foj 1664 Top K different numbers

Problem Description

There are N numbers in an array. Now your question is to select the top K different numbers in the array. The top K different numbers are the largest K different numbers in the array. If there are less than K different numbers, just output “-1” instead.
Input

There are multiply test cases. For each test case, the first line are two integer N and K(1≤N≤10000, 1≤K≤N). In the second line, there are N integers which are separated by a space. It is guaranteed that the element in the array will fit within a 32-bit integer.

Output

For each test case, output the top K different numbers increasingly. Please separate the K numbers by a space. If there are less than K different numbers, just output “-1” instead.

Sample Input

2 1
1 2
3 2
1 1 2
3 3
1 1 2

Sample Output

2
1 2
-1

这题方法很多种，我是使用了set的方法。

方法在代码中体现

#include<iostream>
#include<set>//定义
#include<cstdlib>
#include<cstring>
using namespace std;
int a[10001];
int MyCompare(const void *e1,const void *e2){//快排，从小到大（*p1-*p2)
	int *p1,*p2;
	p1=(int *)e1;
	p2=(int *)e2;
	return *p1-*p2;
}
int main()
{
	int n,i,k,num;
	set<int>p;
	set<int>::iterator it; //迭代器，类似于指针，自我理解，来获得地址
	while(cin>>n){
		cin>>k;
		memset(a,0,sizeof(a));//清空初始化，用到cstring头文件
		p.clear();
		for(i=0;i<n;i++)
		{
			cin>>num;
			p.insert(num);//存到set中，重复的会过滤
		}
		if(k>p.size()){//不重复的个数p.size（）；函数
			cout<<"-1";
		}
		else{
			for(it=p.begin(),i=0;it!=p.end();i++,it++)//起始地址p.begin();末尾p.end();
			{
				a[i]=*it;//赋值到数组中
			}
			qsort(a,len,sizeof(int),MyCompare);//快排
			for(i=p.size()-k;i<p.size();i++)
			{
				cout<< a[i];
				if(i!=p.size()-1) cout<<" ";
			}
		}
		cout<<endl;
	}
	return 0;
}

收获：强化了set使用，对set的一些函数有初步了解