hdu2143 box

box

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7186    Accepted Submission(s): 1452

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!

 

题意：有3个数字abc，五种操作(+-*/%)，如果有一个数字能被另外两个数字用那5种操作之一一次计算出来，则输出oh,lucky!，否则输出what a pity!

加减互逆，乘除互逆，二者有一即可，这道题用除法会出问题,除法不是整除.也不能直接用加法。

#include<stdio.h>#include <iostream>using namespace std;int main(){    long long a,b,c;    while (scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)    {        if(a*b==c|| a*c==b|| b*c==a)        {            printf("oh,lucky!\n");        }        else if((b!=0&& (a%b==c||c%b==a))||(a!=0&&(c%a==b||b%a==c))||(c!=0&&(a%c==b ||b%c==a)))        {           printf("oh,lucky!\n");        }        else        {            printf("what a pity!\n");        }    }    return 0;}