HDU 1005

Number Sequence


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

#include <iostream>  using namespace std;  int f[10000];   int main()  {      int A,B,n;      f[1] = f[2] = 1;      while(cin>>A>>B>>n, A || B || n)      {   int i;        for( i=3; i<10000 ;i++)          {              f[i] = (A*f[i-1] + B*f[i-2]) % 7;              //如果有两个连着 =1，则后面的全部和前面相同，即出现了周期              //这时就没必要再进行下去了，跳出循环, i-2为周期               if(f[i] == 1 && f[i-1] == 1)                              break;          }          n = n % (i-2);          // 把n对周期求模，当n = i-2时, n=0,此时本来应该取arr[i-2]的，所以把arr[0]=arr[i-2]           //也可以这样：          //if(n==0)   n=i-2;          f[0] = f[i-2];         cout << f[n] << endl;      }      return 0;  }