CF 104741A - A+B问题

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Solve time: 51s
Verified: yes

Solution

Problem Understanding

We are given multiple independent queries. Each query contains three numbers written in the same unknown positional numeral system with base $X$, where $X$ is some integer between 2 and 16 inclusive. The digits are written using characters 0-9 and A-F, so they can represent values up to 15.

For each query, we are told that the first number plus the second number equals the third number when interpreted in base $X$. The base itself is unknown, and different queries may correspond to different valid bases. Our task is to recover any base $X$ that makes the equality correct for each query.

The constraints imply up to $10^5$ queries, so each query must be checked in constant or near-constant time. A solution that tries to fully convert strings into large integers using repeated arithmetic for each base independently would still work if carefully bounded, but anything quadratic in digit length per query would be risky under worst-case repetition.

A subtle edge case comes from invalid digits. If a number contains a digit like F, then any base $X \le 15$ is immediately invalid. Another failure mode is ignoring carry propagation: even if digit-wise sums match locally, a missing carry check can incorrectly accept invalid bases.

For example, consider a case where $A = 1F$, $B = 1$, and $S = 20$. In base 16 this works, but in base 15 it is invalid because digit F is not allowed. A naive approach that only checks arithmetic after converting character-by-character without validating digit ranges would incorrectly accept such cases.

Approaches

The brute-force idea is straightforward: for each query, try every base $X$ from 2 to 16, convert $A$, $B$, and $S$ into integers in that base, and check whether the equality holds. Since the base range is small and fixed, this already suggests feasibility.

However, the key cost is conversion. If we convert a number by repeated multiplication and addition for each digit, each conversion is $O(L)$, where $L$ is the number of digits. Doing this three times per base gives $O(3L)$, and across 15 bases we get $O(45L)$ per query. With $10^5$ queries, this is still fine in Python if implemented cleanly, but we can do better by avoiding repeated full conversion.

The key observation is that we do not need full integer values. We only need to check whether the addition holds in a given base. That can be verified digit by digit with carry simulation, exactly like manual addition. This avoids constructing large integers and reduces each check to a linear scan over digits once.

So the optimal strategy is to try each base and validate the equation using digit-wise addition with carry propagation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (full conversion per base)	$O(T \cdot 16 \cdot L)$	$O(1)$ extra	Accepted
Optimal (digit-wise carry check)	$O(T \cdot 16 \cdot L)$	$O(1)$ extra	Accepted

The improvement is mainly in constant factors and memory behavior rather than asymptotic complexity, but it simplifies correctness and avoids unnecessary integer construction.

Algorithm Walkthrough

We process each query independently. For a fixed query, we determine the smallest possible base from its digits and then test all valid bases up to 16.

Extract all digit values appearing in $A$, $B$, and $S$, and compute the maximum digit value. The base must be strictly greater than this value, otherwise the representation is invalid.
For each candidate base $X$ from $\max_digit + 1$ to 16, attempt to verify whether $A + B = S$ holds in base $X$.
To verify a base, simulate addition from the least significant digit to the most significant digit. We reverse all three strings so index 0 corresponds to the units place.
Maintain a carry initialized to 0. At each digit position, compute the sum of corresponding digits from $A$ and $B$, add the carry, and compare against the digit of $S$ modulo $X$. If mismatch occurs, reject this base immediately.
After processing all digits, ensure that any remaining carry matches the remaining digits of $S$. If everything is consistent, accept this base and output it.

We stop at the first valid base since any valid answer is acceptable.

Why it works

The algorithm enforces exact digit-level consistency of addition under base $X$. The carry variable captures all cross-digit dependencies, so every position is validated under the same arithmetic rules that define positional numeral systems. Because we try all possible bases in the only feasible range, completeness is guaranteed, and because each candidate is checked using exact arithmetic rules, soundness is guaranteed.

Python Solution

import sys
input = sys.stdin.readline

def val(c):
    if '0' <= c <= '9':
        return ord(c) - ord('0')
    return ord(c) - ord('A') + 10

def ok(a, b, s, base):
    carry = 0
    i = len(a) - 1
    j = len(b) - 1
    k = len(s) - 1

    while k >= 0:
        av = val(a[i]) if i >= 0 else 0
        bv = val(b[j]) if j >= 0 else 0
        sv = val(s[k])

        total = av + bv + carry
        if total % base != sv:
            return False
        carry = total // base

        i -= 1
        j -= 1
        k -= 1

    while i >= 0 or j >= 0:
        av = val(a[i]) if i >= 0 else 0
        bv = val(b[j]) if j >= 0 else 0
        total = av + bv + carry
        carry = total // base
        if total % base != 0:
            return False
        i -= 1
        j -= 1

    return carry == 0

def solve():
    T = int(input())
    for _ in range(T):
        a, b, s = input().split()

        max_digit = 0
        for ch in a + b + s:
            max_digit = max(max_digit, val(ch))

        for base in range(max_digit + 1, 17):
            if ok(a, b, s, base):
                print(base)
                break

if __name__ == "__main__":
    solve()

The solution starts by converting characters into numeric digit values consistently across the entire input. The ok function is the core verifier: it simulates addition from right to left while enforcing base constraints through modular arithmetic and carry propagation. The second loop handles leftover digits when one number is longer than the other, ensuring no hidden contribution remains.

A common pitfall is forgetting that the base must be strictly greater than every digit present. That check avoids unnecessary simulation in invalid bases and prevents false acceptance.

Worked Examples

Consider the query 1 1 10.

We test possible bases starting from 2. In base 2, the digit 1 + 1 produces 10, which matches the third number.

Position	A digit	B digit	S digit	carry in	sum	mod base	carry out
units	1	1	0	0	2	0	1
next	0	0	1	1	1	1	0

This confirms base 2 is valid.

Now consider F F 1E.

We test base 16. In hexadecimal, F + F = 1E.

Position	A digit	B digit	S digit	carry in	sum	mod 16	carry out
units	15	15	14	0	30	14	1
next	0	0	1	1	1	1	0

This confirms base 16 is valid.

The traces show that correctness depends entirely on consistent carry propagation rather than direct string equality.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(T \cdot 16 \cdot L)$	Each query tries up to 15 bases, each validated by a linear scan over digits
Space	$O(1)$	Only constant extra variables for carry simulation

The constraints allow up to $10^5$ queries, but each query involves only small constant-factor work over short strings, so the solution comfortably fits within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from sys import stdout
    out = io.StringIO()
    backup = sys.stdout
    sys.stdout = out

    solve()

    sys.stdout = backup
    return out.getvalue().strip()

# basic samples
assert run("2\n1 1 10\nF F 1E\n") == "2\n16"

# minimal digits
assert run("1\n0 0 0\n") == "2"

# base boundary digit 15 forces base 16
assert run("1\nF 0 F\n") == "16"

# carry-heavy case
assert run("1\n1F 1 20\n") == "16"

# multiple queries
assert run("3\n1 1 10\nF F 1E\n1 0 1\n") == "2\n16\n2"

Test input	Expected output	What it validates
`1 1 10`	`2`	smallest base with carry
`F F 1E`	`16`	max-base digit handling
`0 0 0`	`2`	trivial zero case
`1F 1 20`	`16`	carry across different lengths
multiple queries	mixed	batch correctness

Edge Cases

A subtle case is when all digits are valid only in base 16, such as involving F. The algorithm correctly sets the minimum base to 16 and avoids unnecessary checking of smaller bases.

For input F 0 F, the maximum digit is 15, so only base 16 is tested. The carry simulation immediately confirms F + 0 = F, producing zero carry and exact digit match, so the output is 16.

Another edge case is when numbers have different lengths. For 1 0 1, the second number contributes nothing beyond its single digit, and the carry logic correctly treats missing positions as zero. The algorithm still produces base 2 without special casing because absent digits are naturally modeled as zeros in positional arithmetic.