The Great XOR


Problem Statement :


Given a long integer x, count the number of values of a satisfying the following conditions:
a + x > x
0 < a < x
where a and x are long integers and ^ is the bitwise XOR operator.

You are given q queries, and each query is in the form of a long integer denoting x. For each query, print the total number of values of a satisfying the conditions above on a new line.

For example, you are given the value x=5. Condition 2 requires that a<x. The following tests are run:

1^5 = 4
 ^ 5 = 7
3 ^ 5 = 6
4 ^ 5 =1

We find that there are 2 values meeting the first condition: 2 and 3.

Function Description

Complete the theGreatXor function in the editor below. It should return an integer that represents the number of values satisfying the constraints.

theGreatXor has the following parameter(s):

x: an integer
Input Format

The first line contains an integer q, the number of queries.
Each of the next q lines contains a long integer describing the value of x for a query.

Constraints
1 <= q <= 10^5
1 <= x <= 10^10


Solution :



title-img 


                            Solution in C :

In C++ :





#include <iostream>

using namespace std;

int main()
{
	long long n,a,n0,nb,p2;
	cin>>n;
	for (;n>0;--n){
		nb=0;
		p2=1;
		cin>>a;
		while (a>0){
			if ((a&1)==0){
				nb=nb+p2;
			}
			a>>=1;
			p2<<=1;
		}
		cout<<nb<<endl;
	}
    return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for(int a0 = 0; a0 < q; a0++){
            long x = in.nextLong();
            // your code goes here
            long b = 1;
            long total = 0;
            while(b < x) {
                if((b & x) == 0) {
                    total += b;
                }
                b = b * 2;
            }
            System.out.println(total);
        }
    }
}









In C :





#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int q; 
    scanf("%d",&q);
    for(int a0 = 0; a0 < q; a0++){
        long x; 
        scanf("%ld",&x);
        long result = 0;
        for (long mask = 1; x > 0; x >>= 1, mask <<= 1) {
           if (!(x & 1)) {
              result += mask; 
           }
        }
        printf("%ld\n", result);
    }
    return 0;
}








In Python3 :





#!/bin/python3

import sys


q = int(input().strip())
for a0 in range(q):
    x = int(input().strip())
    size = 1
    result = 0
    while x > 0 :
        if x & 1 == 0:
            result += size
        size *=2
        x = x >> 1
    # your code goes here
    print (result)








View More Similar Problems

Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

View Solution →

Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

View Solution →

Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

View Solution →

Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

View Solution →

Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

View Solution →

Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →