【ACM】杭电OJ 1005 Number Sequence (for java)

版权声明：MZ21G https://blog.csdn.net/qq_35793285/article/details/84678907

最开始的代码是这样的，没说我超时，直接报错~~

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	
	public static int num(int A,int B,int n) {
			int[] ans = new int[n + 1];
			ans[1] = 1;
			ans[2] = 1;
			for(int i = 3; i <= n; i++) {
				ans[i] = (A * ans[i - 1]  + B * ans[i - 2]) % 7;
			}
			return ans[n];
	}
	

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		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) {
				break;
			}
			
			int res = num(A,B,n);		
			System.out.println(res);

		}
	}	
}

后来通过参考别人的代码发现是有规律的，最后是对7取余数，所以，f(n-1)，f(n-2)最多各有七种情况（0、1、2、3、4、5、6），所以f(n)最多有7*7=49种情况，所以从一开始到49可能不尽相同，但是从50开始则开始循环，f(50)=f(50%49)=f(1)。这是解决超时和内存不够的办法。 

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	
	public static int num(int A,int B,int n) {
			int[] ans = new int[n + 1];
			ans[1] = 1;
			ans[2] = 1;
			for(int i = 3; i <= n; i++) {
				ans[i] = (A * ans[i - 1]  + B * ans[i - 2]) % 7;
			}
			return ans[n];
	}
	

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		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) {
				break;
			}
			
			int res = num(A,B,n);		
			System.out.println(res);

		}
	}	
}

参考资料：

https://blog.csdn.net/CSDN___CSDN/article/details/82933844