Number Sequence

Posted 2020-11-04 给我一个团队，干翻TX

题目是这样的：

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

　　首先，我来说说关于我的思路，我是通过找规律找周期来做的，但是做的时候要到了问题，我总是以为，周期的开头值必须是1，结果，就陷入了错误之中，并且无法自拔，

看了大量的博客之后，才知道如何去修改。（简单说说吧，就是周期的开头值不一定是1）

还不明白的，可以看看我的代码。

先看看我错误的代码吧：

package package1;

import java.util.Scanner;

public class Main
{
public static int A,B,n;
public static final int Maxn = 100;
public static int[] Arr = new int[Maxn];

public static void main(String[] args)
{
Scanner cin = new Scanner(System.in);
while(cin.hasNext())
{
A = cin.nextInt();
B = cin.nextInt();
n = cin.nextInt();
if(A == 0 && B == 0 && n == 0)
{
return;
}
int T;
T = init();
Arr[0] = Arr[T];
System.out.println(Arr[n%T]);
}
}
public static int init()
{
Arr[1] = 1;
Arr[2] = 1;
int i;
for(i = 3 ; i < Maxn ; i++)
{
Arr[i] = (A * Arr[i-1] + B * Arr[i-2])%7;
if(Arr[i-1] == 1 && Arr[i] == 1)
{
break;
}
}
return (i - 2);
}
}

 

修改之后的代码实现如下：