題目鏈接:https://www.luogu.com.cn/problem/P1200

代碼實例:

方法一:

import java.util.Scanner; public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String hx = scanner.next();String xz = scanner.next();char a[] = new char[]{'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};int b = 1,c = 1 ;for (int i = 0; i < hx.length(); i++) {for (int j = 0; j < a.length; j++) {if(hx.charAt(i)==a[j]) {b *= j+1; }}}for (int i = 0; i < xz.length(); i++) {for (int j = 0; j < a.length; j++) {if(xz.charAt(i)==a[j]) {c *= j+1; }}}if(b%47==c%47) {System.out.println("GO");}else {System.out.println("STAY");}scanner.close();}}

方法二:

import java.util.Scanner;public class Main{public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String hx = scanner.next();String xz = scanner.next();int b = 1,c = 1 ;for (int i = 0; i < hx.length(); i++) {b *= (hx.charAt(i)-'@');//這里對應ACSLL碼表------->'@'的ACSLL碼為64.'A'的ACSLL碼為65，所以65-64==1}for (int i = 0; i < xz.length(); i++) {b *= (xz.charAt(i)-'@');}if(b%47==c%47) {System.out.println("GO");}else {System.out.println("STAY");}scanner.close();}}

總結

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