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 31 2 100 0 0

 

 

Sample Output

25
代码：
#include<cstdio>
const int max=52;
int f[max];
int  main()
{
 int a,b,n;
 scanf("%d%d%d",&a,&b,&n);
 a%=7;
 b%=7;
 while(a||b||n)
 {
  f[1]=1,f[2]=1;
  for(int i=3;i<max;i++)
  {
   f[i]=(a*f[i-1]+b*f[i-2])%7;
   if(f[i-1]==f[3]&&f[i]==f[4]&&i>4)
    break;
  }
  int t=i-4;   
        if(n<4)
   printf("%d/n",f[n]);  
        else
   printf("%d/n",f[(n-4)%t+4]);
 /* n=(n-2)%(i-4);
  printf("%d/n",f[2+n]);*/
  scanf("%d%d%d",&a,&b,&n);
 }
 return 0;
}