Number Sequence(快速幂矩阵）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 131753    Accepted Submission(s): 31988

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：简单的快速幂矩阵，加一个f(n-1) = f(n-1)即可

代码：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 struct Mtr{
     int a , b , c , d ;
     Mtr operator * (const Mtr m) const
     {
         Mtr res ;
         res.a = (a*m.a% + b*m.c%)%;
         res.b = (a*m.b%+b*m.d%)%;
         res.c = (c*m.a%+d*m.c%)%;
         res.d = (c*m.b%+d*m.d%)%;
         return res;
     }
 };
 int cal(Mtr a , int n)
 {
     Mtr c;
     c.a = c.d = ;
     c.b = c.c = ;
     while(n>)
     {
         if(n&)
             c = c*a;
         a = (a*a);
         n/=;
     }
     return (c.a+c.b)%;
 }
 int main()
 {
     int a , b , n;
     while(~scanf("%d%d%d",&a,&b,&n)&&(a!=||b!=||n!=))
     {
         Mtr tm;
         tm.a = a;
         tm.b = b;
         tm.c = ;
         tm.d = ;
         int tt = cal(tm,n-);
         printf("%d\n",tt);
     }
     return ;
 }