Xor profit question

#1

sir I could not understand the question.Can you pl explaion it

#2

Before coming to the problem, let’s first understand this term “XOR”:

XOR is a bitwise operator with two operands. It operates on binary representation of a number.
If two bits are same (i.e. either 1 and 1 or 0 and 0) then it’s result is 0 else (i.e. either 1 and 0 or 0 and 1) then it’s result is 1.
It is represented by ^ in C++.

Example:
First no. = 6 (110 is it’s binary representation)
Second no. = 5 (101)
6(110) XOR(^) 5(101) = 3(011)

These are the bitwise operators supported by C++:

  1. & (bitwise AND) Takes two numbers as operands and does AND on every bit of two numbers. The result of AND is 1 only if both bits are 1.

  2. | (bitwise OR) Takes two numbers as operands and does OR on every bit of two numbers. The result of OR is 1 any of the two bits is 1.

  3. ^ (bitwise XOR) Takes two numbers as operands and does XOR on every bit of two numbers. The result of XOR is 1 if the two bits are different.

  4. << (left shift) Takes two numbers, left shifts the bits of the first operand, the second operand decides the number of places to shift.

  5. >> (right shift) Takes two numbers, right shifts the bits of the first operand, the second operand decides the number of places to shift.

  6. ~ (bitwise NOT) Takes one number and inverts all bits of it.

Now, coming to the question:

You are given two numbers x and y.
You have to calculate XOR of all the pairs (a and b) in the range [x, y].
Then, you have to print the maximum value out of all the XOR you have calculated.

Constraints:

  1. Time limit is 1 sec.
  2. a and b can be from x to y considering that a <= b.
#3

Hope, this would help.
If you still have doubts, feel free to ask.
If you are satisfied, give a like. :wink:

#4

Thanks sir understood now.:wink: