stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.164639734622789x):
#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 3;

int n, k, q, freq[N];

int main() {

#ifndef ONLINE_JUDGE

//	freopen("input.in", "r", stdin);

#endif

	scanf("%d %d %d", &n, &k, &q);

	for (int i = 0; i< q; ++i) {

		int a;

		scanf("%d", &a);

		++freq[a];

	}

	for (int i = 1; i <=n; ++i) {

		if (k - (q - freq[i]) > 0) {

			puts("Yes");

		}else {

			puts("No");

		}

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.403048853755527x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;

ll n , k ,q ,a[100001] ,x ;

int main()

{

    cin >> n >> k >> q ;

    for (int  i = 1 ;i <= n ;i++)

    {

        a[i] += k;

        a[i] -= q ;

    }

    while (q--)

    {

     cin >> x ;

        a[x]+=1;

    }

    for (int  i = 1 ;i <= n ; i++)

    {

        if (a[i]  > 0 )cout <<"Yes"<<endl;

        else cout <<"No"<<endl;

    }

    return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.129507647241601x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;

ll n , k ,q ,a[100001] ,x ;

int main()

{

    cin >> n >> k >> q ;

    for (int  i = 1 ;i <= n ;i++)

    {

        a[i] += k;

        a[i] -= q ;

    }

    while (q--)

    {

     cin >> x ;

        a[x]+=1;

    }

    for (int  i = 1 ;i <= n ; i++)

    {

        if (a[i]  > 0 )cout <<"Yes"<<endl;

        else cout <<"No"<<endl;

    }

    return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4106700194408655x):
#include<cstdio>

#include<iostream>

using namespace std;

const int N=100010;

int n,k,q,pl[N];

int main()

{

	scanf("%d%d%d",&n,&k,&q);

	for(int i=1;i<=n;i++) pl[i]=k;

	for(int i=1;i<=q;i++)

	{

		int x;

		scanf("%d",&x);

		pl[x]++;

	}

	for(int i=1;i<=n;i++)

	{

		if(pl[i]-q>0) printf("Yes\n");

		else printf("No\n");

	}

	return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.157370417350568x):
#include<cstdio>

#include<iostream>

using namespace std;

const int N=100010;

int n,k,q,pl[N];

int main()

{

	scanf("%d%d%d",&n,&k,&q);

	for(int i=1;i<=n;i++) pl[i]=k;

	for(int i=1;i<=q;i++)

	{

		int x;

		scanf("%d",&x);

		pl[x]++;

	}

	for(int i=1;i<=n;i++)

	{

		if(pl[i]-q>0) printf("Yes\n");

		else printf("No\n");

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3820926564393996x):
#include <stdio.h>

#include <iostream>

using namespace std;



int N,K,Q,a,nilai;

int main (){

int nilai[100001];

cin >> N >> K >>Q;

	for (int i=1; i<=N; i++){

		nilai[i]=K-Q;

	}

	for (int i=1; i<=Q; i++){

		cin>>a;

		nilai[a]++;

	}	

	for (int i=1; i<=N; i++){

		if (nilai[i]<=0){

			cout << "No"<< endl;

		}else {

			cout <<"Yes"<< endl;

		}

	}

}
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.052892371798805x):
#include <stdio.h>

#include <iostream>

using namespace std;



int N,K,Q,a,nilai;

int main (){

int nilai[100001];

cin >> N >> K >>Q;

	for (int i=1; i<=N; i++){

		nilai[i]=K-Q;

	}

	for (int i=1; i<=Q; i++){

		cin>>a;

		nilai[a]++;

	}	

	for (int i=1; i<=N; i++){

		if (nilai[i]<=0){

			cout << "No"<< endl;

		}else {

			cout <<"Yes"<< endl;

		}

	}

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4029935028878593x):
#include <iostream>

using namespace std;



long long n,k,q,bil[100001],a;



int main(){

    cin>>n>>k>>q;

    for(int h=1;h<=n;h++){

    bil[h]=k-q;

    }

    for(int i=1;i<=q;i++){

        cin>>a;

        bil[a]++;

    }

    for(int i=1;i<=n;i++){

        if(bil[i]<=0){

            cout<<"No"<<endl;

        }else{

            cout<<"Yes"<<endl;

        }

    }

}
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.129305286006479x):
#include <iostream>

using namespace std;



long long n,k,q,bil[100001],a;



int main(){

    cin>>n>>k>>q;

    for(int h=1;h<=n;h++){

    bil[h]=k-q;

    }

    for(int i=1;i<=q;i++){

        cin>>a;

        bil[a]++;

    }

    for(int i=1;i<=n;i++){

        if(bil[i]<=0){

            cout<<"No"<<endl;

        }else{

            cout<<"Yes"<<endl;

        }

    }

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4029868111164416x):
#include <iostream>

#include <algorithm>

#include <cstring>

#include <cmath>

#include <cstdio>

#include <vector>

#include <queue>



using namespace std;



typedef long long ll;

const int maxn=1e5+7;



int a[maxn];



int main()

{

//    freopen("input.txt","r",stdin);

    

    int n,k,q;

    cin>>n>>k>>q;

    for(int i=0;i<n;i++) a[i]=k-q;

    for(int i=0;i<q;i++)

    {

    	int x;

    	cin>>x;

    	a[x-1]++;

	}

	for(int i=0;i<n;i++) 

	{

		if(a[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

    

    return 0;

}

//想着稍微改一改思路来做这道题也不错 

//考虑一下我为什么会把G题的思路想起来
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.129280821076004x):
#include <iostream>

#include <algorithm>

#include <cstring>

#include <cmath>

#include <cstdio>

#include <vector>

#include <queue>



using namespace std;



typedef long long ll;

const int maxn=1e5+7;



int a[maxn];



int main()

{

//    freopen("input.txt","r",stdin);

    

    int n,k,q;

    cin>>n>>k>>q;

    for(int i=0;i<n;i++) a[i]=k-q;

    for(int i=0;i<q;i++)

    {

    	int x;

    	cin>>x;

    	a[x-1]++;

	}

	for(int i=0;i<n;i++) 

	{

		if(a[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

    

    return 0;

}

//想着稍微改一改思路来做这道题也不错 

//考虑一下我为什么会把G题的思路想起来
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.771096767085366x):
#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

	while(cin >> n >> k >> q)

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

		for(long long int i = 1; i <= q; i++)

		{

			scanf("%lld", &m);

			A[m]++;

		}

		for(long long int i = 1; i <= n; i++)

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.389257064973182x):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<stack>

#include<cstring>

#include<cmath>

#include<queue>

#include<set>

using namespace std;

typedef long long ll;

ll a[1111111];

ll b[1111111];

ll c[1111111];



int main()

{

	//ios::sync_with_stdio(false);

	ll n,k,q,x;

	cin>>n>>k>>q;

	for(ll i=0;i<n;i++)

		a[i]=k;

	set<ll> st; 

	for(ll i=0;i<q;i++)

	{

		scanf("%lld",&x);

		a[x-1]++;

	}

	



	for(ll j=0;j<n;j++)

		if(a[j]-q>0)

			cout<<"Yes"<<endl;

		else

			cout<<"No"<<endl;

	return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.07908525044564x):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<stack>

#include<cstring>

#include<cmath>

#include<queue>

#include<set>

using namespace std;

typedef long long ll;

ll a[1111111];

ll b[1111111];

ll c[1111111];



int main()

{

	//ios::sync_with_stdio(false);

	ll n,k,q,x;

	cin>>n>>k>>q;

	for(ll i=0;i<n;i++)

		a[i]=k;

	set<ll> st; 

	for(ll i=0;i<q;i++)

	{

		scanf("%lld",&x);

		a[x-1]++;

	}

	



	for(ll j=0;j<n;j++)

		if(a[j]-q>0)

			cout<<"Yes"<<endl;

		else

			cout<<"No"<<endl;

	return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

	while(cin >> n >> k >> q)

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

		for(long long int i = 1; i <= q; i++)

		{

			scanf("%lld", &m);

			A[m]++;

		}

		for(long long int i = 1; i <= n; i++)

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8328790646267514x):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<stack>

#include<cstring>

#include<cmath>

#include<queue>

#include<set>

using namespace std;

typedef long long ll;

ll a[1111111];

ll b[1111111];

ll c[1111111];



int main()

{

	//ios::sync_with_stdio(false);

	ll n,k,q,x;

	cin>>n>>k>>q;

	for(ll i=0;i<n;i++)

		a[i]=k;

	set<ll> st; 

	for(ll i=0;i<q;i++)

	{

		scanf("%lld",&x);

		a[x-1]++;

	}

	



	for(ll j=0;j<n;j++)

		if(a[j]-q>0)

			cout<<"Yes"<<endl;

		else

			cout<<"No"<<endl;

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

    for(int i=1;i<=q;i++)

    {

        int x;

        cin>>x;

        a[x]++;

    }

    for(int i=1;i<=n;i++)

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 6.792369718453121x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 24.832715069356023x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



const int MAXN = 100005;



int N, K, Q, ai, A[MAXN];

int main(){

	scanf("%d %d %d", &N, &K, &Q);

	for(int i = 0; i < Q; i++){

		scanf("%d", &ai);

		A[ai - 1]++;

	}

	for(int i = 0; i < N; i++){

		if(K - Q + A[i] > 0)printf("Yes\n");

		else printf("No\n");

	}

	

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.813959080538226x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 3;

int n, k, q, freq[N];

int main() {

#ifndef ONLINE_JUDGE

//	freopen("input.in", "r", stdin);

#endif

	scanf("%d %d %d", &n, &k, &q);

	for (int i = 0; i< q; ++i) {

		int a;

		scanf("%d", &a);

		++freq[a];

	}

	for (int i = 1; i <=n; ++i) {

		if (k - (q - freq[i]) > 0) {

			puts("Yes");

		}else {

			puts("No");

		}

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.808218258261481x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define ll long long

using namespace std;

ll n , k ,q ,a[100001] ,x ;

int main()

{

    cin >> n >> k >> q ;

    for (int  i = 1 ;i <= n ;i++)

    {

        a[i] += k;

        a[i] -= q ;

    }

    while (q--)

    {

     cin >> x ;

        a[x]+=1;

    }

    for (int  i = 1 ;i <= n ; i++)

    {

        if (a[i]  > 0 )cout <<"Yes"<<endl;

        else cout <<"No"<<endl;

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.841149829011337x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include<cstdio>

#include<iostream>

using namespace std;

const int N=100010;

int n,k,q,pl[N];

int main()

{

	scanf("%d%d%d",&n,&k,&q);

	for(int i=1;i<=n;i++) pl[i]=k;

	for(int i=1;i<=q;i++)

	{

		int x;

		scanf("%d",&x);

		pl[x]++;

	}

	for(int i=1;i<=n;i++)

	{

		if(pl[i]-q>0) printf("Yes\n");

		else printf("No\n");

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.814995445317079x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include <stdio.h>

#include <iostream>

using namespace std;



int N,K,Q,a,nilai;

int main (){

int nilai[100001];

cin >> N >> K >>Q;

	for (int i=1; i<=N; i++){

		nilai[i]=K-Q;

	}

	for (int i=1; i<=Q; i++){

		cin>>a;

		nilai[a]++;

	}	

	for (int i=1; i<=N; i++){

		if (nilai[i]<=0){

			cout << "No"<< endl;

		}else {

			cout <<"Yes"<< endl;

		}

	}

}

// Faster C++ solution for the same problem (measured speedup: 4.914554524840532x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



long long n,k,q,bil[100001],a;



int main(){

    cin>>n>>k>>q;

    for(int h=1;h<=n;h++){

    bil[h]=k-q;

    }

    for(int i=1;i<=q;i++){

        cin>>a;

        bil[a]++;

    }

    for(int i=1;i<=n;i++){

        if(bil[i]<=0){

            cout<<"No"<<endl;

        }else{

            cout<<"Yes"<<endl;

        }

    }

}

// Faster C++ solution for the same problem (measured speedup: 4.841340821943944x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <algorithm>

#include <cstring>

#include <cmath>

#include <cstdio>

#include <vector>

#include <queue>



using namespace std;



typedef long long ll;

const int maxn=1e5+7;



int a[maxn];



int main()

{

//    freopen("input.txt","r",stdin);

    

    int n,k,q;

    cin>>n>>k>>q;

    for(int i=0;i<n;i++) a[i]=k-q;

    for(int i=0;i<q;i++)

    {

    	int x;

    	cin>>x;

    	a[x-1]++;

	}

	for(int i=0;i<n;i++) 

	{

		if(a[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

    

    return 0;

}

//想着稍微改一改思路来做这道题也不错 

//考虑一下我为什么会把G题的思路想起来 


























// Faster C++ solution for the same problem (measured speedup: 4.841363913498246x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}