hdu 1005 Number Sequence (矩阵快速幂加速线性递推式)
题意：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).
思路：矩阵加速线性递推式。
这题第一次看是2012年11月2333，当时用pascal写的
/* ***********************************************
Author :111qqz
Created Time :Mon 31 Oct 2016 05:03:40 AM CST
File Name :code/hdu/1005.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6struct Mat
7{
8 LL mat[2][2];
9 void clear()
10 {
11 ms(mat,0);
12 }
13}M,M1;
14Mat operator * ( Mat a,Mat b)
15{
16 Mat c;
17 c.clear();
18 for ( int i = 0 ;i < 2 ; i ++)
19 for ( int j = 0 ; j < 2 ; j++)
20 for ( int k = 0 ; k < 2 ; k++)
21 c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j] )%7;
22 return c;
23}
24Mat operator ^ ( Mat a,LL b)
25{
26 Mat res;
27 res.clear();
28 for ( int i = 0 ; i < 2 ; i++)
29 res.mat[i][i] = 1;
30 while (b>0)
31 {
32 if (b&1) res = res * a;
33 b = b >> 1LL;
34 a = a * a;
35 }
36 return res;
37}
38LL A,B,n;
39void init()
40{
41 M.clear();
42 M.mat[0][1] = 1;
43 M.mat[1][0] = B;
44 M.mat[1][1] = A;
45 M1.clear();
46 M1.mat[0][0] = 1;
47 M1.mat[1][0] = 1;
48}
49int main()
50{
51 #ifndef ONLINE_JUDGE
52 freopen("code/in.txt","r",stdin);
53 #endif
54 while (~scanf("%lld%lld%lld",&A,&B,&n))
55 {
56 init();
57 if (A==0&&B==0&&n==0) break;
58 Mat res = (M^(n-2))*M1;
59 printf("%lld\n",res.mat[1][0]);
}
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}