HDOJ 1005

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

/*

*/
#include<iostream>
using namespace std;
int a,b;
long n,t[500];

int main()
{
    int len;
	t[1] = t[2] =1;

	while(scanf("%d %d %ld",&a,&b,&n)!=EOF)
	{
		if(a==0 && b==0 && n==0)
			break;
		if(n>=3)
		{
		//for(len=3; len ==3 || !( t[len-2]==1 && t[len-1] == 1 );len++)
	    for(len =3; len < 200 ; len++)
		{
		    t[len]=(a*t[len-1]+b*t[len-2])%7;
			if(t[len] == 1 && t[len-1] == 1)
				break;
		    //printf("%d ",t[len]);
		}
		 len -= 2;

		n = n%len;
		if(n ==0)
			n = len;
		printf("%ld\n",t[n]);
		}
		else printf("1\n");
		//if(n>len) printf("%d\n",t[(n%len==0)? len : n%len]);
		//else printf("%d\n",t[n]);
	}
	return 0;
}