A. Diverse Team
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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
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.
思路：通过另个for循环，第二个for循环判断输入的数字是否在前面出现过，倘若没有标记，若标记的数大于等于k，则任意输出k个数
#include<bits/stdc++.h>
using namespace std;
int a[108],b[108];
int main()
{
int n,k,m=1,t;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
{
int temp;
scanf("%d",&temp);
a[i]=temp;
t=1;
for(int j=1;j<i;j++)
{
if(temp==a[j])
{
t=0;
}
}
if(t)
{
b[m]=i;
m++;
}
}
m--;
if(m>=k)
{
printf("YES\n");
for(int i=1;i<=k;i++)
{
printf("%d ",b[i]);
}
}
else
{
printf("NO\n");
}
return 0;
}