poj3744 概率

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

这个题想直接转移那肯定是…
必然会t
我确实想到了矩阵优化
但是我很傻比我矩阵到了1e8..
还是t了
发现用1减掉到不了的概率就可以了。。
哇我是真的sb

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cstdio>
using namespace std;
struct qq
{
    double tu[2][2];
};
int tu[20];
qq cheng(qq q, qq w)
{
    qq fanhui = { { { 0,0 },{ 0,0 } } };
    for (int a = 0;a < 2;a++)
    {
        for (int b = 0;b < 2;b++)
        {
            for (int c = 0;c < 2;c++)
            {
                fanhui.tu[a][b] += q.tu[a][c] * w.tu[c][b];
            }
        }
    }
    return fanhui;
}
qq ksm(qq ds, int zhishu)
{
    qq fanhui = { {{1,0},{0,1}} };
    while (zhishu)
    {
        if (zhishu & 1)fanhui = cheng(fanhui, ds);
        zhishu >>= 1;
        ds = cheng(ds, ds);
    }
    return fanhui;
}
int main()
{
    int n;
    double p, fp;
    while (cin >> n >> p)
    {
        fp = 1 - p;
        double qqq = 1;
        for (int a = 1;a <= n;a++)scanf("%d", &tu[a]);
        qq csjz = { { { p,fp },{ 1,0 } } };
        qq jiegg= ksm(csjz,tu[1]-1);
        qqq = 1 - jiegg.tu[0][0];
        sort(tu + 1, tu + n + 1);
        for (int a = 2;a <= n;a++)
        {
            if (tu[a] == tu[a - 1])continue;
            int zss = tu[a]- tu[a-1]-1;
            qq jieguo = ksm(csjz, zss);
            qqq *= (1 - jieguo.tu[0][0]);
        }
        printf("%.7f\n", qqq);
    }
    return 0;
}