There are nn students in a school class, the rating of the ii-th student on Codehorses is aiai. You have to form a team consisting of kk students (1≤k≤n1≤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 kk 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 nn and kk (1≤k≤n≤1001≤k≤n≤100) — the number of students and the size of the team you have to form.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100), where aiai is the rating of ii-th student.
Output
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print kk distinct integers from 11 to nn 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 11 to nn.
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
题意就是让你从n个人当中选取k个人组成一个数值均不相同的球队，让你判断是否可以。
做法比较无脑，可能也是水题的原因，直接标记然后录入就弄好了。
#include<stdio.h>
#include<string.h>
using namespace std;
bool book[110];
int main()
{
memset(book,0,sizeof(book));
int n,m,x;
scanf("%d%d",&n,&m);
int ans=0,a[110];
for(int i=1; i<=n; i++)
{
scanf("%d",&x);
if(book[x]==0)
{
book[x]=1;
a[ans++]=i;
}
}
if(ans<m)
printf("NO\n");
else
{
printf("YES\n");
for(int i=0; i<m-1; i++)
{
printf("%d ",a[i]);
}
printf("%d\n",a[m-1]);
}
return 0;
}