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. 