Diverse Team CodeForces - 988A

最新推荐文章于 2018-08-07 17:22:25 发布

原创最新推荐文章于 2018-08-07 17:22:25 发布 · 290 阅读

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本文介绍了一个算法挑战问题，旨在从具有不同评分的学生中选择一个由k名成员组成的团队，确保每个成员的评分各不相同。文章提供了一段C++代码实现，用于解决这一问题，并展示了几种可能的输入输出示例。

There are n students in a school class, the rating of the i-th student on Codehorses is ai. You have to form a team consisting of k students (1≤k≤n

) such that the ratings of all team members are distinct.

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print k

distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.

Input

The first line contains two integers n

and k (1≤k≤n≤100

) — the number of students and the size of the team you have to form.

The second line contains n

integers a1,a2,…,an (1≤ai≤100), where ai is the rating of i

-th student.

Output

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print k

distinct integers from 1 to n

which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.

Assume that the students are numbered from 1

to n

Examples

Input

5 3
15 13 15 15 12

Output

YES
1 2 5

Input

5 4
15 13 15 15 12

Output

NO

Input

4 4
20 10 40 30

Output

YES
1 2 3 4

Note

All possible answers for the first example:

{1 2 5}
{2 3 5}
{2 4 5}

Note that the order does not matter.

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int a[110],b[110];
int main()
{
    int n,m,i,k,p=0,x;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    cin>>n>>m;
    k=0;
    for(i=0;i<n;i++)
    {
        cin>>x;
        if(a[x]==0)
        {
            k++;
            a[x]=1;
            b[p++] = i+1;
        }
    }
    int f=0;
    if(k >= m)
    {
        printf("YES\n");
        for(i=0;i<m;i++)
        {
            if(f==0) f= 1;
            else printf(" ");
            printf("%d",b[i]);
        }
        printf("\n");
    }
    else printf("NO\n");
    return 0;
}