Codeforces Round #521 (Div. 3) D - Cutting Out(二分答案)

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D. Cutting Out

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array ss consisting of nn integers.

You have to find any array tt of length kk such that you can cut out maximum number of copies of array tt from array ss.

Cutting out the copy of tt means that for each element titi of array tt you have to find titi in ss and remove it from ss. If for some titi you cannot find such element in ss, then you cannot cut out one more copy of tt. The both arrays can contain duplicate elements.

For example, if s=[1,2,3,2,4,3,1]s=[1,2,3,2,4,3,1] and k=3k=3 then one of the possible answers is t=[1,2,3]t=[1,2,3]. This array tt can be cut out 22 times.

  • To cut out the first copy of tt you can use the elements [1,2––,3,2,4,3––,1––][1,2_,3,2,4,3_,1_] (use the highlighted elements). After cutting out the first copy of tt the array ss can look like [1,3,2,4][1,3,2,4].
  • To cut out the second copy of tt you can use the elements [1––,3––,2––,4][1_,3_,2_,4]. After cutting out the second copy of tt the array ss will be [4][4].

Your task is to find such array tt that you can cut out the copy of tt from ss maximum number of times. If there are multiple answers, you may choose any of them.

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of elements in ss and the desired number of elements in tt, respectively.

The second line of the input contains exactly nn integers s1,s2,…,sns1,s2,…,sn (1≤si≤2⋅1051≤si≤2⋅105).

Output

Print kk integers — the elements of array tt such that you can cut out maximum possible number of copies of this array from ss. If there are multiple answers, print any of them. The required array tt can contain duplicate elements. All the elements of tt (t1,t2,…,tkt1,t2,…,tk) should satisfy the following condition: 1≤ti≤2⋅1051≤ti≤2⋅105.

Examples

input

Copy

7 3
1 2 3 2 4 3 1

output

Copy

1 2 3 

input

Copy

10 4
1 3 1 3 10 3 7 7 12 3

output

Copy

7 3 1 3

input

Copy

15 2
1 2 1 1 1 2 1 1 2 1 2 1 1 1 1

output

Copy

1 1 

题意:

给你长度为n的序列,让你从中挑k个数组成序列ans,使得ans序列在原序列中出现次数最多。这里不用管顺序,序列是无序的。

1 2 3 2 4 3 1

出现次数最多的序列是[1,2,3]

解析:

直接二分答案,答案要求最大化出现次数,那么我们就二分次数mid,把出现次数>=mid的数都加进答案序列,看这个

序列最终能不能>=k

一般思维题、构造题,使你构造一个XXX来最大化或最小化答案,并且未知量只有这个答案,

那么我们可以直接二分这个答案使所有条件已知,来尝试构造这个序列,能不能成功.

思维题二分的一般问题都是最大化/最小化某一个值,尤其是最大化最小值/最小化最大值

还有就是最大/小化某一个值来构造序列

#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
const int MAXN = 2e5+100;

int a[MAXN];
int ans[2][MAXN];
int to,k;
bool check(int x,int now)
{
    int cnt=0;
    for(int i=1;i<=to;i++)
    {
        if(a[i]>=x)
        {
            int tmp=a[i]/x;
            for(int j=0;j<tmp;j++) ans[now][cnt++]=i;
        }
        if(cnt>=k) return true;
    }
    return false;
}

int main()
{
    int n;
    scanf("%d%d",&n,&k);

    for(int i=1;i<=n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        a[tmp]++;
        to=max(to,tmp);
    }
    int l=1;
    int r=n;
    int now=0;
    while(l<r)
    {
        int mid=(l+r)>>1;
        if(check(mid,now^1)) now=now^1,l=mid+1;
        else r=mid;
    }
    if(check(l,now^1)) now=now^1;
    for(int i=0;i<k;i++)
    {
        if(i==0) printf("%d",ans[now][i]);
        else printf(" %d",ans[now][i]);
    }
    printf("\n");
}

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转载自blog.csdn.net/qq_37025443/article/details/84173801