Number Sequence_hdu_1005(规律)

9329854 2013-10-13 14:36:41 Accepted 1005 171MS 5072K 654 B Java zhangyi

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

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 86162 Accepted Submission(s): 20434

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

*/

import java.util.Scanner;

public class Main{
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
while(true){
int a=input.nextInt();
int b=input.nextInt();
long n=input.nextInt();
if(a==0&&b==0&&n==0)
break;
int f[]=new int[1000];
f[1]=f[2]=1;
int s[][]=new int[7][7];
s[1][1]=2;
int i=0;
for(i=3;i<60;i++){
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(s[f[i-1]][f[i]]!=0){
break;
}
s[f[i-1]][f[i]]=i;
}
int d=i-s[f[i-1]][f[i]];
n-=s[f[i-1]][f[i]];
n%=d;
if(n==0)n+=d;
n+=s[f[i-1]][f[i]];
System.out.println(f[(int)n]);
}
}
}