stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 1.8342124117572256x):




#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;

 

#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'

#define mset(d,val) memset(d,val,sizeof(d))

#define setp(x) cout<<fixed<<setprecision(x)

#define forn(i,a,b) for(int i=a;i<b;i++)

#define fore(i,a,b) for(int i=a;i<=b;i++)

#define pb push_back

#define mp make_pair

#define F first

#define S second

#define INF ll(1e18)

#define MOD 998244353

#define pqueue priority_queue

#define fbo find_by_order

#define ook order_of_key

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

typedef long double ld;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

 

#define MAXN 1000000007

 

 

int main() {

	ios_base::sync_with_stdio(0); cin.tie(0);

	ll n,k,q,t;

	cin>>n>>k>>q;

	ll arr[n];

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

		arr[i]=0;

	}

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

		cin>>t;

		arr[t-1]+=1;

	}

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

		if (arr[i]<=q-k) {

			cout<<"No"<<endl;

		}

		else {

			cout<<"Yes"<<endl;

		}

	}

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

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

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

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

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

    int m=q-k;

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

    {

        sf("%d",&s);

        ++ans[s];

    }

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

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3934215270753076x):
/*

    Author: Alam Khan

    AUST CSE 40th Batch



*/



#include<bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/pb_ds/detail/standard_policies.hpp>

using namespace __gnu_pbds;



typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<ll> vl;

typedef deque<ll> dl;

typedef stack<ll> stl;

typedef set<ll> sl;

typedef map<string, ll> msl;

typedef map<ll,ll> mll;

typedef tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;

#define sf(n)     scanf("%lld",&n)

#define sff(n,m)     scanf("%lld %lld",&n,&m)

#define sfff(n,m,r)     scanf("%lld %lld %lld",&n,&m,&r)

#define sfs(n)     scanf("%s",n)

#define pf(n)     printf("%lld\n",n)

#define pff(n,m)     printf("%lld %lld\n",n,m)

#define pfff(n,m,r)     printf("%lld %lld %lld\n",n,m,r)

#define pfs(n)     printf("%s\n",n)

#define pfcs(i,n)     printf("Case %lld: %lld\n",i,n)

#define pb     push_back

#define prf     printf

#define inf     2e18

#define low     -1000000000000

#define PI        acos(-1.0)

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

#define rep0(i,n) for(i=0;i<n;i++)

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

#define repe1(i,n) for(i=1;i<=n;i++)

#define repe0(i,n) for(i=0;i<=n;i++)

#define repe(i,a,n) for(i=a;i<=n;i++)

#define endl  "\n"

#define FAST  ios_base::sync_with_stdio(false);cin.tie(NULL);

#define fr freopen("input.txt","r",stdin)

#define fw freopen("output.txt","w",stdout)

ll a[100009];

int main()

{

    ll i,n,t,k,j,x=0,y=0,m;

    cin>>n>>k>>m;

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

    {

        cin>>x;

        a[x]++;

    }

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

    {

        y= m-a[i];

        a[i] = -y;

        a[i]+= k;



    }

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

    {

        if(a[i]>0)

            cout<<"Yes"<<endl;

        else cout<<"No"<<endl;

    }

    return 0;

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

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.41346125907886x):
/*

    Author: Alam Khan

    AUST CSE 40th Batch



*/



#include<bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/pb_ds/detail/standard_policies.hpp>

using namespace __gnu_pbds;



typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<ll> vl;

typedef deque<ll> dl;

typedef stack<ll> stl;

typedef set<ll> sl;

typedef map<string, ll> msl;

typedef map<ll,ll> mll;

typedef tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;

#define sf(n)     scanf("%lld",&n)

#define sff(n,m)     scanf("%lld %lld",&n,&m)

#define sfff(n,m,r)     scanf("%lld %lld %lld",&n,&m,&r)

#define sfs(n)     scanf("%s",n)

#define pf(n)     printf("%lld\n",n)

#define pff(n,m)     printf("%lld %lld\n",n,m)

#define pfff(n,m,r)     printf("%lld %lld %lld\n",n,m,r)

#define pfs(n)     printf("%s\n",n)

#define pfcs(i,n)     printf("Case %lld: %lld\n",i,n)

#define pb     push_back

#define prf     printf

#define inf     2e18

#define low     -1000000000000

#define PI        acos(-1.0)

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

#define rep0(i,n) for(i=0;i<n;i++)

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

#define repe1(i,n) for(i=1;i<=n;i++)

#define repe0(i,n) for(i=0;i<=n;i++)

#define repe(i,a,n) for(i=a;i<=n;i++)

#define endl  "\n"

#define FAST  ios_base::sync_with_stdio(false);cin.tie(NULL);

#define fr freopen("input.txt","r",stdin)

#define fw freopen("output.txt","w",stdout)

ll a[100009];

int main()

{

    ll i,n,t,k,j,x=0,y=0,m;

    cin>>n>>k>>m;

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

    {

        cin>>x;

        a[x]++;

    }

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

    {

        y= m-a[i];

        a[i] = -y;

        a[i]+= k;



    }

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

    {

        if(a[i]>0)

            cout<<"Yes"<<endl;

        else cout<<"No"<<endl;

    }

    return 0;

}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 1.8604658089457082x):
/*

    Author: Alam Khan

    AUST CSE 40th Batch



*/



#include<bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/pb_ds/detail/standard_policies.hpp>

using namespace __gnu_pbds;



typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<ll> vl;

typedef deque<ll> dl;

typedef stack<ll> stl;

typedef set<ll> sl;

typedef map<string, ll> msl;

typedef map<ll,ll> mll;

typedef tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;

#define sf(n)     scanf("%lld",&n)

#define sff(n,m)     scanf("%lld %lld",&n,&m)

#define sfff(n,m,r)     scanf("%lld %lld %lld",&n,&m,&r)

#define sfs(n)     scanf("%s",n)

#define pf(n)     printf("%lld\n",n)

#define pff(n,m)     printf("%lld %lld\n",n,m)

#define pfff(n,m,r)     printf("%lld %lld %lld\n",n,m,r)

#define pfs(n)     printf("%s\n",n)

#define pfcs(i,n)     printf("Case %lld: %lld\n",i,n)

#define pb     push_back

#define prf     printf

#define inf     2e18

#define low     -1000000000000

#define PI        acos(-1.0)

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

#define rep0(i,n) for(i=0;i<n;i++)

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

#define repe1(i,n) for(i=1;i<=n;i++)

#define repe0(i,n) for(i=0;i<=n;i++)

#define repe(i,a,n) for(i=a;i<=n;i++)

#define endl  "\n"

#define FAST  ios_base::sync_with_stdio(false);cin.tie(NULL);

#define fr freopen("input.txt","r",stdin)

#define fw freopen("output.txt","w",stdout)

ll a[100009];

int main()

{

    ll i,n,t,k,j,x=0,y=0,m;

    cin>>n>>k>>m;

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

    {

        cin>>x;

        a[x]++;

    }

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

    {

        y= m-a[i];

        a[i] = -y;

        a[i]+= k;



    }

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

    {

        if(a[i]>0)

            cout<<"Yes"<<endl;

        else cout<<"No"<<endl;

    }

    return 0;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.422619142781979x):
#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;

int a[100005];

int main()

{

	int N,K,Q,i,j,m,n;

	cin>>N>>K>>Q;

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

	{

		a[i]=K-Q;

	}

	for(j=1;j<=Q;j++)

	{

		cin>>m;

		a[m]++;

	}

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

	{

		if(a[i]<=0)

		cout<<"No\n";

		else

		cout<<"Yes\n";

	}

	return 0;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.4267701179134875x):
#include <bits/stdc++.h>



using namespace std;



const int MAXN = 1e5 + 15;



int N, K, Q;

int cnt[MAXN];



int main() {



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



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

		scanf("%d", &player);

		cnt[player]++;

	}



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

		if ( K - Q + cnt[i] > 0 ) {

			puts("Yes");

		} else {

			puts("No");

		}

	}



	return 0;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.3952989875374544x):
#include <bits/stdc++.h>

#define pb push_back

#define mp make_pair

#define pii pair <int, int>

#define fi first

#define sc second

#define ALL(x) x.begin(), x.end()

#define RALL(X) x.begin(), x.end()

#define FOR(i, n, k) for(i=0; i<n; i+=k)

#define FO(i, n, k) for(i=1; i<=n; i+=k)

#define CLEAR(a, b) memset(a, b, sizeof(a))

#define N 100005

#define mid ((l+r)/2)

#define dbg(x) (cerr << #x << " : " << x << endl)

#define endl "\n"

#define MOD 100000009



using namespace std;



typedef long long lli;

#define int long long



signed main(){

	ios_base::sync_with_stdio(false);

	int n, k, q;

	cin >> n >> k >> q;

	map <int, int> ar;

	

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

		int x;

		cin >> x;

		ar[x]++;

	}



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

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

			cout << "Yes" << endl;

		}else

			cout << "No" << endl;

	}	

	return 0;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.40540842879706x):
/*

the top of the losers

*/

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#define sz(v)   ((int)(v).size())

#define  all(v)    ((v).begin()),((v).end())

#define  allr(v)    ((v).rbegin()),((v).rend())

#define   pb         push_back

#define   mp         make_pair

#define    Lcm(a,b)  ((a*b)/(__gcd(a,b)))

#define   length(a)      hypot( (a.imag()),(a.real()) )

#define    x            real()

#define    y            imag()

#define    clr(v,d)      memset( v, d ,sizeof(v))

#define     point        complex<double>

#define    dp(a,b)          (((conj(a))*(b)).real())

#define    cp(a,b)          (((conj(a))*(b)).imag())

typedef  long long     ll ;

typedef  unsigned long long     ull ;

const double eps= (1e-9);

using namespace std;

using namespace __gnu_pbds;

int getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1;}

int setBit1(int num, int idx) {return num | (1<<idx);}

ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}

int dcmp(long double a,long double b){   return fabsl(a-b)<=eps ? 0: (a>b)? 1:-1  ;}

int countNumBit1(int mask) {int ret=0; while (mask) 	{mask &= (mask-1);++ret;	}return ret;}

void GO(){ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);}

ll arr[100009];

int main(){

    GO();

    int n,k,q;

    cin>>n>>k>>q;

    arr[1]+=k;

    arr[n+1]-=k;

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

        int no;

        cin>>no;

        arr[1]+=-1;

        arr[n+1]-=-1;

        arr[no]+=1;

        arr[no+1]-=1;

    }



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

        arr[i]+=arr[i-1];

    }

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

        if(arr[i]<=0){

            cout<<"No\n";

        }

        else{

            cout<<"Yes\n";

        }

    }

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.401621147636554x):
#include<bits/stdc++.h>

using namespace std;



int a[100005], b[100005];

int n,k,q,tmp;



int main()

{

	ios::sync_with_stdio(0);

	cin>>n>>k>>q;



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

		{

			cin>>tmp;

			b[tmp]++;

		}



	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;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.417397673791271x):
#include <bits/stdc++.h>

using namespace std;



int skor[100010], n, q, k, a;

int main()

{

    cin >> n >> k >> q;

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

    {

        skor[i] = k - q;

    }

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

    {

        cin >> a;

        skor[a]++;

    }

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

    {

        if (skor[i] > 0)

        {

            cout << "Yes" << endl;

        }

        else

        {

            cout << "No" << endl;

        }

        

    }

    

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.42154413346706x):
#include <iostream>

#include <vector>

#include <cstdio>

#include <algorithm>

#include <queue>

#include <map>

#include <cstring>

using namespace std;

typedef long long ll;

const ll maxn = 1e6+10;

const ll inf = 1e16;

const ll mod = 998244353;



int N,K,Q;

int arr[maxn];

int main(){

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

//    ios::sync_with_stdio(0);

//    cin.tie(0);cout.tie(0);

    cin>>N>>K>>Q;

    int t;

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

        scanf("%d",&t);

        arr[t]++;

    }

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

        if(Q-arr[i]>=K) puts("No");

        else puts("Yes");

    }

    

    return 0;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.4065926550957473x):
#include<bits/stdc++.h>



using namespace std;

map<int,int> a;

int main() {

	int n, k, q, t;

	cin >> n >> k >> q;

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

		cin >> t;

		a[t] ++;

	}

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

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

		else cout << "No" << endl;

	}

	return 0;

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.4171170910319788x):
#include <bits/stdc++.h>

using namespace std;

long long a[100000+10];

long long aa[100000+10];

int main()

{

	int n,k,q;

	cin>>n>>k>>q;

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

	{

		a[i]=k;

	}

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

	{

		int m;

		cin>>m;

		aa[m]++;

	}

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

	{

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

		{

			cout<<"Yes"<<endl;

		}

		else

		{

			cout<<"No"<<endl;

		}

	}

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 1.4374331006330818x):
#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; 

}