# 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 :

```                            ```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();
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)) {
}
}
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
print (result)```
```

## 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

## 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

## 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