//怀疑hdu的数据有点弱啊，下面这个代码竟AC了，明明这题是考hash嘛
//不过用时花了921ms,差那么一点点就TLE了，不能侥幸啊，还是该用hash做
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int hash[1000001];
bool compare(int a,int b){
     return a>b;
}
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
         for(int i=0;i<n;i++)
              scanf("%d",&hash[i]);
         sort(hash,hash+n,compare);
         for(int i=0;i<m-1;i++)
               printf("%d ",hash[i]);
         printf("%d\n",hash[m-1]);
    }
    return 0;
}

下面是hash做法

//题目中的数据范围都是【-500000，500000】，并且总数不超过1000000
//那么值加上500000即为存储的下标，下标越大值越大，这样就不用排序了
//用时671ms
#include<iostream>
int hash[1000001];
int main(){
    int n,m,temp,cnt;
    while(scanf("%d%d",&n,&m)!=EOF){
          memset(hash,0,sizeof(hash));
          for(int i=0;i<n;i++){
                scanf("%d",&temp);
                hash[temp+500000]=1; 
          }
          cnt=0;
          for(int i=1000000;i>=0;i--){
                if(hash[i]==1){
                      if(cnt==m-1){
                             printf("%d\n",i-500000);            
                             break;     
                      }
                      printf("%d ",i-500000);
                      cnt++;                                            
                }
          }  
    }
    return 0;
}

 如果允许其中有数据相同，相应的代码如下：

#include<iostream>
int hash[1000001];
int main(){
    int n,m,temp,cnt;
    while(scanf("%d%d",&n,&m)!=EOF){
          memset(hash,0,sizeof(hash));
          for(int i=0;i<n;i++){
                scanf("%d",&temp);
                hash[temp+500000]++; //存放这个数的数量
          }
          cnt=0;
          for(int i=1000000;i>=0;i--){
                if(hash[i]>0){
                      for(int j=0;j<hash[i];j++){
                           if(cnt==m-1){
                                 printf("%d\n",i-500000);
                                 cnt++;           
                                 break;     
                           }
                           printf("%d ",i-500000);
                           cnt++;      
                      } 
                      if(cnt==m)
                           break;                                   
                }
          }  
    }
    return 0;
}