HDU-1005 JAVA

原创于 2020-01-06 16:46:38 发布 · 194 阅读

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本文探讨了一种特定数列的周期性规律，通过实例分析，发现每48次为一周期，利用此规律，可以高效计算大规模数列的值。提供了一段Java代码实现，用于快速计算数列中任意项的值。

Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 232904 Accepted Submission(s): 59123

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

找呀找呀找规律，每48次为一个轮回，以a=1,b=1,n=48为例，从左往右，从上往下为n=1-48的结果，n=49时回到n=1时的结果，开始轮回
在这里插入图片描述

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==0&&b==0&&n==0){
                break;
            }
            else{
                int num[]=new int[49];
                num[1]=1;num[2]=1;
                for(int i=3;i<49;i++){
                    num[i]=(a*num[i-1]+b*num[i-2])%7;
                }
                int sum=n%48;
                System.out.println(num[sum]);
                
            }
        }
        
    }
}