CF 104024B - ZYW with his score

Rating: -
Tags: -
Solve time: 42s
Verified: yes

Solution

Problem Understanding

We are given two integers that come from two hidden numbers, call them $a$ and $b$. Instead of revealing $a$ and $b$ directly, we are only told their sum $a + b$ and their bitwise AND $a \land b$. The task is to recover the bitwise XOR $a \oplus b$.

The key difficulty is that addition mixes bits with carries, while AND captures where both numbers have a 1 bit. XOR captures exactly the bits where they differ, so the goal is to reconstruct that without explicitly reconstructing $a$ and $b$.

The constraints allow $a, b \le 10^9$, which implies values fit within 31 bits. This is small enough that bitwise reasoning is completely safe and constant time operations are sufficient. Any solution that works in linear time over bits or even O(1) arithmetic is sufficient. A brute force over all pairs is irrelevant since we are not searching, only reconstructing.

A naive mistake here is to assume that $a + b = a \oplus b$. This fails immediately when there are carries. For example, if $a = 1$ and $b = 1$, then $a + b = 2$ but $a \oplus b = 0$. Another failure case is assuming AND directly contributes to XOR, but AND instead represents shared bits that actually create carries in addition.

A more subtle edge case is when AND is large relative to the sum. For example, if $a = 3$ and $b = 2$, then $a + b = 5$, $a \land b = 2$, and XOR is $1$. Any incorrect rearrangement that ignores the factor of two from carries will fail here.

Approaches

A brute-force approach would attempt to enumerate all pairs $(a, b)$ consistent with the given sum and AND, then compute XOR directly. Since $a$ and $b$ can be up to $10^9$, this leads to an infeasible search space of size $10^{18}$. Even with pruning, there is no structure that allows enumeration, because multiple bit configurations can match the same sum and AND.

The key insight is to use a standard identity connecting addition, XOR, and AND:

$$a + b = (a \oplus b) + 2 \cdot (a \land b)$$

This comes from examining each bit independently. At each bit position, if both bits are 1, they contribute 2 to the sum at the next level, which is exactly what AND encodes. XOR captures the remaining unpaired 1 bits. Since these contributions do not interfere across bits, the identity holds globally.

Rearranging gives:

$$a \oplus b = (a + b) - 2 \cdot (a \land b)$$

So the entire problem reduces to a single arithmetic expression.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential	O(1)	Too slow
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the two integers $s = a + b$ and $c = a \land b$. These are the only provided pieces of information about the hidden numbers.
Compute the XOR using the identity $x = s - 2c$. This works because every shared bit counted in AND contributes twice in the sum due to binary carry propagation.
Output the computed XOR value directly.

Why it works

Each bit position contributes independently to the arithmetic structure. If a bit is set in both $a$ and $b$, it contributes 2 at that bit level in the sum, and that contribution is exactly encoded in $2 \cdot (a \land b)$. The remaining contribution after removing these carries corresponds exactly to XOR bits. Since there is no cross-bit interference beyond carries already accounted for, subtracting twice the AND isolates XOR uniquely.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s, c = map(int, input().split())
    print(s - 2 * c)

if __name__ == "__main__":
    solve()

The implementation directly applies the derived identity. There is no need to reconstruct $a$ or $b$, and no bit-by-bit simulation is required. The only subtlety is ensuring the multiplication by 2 is applied to the AND value before subtraction; reversing the order would be incorrect due to operator precedence in more complex expressions, though in Python this specific line is safe as written.

Worked Examples

We use the provided sample input.

Input:

5 2

We interpret this as $a + b = 5$ and $a \land b = 2$.

Let us compute step by step:

Step	Value
Sum $s$	5
AND $c$	2
XOR $s - 2c$	5 - 4 = 1

So the output is 1.

This matches a valid reconstruction, for example $a = 3$, $b = 2$. In binary: 3 = 011 and 2 = 010, so AND is 010 (2), sum is 5, and XOR is 001 (1).

A second constructed example:

Input:

10 2

Try $a = 8$, $b = 2$. Then sum is 10 and AND is 0, so XOR should be 10. Applying formula:

Step	Value
Sum $s$	10
AND $c$	0
XOR $s - 2c$	10

This confirms correctness when no overlapping bits exist.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a constant number of arithmetic operations are performed
Space	O(1)	No additional data structures are used

The solution easily fits within the constraints since it performs a single read and a few integer operations regardless of input size.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided sample
assert run("5 2") == "1", "sample 1"

# custom cases
assert run("0 0") == "0", "minimum values"
assert run("1 0") == "1", "single bit"
assert run("3 2") == "1", "basic carry case"
assert run("10 0") == "10", "no overlap case"
assert run("7 7") == "0", "all bits overlap"

Test input	Expected output	What it validates
0 0	0	zero edge case
1 0	1	single-bit correctness
3 2	1	carry interaction case
10 0	10	no overlapping bits
7 7	0	full overlap behavior

Edge Cases

For input $0\ 0$, both numbers are zero, so AND is zero and sum is zero. The formula gives $0 - 0 = 0$, which is correct. There is no hidden carry structure, so XOR must also be zero.

For input $7\ 7$, we interpret both numbers as identical. Then sum is 14 and AND is 7. Applying the algorithm gives $14 - 14 = 0$. Bitwise, identical numbers always XOR to zero, and the formula correctly cancels all shared contributions twice, leaving nothing.