stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std; 

int main(){

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

//  freopen("test.out","w",stdout);

	int n,q;

	long long k;

	cin>>n>>k>>q;

	int score0[n],score1[n],player[q];

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

		score1[i]=0; 

		score0[i]=k;

	}

	memset(player,0,sizeof(player));

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

		cin>>player[i];

        score1[player[i]-1]++;

	}

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

		score0[i]-=q-score1[i];

		if(score0[i]>0) 	cout<<"Yes"<<endl;

		else 	cout<<"No"<<endl;

	} 

	

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int n,k,q,i,j;

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

	int a[n],b[q];

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

		a[i]=k;

	}

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

		scanf("%d",&b[i]);

	}

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

		a[b[i]-1]=a[b[i]-1]+1;

	}

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

		a[j]=a[j]-q;

		if(a[j]>0)

		printf("Yes\n");

		else

		printf("No\n");

	}

	return 0;

} 

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;



ll s[MAXN];



int main() {

	ios::sync_with_stdio(false);

	ll n, k, q, a;

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

		memset(s, 0, sizeof(s));

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

			cin >> a;

			s[a]++;

		}

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

			if(s[i] > q - k) {

				cout << "Yes\n";

			} else {

				cout << "No\n";

			}

		}

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>

#include<iostream>

#include<math.h> 

#include<string.h>

#include<algorithm>

using namespace std;

typedef long long ll;

ll n,k,q,b;

ll a[100000+100];

int main()

{

	cin>>n>>k>>q;

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

	{

		a[i]=k;

	}

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

	{

		cin>>b;

		a[b]++;

	}

	if(q<k)

	{

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

		printf("Yes\n");

	}

	else

	{

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

		{

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

			printf("No\n");

			else

			printf("Yes\n");

		}	

	}

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

const int N=1e6;

typedef long long ll;

ll arr[N];

int main(){

	int n,k,q,a;

	int flag=0;

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

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

		arr[i]=k;

	} 

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

		scanf("%d",&a);

		arr[a]++;

		flag++;

	} 

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

		if(arr[i]-flag<=0){

			puts("No");

		}

		else{

			puts("Yes");

		}

	}

	 

	

	

	return 0; 

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

using namespace std;

const int MAX=1e5;

int a[MAX];

int main()

{

    memset(a,0,sizeof(a));

    int n,k,q;

    cin>>n>>k>>q;

    int t=q;

    int b;

    while(q--){

        cin>>b;

        a[b-1]++;

    }

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

        if(a[i]>t-k){

            cout<<"Yes"<<endl;

        }

        else{

            cout<<"No"<<endl;

        }

    }

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define mp make_pair

#define ff first

#define ss second

#define pb push_back

#define w(x) scanf("%d",&x);while(x--)

#define ws(x) scanf("%d",&x);getchar();while(x--)

#define rep(i,b) for(i=0;i<=b;i++)

#define ios ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define pi 2*acos(0.0)

#define endl "\n"

using namespace std;

typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<int> vi;



const int N=1e6+2;

long long  mod=1e9+9;

int t,j,cs=0,n,a,b,i,k,x;

string s;



int main()

{

    cin>>n>>a>>b;

    k=b-a;

    int ar[n]={};

    for(i=0;i<b;i++){

        cin>>x;

        ar[x-1]++;

    }

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

        if(ar[i]>k)

            cout<<"Yes";

        else

            cout<<"No";

        cout<<endl;

    }

}




// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please





int main() {

	cin.tie(0);

	ios::sync_with_stdio(false);





	int N, K, Q;

	cin >> N >> K >> Q;



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		int a;

		cin >> a;

		P[a]++;

	}

	rep1(i, N) {

		if (P[i] > 0) co("Yes");

		else co("No");

	}



	Would you please return 0;

}

// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 19;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//#include<bits/stdc++.h>

#include<cstdio>

#include<string>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	string S;

	rep1(i, N) {

		if (P[i] > 0) S += "Yes\n";

		else S += "No\n";

	}

	printf("%s", S.c_str());



	Would you please return 0;

}

// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 12;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

		if (i % 100 == 0) {

			fwrite(dn, 1, di - dn, stdout);

			di = dn;

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//#include<bits/stdc++.h>

#include<cstdio>

#include<string>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	string S;

	rep1(i, N) {

		if (P[i] > 0) S += "Yes\n";

		else S += "No\n";

		if (i % 100 == 0) {

			printf("%s", S.c_str());

			S = "";

		}

	}

	printf("%s", S.c_str());



	Would you please return 0;

}

// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 12;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

		if (i % 100 == 0) {

			fwrite(dn, 1, di - dn, stdout);

			di = dn;

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

    scanf("%d%d%d",&n,&m,&T);

    for(i=0;i<T;i++)

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	int n,k,q,i;

    int a;

	int player[100001]={0};



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

    

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

		player[i]=k-q;

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

	{	scanf("%d",&a);

	    player[a]+=1;

	}

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

	{	

		if(player[i]<=0)

	        printf("No\n");

		else

			printf("Yes\n");

	}

return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define maxn 100005

using namespace std;

int a[maxn];

int main()

{

    int n,k,q;

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

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

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

        int num;

        scanf("%d",&num);

        a[num]++;

    }

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

        if(a[i]>0) puts("Yes");

        else puts("No");

    }

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	int n,k,q,i;

    int a;

	int player[100001]={0};



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

    

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

		player[i]=k-q;

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

	{	scanf("%d",&a);

	    player[a]+=1;

	}

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

	{	

		if(player[i]<=0)

	        printf("No\n");

		else

			printf("Yes\n");

	}

return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<iostream>

#include<set>

#include<queue>

using namespace std;



int value[555555] = {0};



int main() 

{

	int n, score,m,mark;

	cin >> n >> score >> m;

	int t = m;

	while (t--)

	{

		cin >> mark;

		value[mark-1]++;

	}

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

	{

		if (score > m - value[i])printf("Yes\n");

		else printf("No\n");

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	int n,k,q,i;

    int a;

	int player[100001]={0};



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

    

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

		player[i]=k-q;

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

	{	scanf("%d",&a);

	    player[a]+=1;

	}

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

	{	

		if(player[i]<=0)

	        printf("No\n");

		else

			printf("Yes\n");

	}

return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<string>

using namespace std;

typedef long long ll;

const int maxn=1e5+100;

int main()

{

	ll n,k,q;

	ll x;

	ll a[maxn];

	memset(a,0,sizeof(a));

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

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

	{

		scanf("%d",&x);

		a[x]++;

	}

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

	{

		if(a[i]+k-q>0)

			printf("Yes");

		else

			printf("No");

		if(i!=n)

			printf("\n");

//		printf("%d ",a[i]+k-q);

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<vector>

#include<queue>

#include<cmath>

#include<algorithm>

#include<iomanip>

#include<functional>

#include<set>

#include<map>

#include<climits>

#include<numeric>

#include<utility>

#define ul unsigned long long int

#define ll long long int

#define MAX 1000000+1

using namespace std;

#define F first

#define S second

#define pi 2*acos(0.0)

#define PB push_back

#define CL clear()







int main()

{

   int n, k, q;

   int c;

   cin >> n >> k>> q;

   c = q;

   int arr[n+1];

   for(int i = 0; i <= n; i++) arr[i] = 0;

   int a;

   while(q--)

   {

       cin >> a;



       arr[a]++;

   }

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

   {

       if(k-c+arr[i]>0) cout <<"Yes\n";

       else cout <<"No\n";

   }



return 0;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn=2e5+6;

int main()

{

	int n,k,q,a,vis[maxn],flag[maxn];

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

	memset(vis,0,sizeof(vis));

	memset(flag,0,sizeof(flag));

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

	{

		scanf("%d",&a);

		vis[a]++;

	}

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

	{

		if(q-vis[j]<k)

		{

			flag[j]=1;

		}

		else

		{

			flag[j]=0;

		}

	}

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

	{

		if(flag[i]==1)

		{

			printf("Yes\n");

		}

		else

		{

			printf("No\n");

		}

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<vector>

#include<queue>

#include<cmath>

#include<algorithm>

#include<iomanip>

#include<functional>

#include<set>

#include<map>

#include<climits>

#include<numeric>

#include<utility>

#define ul unsigned long long int

#define ll long long int

#define MAX 1000000+1

using namespace std;

#define F first

#define S second

#define pi 2*acos(0.0)

#define PB push_back

#define CL clear()







int main()

{

   int n, k, q;

   int c;

   cin >> n >> k>> q;

   c = q;

   int arr[n+1];

   for(int i = 0; i <= n; i++) arr[i] = 0;

   int a;

   while(q--)

   {

       cin >> a;



       arr[a]++;

   }

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

   {

       if(k-c+arr[i]>0) cout <<"Yes\n";

       else cout <<"No\n";

   }



return 0;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<stdio.h>

#include<string.h>

#include<iostream>

using namespace std;

int main()

{

    int n,k,q,x;

    int a[100002]={0};

    cin>>n>>k>>q;

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(a[i]+k-q>0)

            printf("Yes\n");

        else

            printf("No\n");

    }

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

long long n,k,q;

long long a[200005],cnt[200005];

int main()

{

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

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

        scanf("%lld",&a[i]),cnt[a[i]] ++;

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

    {

    	if(k-(q-cnt[i]) <= 0) printf("No\n");

    	else printf("Yes\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int maxn=200000+100;



ll n,m,q,t[maxn<<2],x,y,z;

/*

ll low(ll x){

	return x&(-x);

}

ll query(ll x){

	ll res=0;

	for(;x>0;x-=low(x))res+=t[x];

	return res;

}

void updata(ll x,ll k){

	for(;x<=n;x+=low(x))t[x]+=k;

}*/



int main(){

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

	memset(t,0,sizeof(t));

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

		scanf("%lld",&x);

		t[x]++;

	}

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

		int k=t[i]+m-q;

		if(k<=0)printf("No\n");

		else printf("Yes\n");

	}

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

long long n,k,q;

long long a[200005],cnt[200005];

int main()

{

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

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

        scanf("%lld",&a[i]),cnt[a[i]] ++;

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

    {

    	if(k-(q-cnt[i]) <= 0) printf("No\n");

    	else printf("Yes\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define maxn 100005

using namespace std;



int a[maxn],b[maxn];



int main(){

	int n,k,q;

	cin>>n>>k>>q;

	

	memset(a,0,sizeof(a));

	memset(b,0,sizeof(b));

	

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

		int flag;

		cin>>flag;

		b[flag]++;

	}

	

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

	

		a[i] = k-(q-b[i]);

		

		if(a[i]<=0)

		 cout<<"No"<<endl;

		else

		  cout<<"Yes"<<endl;

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

long long n,k,q;

long long a[200005],cnt[200005];

int main()

{

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

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

        scanf("%lld",&a[i]),cnt[a[i]] ++;

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

    {

    	if(k-(q-cnt[i]) <= 0) printf("No\n");

    	else printf("Yes\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

#include<string>

#include<cstring>

int a[100010];

using namespace std;

int main(){

	memset(a,0,sizeof a);

	int n,k,q,b;

	cin>>n>>k>>q;

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

		cin>>b;

		a[b]++;

	}

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

		if((q-a[i])>=k) cout<<"No"<<endl;

		else cout<<"Yes"<<endl;

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

long long n,k,q;

long long a[200005],cnt[200005];

int main()

{

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

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

        scanf("%lld",&a[i]),cnt[a[i]] ++;

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

    {

    	if(k-(q-cnt[i]) <= 0) printf("No\n");

    	else printf("Yes\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn=1e5+100;

int main()

{

	int c[maxn],num[maxn],n,q,k,x;

	memset(num,0,sizeof num);

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

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

	{

		scanf("%d",&x);

		num[x]++;

	}

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

	{

		if(num[i]<=q-k) puts("No");

		else puts("Yes");

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

long long n,k,q;

long long a[200005],cnt[200005];

int main()

{

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

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

        scanf("%lld",&a[i]),cnt[a[i]] ++;

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

    {

    	if(k-(q-cnt[i]) <= 0) printf("No\n");

    	else printf("Yes\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>



using namespace std;

const int N =100010;

int b[N];

int vis[N];

int main()

{

	int n,k,q;

	cin>>n>>k>>q;

	memset(vis,0,sizeof(vis));

	int id;

	for(int i=1;i<=q;i++) cin>>id,vis[id]++;

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

	{

		b[i]=k;

		b[i]=b[i]-(q-vis[i]);

		if(b[i]>=1)	puts("Yes");

		else puts("No");	

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

long long n,k,q;

long long a[200005],cnt[200005];

int main()

{

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

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

        scanf("%lld",&a[i]),cnt[a[i]] ++;

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

    {

    	if(k-(q-cnt[i]) <= 0) printf("No\n");

    	else printf("Yes\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}


// Optimized solution to the same problem (faster):
#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;

}

// Fastest known solution for this problem: