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 implementation 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 |
Javascript
<script> // Javascript code queries on XOR of // greatest odd divisor of the range // Precompute the prefix XOR of greatest // odd divisor function prefixXOR(arr, preXOR, n)
{ // Finding the Greatest Odd divisor
for (let i = 0; i < n; i++)
{
while (arr[i] % 2 != 1)
arr[i] = parseInt(arr[i] / 2);
preXOR[i] = arr[i];
}
// Finding prefix XOR
for (let i = 1; i < n; i++)
preXOR[i] = preXOR[i - 1] ^ preXOR[i];
} // Return XOR of the range function query(preXOR, l, r)
{ if (l == 0)
return preXOR[r];
else
return preXOR[r] ^ preXOR[l - 1];
} // Driver code let arr = [ 3, 4, 5 ]; let n = arr.length; let preXOR = new Array(n);
prefixXOR(arr, preXOR, n); document.write(query(preXOR, 0, 2) + "<br>" );
document.write(query(preXOR, 1, 2) + "<br>" );
// This code is contributed by subham348 </script> |
Output
7 4
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)