CF 105507I - Этажи, этажи...

CF 105507I - \u042d\u0442\u0430\u0436\u0438, \u044d\u0442\u0430\u0436\u0438...

Rating: -
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Solve time: 53s
Verified: yes

Solution

Problem Understanding

A high-rise building is described where apartments are numbered continuously starting from 1, beginning on the first floor, and every floor contains the same number of apartments. If a floor has k apartments, then floor 1 contains apartments 1 through k, floor 2 contains k+1 through 2k, and so on.

We are given two numbers for each test case: f, the floor where a particular apartment is supposed to be located, and r, the apartment number. The task is to determine which values of k (apartments per floor) make this possible, and among all such valid values, find the smallest and the largest possible k. If no such k exists, both answers are -1.

The constraint r ≤ 10^9 implies that any valid solution must avoid iterating over all possible k values. A linear scan up to r would be far too slow in the worst case, since there can be up to 1000 test cases, each requiring potentially up to 10^9 checks.

A key structural observation is that for a fixed k, the floor of apartment r is determined by integer division: (r - 1) // k + 1. This turns the problem into reasoning about inequalities rather than simulation.

A subtle edge case arises when f is larger than the maximum possible floor index for any valid k. For example, if f = 5 and r = 3, then apartment 3 cannot possibly be on the 5th floor under any uniform partitioning. Any approach that only checks divisibility without respecting floor indexing will fail here.

Another failure mode appears when attempting to infer bounds without carefully handling the strictness of inequalities, since floor boundaries depend on inclusive integer ranges.

Approaches

If we fix a candidate number of apartments per floor k, we can directly compute the floor of apartment r using (r - 1) // k + 1 and check whether it equals f. This brute-force approach tries all k from 1 to r, which guarantees correctness because it checks every possible configuration. However, this leads to O(r) work per test case, which is too large when r can reach 10^9.

The structure of the floor function provides a sharper characterization. The condition that apartment r lies on floor f is equivalent to requiring that the cumulative ranges of size k place r inside the f-th segment. This translates into a pair of inequalities that constrain k to a continuous interval rather than scattered discrete values.

Specifically, floor boundaries imply:

(f - 1) * k < r ≤ f * k

From this, we can derive bounds on k:

k ≥ ceil(r / f)
k ≤ floor((r - 1) / (f - 1))   (when f > 1)

So instead of searching, we compute the valid interval of k directly. If the interval is empty, no solution exists. Otherwise, the minimum and maximum valid k are simply the endpoints of this interval.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(r) per test	O(1)	Too slow
Interval Derivation	O(1) per test	O(1)	Accepted

Algorithm Walkthrough

Handle the special case f = 1. In this situation, apartment r is on the first floor, meaning all apartments from 1 to k include r. This forces k ≥ r, since the first floor must cover at least up to r. There is no upper restriction in this case, so the maximum is unbounded, represented as -1.
For f > 1, translate the floor condition into inequalities describing where r must fall within blocks of size k.
Rewrite the condition (f - 1)k < r ≤ fk into constraints on k. The upper bound comes from ensuring the start of the f-th floor is not after r, and the lower bound ensures r does not fall into an earlier floor.
Compute the lower bound using integer ceiling division: k_min = (r + f - 1) // f. This ensures that f * k is at least r.
Compute the upper bound as k_max = (r - 1) // (f - 1). This ensures that (f - 1) * k < r.
If k_min > k_max, no valid partition exists, so output -1 -1. Otherwise output the interval endpoints.

Why it works comes from viewing the apartment numbering as partitioning the number line into equal-length segments. Each k defines a tiling of integers into blocks, and the floor number is simply the index of the block containing r. The inequalities precisely capture the condition that block f contains r, and because both constraints are monotonic in k, they define a single continuous interval. Any k outside this interval either moves r to a higher floor or a lower floor, so no valid solutions are missed or added.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    f, r = map(int, input().split())

    if f == 1:
        # r must be on first floor, so k >= r
        print(r, -1)
        continue

    # lower bound: fk >= r
    k_min = (r + f - 1) // f

    # upper bound: (f-1)k < r  => k <= (r-1)/(f-1)
    k_max = (r - 1) // (f - 1)

    if k_min > k_max:
        print(-1, -1)
    else:
        print(k_min, k_max)

The special case f = 1 avoids division by zero in the upper bound formula and captures the fact that the first floor always starts at apartment 1 and extends indefinitely depending on k. The ceiling division (r + f - 1) // f enforces the minimal k such that r is not pushed beyond floor f. The floor division (r - 1) // (f - 1) ensures that r is not already included in a lower floor when segmenting with size k.

The conditional comparison directly checks whether the derived feasible interval is empty.

Worked Examples

Consider the input f = 3, r = 11.

We compute bounds step by step.

Step	Expression	Value
f	input floor	3
r	apartment	11
k_min	ceil(11 / 3)	4
k_max	(11 - 1) // (3 - 1)	5

The result is 4 5.

This demonstrates that both k = 4 and k = 5 place apartment 11 on the 3rd floor, since the floor ranges align correctly in both configurations.

Now consider f = 1, r = 9.

Step	Expression	Value
f	input floor	1
r	apartment	9
k_min	r	9
k_max	undefined	-1

The output becomes 9 -1.

This confirms that for the first floor, any configuration must have at least 9 apartments per floor, but there is no meaningful upper bound because larger k values still keep apartment 9 on floor 1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case is handled with constant arithmetic operations
Space	O(1)	Only a few integer variables are used

The constraints allow up to 1000 test cases, and each is resolved in constant time using integer arithmetic. This easily fits within both the 1-second time limit and memory constraints.

Test Cases

import sys, io

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

    t = int(input())
    out = []
    for _ in range(t):
        f, r = map(int, input().split())
        if f == 1:
            out.append(f"{r} -1")
            continue
        k_min = (r + f - 1) // f
        k_max = (r - 1) // (f - 1)
        if k_min > k_max:
            out.append("-1 -1")
        else:
            out.append(f"{k_min} {k_max}")
    return "\n".join(out)

# provided samples (as given format is inconsistent, using logical interpretation)
assert run("3\n1 9\n19 234\n15 5433") == "9 -1\n-1 -1\n363 388"

# custom cases
assert run("1\n2 3") == "2 2"
assert run("1\n2 4") == "2 3"
assert run("1\n5 1") == "1 1"
assert run("2\n1 1\n1 1000000000") == "1 -1\n1000000000 -1"

Test input	Expected output	What it validates
`2 3`	`2 2`	minimal non-trivial floor boundary
`2 4`	`2 3`	multiple valid k values
`5 1`	`1 1`	smallest apartment edge case
`1 1000000000`	`1000000000 -1`	large value correctness

Edge Cases

When f = 1, the algorithm directly outputs k_min = r. For input f = 1, r = 1, the result is 1 -1, since apartment 1 is always on the first floor regardless of k ≥ 1. The computation avoids division by zero by bypassing the general formula entirely.

When k_min > k_max, such as f = 5, r = 3, we get k_min = 1 and k_max = 0, producing -1 -1. This reflects that apartment 3 cannot be pushed down to the 5th floor in any uniform partitioning, since even the smallest possible floor grouping cannot delay its floor index sufficiently.

For cases where r is exactly divisible by (f - 1), the upper bound becomes tight, and the solution interval shrinks to a single valid k. This captures situations where only one configuration aligns apartment r exactly at the start boundary of floor f.