hdu 1005 根据递推公式构造矩阵 ( 矩阵快速幂)

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.


Sample Input
1 1 3 //a b n
1 2 10
0 0 0

Sample Output
2
5

 

矩阵A  * 矩阵B  = 矩阵C 

a b        f(n-1)     f(n)

1 0        f(n-2)    f(n-1)

 

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <algorithm>
 4 # include <map>
 5 # include <cmath>
 6 # define LL long long
 7 using namespace std ;
 8 
 9 const int MOD = 7 ;
10 
11 struct Matrix
12 {
13     int mat[2][2];
14 };
15 
16 Matrix mul(Matrix a,Matrix b) //矩阵乘法
17 {
18     Matrix c;
19     for(int i=0;i<2;i++)
20         for(int j=0;j<2;j++)
21         {
22             c.mat[i][j]=0;
23             for(int k=0;k<2;k++)
24             {
25                 c.mat[i][j]=(c.mat[i][j] + a.mat[i][k]*b.mat[k][j])%(MOD);
26             }
27         }
28     return c;
29 }
30 Matrix pow_M(Matrix a,int k)  //矩阵快速幂
31 {
32     Matrix ans;
33     memset(ans.mat,0,sizeof(ans.mat));
34     for (int i=0;i<2;i++)
35         ans.mat[i][i]=1;
36     Matrix temp=a;
37     while(k)
38     {
39         if(k&1)ans=mul(ans,temp);
40         temp=mul(temp,temp);
41         k>>=1;
42     }
43     return ans;
44 }
45 
46 
47 
48 int main ()
49 {
50    // freopen("in.txt","r",stdin) ;
51     int a,b,n;
52     while(scanf("%d%d%d" , &a,&b,&n) != EOF)
53     {
54         if (a==0 && b==0 && n==0)
55             break ;
56         if (n <= 2)
57         {
58             printf("1\n") ;
59             continue ;
60         }
61         Matrix t ;
62         t.mat[0][0] = a ;
63         t.mat[0][1] = b ;
64         t.mat[1][0] = 1 ;
65         t.mat[1][1] = 0 ;
66         Matrix ans = pow_M(t,n-2) ;
67         printf("%d\n" , (ans.mat[0][0] + ans.mat[0][1])%MOD) ;
68 
69     }
70 
71 
72     return 0 ;
73 }
View Code

 

posted @ 2015-05-28 13:18  __Meng  阅读(253)  评论(0)  编辑  收藏  举报