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 5

 

这道题完全不知道如何找规律，直接用递归会爆栈，最后是看的网上大神总结的周期

AC代码：

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int n = sc.nextInt();
            if (a < 1 & a > 1000 & b < 1 & b > 1000 & b < 1 & a > 100000000)
                break;
            if (a == 0 & b == 0 & n == 0)
                break;
            int f[] = new int[49];
            for (int i = 1; i < 49; i++) {  //通过大神总结出来的周期，直接用递归会爆栈
                if (i == 1 || i == 2) {
                    f[i] = 1;
                } else {
                    f[i] = (a * f[i - 1] + b * f[i - 2]) % 7;
                }
            }
            System.out.println(f[n % 49]);
        }

    }
}