Minimum Cars required CodeChef Solution

Problem – Minimum Cars required CodeChef Solution

A single car can accommodate at most 4 people.

N friends want to go to a restaurant for a party. Find the minimum number of cars required to accommodate all the friends.

Input Format

  • The first line contains a single integer T – the number of test cases. Then the test cases follow.
  • The first and only line of each test case contains an integer N – denoting the number of friends.

Output Format

For each test case, output the minimum number of cars required to accommodate all the friends.

Constraints

  • 1≤T≤1000
  • 2≤N≤1000

Sample 1:

Input: 4
4
2
7
98
Output: 1
1
2
25

Explanation:

Test Case 1: There are only 4 friends and a single car can accommodate 4 people. Thus, only 1 car is required.

Test Case 2: There are only 2 friends and a single car can accommodate 4 people. Thus, only 1 car is required

Test Case 3: There are 7 friends and 2 cars can accommodate 8 people. Thus, 2 cars are required.

Minimum Cars required CodeChef Solution in Java

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
	    
	    Scanner sc=new Scanner(System.in);
	    int T=sc.nextInt();
	    for(int i=0;i<T;i++)
	    {
		int x=sc.nextInt();
		int a=x%4;
		int b=x/4;
		if(a==0)
		System.out.println(b);
		else
		System.out.println(b+1);
	    }
	}
}

Minimum Cars required CodeChef Solution in Pyth 3

# cook your dish here
n=int(input())
for i in range(0,n):
    e=int(input())
    d=e%4
    if(d<4 and d!=0):
        print((e//4)+1)
    elif(d==0):
        print(e//4)
        

Minimum Cars required CodeChef Solution in C++17

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--){
	    int n;
	    cin>>n;
	   // cout<<n*2<<endl;
	   if(n%4==0)
	    {
	        cout<<(n/4)<<endl;
	    }
	    else{
	        cout<<(n/4)+1<<endl;
	    }
	}

	return 0;
}
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