思路:數值範圍就那麼大,看個數組,一個一個的放,然後從後面輸出
代碼:
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
#define mst(a,b) memset(a,b,sizeof(a))
#define REP(a,b,c) for(int a = b; a < c; ++a)
#define eps 10e-8
const int MAX_ = 1000010;
const int N = 500000;
const int INF = 0x7fffffff;
char str[MAX_];
int hash[MAX_];
int a[MAX_];
int Minimize(char str[])
{
int i= 0, j= 1, len, k =0;
len = strlen(str);
while(i < len && j < len && k < len){
int t = str[(i+k)%len] - str[(j+k)%len];
if(!t)++k;
else {
if(t > 0)i = i+k+1;
else j = j+ k + 1;
if(i == j)++j;
k = 0;
}
}
return i>j?j:i;
}
int main(){
int T, n, m;
//scanf("%d", &T);
while(~scanf("%d%d", &n,&m)){
mst(hash, 0);
REP(i, 0 , n){
scanf("%d", &a[i]);
hash[a[i] + N]++;
}
for(int i = 1000000; i > -1 && m; --i){
if(hash[i]){
printf("%d", i-N);
--m;
if(m)printf(" ");
}
}
printf("\n");
}
return 0;
}