Codeforces Round #441 (Div. 2, by Moscow Team Olympiad) B. Divisiblity of Differences

http://codeforces.com/contest/876/problem/B

题意：

给出n个数，要求从里面选出k个数使得这k个数中任意两个的差能够被m整除，若不能则输出no。

思路：

差能够被m整除，其实就是对m取余的余数相同。那么就统计n个数的余数丢到一个map里面，最后判断是否有某个数的数量大于等于k。

代码：

 #include <stdio.h>
 #include <map>
 #include <vector>
 using namespace std;

 map<int,int> mmp;

 int a[];
 int b[];

 vector<int> v;

 int main()
 {
     int n,k,m;

     scanf("%d%d%d",&n,&k,&m);

     for (int i = ;i < n;i++)
     {
         int t;

         scanf("%d",&t);

         b[i] = t;

         t %= m;

         a[i] = t;
     }

     for (int i = ;i < n;i++) mmp[a[i]]++;

     map<int,int>::iterator it;

     int ans = -;

     for (it = mmp.begin();it != mmp.end();++it)
     {
         if (it -> second >= k)
         {
             ans = it -> first;
             break;
         }
     }

     if (ans == -) printf("No\n");
     else
     {
         printf("Yes\n");

         int cnt = ;

         for (int i = ;i < n;i++)
         {
             if (b[i] % m == ans)
             {
                 v.push_back(b[i]);
                 cnt++;
             }

             if (cnt >= k) break;
         }

         for (int i = ;i < v.size();i++)
         {
             printf("%d%c",v[i],i == v.size() -  ? '\n' :' ');
         }
     }

     return ;
 }