        A.data[][] = ;

        A = A * T;

    }

    printf("%.7f\n",A.data[][]);

  }

}