hdu2143

转自 http://blog.csdn.net/ysc504/article/details/8505630

box

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5323    Accepted Submission(s): 1023

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!

 

Author

kiki

 

Source

HDU 2007-11 Programming Contest

 

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威士忌

 

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1. #include <stdio.h>  
2. int fun(__int64 a, __int64 b, __int64 c)  
3. {  
4.     if(a + b == c)  
5.         return 1;  
6.     if(a * b == c)  
7.         return 1;  
8.     if(a != 0 && b % a == c)  
9.         return 1;  
10.     return 0;  
11. }  
12. int main()  
13. {  
14.     __int64 a, b, c, flag;  
15.     while (~scanf("%I64d%I64d%I64d", &a, &b, &c))  
16.     {  
17.           
18.         flag = 0;  
19.         if(fun(a, b, c))  
20.             flag = 1;  
21.         else if(fun(a, c, b))  
22.             flag = 1;  
23.         else if(fun(b, a, c))  
24.             flag = 1;  
25.         else if(fun(b, c, a))  
26.             flag = 1;  
27.         else if(fun(c, b, a))  
28.             flag = 1;  
29.         else if(fun(c, a, b))  
30.             flag = 1;  
31.         if(flag)  
32.             printf("oh,lucky!\n");  
33.         else  
34.             printf("what a pity!\n");  
35.     }  
36.     return 0;  
37. }