#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
int a[N];
signed main()
{
	int n,k,q;
	cin>>n>>k>>q;
	for(int i=1;i<=n;i++)a[i]=k-q;
	while(q--)
	{
		int x;
		cin>>x;
		a[x]++;
	}
	for(int i=1;i<=n;i++)
	{
		if(a[i]>0)cout<<"Yes"<<'\n';
		else cout<<"No"<<'\n';
	}
}

先让大家都扣q分如果你回答正确了那么你就多扣了分,給它加回去即可

s = list(map(int, input().split())) n, k, q = s[0], s[1], s[2] a_list = [int(input()) for _ in range(q)]

count_dict = {} for a in a_list: if a in count_dict: count_dict[a] += 1 else: count_dict[a] = 1

for i in range(1, n + 1): values = k - q if i in count_dict: values += count_dict[i] result = 'Yes' if values > 0 else 'No' print(result)

#include <bits/stdc++.h>

using namespace std; int t=1; const int N=1e5+10,INF=0x3f3f3f3f; int a[N];

void solve() { int n,k,q; cin>>n>>k>>q; for(int i=1;i<=n;i++) a[i]=k-q; for(int i=1;i<=q;i++) { int md; cin>>md; a[md]++; }

for(int i=1;i<=n;i++)
{
	if(a[i]>0)
	cout<<"Yes"<<endl;
	else
	cout<<"No"<<endl; 
}

} int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); while(t--) { solve(); }

return 0;

}

#include<bits/stdc++.h> using namespace std;

long long n,a[100006],k,q,d; int main(){ cin>>n>>k>>q; for(int i=1;i<=q;i++){ cin>>d; a[d]++; } for(int i=1;i<=n;i++){ if(k-(q-a[i])>0)cout<<"Yes"<<endl; else cout<<"No"<<endl; } return 0; }

#include<bits/stdc++.h>
using namespace std;

int t,N,Q,K;

void callFinalre(vector<int> &re) {
	for(int i=1;i<=N;i++)
	{
		if(re[i] > 0)  cout<<"Yes"<<endl;
		else{
			cout<<"No"<<endl;
		}		
	}
}

int main()
{
	cin>>N>>K>>Q;
	
	vector<int> re(N+1,K-Q);
	
	for(int i=0;i<Q;i++){
		cin>>t;
		re[t]++;	
	}
	
	callFinalre(re);
	return 0;
}

首先，本题的纯模拟代码：

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<cstring>
#include<string>
#include<array>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<sstream>
#include<unordered_set>
#include<unordered_map>
using namespace std;
using i64=long long;
using ui=unsigned int;
//using i128=__int128;
const int INF=1e9;
//const i64 INT=F=1e18;
//const int mod=998244353;
const int mod=80112002;
//const int N=1e9+7;
void solve(){
    int n,k,q;
    cin>>n>>k>>q;
    vector<int>a(n+1);
    for(int i=1;i<=n;i++)a[i]=k;
    while(q--){
        int x;
        cin>>x;
        for(int i=1;i<=n;i++){
            if(i!=x)a[i]--;
        }
    }
    cout<<'\n';
    for(int i=1;i<=n;i++){
        if(a[i]>0)cout<<"Yes"<<'\n';
        else cout<<"No"<<'\n';
    }
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int _=1;
    //cin>>_;
    while(_--)solve();
    return 0;
}```

优化思路：可以将q次查询的数，每一个数都加1，之后再将它们减去q次查询判断即可 代码：

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<cstring>
#include<string>
#include<array>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<sstream>
#include<unordered_set>
#include<unordered_map>
using namespace std;
using i64=long long;
using ui=unsigned int;
//using i128=__int128;
const int INF=1e9;
//const i64 INT=F=1e18;
//const int mod=998244353;
const int mod=80112002;
//const int N=1e9+7;
void solve(){
    int n,k,q;
    cin>>n>>k>>q;
    int q1=q;
    vector<int>a(n+1);
    for(int i=1;i<=n;i++)a[i]=k;
    while(q--){
        int x;
        cin>>x;
        a[x]++;
    }
    for(int i=1;i<=n;i++){
        a[i]-=q1;
        if(a[i]>0)cout<<"Yes"<<'\n';
        else cout<<"No"<<'\n';
    }
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int _=1;
    //cin>>_;
    while(_--)solve();
    return 0;
}