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

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

#define debug  freopen("in.txt","r",stdin),freopen("out.txt","w",stdout);

#define pb push_back

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

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int maxn = 1e6+10;

const int maxM = 1e6+10;

const int inf = 0x3f3f3f3f;



int K,A,B;

int main(){

	// debug;

	ios;



	cin>>K>>A>>B;

	if(B/K - (A-1)/K > 0) cout<<"OK\n";

	else cout<<"NG\n";



	return 0;

}


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



int main ()

{

    int K,A,B;

    scanf("%d%d%d",&K,&A,&B);

    if((B/K)*K<A){

        printf("NG\n");

    }

    else{

        printf("OK\n");



    }



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h> //万能头文件 

using namespace std;

int main(){

	int a, b, i, j, k;//定义变量，题目给的是a, b, k就最好也定义成一样的 

	scanf("%d%d%d", &k, &a, &b);//读入 

	for(int i = a; i<=b; i++){//查找 

		if(i % k == 0){//判断是否在范围中 ,寻找倍数 

			printf("OK");

			return 0;//结束循环 

		}

	}

	printf("NG");//如果没有，输出NG 

}

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

using namespace std;

int dis,ud,dd;

int main ()

{

	scanf("%d%d%d",&dis,&dd,&ud);

	(dis<=(ud-dd+1+(dd-1)%dis))?printf("%s\n","OK"):printf("%s\n","NG");

	return 0;

}

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



using namespace std;



int main() {

    int k,a,b,v;

    cin>>k>>a>>b;

    for(v=0;v<=b;v=v+1){

        if(v%k==0){

            if(v>=a && v<=b){

                cout<<"OK"<<"\n";

                return 0;

            }

        }

    }

    cout<<"NG"<<"\n";

    return 0;

}

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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

using namespace std;

typedef long long int  ll;

const ll mod = 1e9+7;

const ll mod1 = 998244353;

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



void swap(ll &a , ll &b){

    ll t = a;

    a = b;

    b = t;

}



ll min(ll a, ll b){

    if(a<b)return a;

    return b;

}



ll max(ll a, ll b){

    if(a>b)return a;

    return b;

}



 ll pow(ll a, ll b) {

    a %= mod;

    ll ans = 1;

    while (b > 0) {

        if (b & 1)

            ans = ans * a % mod;

        a = a * a % mod;

        b /= 2;

    }

    return ans;

 }



 int n;

 int visited[102] = {0};

int x[102], y[102];



void dfs(int i)

{

    visited[i] = 1;

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

    {

        if(x[j] == x[i] || y[j]==y[i] )

        {

            if(visited[j] == 0)

            {

                dfs(j);

            }

        }

    }

}



int main(){



   ios_base :: sync_with_stdio(0);

   cin.tie(0);



   int a,b,k;

   cin>>k>>a>>b;



   int flag = 0;

   for(int i=a; i<=b; i++)

   {

        if(i%k==0)

        {

            flag = 1;

            cout<<"OK\n";

            break;

        }

   }



    if(flag == 0)

        cout<<"NG\n";



    return 0;



}


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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

using namespace std;



bool solve(int k, int a, int b){

	for (int i = a; i <=b; i++)

		if (i % k == 0)

			return true;

	return false;

}



int main(){

	int k, a, b;

	cin >> k;

	cin >> a >> b;

	cout << (solve(k,a,b) ? "OK" : "NG");

}


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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

using namespace std;

int main()

{

    int k,a,b,ans;

    cin>>k>>a>>b;

    ans = b/k - (a-1)/k;

    if(ans) cout<<"OK"<<endl;

    else cout<<"NG"<<endl;

}


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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

using namespace std;

int main()

{

    int K,A,B,i;

    cin>>K>>A>>B;

    for(i=A;i<=B;i++)

    {

        if(i%K==0)

        {

            cout<<"OK";

            return 0;

        }



    }

    cout<<"NG";

    return 0;

}


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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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



using namespace std;

int main(void)

{

    short int K,A,B,count=0;

    cin>>K;

    cin>>A>>B;



    for(int i=A;i<=B;i++)

    {

        if(i%K==0)

        {

            count=1;

            break;

        }

    }





    if(count==1) cout<<"OK"<<endl;

    else cout<<"NG"<<endl;



}

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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;

using namespace std;



typedef long long ll;

typedef pair<int,int> ii;

typedef tree<ii,null_type,less<ii>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;



int main() {

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

	int k,a,b; cin>>k>>a>>b;

	for (int i=a; i<=b; i++) {

		if (i%k==0) {   

			cout<<"OK"<<endl;

			return 0;

		}

	}

	cout<<"NG"<<endl;

	return 0;

}


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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

//	Created by shamim_nik

//



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;

using namespace std;



typedef long long ll;

typedef tree<int,null_type,less<int>,rb_tree_tag,

		tree_order_statistics_node_update> indexed_set;



#define br printf("\n")



int main () {

	ios::sync_with_stdio(0);

	cin.tie(0);



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

//	freopen("output.txt", "w", stdout);



	int a, b, k;

	cin >> k >> a >> b;

	for(int i = a; i <= b; i++) {

		if(i % k == 0) {

			cout << "OK" << endl;

			return 0;

		}

	}

	cout << "NG" << endl;

	return 0;

}

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



int main(){

	int n;

	scanf("%d", &n);

	int T, K;

	scanf ("%d %d", &T, &K);

	

	for (int i = T; i <= K; i++){

		if (i % n == 0){

			puts("OK");

			return 0;

		}

		else continue;

	}

	puts("NG");

}

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

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<cstring>

#include<string>

typedef long long ll;

template <typename T>void read(T &x){x=0;char ch=getchar();ll f=1;while(!isdigit(ch)){if(ch=='-')f*=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}x*=f;}

#define IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)

using namespace std;

#pragma endregion



int main() {

    IO;

    int k, a, b;

    cin >> k >> a >> b;

    if (a%k==0||b%k==0) 

        cout << "OK\n";

    else

        cout << (a / k - b / k ? "OK\n" : "NG\n");

}

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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

using namespace std;

int main(){

    int k,a,b;

    cin>>k>>a>>b;

    for(int i=1;k*i<=b;i++){

        if(k*i>=a&&k*i<=b){

            cout<<"OK";

            return 0;

        }

    }

    cout<<"NG";

}


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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

using namespace std;

int main(){

	int k,a,b;

	cin>>k>>a>>b;

	if(a%k==0||b%k==0){

		cout<<"OK"<<endl;

	}else{

		int x=b-a;

		int y=a%k;

		if(x+y>=k)cout<<"OK"<<endl;

		else cout<<"NG"<<endl;

	}

} 

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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

// power code taken from geeks for geeks

// */



#include <bits/stdc++.h>

#define MODD 1000000007LL

#define MAX_SIZE 100000000

#define MAX_N 200001

using namespace std;

typedef long long ll;

typedef long double ld;



ll max(ll a,ll b){return a>b?a:b;}

ll min(ll a,ll b){return a<b?a:b;}

double max(double a,double b){return a>b?a:b;}

double min(double a,double b){return a<b?a:b;}



ll modd(ll a){if(a<0LL) return a*-1LL;return a;}

double modd(double a){if(a<0.0) return a*-1.0;return a;}



ll modulo(ll a){return a % MODD; }



ll power(ll k, ll p){

    if (p == 0)

        return 1LL;

 

    ll P = power(k, p/2LL);

    P = (P * P) % MODD;

 

    if (p % 2 == 1)

        P = (P * k) % MODD;

 

    return P;

}





struct p_str{

    ll a,b,c;

};





void swap(ll& s1, ll &s2){

    ll temp=s1;

    s1=s2;

    s2=temp;

}

ll gcd(ll a, ll b){

    if(b==0){

        return a;

    }

    return gcd(b, a%b);

}



void solve(){

    ll k;

    cin>>k;

    ll a, b;

    cin>>a>>b;

    ll i=a;

    while(i<=b){

        if(i%k==0){

            cout<<"OK"<<endl;

            return;

        }

        i++;

    }

    cout<<"NG"<<endl;

}





int main(){





    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    

    ll t=1;



    // cin>>t;

    

    while(t--){

        solve();

    }

    return 0;

}


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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

using namespace std;



#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)

template <typename Arg1>

void __f(const char* name, Arg1&& arg1) {

  cerr << name << " : " << arg1 << "\n";

}

template <typename Arg1, typename... Args>

void __f(const char* na, Arg1&& arg1, Args&&... args) {

  const char* c = strchr(na + 1, ',');

  cerr.write(na, c - na) << " : " << arg1 << " , ";

  __f(c + 1, args...);

}

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

#define endl "\n"

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

#define UNIQUE(v) sort(ALL(v)),v.erase(unique(ALL(v)),v.end())

typedef long long LL;

typedef pair <int,int> pii;

const double pi = acos(-1.0);

const double eps = 1e-8;

const int inf = 1e9;

const LL INF = 1e18;

const int mod = 1e9 + 7;

const int N = 1e5 + 5;



void solve() {

  int k, a, b;

  cin >> k >> a >> b;

  if (b/k - (a - 1)/k) cout << "OK" << endl;

  else cout << "NG" << endl;

}



int main() {

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

  #ifdef LOCAL

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

  #endif

  int t = 1;

  for (int tc = 1; tc <= t; tc++) solve();

  return 0;

}




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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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


#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,x,y,i=0,m=1;

    cin>>n>>x>>y;

    while(1)

    {

        i++;

        m=n*i;

        if(m>=x && m<=y)

            {

                cout<<"OK";

            return 0;

            }

        if(m>y)

            {

                cout<<"NG";

            return 0;

            }

    }



   return 0;

}


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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

using namespace std;

int main()

{

    int a,b,k,c=0;

    cin>>k>>a>>b;

    for(int i=a;i<=b;i++)

    {

        if(i%k==0)

        {

            c=1;

            break;

        }

    }

    if(c==1)

        cout<<"OK\n";

    else

        cout<<"NG\n";

}


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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

using namespace std;

int main()

{

  int a,b,k,x=0;

  cin>>k>>a>>b;

  for(int i=a;i<=b;i++)

  {

      if(i%k==0){

        x=1;

        break;

        }

  }

if(x==1){

cout<<"OK";

}

else{

cout<<"NG";

}

}

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



int main(){

	int kelipatan,batas1,batas2;

	scanf("%d %d %d", &kelipatan,&batas1,&batas2);

	int temp=batas2/kelipatan;

	if(temp*kelipatan>=batas1){

		printf("OK\n");

	}

	else printf("NG\n");

	return 0;

}

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

using namespace std;

int main()

{

	int k,a,b;

	cin>>k>>a>>b;

	if(b/k-a/k!=0||a%k==0)

		puts("OK");

	else

		puts("NG");

}

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



int main() {

  int K, A, B, objective = 0;

  scanf("%d %d %d", &K, &A, &B);

  

  for(int i = A; i <= B; i++) {

    if(i % K == 0) {

      objective = 1;

      break;

    }

  }



  (objective) ? puts("OK") : puts("NG");

  return 0;

}

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

using namespace std;



#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define pop_b pop_back

#define pf push_front

#define pop_f pop_front



typedef long long ll;

typedef pair<ll,ll> ii;

typedef long double ld; 

typedef vector<int> vi;



#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

void err(istream_iterator<string> it) {}

template<typename T, typename... Args>

void err(istream_iterator<string> it, T a, Args... args) {

	cerr << *it << " = " << a << endl;

	err(++it, args...);

}



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

#define repr(i, n) for (ll i = (n) - 1; i >= 0; i--)

#define rep2(i, l, r) for (ll i = (l); i < (r); i++)

#define rep2r(i, l, r) for (ll i = (r) - 1; i >= (l); i--)



int main()

{

	int k, a, b;

	cin>>k>>a>>b;

	int p = (a)/k;

	cout<<(k*(p + (a%k ? 1 : 0)) <= b ? "OK" : "NG")<<endl;

}

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



int main() {

  int K, A, B, objective = 0;

  scanf("%d %d %d", &K, &A, &B);

  

  for(int i = A; i <= B; i++) {

    if(i % K == 0) {

      objective = 1;

      break;

    }

  }



  (objective) ? puts("OK") : puts("NG");

  return 0;

}

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

using namespace std;

int main()

{

    int a, b, k, i;

    cin>>k>>a>>b;

    for(i=a; i<=b; i++)

    {

        if(i%k==0)

        {

             cout<<"OK";

           return 0;

        }

    }

     cout<< "NG"<<endl;

     return 0;

}


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



int main() {

  int K, A, B, objective = 0;

  scanf("%d %d %d", &K, &A, &B);

  

  for(int i = A; i <= B; i++) {

    if(i % K == 0) {

      objective = 1;

      break;

    }

  }



  (objective) ? puts("OK") : puts("NG");

  return 0;

}

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

#include<math.h>

#define pi 2*acos(0)

int main()

{

    int a,b,c;

    int x=0;

    std::cin>>a>>b>>c;



    x = c/a;

    if(a*x>= b){

        std::cout<<"OK"<<std::endl;

    }

    else{

        

        std::cout<<"NG"<<std::endl;

    }

    return 0;

}

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



int main() {

  int K, A, B, objective = 0;

  scanf("%d %d %d", &K, &A, &B);

  

  for(int i = A; i <= B; i++) {

    if(i % K == 0) {

      objective = 1;

      break;

    }

  }



  (objective) ? puts("OK") : puts("NG");

  return 0;

}

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

using namespace std;



typedef long long ll;



main()

{

    int a, b, k, i;

    cin >> k >> a >> b;

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

        if(i % k == 0){

            cout << "OK";

            break;

        }

    }

    if(i == b+1){

        cout << "NG";

    }







}











///promy_pompom

//hello world


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



int main() {

  int K, A, B, objective = 0;

  scanf("%d %d %d", &K, &A, &B);

  

  for(int i = A; i <= B; i++) {

    if(i % K == 0) {

      objective = 1;

      break;

    }

  }



  (objective) ? puts("OK") : puts("NG");

  return 0;

}

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

#include <iostream>

#include <cstdio>

#include <cctype>

using namespace std;

 

int main()

{

  char ch;

  while(1){

    scanf("%c",&ch); //scanf???' '???????????????

    if(ch=='\n') break;

    if(islower(ch)!=0){ //?°???????????????°true

      ch=toupper(ch); //?°?????????? ??§??????

      cout << ch;

    }else if(isupper(ch)!=0){ //??§??????????????°true

      ch=tolower(ch); //??§????????? ?°???????

      cout << ch;

    }else{

      cout << ch;

    }

  }

  cout << endl;

  return 0;

}

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

#include<cctype>

int main()

{

    char ch;

    while(1)

    {

        scanf("%c", &ch);

        if( isalpha(ch) )

        {

            if( islower(ch) ) printf("%c", toupper(ch));

            else printf("%c", tolower(ch));

        }

        else printf("%c", ch);

    if ( ch == '\n' )break;

    }

    return 0;

}

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

#include <cctype>

#include <cstdio>

using namespace std;



int main()

{

	char ch;//ch:?????????

	

	while(1){

		scanf("%c",&ch);

		if(ch == '\n') break;

		

		if(islower(ch) != 0){ //ch????°???????????????\????°????????????????

			ch = toupper(ch); //?°????????????§??????

			cout << ch;

		}else if(isupper(ch) != 0){ //??§???????????????????????\?????§???????????????

			ch = tolower(ch); //o??????????°???????

			cout << ch;

		}else{ //??±????????\???

			cout << ch;

		}

	}

	cout << endl;

	return 0;

}

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

#include<cctype>

int main()

{

    char ch;

    while(1)

    {

        scanf("%c", &ch);

        if( isalpha(ch) )

        {

            if( islower(ch) ) printf("%c", toupper(ch));

            else printf("%c", tolower(ch));

        }

        else printf("%c", ch);

    if ( ch == '\n' )break;

    }

    return 0;

}

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

using namespace std;

int main(){

	

		double h,b;

		cin>>h>>b;

		cout<<ceil(h/b);

	return 0;

}

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

int main() {

 int h,a;

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

 int x = h/a;

 int z = h%a;

 if (z==0)

 		printf("%d",x);

	else 

		printf("%d", ++x);

return 0;

}


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

using namespace std;







int main()

{

   int H , A,coun = 0;

   cin>>H>>A;



   for( int i =0; H>0;i++)

   {

       H = H - A;

       coun+=1;

   }

   cout<<coun;





    return 0;

}

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



int main()

{	

	int x,y,z;

	scanf("%d %d", &x, &y);

	if(x<y){

		printf("1");

	}

	else if(x>=y){

		z=(x-1)/y+1;

		printf("%d", z);

	} 

	return 0;

}

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

# include <cmath>

using namespace std;



int main()

{

	float x,y; cin>>x>>y;

	cout<<ceil(x/y);

}


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

int main()

{

	int H,A;

	scanf("%d%d",&H,&A);

	H%A<=0?printf("%d",H/A):printf("%d",H/A +1);//kalo yang pertama kondisi contoh 10/5 jdi habis kalo kondisi kedua contoh 10/4 jdi sisa dan harus ditambah 1 kali atatck

	return 0;

}


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



using namespace std;



int main()

{

    int h, a ;

    int counter =0;

    cin>>h>>a;

    for(int i=h ; i>0 ; i-=a)

    {

        counter++;

    }

    cout<<counter;

    return 0;

}


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

int main()

{

	int H,A;

	scanf("%d%d",&H,&A);

	H%A<=0?printf("%d",H/A):printf("%d",H/A +1);//kalo yang pertama kondisi contoh 10/5 jdi habis kalo kondisi kedua contoh 10/4 jdi sisa dan harus ditambah 1 kali atatck

	return 0;

}


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



using namespace std;

int main()

{

    int a,b;

    cin>>a>>b;



    if(a%b==0)

    {

        cout<<a/b<<endl;

    }

    else

    {

        cout<<a/b+1<<endl;

    }

    return 0;

}


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

int main()

{

	int H,A;

	scanf("%d%d",&H,&A);

	H%A<=0?printf("%d",H/A):printf("%d",H/A +1);//kalo yang pertama kondisi contoh 10/5 jdi habis kalo kondisi kedua contoh 10/4 jdi sisa dan harus ditambah 1 kali atatck

	return 0;

}


// Fastest known solution for this problem: