988A Diverse Team

A. Diverse Team

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1\le k\le n$) 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 $k$ ($1\le k\le n\le 100$) — the number of students and the size of the team you have to form.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 100$), where $a_i$ 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

Copy

5 3
15 13 15 15 12

output

Copy

YES
1 2 5 

input

Copy

5 4
15 13 15 15 12

output

Copy

NO

input

Copy

4 4
20 10 40 30

output

Copy

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.

题意：在一所学校里有n个学生，每个学生有一个rating分，你需要组建一个团队，有m个人，每个人的rating分不一样。能不能组建一个这样的团队，能就输出YES，然后输出组建团队的下标，不能就输出NO。

题解：暴力 模拟  首先判断不重复的分数个数事是否大于m，我用的是set集合判断不重复的个数，大于输出YES，否则就NO。然后开个标记数组对输出的数组进行标记，不让重复输出。

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C++:

#include<bits/stdc++.h>
using namespace std;
set<int>se;
int vis[110];
int main()
{
    int n,m,a[110];
    cin>>n>>m;
    for(int i=1; i<=n; i++)
        cin>>a[i],se.insert(a[i]);
    if(se.size()<m) puts("NO");
    else
    {
        puts("YES");
   int cnt=0;
        for(int i=1; i<=n; i++)
            if(!vis[a[i]]&&cnt<m)
                vis[a[i]]=1,cnt++,cout<<i<<" ";
    }
    return 0;
}

python:

n,m=map(int,input().split(" "))
a=input().split(" ")
b=list(set(a))
if len(b)<m:
    print("NO")
else:
    print("YES")
    for i in range(m):
        print(a.index(b[i])+1,end=' ')