CF 102411A - Accurate Movement

Rating: -
Tags: -
Solve time: 5m 32s
Verified: yes

Solution

Problem Understanding

We have a box of length n containing two bars on parallel rails. The short bar has length a, the long bar has length b, with a < b. The long bar carries a stopper at each end, and the short bar must remain completely between those two stoppers.

Initially, both bars start flush with the left side of the box. The goal is to move both bars until they are flush with the right side, using as few moves as possible. During one move, exactly one bar may be repositioned, while the other stays fixed. The only restriction is that after every move the short bar must still lie between the long bar's two stoppers.

The input contains a, b, and n, where 1 ≤ a < b ≤ n ≤ 10^7. The upper bound of 10^7 tells us that an algorithm performing a constant amount of work is ideal, while even an algorithm proportional to n^2 is completely infeasible. A linear simulation can reach about ten million iterations in the worst case, which is potentially borderline under a two second limit in Python. The direct formula is preferable because it reduces the computation to a few arithmetic operations. The official limit is two seconds and the memory limit is 512 MB.

The first edge case is when the long bar already fills the entire box. For example, with input 1 3 3, the long bar has length 3 and the box also has length 3. It cannot move at all, while the short bar can move directly from the left stopper to the right stopper, so the answer is 1. A careless solution that always assumes both bars have to move would return at least 2.

The second edge case occurs when b - a = 1. For example, 2 3 5 has only one unit of free space inside the long bar around the short bar. The long bar must travel 5 - 3 = 2 units, and each pair of moves can advance it by only one unit, so the answer is 2 * 2 + 1 = 5. A formula using integer division instead of ceiling would incorrectly calculate only 3 moves.

The third edge case is when the required movement of the long bar is not divisible by b - a. For example, 2 5 8 requires the long bar to travel 3 units, while the short bar can create a gap of at most 3 units. Here one full pair of moves advances the long bar by 3, giving 3 moves in total. With 2 4 9, however, the long bar must travel 5 units and each pair can advance it by at most 2, so three pairs are needed and the answer is 7. The ceiling operation is essential.

Approaches

A brute-force approach can model the positions of both bars explicitly. Let x be the left endpoint of the long bar and y the left endpoint of the short bar. A legal state satisfies

0 ≤ x ≤ n - b

and

x ≤ y ≤ x + b - a.

Starting from (0, 0), we could perform a shortest-path search over all possible integer states. There are O(n^2) possible pairs (x, y), and from one state we could try many possible new positions for the selected bar. That gives up to O(n^3) work in the most direct implementation. With n = 10^7, this is on the order of 10^21 operations, so it is not remotely viable.

The brute force works because it explicitly represents every legal relative position of the bars. The key observation is that we do not actually need to consider all those positions. Let

d = b - a.

Suppose the long bar starts at position x. Since the short bar must remain between the two stoppers, its left endpoint can be at most x + d. Thus, when the long bar is fixed, the short bar can move right by at most d.

Now reverse the roles. If the short bar is at y, the long bar can move right until its left endpoint reaches y. If we first place the short bar at x + d, the long bar can then move from x to x + d. So one short-bar move followed by one long-bar move can advance the long bar by exactly d.

There is never a benefit to stopping either move before its maximum possible position. Moving farther right cannot make a future state worse, because every later legal position is also farther to the right. This lets us greedily alternate the bars, with every move except the final short-bar move contributing exactly d units of progress for the long bar. The official tutorial gives the same observation and resulting formula.

The long bar must move from position 0 to position n - b, so the number of short/long move pairs required is

ceil((n - b) / (b - a)).

Each pair uses two moves, and after the long bar reaches its final position, one final move places the short bar against the right side of the box. Hence the answer is

2 * ceil((n - b) / (b - a)) + 1.

We can compute the ceiling for positive integers with

ceil(x / d) = (x + d - 1) // d.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n³)	O(n²)	Too slow
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read a, b, and n. The only quantities that matter for the movement count are the distance the long bar must travel, n - b, and the maximum relative shift available between the bars, b - a.
Compute distance = n - b. This is the total distance that the long bar's left endpoint has to travel before the long bar reaches the right side of the box.
Compute gap = b - a. This is the largest distance by which the short bar can be placed ahead of the long bar while still remaining between the long bar's two stoppers.
Compute pairs = (distance + gap - 1) // gap. Every pair consists of moving the short bar as far right as possible and then moving the long bar as far right as possible. Such a pair advances the long bar by gap, except possibly the final pair, which may advance it by less because the long bar has reached the boundary.
Multiply pairs by two because every pair contains one short-bar move and one long-bar move.
Add one final move for the short bar. After the long bar has reached its target position, the short bar can be moved directly to its target position while staying between the final stoppers.

Why it works

The invariant is that immediately before every long-bar move, the short bar can be placed at most b - a units ahead of the long bar. Consequently, a single long-bar move can advance its left endpoint by at most b - a. The greedy strategy reaches exactly that maximum whenever more progress is needed, so no strategy can move the long bar faster in terms of number of moves.

The long bar needs to travel n - b units, so at least ceil((n - b) / (b - a)) long-bar moves are necessary. Before every such long-bar move, except for the initial configuration, a short-bar move is needed to create the required gap. Once the long bar reaches its target, exactly one short-bar move remains. The constructed sequence achieves this lower bound, giving 2 * ceil((n - b) / (b - a)) + 1.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    a, b, n = map(int, input().split())

    distance = n - b
    gap = b - a

    pairs = (distance + gap - 1) // gap
    answer = 2 * pairs + 1

    print(answer)

if __name__ == "__main__":
    solve()

The first line reads the three dimensions directly from standard input. There is only one test case in this problem, so no test-case loop is needed.

distance = n - b is the amount by which the long bar's left endpoint must move. If n == b, this becomes zero, which correctly leads to zero pairs and an answer of 1.

gap = b - a is always positive because the statement guarantees a < b. This means the ceiling division is safe and never divides by zero.

The expression (distance + gap - 1) // gap performs ceiling division without floating point arithmetic. Floating point is unnecessary here and could introduce avoidable precision issues, while integer arithmetic is exact for the entire input range.

Finally, 2 * pairs + 1 counts the two moves in each short/long pair and the one final short-bar move. Python integers have arbitrary precision, although the answer here is easily within ordinary 64-bit integer range.

Worked Examples

Sample 1

The input is 1 3 6. The official sample gives the answer 5.

Here the long bar has to move 6 - 3 = 3 units, while the maximum gap between the two bars is 3 - 1 = 2.

Step	`a`	`b`	`n`	`distance`	`gap`	`pairs`	`answer`
Read input	1	3	6	3	2	2	5
Ceiling division	1	3	6	3	2	2	5
Final formula	1	3	6	3	2	2	5

The first pair moves the short bar forward by two units and then moves the long bar forward by two units. The second pair finishes the remaining one unit of long-bar movement. The final short-bar move places it against the right side. Thus the sequence requires 2 * 2 + 1 = 5 moves.

Sample 2

The input is 2 4 9. The official sample gives the answer 7.

The long bar must travel 9 - 4 = 5 units, while the maximum gap is 4 - 2 = 2.

Step	`a`	`b`	`n`	`distance`	`gap`	`pairs`	`answer`
Read input	2	4	9	5	2	3	7
Ceiling division	2	4	9	5	2	3	7
Final formula	2	4	9	5	2	3	7

Two pairs advance the long bar by four units. A third pair handles the remaining one unit, and the final short-bar move completes the configuration. The ceiling is what changes 5 / 2 from two pairs to three.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a constant number of arithmetic operations are performed
Space	O(1)	Only a few integer variables are stored

The maximum value of n is 10^7, but the algorithm never loops up to n. It performs only integer subtraction, addition, multiplication, and division, so it is comfortably within the two second time limit and uses negligible memory.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve(inp: str) -> str:
    data = list(map(int, inp.split()))
    a, b, n = data

    distance = n - b
    gap = b - a
    pairs = (distance + gap - 1) // gap

    return str(2 * pairs + 1)

# provided samples
assert solve("1 3 6\n") == "5\n"[:-1], "sample 1"
assert solve("2 4 9\n") == "7", "sample 2"

# minimum-size input
assert solve("1 2 2\n") == "1", "long bar already fills the box"

# all possible movement is by one unit
assert solve("1 2 5\n") == "7", "gap is one"

# exact divisibility
assert solve("2 5 8\n") == "3", "distance is exactly one gap"

# non-divisible boundary case
assert solve("2 4 8\n") == "5", "distance is not divisible by gap"

# maximum-size input
assert solve("1 10000000 10000000\n") == "1", "maximum n with b = n"

The custom cases cover the configuration where no long-bar movement is needed, the smallest possible gap, exact divisibility, non-divisible movement, and the largest permitted input value.

Test input	Expected output	What it validates
`1 2 2`	`1`	Long bar already fills the box
`1 2 5`	`7`	Smallest possible gap, repeated moves
`2 5 8`	`3`	Exact divisibility
`2 4 8`	`5`	Ceiling division boundary
`1 10000000 10000000`	`1`	Maximum input size and zero required long-bar movement

Edge Cases

When n = b, the long bar already spans the complete box. For input 1 3 3, we have distance = 3 - 3 = 0 and gap = 3 - 1 = 2. The ceiling division gives pairs = 0, so the answer is 2 * 0 + 1 = 1. The only required move is the short bar moving from the left side to the right side.

When b - a = 1, the short bar has almost the same length as the long bar, so it can create only a one-unit gap. For input 1 2 5, the long bar must move 3 units. Since each long-bar move can advance it by only one unit, three pairs are required, followed by the final short-bar move. The formula gives 2 * ceil(3 / 1) + 1 = 7.

When the distance is exactly divisible by the gap, no partially completed pair is necessary. For input 2 5 8, the long bar travels 3 units and the available gap is also 3. Thus one short-bar move and one long-bar move put the long bar exactly at its target, followed by one final short-bar move. The answer is 3.

When the distance is not divisible by the gap, the final long-bar move uses only part of the available gap. For input 2 4 8, the long bar must travel 4 units while each pair can provide only 2 units. Two pairs are needed, giving 2 * 2 + 1 = 5. Replacing the ceiling with ordinary integer division would happen to work for this exact-divisibility case, but it would fail on inputs such as 2 4 9, where 5 / 2 requires three pairs rather than two.

If you want, I can also turn this into a shorter Codeforces-style editorial that keeps the proof but removes the deliberately detailed brute-force discussion.