stl大法好

#include<iostream>
#include<set>
using namespace std;
set<int>a;
int main(){
    int n,k,cnt=0;
    cin>>n>>k;
    for(int i=1;i<=n;++i){
        int t;
        cin>>t;
        if(a.count(t)==1);
        else a.insert(t),++cnt;
    }
    int t=n-cnt-k;
    if(t>0)t=0;
    cout<<cnt+t;
    return 0;
}

筛掉重复的，计数，取\(min\{ n-k,totleft \}\)

#include<cstdio>
#include<algorithm>
const int MAXN=100000+5;
int n,k,totrep,nums[MAXN],totleft;bool disabled[MAXN];
int main()
{
    scanf("%d %d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&nums[i]);
    std::sort(nums+1,nums+n+1);
    for(int i=1;i<n;i++)
        if(nums[i]==nums[i+1])
            disabled[i+1]=1;
    for(int i=1;i<=n;i++)
        if(!disabled[i]) totleft++;
    if(totleft>n-k) printf("%d",n-k); else printf("%d",totleft);
    return 0;
}

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;

const int max_n = 1e5 + 10;
int n, k, a[max_n];

int main() {

    
    cin >> n >> k;  
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    
    sort(a + 1, a + 1 + n);
    
    int multi = 0;
    for (int i = 2; i <= n; i++)
        if (a[i] == a[i - 1]) multi++;
    if (k <= multi) cout << n - multi << endl;
    else cout << n - multi - (k - multi) << endl;
}