Codechef Divisible by 3 problem solution

In this Codechef Divisible by 3 problem-solution Stack likes the number 3 a lot.

He has two non-negative integers AA and BB.

In one operation, Stack can do either of the following:

A:=|A-B|A:=∣A−B∣ (change AA to |A-B|∣A−B∣)

B:=|A-B|B:=∣A−B∣ (change BB to |A-B|∣A−B∣)

Note that |X|∣X∣ denotes absolute value of XX. For example |-7| = 7∣−7∣=7 and |4| = 4∣4∣=4.

Find the minimum number of operations after which at least one integer out of A and B becomes divisible by 3.

Problem solution in Python.

# cook your dish here
for i in range(int(input())):
    A, B = [int(i) for i in input().split()]
    if A % 3 ==0 or B % 3 == 0:
        print(0)
    elif A <= B:
        if (B-A) % 3 == 0:
            print(1)
        else:
            print(2)
    else:
        if (A-B) % 3 == 0:
            print(1)
        else:
            print(2)

Problem solution in Java.

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 a = sc.nextInt();
		    int b = sc.nextInt();
		    
		    if(a%3==0||b%3==0){
		        System.out.println(0);
		    }else if(Math.abs(a-b)%3==0){
		        System.out.println(1);
		    }else{
		        System.out.println(2);
		    }
		}
		// your code goes here
	}
}

Problem solution in C++.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define loop(i,a,b) for(int i=a;i<b;i++)
#define range(a)    a.begin(),a.end()
#define p(a)        push_back(a)

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--){
	    ll a,b;
	    cin>>a>>b;
	    int count=0;
	    if(a%3==0 || b%3==0){
	        cout<<0<<endl;
	    }
	    else if(a==b){
	        cout<<1<<endl;
	    }
	    else{
	        int ans1=a%3;
	        int ans2=b%3;
	        if(ans1==ans2){
	            cout<<1<<endl;
	        }
	        else{
	            cout<<2<<endl;
	        }
	    }
	}
	return 0;
}

Problem solution in C.

#include<stdio.h>
int main()
{
    int t,i,a,b,p=0,q=0;
    scanf("%d",&t);
    for(i=0;i<t;i++){
        scanf("%d %d",&a,&b);
        while(a%3!=0&&b%3!=0){
            if(a>b){
            a=a-b;
            p++;
            }
        else{
            b=b-a;
            p++;
        }
        if(a%3==0||b%3==0)
            break;
        }
        q=p;
        p=0;
        printf("%d\n",q);

    }
    return 0;
}