HDOJ2143 box

box

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9731    Accepted Submission(s): 2155

Problem Description

One day, winnie received a box and a letter. In the letter, there are three integers and five operations(+,-,*,/,%). If one of the three integers can be calculated by the other two integers using any operations only once.. He can open that mysterious box. Or that box will never be open.

 

Input

The input contains several test cases.Each test case consists of three non-negative integers.

 

Output

If winnie can open that box.print "oh,lucky!".else print "what a pity!"

 

Sample Input

1 2 3

 

Sample Output

oh,lucky!

 简单题，十分简单。

加和减是相对应的，乘和除是相对的，另外取余的时候a%b，b不能是0，需要防护。

import java.util.Scanner;public class Main{private static Scanner scanner;public static void main(String[] args) {scanner = new Scanner(System.in);while (scanner.hasNext()) {//没讲多大，使用longlong a = scanner.nextLong();long b = scanner.nextLong();long c = scanner.nextLong();boolean boo = false;// 加和减相对，乘和除相对if (a + b == c || b + c == a || a + c == b) {boo = true;} else if (a * b == c || b * c == a || a * c == b) {boo = true;}// 防范//对于：a%0出现的错误是java.lang.ArithmeticException: / by zeroelse if (a != 0 && (b % a == c || c % a == b)) {boo = true;} else if (b != 0 && (a % b == c || c % b == a)) {boo = true;} else if (c != 0 && (b % c == a || a % c == b)) {boo = true;}if (boo) {System.out.println("oh,lucky!");} else {System.out.println("what a pity!");}}}}