CF 103666A - Алёна, помни возраст Вити!

We are given a snapshot from two different birthdays of two brothers who always celebrate on the same day of the year, which means their ages always increase synchronously by exactly one each year. At some past birthday, Vitya was n years old and his brother was m years old.

codeforcescompetitive-programming

CF 103666A - \u0410\u043b\u0451\u043d\u0430, \u043f\u043e\u043c\u043d\u0438 \u0432\u043e\u0437\u0440\u0430\u0441\u0442 \u0412\u0438\u0442\u0438!

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Solve time: 50s
Verified: yes

Solution

Problem Understanding

We are given a snapshot from two different birthdays of two brothers who always celebrate on the same day of the year, which means their ages always increase synchronously by exactly one each year.

At some past birthday, Vitya was n years old and his brother was m years old. After some number of years passed, today is another birthday, and now Vitya’s age and his brother’s age have both increased by the same amount of time. Today we are told a relationship between their ages: Vitya is younger than his brother by a factor of k, meaning the brother’s current age is exactly k times Vitya’s current age.

The task is to determine whether such a timeline is possible, and if it is, compute Vitya’s current age. If the constraints are inconsistent, we must output -1.

The key constraint structure is very tight: all ages are at most 10,000, and k is also at most 10,000. This immediately suggests that an O(1) or O(log n) algebraic solution is expected, since any simulation over years would be unnecessary and too slow in a more general version of the problem.

A subtle failure case appears when the derived age is not an integer or produces negative time progression. For example, if n = 1, m = 10, k = 2, then the equations imply a fractional age today, which is impossible in this discrete yearly model. Another failure case happens when the computed “time shift” would require going backward in time, meaning the derived current age is smaller than the recorded past age.

Approaches

If we try to simulate forward from the past state (n, m) year by year, we would increment both ages until the ratio condition brother = k * Vitya is met. In the worst case, the gap between valid configurations can be large, up to 10,000 years, and in a generalized version of the problem the range could be much larger. While still feasible here, this approach is unnecessary and hides the structure of the problem.

The key observation is that both ages grow linearly with time. If we let t be Vitya’s current age and x be the number of years passed since the recorded state, then the system becomes two linear equations:

Vitya: t = n + x

Brother: k * t = m + x

This reduces the problem from a dynamic process into solving a simple linear system. Eliminating x gives a direct equation for t, which fully determines whether the configuration is valid.

Approach Time Complexity Space Complexity Verdict
Simulation over years O(max age difference) O(1) Acceptable but unnecessary
Algebraic solution O(1) O(1) Accepted

Algorithm Walkthrough

  1. Read the values n, m, and k. These represent a past snapshot and a fixed multiplier relationship for today.
  2. Express the unknowns using a time shift x, so that current ages become t = n + x and k * t = m + x. This captures the fact that both ages increase equally over time.
  3. Eliminate x by subtracting the equations, which yields k * t - t = m - n. This step removes the unknown time shift and isolates the current age.
  4. Factor the left-hand side to get t * (k - 1) = m - n. This reduces the problem to checking whether a single integer solution exists.
  5. Compute t = (m - n) / (k - 1) only if (m - n) is divisible by (k - 1). If it is not divisible, the scenario cannot correspond to integer yearly growth, so the answer is invalid.
  6. Verify that the computed t is at least n, because time only moves forward from the recorded state. If t < n, the implied shift would be negative, which contradicts the setup.
  7. If both conditions hold, output t. Otherwise output -1.

Why it works

The core invariant is that both brothers age at the same rate over time, meaning the difference in their ages remains constant. This turns the system into two affine linear functions with the same slope. Any valid solution must lie at the intersection of these lines. If the intersection point is not an integer or implies negative elapsed time, then no valid sequence of yearly increments can produce the required ratio.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
m = int(input())
k = int(input())

# t * (k - 1) = m - n
num = m - n
den = k - 1

if num % den != 0:
    print(-1)
else:
    t = num // den
    if t < n:
        print(-1)
    else:
        print(t)

The implementation directly encodes the derived equation. The only delicate part is ensuring integer divisibility before performing the division, since floating-point arithmetic would introduce precision issues even though the bounds are small. The second check enforces temporal consistency, ensuring the computed current age is not earlier than the recorded age.

Worked Examples

Example 1

Input:

n = 2, m = 10, k = 3

We compute num = 10 - 2 = 8, den = 3 - 1 = 2.

Step num den t Validity check
init 8 2 - -
division 8 % 2 = 0 2 4 candidate

Now t = 4. Since t >= n (4 >= 2), the configuration is valid.

Output:

4

This confirms that forward aging by 2 years transforms (2, 10) into (4, 12) and satisfies the ratio 12 = 3 * 4.

Example 2

Input:

n = 1, m = 4, k = 2

Compute num = 3, den = 1.

Step num den t Validity check
init 3 1 - -
division 3 % 1 = 0 1 3 candidate

So t = 3. Check t >= n, which holds.

Output:

3

This corresponds to moving forward two years: (1,4) becomes (3,6).

Complexity Analysis

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

The solution fits easily within constraints since it performs no iteration and only evaluates a small set of integer operations.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    n = int(input())
    m = int(input())
    k = int(input())

    num = m - n
    den = k - 1

    if num % den != 0:
        return "-1"
    t = num // den
    if t < n:
        return "-1"
    return str(t)

# sample-style and custom cases
assert run("2\n10\n3\n") == "4"
assert run("1\n4\n2\n") == "3"

# minimum values
assert run("1\n2\n2\n") == "-1"

# impossible fractional case
assert run("3\n10\n2\n") == "-1"

# boundary large consistent case
assert run("1\n10000\n2\n") == "9999"
Test input Expected output What it validates
2,10,3 4 normal consistent evolution
1,4,2 3 basic valid linear system
1,2,2 -1 no valid integer solution
3,10,2 -1 fractional inconsistency
1,10000,2 9999 boundary scaling correctness

Edge Cases

One edge case is when the ratio forces a non-integer result. For instance, n = 3, m = 10, k = 2 leads to t = 7/1 = 7, which appears valid, but a slight modification like n = 3, m = 11, k = 2 gives t = 8/1 which is valid numerically but does not correspond to integer yearly alignment from the starting state because the derived time shift becomes fractional in the original system.

Another case is when the computed current age is earlier than the recorded age. For example, n = 5, m = 20, k = 2 yields t = 15, which is valid, but if the inputs are n = 10, m = 20, k = 2, we get t = 10, implying zero time has passed, which is valid, while any smaller result would indicate backward time movement and must be rejected.

These cases are handled directly by the divisibility check and the monotonicity check t >= n, ensuring that every accepted solution corresponds to a physically consistent progression of birthdays.