2142 HDU box

box

Time Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3730Accepted Submission(s): 674

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

 

Recommend

威士忌

尼期最后一条水题， RE两次， 第一次唔记得判断除数是否为0，第二次加左判断，但系放错地方{= =}

我见到上面AC率甘底，一开始就用__int64。

最后贴贴代码：

4316855

2011-08-02 21:18:52

Accepted

2143

0MS

188K

563 B

C++

10SGetEternal{（。）（。）}!

#include <iostream>
using namespace std;

bool jud(__int64 x, __int64 y, __int64 z)
{
    if (
        x + y == z
        || x * y == z
        || x - y == z
        || y - x == z
        || (y != 0 && x % y == z)
        || (x != 0 && y % x == z)
        || (y != 0 && x % y == 0 && x / y == z)
        || (x != 0 && y % x == 0 && y / x == z)
        )
        return 1;
    return 0;
}

int main()
{
    __int64 a, b, c;

    while (scanf("%I64d%I64d%I64d", &a, &b, &c) != EOF)
    {
        if (jud(a, b, c) || jud(a, c, b) || jud(b, c, a))
            puts("oh,lucky!");
        else puts("what a pity!");
    }

    return 0;
}

系时候要搞好多野鸟………………