PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4
Author: yash5507
Tester: jay_1048576
Editorialist: iceknight1093
DIFFICULTY:
2028
PREREQUISITES:
Familiarity with bitwise operations
PROBLEM:
Let F(x, y, z) = (x\mid y)  (x\ \& \ z).
Given A, B, C, find the number of integers 0 \leq x \lt 2^C such that F(x, A, B) is maximum.
EXPLANATION:
Almost everything in this problem is a bitwise operation, except the subtraction in F(x, y, z).
However, notice that (x\mid y) will contain all the set bits of x, while (x \ \& \ z) cannot contain any bits that aren’t set in x.
So, (x\ \& \ z) is a submask of (x\mid y), and this means the subtraction operation is really just bitwise XOR.
That is, we can rewrite the function as F(x, y, z) = (x\mid y) \oplus (x\ \& \ z).
Now that everything we’re dealing with is bitwise operation, it helps to look at what’s going on bitbybit.
Let’s fix a bit k and see what happens.
Let x_k denote the value of the kth bit of x (which is either 0 or 1). Similarly define A_k, B_k, F_k(x, A, B).
We already know A_k and B_k, our aim is to find out what values x_k can possibly take.
We have F_k(x, A, B) = (x_k\mid A_k) \oplus (x_k\ \& \ B_k), and we’d like to maximize F_k(x, A, B).
There are 4 cases depending on their values:

A_k = B_k = 1
Here, (x_k \mid A_k) = 1, and (x_k \ \& \ B_k) = x_k.
So, F_k(x, A, B) =1 \oplus x_k, which is maximized when x_k = 0. 
A_k = B_k = 0
Here, (x_k \mid A_k) = x_k, and (x_k \ \& \ B_k) = 0.
So, F_k(x, A, B) = x_k \oplus 0, which is maximized when x_k = 1. 
A_k = 1 and B_k = 0
Here, F_k(x, A, B) = 1\oplus 0 = 1, and is independent of x_k. So, we can choose x_k to be either 0 or 1 
A_k = 0 and B_k = 1
Here, F_k(x, A, B) = x_k\oplus x_k = 0, once again we can choose x_k freely.
Putting the above together:
 If A_k = B_k, x_k is fixed and we have no choice.
 If A_k \neq B_k, x_k can be chosen freely.
So, the answer is simply 2^d, where d is the number of bits where A and B differ in their binary representations.
d can be computed by iterating over each bit independently; or you can notice that it’s just the number of set bits in (A\oplus B) for an \mathcal{O}(1) solution.
TIME COMPLEXITY
\mathcal{O}(C) or \mathcal{O}(1) per test case.
CODE:
Editorialist's code (Python)
for _ in range(int(input())):
a, b, c = map(int, input().split())
dif = bin(a ^ b).count('1')
print(2 ** dif)