• Problem Statement
• Key Insight
• Code
• Conclusion

Codeforces Problem 1869A - Make It Zero

Today, we're diving into an interesting problem from the Zhongkao examination that involves manipulating an array using bitwise XOR operations. This problem not only tests your understanding of XOR but also your ability to think algorithmically under time constraints. Let's break it down step by step!

problem link: Codeforces Problem 1869A - Make It Zero from Codeforces.

Problem Statement

You are given an array a of length n (2 ≤ n ≤ 100), and your task is to convert all elements of the array into 0 by performing a specific operation up to 8 times. The operation is as follows:

• Select two indices l and r (1 ≤ l ≤ r ≤ n).
• Compute s = a[l] ⊕ a[l+1] ⊕ ... ⊕ a[r], where ⊕ represents the XOR operation.
• Replace all elements a[i] (for l ≤ i ≤ r) with s.

The objective is to determine a sequence of operations that results in all elements of a becoming 0.

Main Concept: How XOR Works

To solve this problem, understanding the properties of XOR is critical:

• Identity Property: x ⊕ 0 = x
• Self-Inverse Property: x ⊕ x = 0
• Commutativity: x ⊕ y = y ⊕ x
• Associativity: (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

The self-inverse property (x ⊕ x = 0) is particularly crucial, as it allows us to neutralize values systematically. By applying XOR operations strategically, we can ensure all elements of the array are reduced to 0.

Intuition and Strategy

The solution varies based on whether the length of the array (n) is even or odd:

Even Length Arrays (n is even):

• Select the entire array (l = 1, r = n).
• Perform the XOR operation twice:
  • First operation sets all elements to the same value s.
  • Second operation reduces them all to 0 (s ⊕ s = 0).

Odd Length Arrays (n is odd):

• Divide the problem into two parts:
  • Apply the operation twice to the first n-1 elements to set them to 0.
  • Apply the operation twice to the last two elements to ensure the entire array becomes 0.

Example Walkthrough

Input Example

n = 5, a = [3, 5, 1, 7, 9]
    

Steps for n = 5 (Odd Length Array):

1. Step 1: Select l = 1, r = 4 ([3, 5, 1, 7]):
   • Compute s = 3 ⊕ 5 ⊕ 1 ⊕ 7 = 4.
   • Replace [3, 5, 1, 7] with [4, 4, 4, 4].
2. Step 2: Select l = 1, r = 4 again:
   • Compute s = 4 ⊕ 4 ⊕ 4 ⊕ 4 = 0.
   • Replace [4, 4, 4, 4] with [0, 0, 0, 0].
3. Step 3: Select l = 4, r = 5 ([0, 9]):
   • Compute s = 0 ⊕ 9 = 9.
   • Replace [0, 9] with [9, 9].
4. Step 4: Select l = 4, r = 5 again:
   • Compute s = 9 ⊕ 9 = 0.
   • Replace [9, 9] with [0, 0].

Output:

k = 4
Operations: (1, 4), (1, 4), (4, 5), (4, 5)
    

Code Implementations

Below are the implementations in Java and C++.

C++

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int t; // Number of test cases
    cin >> t;

    while (t-- > 0) {
        int n; // Length of the array
        cin >> n;
        vector<int> arr(n);

        for (int i = 0; i < n; i++) {
            cin >> arr[i];
        }

        if (n % 2 == 0) {
            cout << 2 << endl;
            cout << 1 << " " << n << endl;
            cout << 1 << " " << n << endl;
        } else {
            cout << 4 << endl;
            cout << 1 << " " << (n - 1) << endl;
            cout << 1 << " " << (n - 1) << endl;
            cout << (n - 1) << " " << n << endl;
            cout << (n - 1) << " " << n << endl;
        }
    }

    return 0;
}
    
        

JAVA

import java.util.*;

public class MakeItZero {
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();  // Number of test cases

        while (t-- > 0) {
            int n = scn.nextInt(); // Length of the array
            int[] arr = new int[n];

            for (int i = 0; i < n; i++) {
                arr[i] = scn.nextInt();
            }

            if (n % 2 == 0) {
                System.out.println(2);
                System.out.println(1 + " " + n);
                System.out.println(1 + " " + n);
            } else {
                System.out.println(4);
                System.out.println(1 + " " + (n - 1));
                System.out.println(1 + " " + (n - 1));
                System.out.println((n - 1) + " " + n);
                System.out.println((n - 1) + " " + n);
            }
        }
        scn.close();
    } }
        

Conclusion

This problem demonstrates the versatility of XOR and its ability to simplify computational tasks. By dividing the array into manageable sections and applying XOR systematically, we can always reduce the array to all zeroes within 8 operations. Whether you're a beginner or an experienced coder, understanding XOR is a powerful tool in your problem-solving arsenal. Happy coding!

• Finding the range of achievable sums using the arithmetic progression formula.
• Proving that every sum between the minimum sum and the maximum sum is achievable by incrementing pointers systematically.

This approach ensures an efficient solution to the problem, making it easier to determine if a specific sum is achievable within the given constraints.