Number Sequence

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

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

Author

CHEN, Shunbao

Source

ZJCPC2004

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有时候会怀疑出这种题目的作者是不是脑子有问题？只有20%的通过率,12万多的acmer提交只有3万不到的通过率!

你想题目一定是很难吧，其实不是，题目是很坑！(1 <= A, B <= 1000, 1 <= n <= 100,000,000)

里面n的取值范围是一个亿,如果直接递归来代入的话！肯定堆栈溢出.....（因为我试过!）

最后在网上查了一下思路：因为f(n)的结果是模7的，那么f(n)的结果只可能在{0,1,2,3,4,5,6,}中，这样f(n)与f(n-1)的结果只可能是7*7 = 49种，合理利用周期性，便可以大大提高程序运行时间。

不想说了，这题多说一句话都make me sick!!

感谢金永庆博友的解答！

http://blog.csdn.net/jinyongqing/article/details/21537175

import java.io.*;
import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        int a, b;
        int n;
        while (input.hasNext())
        {
            a = input.nextInt();
            b = input.nextInt();
            n = input.nextInt();
            if (a == 0 & b == 0 & n == 0)
                System.exit(0);
            int f[] = new int[50];
            for (int i = 1; i < 50; i++)
            {
                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]);
        }
    }  

}