MINDAYSRET - Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Practice

Setter: Utkarsh Gupta
Tester: Lavish Gupta and Abhinav Sharma
Editorialist: Taranpreet Singh

DIFFICULTY

Cakewalk

PREREQUISITES

None

PROBLEM

There are N students in a college. The school can only screen K students per day. What’s the minimum number of days required for all students to be screened?

QUICK EXPLANATION

The number of days required is \displaystyle\left\lceil \frac{N}{K}\right\rceil

EXPLANATION

Simulation solution

In this solution, we can simulate the actual process of the screen and maintain a counter storing number of days taken.

Following pseudo-code represents the simulation solution.

Read N, K
countDays = 0
while N > 0:
    N -= min(N, K)
    countDays++

print(countDays) 

The time complexity of this solution is O(N) in the worst case when K = 1

Faster solution

We can visualize dividing all the students into groups of size K, where the last group may be smaller than K if N \bmod K \neq 0. Each group can be screened in one day. So the number of days required is the number of groups formed.

The number of groups formed is \displaystyle\left\lceil \frac{N}{K}\right\rceil. For example, if N = 7 and K = 3, three days are needed. For N = 9, K = 3, three days are needed.

The time complexity of this solution is O(1) per test case.

Implementation tip: \displaystyle\left\lceil \frac{N}{K}\right\rceil is same as \displaystyle\left\lfloor \frac{N+K-1}{K}\right\rfloor, so you can directly print (N+K-1)/K with floor division. Why this works is an exercize.

TIME COMPLEXITY

The time complexity is O(1) or O(N) per test case.

SOLUTIONS

Setter's Solution
Tester's solution 1
#include <bits/stdc++.h>
using namespace std;
 
 
/*
------------------------Input Checker----------------------------------
*/
 
long long readInt(long long l,long long r,char endd){
    long long x=0;
    int cnt=0;
    int fi=-1;
    bool is_neg=false;
    while(true){
        char g=getchar();
        if(g=='-'){
            assert(fi==-1);
            is_neg=true;
            continue;
        }
        if('0'<=g && g<='9'){
            x*=10;
            x+=g-'0';
            if(cnt==0){
                fi=g-'0';
            }
            cnt++;
            assert(fi!=0 || cnt==1);
            assert(fi!=0 || is_neg==false);
 
            assert(!(cnt>19 || ( cnt==19 && fi>1) ));
        } else if(g==endd){
            if(is_neg){
                x= -x;
            }
 
            if(!(l <= x && x <= r))
            {
                cerr << l << ' ' << r << ' ' << x << '\n';
                assert(1 == 0);
            }
 
            return x;
        } else {
            assert(false);
        }
    }
}
string readString(int l,int r,char endd){
    string ret="";
    int cnt=0;
    while(true){
        char g=getchar();
        assert(g!=-1);
        if(g==endd){
            break;
        }
        cnt++;
        ret+=g;
    }
    assert(l<=cnt && cnt<=r);
    return ret;
}
long long readIntSp(long long l,long long r){
    return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
    return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
    return readString(l,r,'\n');
}
string readStringSp(int l,int r){
    return readString(l,r,' ');
}
 
 
/*
------------------------Main code starts here----------------------------------
*/
 
const int MAX_T = 10000;
const int MAX_LEN = 1e5;
const int MAX_SUM_LEN = 1e5;
 
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
 
int sum_len = 0;
int max_n = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;
 
int n,k;
string s;
 
 
void solve()
{
   n = readIntSp(1,100);
   k = readIntLn(1,100);

   long long ans = 0 , curr = 0 ;
   while(curr < n)
   {
        curr += k ;
        ans++ ;
   }
   cout << ans << endl ;
   return ;
}
 
signed main()
{

    // fast;
    
    int t = 1;
    
    t = readIntLn(1,MAX_T);
    assert(1<=t && t<=10000);
    
    for(int i=1;i<=t;i++)
    {     
       solve();
    }
    
    assert(getchar() == -1);
 
    cerr<<"SUCCESS\n";
    cerr<<"Tests : " << t << '\n';
    // cerr<<"Sum of lengths : " << sum_len << '\n';
    // cerr<<"Maximum length : " << max_n << '\n';
    // cerr<<"Total operations : " << total_ops << '\n';
    // cerr<<"Answered yes : " << yess << '\n';
    // cerr<<"Answered no : " << nos << '\n';
}
Tester's Solution 2
#include <bits/stdc++.h>
using namespace std;
 
 
/*
------------------------Input Checker----------------------------------
*/
 
long long readInt(long long l,long long r,char endd){
    long long x=0;
    int cnt=0;
    int fi=-1;
    bool is_neg=false;
    while(true){
        char g=getchar();
        if(g=='-'){
            assert(fi==-1);
            is_neg=true;
            continue;
        }
        if('0'<=g && g<='9'){
            x*=10;
            x+=g-'0';
            if(cnt==0){
                fi=g-'0';
            }
            cnt++;
            assert(fi!=0 || cnt==1);
            assert(fi!=0 || is_neg==false);
 
            assert(!(cnt>19 || ( cnt==19 && fi>1) ));
        } else if(g==endd){
            if(is_neg){
                x= -x;
            }
 
            if(!(l <= x && x <= r))
            {
                cerr << l << ' ' << r << ' ' << x << '\n';
                assert(1 == 0);
            }
 
            return x;
        } else {
            assert(false);
        }
    }
}
string readString(int l,int r,char endd){
    string ret="";
    int cnt=0;
    while(true){
        char g=getchar();
        assert(g!=-1);
        if(g==endd){
            break;
        }
        cnt++;
        ret+=g;
    }
    assert(l<=cnt && cnt<=r);
    return ret;
}
long long readIntSp(long long l,long long r){
    return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
    return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
    return readString(l,r,'\n');
}
string readStringSp(int l,int r){
    return readString(l,r,' ');
}
 
 
/*
------------------------Main code starts here----------------------------------
*/
 
const int MAX_T = 1000000;
const int MAX_LEN = 1e5;
const int MAX_SUM_LEN = 1e5;
 
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
 
int sum_len = 0;
int max_n = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;
 
int n,k;
string s;
 
 
void solve()
{

    
   n = readIntSp(1,1000);
   k = readIntLn(1,1000);

   cout<<(n+k-1)/k<<"\n";
}
 
signed main()
{

    fast;
    
    int t = 1;
    
    t = readIntLn(1,MAX_T);
    assert(1<=t && t<=1000000);
    
    for(int i=1;i<=t;i++)
    {     
       solve();
    }
    
    assert(getchar() == -1);
 
    cerr<<"SUCCESS\n";
    cerr<<"Tests : " << t << '\n';
    // cerr<<"Sum of lengths : " << sum_len << '\n';
    // cerr<<"Maximum length : " << max_n << '\n';
    // cerr<<"Total operations : " << total_ops << '\n';
    // cerr<<"Answered yes : " << yess << '\n';
    // cerr<<"Answered no : " << nos << '\n';
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class MINDAYSRET{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int N = ni(), K = ni();
        pn((N+K-1)/K);
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    static boolean multipleTC = true;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        new MINDAYSRET().run();
    }
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }
}

Feel free to share your approach. Suggestions are welcomed as always. :slight_smile:

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