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Practice
Setter: Daanish Mahajan
Tester & Editorialist: Taranpreet Singh
DIFFICULTY
Cakewalk
PREREQUISITES
None
PROBLEM
Given a song of duration S, find out the number of times you can play it completely during M minutes ride.
QUICK EXPLANATION
The song can be played at most M/S times.
EXPLANATION
Let’s simulate the process for M = 14, S = 3.
- Song plays for first time during time [1, 3]
- Song plays for second time during time [4, 6]
- Song plays for third time during time [7, 9]
- Song plays for fourth time during time [10,12]
- Song plays for fifth time during time [13,15], but the ride got over at the end of 14-th minute, so this song is not counted.
Hence, for M = 14 and S = 3, we can play song 4 times.
It is easy to write a solution simulating above process, keeping count of the number of times song is played.
count = 0
while M >= S:
count++
M -= S
The count shall store the final number of times song is played completely.
O(1) solution
What we are doing above is repeated subtraction while M >= S. We can see that above loop continues for \lfloor M/S\rfloor iterations. Indeed, \lfloor M/S\rfloor is the number of times the song is played, hence we can directly output \lfloor M/S\rfloor as the answer in O(1) time.
Bonus
Given a song of duration S, find out the number of times you can play it during M minutes ride. If a song is in going on at the end of ride, it is counted as played.
TIME COMPLEXITY
The time complexity is O(1) per test case.
SOLUTIONS
Setter's Solution
#include<bits/stdc++.h>
# define pb push_back
#define pii pair<int, int>
#define mp make_pair
# define ll long long int
using namespace std;
const int maxt = 1000, maxm = 100, maxs = 10;
int main()
{
int t; cin >> t;
while(t--){
int m, s; cin >> m >> s;
cout << (int)(m / s) << endl;
}
}
Tester's Solution
import java.util.*;
import java.io.*;
class SLOOP{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int M = ni(), S = ni();
pn(M/S);
}
//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 SLOOP().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.