Recommendation system predicts the preference that a user would give to an item. Now you are asked to program a very simple recommendation system that rates the user's preference by the number of times that an item has been accessed by this user.

Input Specification:

Each input file contains one test case. For each test case, the first line contains two positive integers: N (≤ 50,000), the total number of queries, and K (≤ 10), the maximum number of recommendations the system must show to the user. Then given in the second line are the indices of items that the user is accessing -- for the sake of simplicity, all the items are indexed from 1 to N. All the numbers in a line are separated by a space.

Output Specification:

For each case, process the queries one by one. Output the recommendations for each query in a line in the format:

query: rec[1] rec[2] ... rec[K]

where query is the item that the user is accessing, and rec[i] (i=1, ... K) is the i-th item that the system recommends to the user. The first K items that have been accessed most frequently are supposed to be recommended in non-increasing order of their frequencies. If there is a tie, the items will be ordered by their indices in increasing order.

Note: there is no output for the first item since it is impossible to give any recommendation at the time. It is guaranteed to have the output for at least one query.

Sample Input:

12 3
3 5 7 5 5 3 2 1 8 3 8 12

Sample Output:

5: 3
7: 3 5
5: 3 5 7
5: 5 3 7
3: 5 3 7
2: 5 3 7
1: 5 3 2
8: 5 3 1
3: 5 3 1
8: 3 5 1
12: 3 5 8
  
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <set>
using namespace std;
const int maxn=50010;
struct node{
    int num;
    int cnt;
    node(int num,int cnt):num(num),cnt(cnt){};
    bool operator < (const node& a) const{
        return cnt!=a.cnt?cnt>a.cnt:num<a.num;
    }
};
set<node> s;
int times[50010];
int main(){
  int n,k;
  scanf("%d %d",&n,&k);
  for(int i=0;i<n;i++){
    int tmp;
    scanf("%d",&tmp);
    if(i!=0){
        printf("%d:",tmp);
           int j=0;
           for(auto it=s.begin();it!=s.end()&&j<k;it++,j++){
            printf(" %d",it->num);
        }
        printf("\n");
    }
    if(s.find(node(tmp,times[tmp]))!=s.end()){
        s.erase(s.find(node(tmp,times[tmp])));
    }
    times[tmp]++;
    s.insert(node(tmp,times[tmp]));
  }
}

注意点：用数组，然后再复制排序，后面3个测试点都会超时，这里要用set。虽然一开始想到的也是要用set，但set只能对数字自动排序，咋办，看了大佬的，原来可以重载<符号，这样set可以自动对结构体排序了，还有就是结构体的构造函数的写法，这样可以直接构造一个node了。之前看的primec++全忘完了。。。

超时代码：虽然写的时候就感觉会超时，但想不到别的办法，只好硬着头皮写了，考试的时候这样也能拿个16分，就酱吧，花不到30分钟还行

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn=50010;
struct node{
    int num;
    int cnt;
}nodes[maxn];
bool cmp(node n1,node n2){
    return n1.cnt==n2.cnt?n1.num<n2.num:n1.cnt>n2.cnt;
}
node res[maxn];
int main(){
  int n,k;
  int count=0;
  scanf("%d %d",&n,&k);
  for(int i=0;i<n;i++){
    int tmp;
    scanf("%d",&tmp);
    if(i!=0){
        printf("%d:",tmp);
        memcpy(res,nodes,sizeof(nodes));
           sort(res,res+maxn,cmp);
           for(int j=0;j<k;j++){
            if(res[j].cnt!=0){
                printf(" %d",res[j].num);
            }
        }
        printf("\n");
    }
    nodes[tmp].num=tmp;
    nodes[tmp].cnt++;
  }
}