HDU 1005 矩阵快速幂

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 169273 Accepted Submission(s): 41762

Problem Description

A number sequence is defined as follows:

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

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 3 1 2 10 0 0 0

Sample Output

2 5

Author

CHEN, Shunbao

Source

ZJCPC2004

题意：给你A B 求f(n)

题解：构造矩阵矩阵快速幂

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 #define mod 7
 4 using namespace std;
 5 int A,B,n;
 6 struct node
 7 {
 8     int a[3][4];
 9 }exm,ans;
10 struct node matrix_mulit(struct node aa,struct node bb)
11 {
12     struct node there;
13     for(int i=0;i<2;i++)
14     {
15         for(int j=0;j<2;j++)
16         {
17             there.a[i][j]=0;
18             for(int k=0;k<2;k++)
19             {
20                 there.a[i][j]=(there.a[i][j]+(aa.a[i][k]*bb.a[k][j]%mod))%mod;
21             }
22         }
23     }
24     return there;
25 }
26 int matrix_quick(int gg)
27 {
28     exm.a[0][0]=A;exm.a[0][1]=B;
29     exm.a[1][0]=1;exm.a[1][1]=0;
30     ans.a[0][0]=1; ans.a[0][1]=0;
31     ans.a[1][0]=1; ans.a[1][1]=0;
32     while(gg)
33     {
34         if(gg&1)
35             ans=matrix_mulit(exm,ans);
36         exm=matrix_mulit(exm,exm);
37         gg>>=1;
38     }
39     return ans.a[0][0];
40 }
41 int main()
42 {
43     while(scanf("%d %d %d",&A,&B,&n)!=EOF)
44     {
45     if(A==0&&B==0&&n==0)
46         break;
47     if(n==1||n==2)
48         printf("1
");
49     else
50         printf("%d
",matrix_quick(n-2));
51     }
52     return 0;
53 }