Given an array of N positive integers. There are Q queries, each include a range [L, R]. For each query output the xor of greatest odd divisor of each number in that range.
Examples:
Input : arr[] = { 3, 4, 5 }
query 1: [0, 2]
query 2: [1, 2]
Output : 7 4
Greatest odd divisor are: { 3, 1, 5 }
XOR of 3, 1, 5 is 7
XOR of 1, 5 is 4
Input : arr[] = { 2, 1, 2 }
query 1: [0, 2]
Output : 1
The idea is to precompute the greatest odd divisor of the array and store it in an array, say preXOR[]. Now, precompute and store prefix XOR of the array preXOR[]. To answer each query, return (preXOR[r] xor preXOR[l-1]).
Below is the implementatin of this approach:
C++
#include <bits/stdc++.h> using namespace std; // Precompute the prefix XOR of greatest // odd divisor void prefixXOR(int arr[], int preXOR[], int n) { // Finding the Greatest Odd divisor for (int i = 0; i < n; i++) { while (arr[i] % 2 != 1) arr[i] /= 2; preXOR[i] = arr[i]; } // Finding prefix XOR for (int i = 1; i < n; i++) preXOR[i] = preXOR[i - 1] ^ preXOR[i]; } // Return XOR of the range int query(int preXOR[], int l, int r) { if (l == 0) return preXOR[r]; else return preXOR[r] ^ preXOR[l - 1]; } // Driven Program int main() { int arr[] = { 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); int preXOR[n]; prefixXOR(arr, preXOR, n); cout << query(preXOR, 0, 2) << endl; cout << query(preXOR, 1, 2) << endl; return 0; } |
Java
// Java code Queries on XOR of // greatest odd divisor of the range import java.io.*; class GFG { // Precompute the prefix XOR of greatest // odd divisor static void prefixXOR(int arr[], int preXOR[], int n) { // Finding the Greatest Odd divisor for (int i = 0; i < n; i++) { while (arr[i] % 2 != 1) arr[i] /= 2; preXOR[i] = arr[i]; } // Finding prefix XOR for (int i = 1; i < n; i++) preXOR[i] = preXOR[i - 1] ^ preXOR[i]; } // Return XOR of the range static int query(int preXOR[], int l, int r) { if (l == 0) return preXOR[r]; else return preXOR[r] ^ preXOR[l - 1]; } // Driven Program public static void main (String[] args) { int arr[] = { 3, 4, 5 }; int n = arr.length; int preXOR[] = new int[n]; prefixXOR(arr, preXOR, n); System.out.println(query(preXOR, 0, 2)) ; System.out.println (query(preXOR, 1, 2)); } } // This article is contributed by vt_m |
Python3
# Precompute the prefix XOR of greatest # odd divisor def prefixXOR(arr, preXOR, n): # Finding the Greatest Odd divisor for i in range(0, n, 1): while (arr[i] % 2 != 1): arr[i] = int(arr[i] / 2) preXOR[i] = arr[i] # Finding prefix XOR for i in range(1, n, 1): preXOR[i] = preXOR[i - 1] ^ preXOR[i] # Return XOR of the range def query(preXOR, l, r): if (l == 0): return preXOR[r] else: return preXOR[r] ^ preXOR[l - 1] # Driver Code if __name__ == '__main__': arr = [3, 4, 5] n = len(arr) preXOR = [0 for i in range(n)] prefixXOR(arr, preXOR, n) print(query(preXOR, 0, 2)) print(query(preXOR, 1, 2)) # This code is contributed by # Sahil_shelangia |
C#
// C# code Queries on XOR of // greatest odd divisor of the range using System; class GFG { // Precompute the prefix XOR of greatest // odd divisor static void prefixXOR(int []arr, int []preXOR, int n) { // Finding the Greatest Odd divisor for (int i = 0; i < n; i++) { while (arr[i] % 2 != 1) arr[i] /= 2; preXOR[i] = arr[i]; } // Finding prefix XOR for (int i = 1; i < n; i++) preXOR[i] = preXOR[i - 1] ^ preXOR[i]; } // Return XOR of the range static int query(int [] preXOR, int l, int r) { if (l == 0) return preXOR[r]; else return preXOR[r] ^ preXOR[l - 1]; } // Driven Program public static void Main () { int []arr = { 3, 4, 5 }; int n = arr.Length; int []preXOR = new int[n]; prefixXOR(arr, preXOR, n); Console.WriteLine(query(preXOR, 0, 2)) ; Console.WriteLine (query(preXOR, 1, 2)); } } // This code is contributed by vt_m |
Output:
7 4
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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