Poor God Water（ACM-ICPC 2018 焦作赛区网络预赛 矩阵快速幂）

Posted 2022-06-05 f-mu

题目描述

God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him that some sequence of eating will make them poisonous.
Every hour, God Water will eat one kind of food among meat, fish and chocolate. If there are 3 continuous hours when he eats only one kind of food, he will be unhappy. Besides, if there are 3 continuous hours when he eats all kinds of those, with chocolate at the middle hour, it will be dangerous. Moreover, if there are 3 continuous hours when he eats meat or fish at the middle hour, with chocolate at other two hours, it will also be dangerous.
Now, you are the doctor. Can you find out how many different kinds of diet that can make God Water happy and safe during N hours? Two kinds of diet are considered the same if they share the same kind of food at the same hour. The answer may be very large, so you only need to give out the answer module 1000000007.

输入

The fist line puts an integer T that shows the number of test cases. (T≤1000)
Each of the next T lines contains an integer N that shows the number of hours. (1≤N≤10^10)

输出

For each test case, output a single line containing the answer.

 

题意：有肉，鱼，巧克力三种食物，有几种禁忌，对于连续的三个食物，1.这三个食物不能都相同；2.若三种食物都有的情况，巧克力不能在中间；3.如果两边是巧克力，中间不能是肉或鱼，求方案数。

分析：设巧克力是1，肉是2，鱼是3。则对于n个小时来说，n-2和n-1的食物可以推出n-1和n的食物，设i=(k-1)*3+j，其中k是第n-2小时吃的食物编号，j是第n-1小时吃的食物编号，例如第n-2小时吃了肉，第n-1小时吃了鱼，则a6是这种方案的种数。通过枚举可算出递推：a1=a4+a7;a2=a1+a4;a3=a1+a7;a4=a5+a8;a5=a2+a8;a6=a2+a5+a8;a7=a6+a9;a8=a3+a6+a9;a9=a3+a6。（左式为第n-1小时和n小时吃某两类事物的方案数，右式为第n-2小时和第n-1小时吃某两类食物的方案数）。则答案为一个9*1的全为1的矩阵（代表a1~a9）与一个9*9的矩阵的n-2（n>=3）次幂相乘（若ai=aj+ak，则res[i][j]和res[i][k]都为1）。

 1 #include <bits/stdc++.h>
 2 
 3 #define maxn 10
 4 #define mod 1000000007
 5 #define inf 0x3f3f3f3f
 6 #define start ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 7 #define ll long long
 8 #define LL long long
 9 using namespace std;
10 
11 struct Mat 
12     ll mat[maxn][maxn];
13 
14     Mat() 
15         memset(mat, 0, sizeof(mat));
16     
17 ;
18 
19 int n = 9;
20 
21 Mat operator*(Mat a, Mat b) 
22     Mat c;
23     memset(c.mat, 0, sizeof(c.mat));
24     int i, j, k;
25     for (k = 1; k <= n; k++) 
26         for (i = 1; i <= n; i++) 
27             if (a.mat[i][k] == 0) continue;//优化
28             for (j = 1; j <= n; j++) 
29                 if (b.mat[k][j] == 0) continue;//优化
30                 c.mat[i][j] = (c.mat[i][j] + (a.mat[i][k] * b.mat[k][j]) % mod) % mod;
31             
32         
33     
34     return c;
35 
36 
37 Mat operator^(Mat a, ll k) 
38     Mat c;
39     int i, j;
40     for (i = 1; i <= n; i++)
41         for (j = 1; j <= n; j++)
42             c.mat[i][j] = (i == j);
43     for (; k; k >>= 1) 
44         if (k & 1)
45             c = c * a;
46         a = a * a;
47     
48     return c;
49 
50 
51 int main() 
52     start;
53     int T;
54     cin >> T;
55     while (T--) 
56         ll n;
57         cin >> n;
58         if (n == 1)
59             cout << 3 << endl;
60         else if (n == 2)
61             cout << 9 << endl;
62         else 
63             Mat res;
64             res.mat[1][4] = res.mat[1][7] = 1;
65             res.mat[2][1] = res.mat[2][4] = 1;
66             res.mat[3][1] = res.mat[3][7] = 1;
67             res.mat[4][5] = res.mat[4][8] = 1;
68             res.mat[5][2] = res.mat[5][8] = 1;
69             res.mat[6][2] = res.mat[6][5] = res.mat[6][8] = 1;
70             res.mat[7][6] = res.mat[7][9] = 1;
71             res.mat[8][3] = res.mat[8][6] = res.mat[8][9] = 1;
72             res.mat[9][3] = res.mat[9][6] = 1;
73             Mat cnt = res ^(n - 2);
74             ll t = 0;
75             for (int i = 1; i <= 9; ++i)
76                 for (int j = 1; j <= 9; ++j)
77                     t = (t + cnt.mat[j][i]) % mod;
78             cout << t << endl;
79         
80     
81     return 0;
82 

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