Dazzling AXNODR Challenge AXNODR Codechef Solution

Dazzling AXNODR Challenge AXNODR Solution

Dazzler has a blank canvas and (N−1) colours numbered from 2 to N.
Let B denote the beauty of the canvas. The beauty of a blank canvas is 1.

Dazzler paints the canvas by using all the (N−1) colours exactly once. On applying the ith colour (2≤i≤N):

If i is odd, B=B & i.
If i is even, B=B⊕i.
Find the beauty of the canvas after applying all (N−1) colours.

Note: The colours are applied in ascending order. Colour number 2 is applied first. The ith numbered colour is applied after (i−1)th numbered colour for all i>2.

Here & and ⊕ denote the bitwise AND and bitwise XOR operations respectively.

Input Format
First line will contain T, the number of test cases. Then the test cases follow.
Each test case contains of a single line of input, a single integer N.

Output Format
For each test case, output a single integer, the beauty of the canvas after applying all (N−1) colours.

Constraints
1≤T≤105
2≤N≤1016

Sample Input 1
2
4
10

Sample Output 1
7
3

Explanation
Initially, B=1.

On applying colour 2: Since 2 is even, B=B⊕2=1⊕2=3.
On applying colour 3: Since 3 is odd, B=B&3=3&3=3.
On applying colour 4: Since 4 is even, B=B⊕4=3⊕4=7.
On applying colour 5: Since 5 is odd, B=B&5=7&5=5.
On applying colour 6: Since 6 is even, B=B⊕6=5⊕6=3.
On applying colour 7: Since 7 is odd, B=B&7=3&7=3.
On applying colour 8: Since 8 is even, B=B⊕8=3⊕8=11.
On applying colour 9: Since 9 is odd, B=B&9=11&9=9.
On applying colour 10: Since 10 is even, B=B⊕10=9⊕10=3.

Test case 1: There are 3 colours numbered 2,3, and 4. Initially, B=1.
The final beauty of the canvas is 7.

Test case 2: There are 9 colours numbered 2,3,4,5,6,7,8,9, and 10. Initially, B=1.
The final beauty of the canvas is 3.

SOLUTION

Program: Dazzling AXNODR Challenge AXNODR Solution in Python

for _ in range (int(input())):
    a=int(input())
    if(a%2==0):
        if(a%4==0):
            t=3^a
        else:
            t=3
    else:
        if ((a-1)%4==0):
            
            t= (3^a-1) and a
        else:
            t=3
        
    print(tmp)

Program: Dazzling AXNODR Challenge AXNODR Solution in C++

#include <iostream>
using namespace std;

int main() {
	long long int t;
	cin>>t;
	long long int n;
	while(t--){
	    cin>>n;
	  if(n%4==0)
	  cout<<n+3<<endl;
	  else if(n%4==1)
	  cout<<n<<endl;
	  else 
	  cout<<"3"<<endl;
	}
	return 0;
}

Program: Dazzling AXNODR Challenge AXNODR Solution in Java

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

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++){
            long n = sc.nextLong();
            if(n%4==0){
                System.out.println(n+3);
            }else if(n%4==1){
                System.out.println(n);
            }else System.out.println(3);
        }
	}
}

Related:

Leave a Comment

five × 4 =