Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124651    Accepted Submission(s): 30286

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

　　这题可以找规律，但是首先想到的还是矩阵快速幂。

代码如下：

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cmath>
 5
 6 using namespace std;
 7
 8 #define MOD 7
 9 #define MAXN 10
10
11 typedef struct MAT
12 {
13     int d[MAXN][MAXN];
14     int r, c;
15     MAT()
16     {
17         r = c = 0;
18         memset(d, 0, sizeof(d));
19     }
20 }MAT;
21
22 MAT mul(MAT m1, MAT m2, int mod)
23 {
24     MAT ans = MAT();
25     ans.r = m1.r;
26     ans.c = m2.c;
27     for(int i = 0; i < m1.r; i++)
28     {
29         for(int j = 0; j < m2.r; j++)
30         {
31             if(m1.d[i][j])
32             {
33                 for(int k = 0; k < m2.c; k++)
34                 {
35                     ans.d[i][k] = (ans.d[i][k] + m1.d[i][j] * m2.d[j][k]) % mod;
36                 }
37             }
38         }
39     }
40     return ans;
41 }
42
43 MAT quickmul(MAT m, int n, int mod)
44 {
45     MAT ans = MAT();
46     for(int i = 0; i < m.r; i++)
47     {
48         ans.d[i][i] = 1;
49     }
50     ans.r = m.r;
51     ans.c = m.c;
52     while(n)
53     {
54         if(n & 1)
55         {
56             ans = mul(m, ans, mod);
57         }
58         m = mul(m, m, mod);
59         n >>= 1;
60     }
61     return ans;
62 }
63
64 int main() {
65     int n, m, t, a, b, k;
66     while(scanf("%d %d %d", &a, &b, &n) && a + b + n) {
67         MAT tmp, ans;
68         tmp.r = 2, tmp.c = 2;
69         ans.r = 2, ans.c = 1;
70         if(n == 1 || n == 2) {
71             printf("%d\n", 1);
72         }
73         else {
74             tmp.d[0][0] = a;
75             tmp.d[0][1] = b;
76             tmp.d[1][0] = 1;
77             tmp.d[1][1] = 0;
78             ans = quickmul(tmp, n - 2, MOD);
79             printf("%d\n", (ans.d[0][0]+ans.d[0][1]) % MOD);
80         }
81     }
82 }