# SLOOP - Editorial

Setter: Daanish Mahajan
Tester & Editorialist: Taranpreet Singh

Cakewalk

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;
void run() throws Exception{
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());}

StringTokenizer st;
}

}

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

String nextLine() throws Exception{
String str = "";
try{
Feel free to share your approach. Suggestions are welcomed as always. 