Number Sequence

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

Total Submission(s): 188565    Accepted Submission(s): 47058

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

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由于有一个模7，所以我们猜想这个数列会循环，什么时候循环一次？当1 1第二次出现的时候就证明开始循环了，所以我们要找第二个1 1

注意：数组大小，循环的条件不要设的太大，会出错，为什么会出错？

我也不知道= =！

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源码;

import java.util.Arrays;

import java.util.Scanner;

public class Main {

    public static void main(String[] args) 

    {

        Scanner scanner = new Scanner(System.in);

        while(scanner.hasNext())

        {

            int a = scanner.nextInt();

            int b = scanner.nextInt();

            int n = scanner.nextInt();

            if(a==0 && b==0 && n==0)

                System.exit(0);

            if(a==0 && b==0)

            {

                System.out.println(0);

                continue;

            }

            int i =0;

            int[] result = new int[100000];

            result[0]=1;

            result[1]=1;

            for(i = 2 ;i < 1000 ; i++)

            {

                   result[i]=(a*result[i-1]+b*result[i-2])%7;

                   if(result[i-2] == 1 && result[i-1] == 1 && i != 2) break;

            }

            System.out.println(result[(n-1)%(i-2)]);

        }

    }

}