hduoj 2143 box_one day, winnie received a box and a letter. in th-优快云博客

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本文介绍了一个简单的编程挑战——通过数学运算判断是否能打开一个神秘盒子。文章提供了完整的代码实现，并强调了处理边界情况的重要性。

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box

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8574 Accepted Submission(s): 1830

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 C

题意很简单，就是有细节问题会注意不到

a.b.c均为int型所以相加或相乘可能超int 要用long long

另外要注意 a%b时要先判定b！=0 不然提交就会出现RE的问题

判定了+就不用判定-，判定了*就不用判定/（题上/是除而不是取整）

#include<stdio.h>
int main()
{
  long long a,b,c;
  while(~scanf("%lld%lld%lld",&a,&b,&c))
  {
      int flag=0;
      if(a+b==c||a+c==b||b+c==a)
      {
          flag=1;
      }
      if(a*b==c||a*c==b||b*c==a)
      {
          flag=1;
      }
      if((b!=0&&a%b==c)||(c!=0&&a%c==b)||(c!=0&&b%c==a)||(a!=0&&b%a==c)||(b!=0&&c%b==a)||(a!=0&&c%a==b))
      {
          flag=1;
      }
      if(flag)
      {
          printf("oh,lucky!\n");
      }
      else
      {
          printf("what a pity!\n");
      }
  }
  return 0;
}