杭电1005

    貌似这道题挺难得，我现在还是不太懂得，还是搞不明白为什么一定会存在循环节，等今天比赛完了，问一下队长吧，这个让我真的很费解。。。。题目：

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
   
   
   
   

ac代码，，，，，这代码也不是我自己写的，唉！失败。。。。

#include <iostream>
using namespace std;
int main()
{
  int a,b,n,f[55];
  while(cin>>a>>b>>n&&a&&b&&n)
  {
    f[1]=1,f[2]=1;
	int i;
	for(i=3;i<50;++i)
	{
	  f[i]=((a*f[i-1])%7+(b*f[i-2])%7)%7;
	  //cout<<f[i]<<" "<<f[i-1]<<" "<<f[i-2]<<endl;
	  if(f[i]==f[i-1]&&f[i-1]==1)
	  {break;}
	}
	i-=2;
	n=n%i;
	//cout<<k<<endl;
    if(n)
		cout<<f[n]<<endl;
	else
		cout<<f[i]<<endl;
  }
  return 0;
}