hdu 2101 A + B Problem Too(数论)

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于 2013-01-13 19:00:50 发布

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分类专栏： ACM--数论

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本文介绍了一种特殊的A+B问题，需要判断两数之和是否能被86整除。通过示例输入输出展示了问题的具体解决过程，并提供了一个C++实现代码示例。

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A + B Problem Too

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8373 Accepted Submission(s): 6277

Problem Description

This problem is also a A + B problem,but it has a little difference,you should determine does (a+b) could be divided with 86.For example ,if (A+B)=98,you should output no for result.

Input

Each line will contain two integers A and B. Process to end of file.

Output

For each case, if(A+B)%86=0,output yes in one line,else output no in one line.

Sample Input

Sample Output

  
  
   
   no
yes

Source

HDU 2007-6 Programming Contest

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思路：(10*a+b)%c=((a%c)*10+b)%c;

#include<iostream>
using namespace std;
const int mm=100010;
char s[mm],ss[mm];
int main()
{ int ans,anss;
  while(cin>>s>>ss)
  {
    ans=anss=0;
    for(int i=0;s[i]!='\0';i++)
    {
      ans=(ans*10+s[i]-'0')%86;
    }
    for(int i=0;ss[i]!='\0';i++)
      anss=(anss*10+ss[i]-'0')%86;
    ans=(ans+anss)%86;
    if(ans)cout<<"no\n";
    else cout<<"yes\n";
  }
}