HDU-1005 JAVA

妖狐艹你老母 2023-06-27 12:35 8阅读 0赞

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][20200106164443343.png]

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]);
                    
                }
            }
            
        }
    }

[20200106164443343.png]: https://img-blog.csdnimg.cn/20200106164443343.png