CF 102591I - Громкость динамика

The computer's speaker volume is currently set to X, and we want to change it to Y. Every second we may perform exactly one operation. An operation increases or decreases the volume either by 1 or by Z.

codeforcescompetitive-programming

CF 102591I - \u0413\u0440\u043e\u043c\u043a\u043e\u0441\u0442\u044c \u0434\u0438\u043d\u0430\u043c\u0438\u043a\u0430

Rating: -
Tags: -
Solve time: 5m 56s
Verified: yes

Solution

Problem Understanding

The computer's speaker volume is currently set to X, and we want to change it to Y. Every second we may perform exactly one operation. An operation increases or decreases the volume either by 1 or by Z. The volume must always stay within the valid range from 0 to 100, including the intermediate values reached after every operation. The task is to compute the minimum number of seconds required to reach the target volume.

The constraints are extremely small. Every possible volume is an integer between 0 and 100, so the entire state space contains only 101 vertices. Such a small graph allows us to explore every reachable state directly. Even if every state generated four transitions, the total amount of work would stay well below one thousand operations. Any graph algorithm such as Breadth First Search comfortably fits within the limits.

Several edge cases are easy to miss when trying to derive a mathematical formula instead of searching the state space.

If Z = 0, the large move does nothing. For example:

Input:
5 8 0

The correct answer is 3, because only +1 changes the volume. A formula based on dividing by Z would fail because division by zero is undefined.

Another subtle case appears when temporarily moving away from the target is beneficial. Consider:

Input:
1 4 5

The answer is 3. One optimal sequence is 1 -> 0 -> 5 -> 4. A greedy strategy that only moves toward the target never discovers this path.

The boundary values also matter. For example:

Input:
100 99 10

The answer is 1. From volume 100 we cannot increase by 10, because intermediate states must remain inside the valid range. Any implementation must reject moves leaving the interval [0, 100].

Approaches

The most direct solution is to view every possible volume as a state. From one volume we may try four operations: add 1, subtract 1, add Z, and subtract Z. Every operation costs one second, so every edge has equal weight. Running Breadth First Search from the starting volume visits states in increasing order of required time. The first time we reach the target, we have found the minimum number of seconds.

Even a brute force search is already fast because there are only 101 states. In the worst case, Breadth First Search processes each state once and inspects four outgoing transitions, resulting in about 404 transition checks.

Trying to build a closed form formula is much harder than it first appears. The ability to move in both directions, the restriction that every intermediate volume stays inside [0, 100], and the special case Z = 0 create many corner cases. Treating the problem as a shortest path problem removes all of these complications. Breadth First Search naturally handles detours, boundaries, and repeated operations without requiring separate reasoning for each situation.

Approach Time Complexity Space Complexity Verdict
Brute Force, Breadth First Search over all states O(101) O(101) Accepted
Optimal, Breadth First Search over all states O(101) O(101) Accepted

Algorithm Walkthrough

  1. Create a distance array of size 101 and initialize every value to -1. This array records the minimum number of seconds needed to reach each volume.
  2. Set the distance of the starting volume X to 0 and push X into a queue. Breadth First Search always expands states in order of increasing distance.
  3. Repeatedly remove the front volume from the queue.
  4. Generate the four possible next volumes by applying +1, -1, +Z, and -Z.
  5. Ignore every generated volume that lies outside the interval [0, 100], because such a move is not allowed.
  6. If a valid volume has not been visited before, assign its distance as the current distance plus one and push it into the queue. The first visit is always optimal because Breadth First Search explores shorter paths before longer ones.
  7. Continue until the queue becomes empty or the target volume has been reached.
  8. Output the recorded distance for volume Y.

Why it works

Breadth First Search maintains the invariant that every state removed from the queue already has its minimum possible distance from the starting state. Since every allowed operation costs exactly one second, all graph edges have equal weight. When a new state is discovered, it is reached through the shortest possible sequence of operations. Every legal sequence of button presses corresponds to a path in this graph, and every graph path corresponds to a legal sequence of operations. Because Breadth First Search computes shortest paths in unweighted graphs, the recorded distance for Y is exactly the minimum required time.

Python Solution

import sys
from collections import deque

input = sys.stdin.readline

def solve():
    x, y, z = map(int, input().split())

    dist = [-1] * 101
    dist[x] = 0

    q = deque([x])

    while q:
        cur = q.popleft()
        if cur == y:
            break

        for nxt in (cur + 1, cur - 1, cur + z, cur - z):
            if 0 <= nxt <= 100 and dist[nxt] == -1:
                dist[nxt] = dist[cur] + 1
                q.append(nxt)

    print(dist[y])

if __name__ == "__main__":
    solve()

The distance array serves two purposes. It stores the shortest known time and also marks whether a volume has already been visited. This avoids processing the same state multiple times.

Every iteration considers exactly four candidate moves. The boundary check comes before accessing the distance array so that invalid volumes such as -1 or 101 are ignored safely.

The early exit when cur == y is optional but avoids unnecessary work after the optimal answer has already been found.

The implementation also handles Z = 0 naturally. In that case the transitions cur + z and cur - z both equal cur, which has already been visited, so no infinite loop occurs.

Worked Examples

Sample 1

Input:

0 10 3
Step Current Volume Newly Reached Distance
1 0 1, 3 1
2 3 6 2
3 6 9 3
4 9 10 4

The shortest sequence is 0 -> 3 -> 6 -> 9 -> 10. Breadth First Search reaches volume 10 after four operations, matching the sample answer.

Sample 2

Input:

10 0 3
Step Current Volume Newly Reached Distance
1 10 9, 7, 13 1
2 7 6, 4 2
3 4 3, 1 3
4 1 0 4

One shortest path is 10 -> 7 -> 4 -> 1 -> 0. The algorithm finds the target after four seconds.

Complexity Analysis

Measure Complexity Explanation
Time O(101) Every volume is processed at most once, with four transitions checked from each state.
Space O(101) The distance array and queue store at most 101 states.

Since there are only 101 possible volumes, the running time is effectively constant. The solution easily satisfies any reasonable contest limits.

Test Cases

import sys
import io
from collections import deque

def solve():
    input = sys.stdin.readline
    x, y, z = map(int, input().split())

    dist = [-1] * 101
    dist[x] = 0
    q = deque([x])

    while q:
        cur = q.popleft()
        if cur == y:
            break
        for nxt in (cur + 1, cur - 1, cur + z, cur - z):
            if 0 <= nxt <= 100 and dist[nxt] == -1:
                dist[nxt] = dist[cur] + 1
                q.append(nxt)

    print(dist[y])

def run(inp: str) -> str:
    backup_stdin = sys.stdin
    backup_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    solve()
    sys.stdin = backup_stdin
    sys.stdout = backup_stdout
    return out.getvalue()

assert run("0 10 3\n") == "4\n", "sample 1"
assert run("10 0 3\n") == "4\n", "sample 2"
assert run("0 0 5\n") == "0\n", "already at target"
assert run("5 8 0\n") == "3\n", "Z equals zero"
assert run("100 99 10\n") == "1\n", "upper boundary"
assert run("1 4 5\n") == "3\n", "temporary detour"
Test input Expected output What it validates
0 0 5 0 Starting volume already equals the target.
5 8 0 3 Large move has zero length.
100 99 10 1 Moves outside the valid range are rejected.
1 4 5 3 The shortest path may temporarily move away from the target.

Edge Cases

When Z = 0, the graph contains self loops produced by +Z and -Z. Consider:

Input:
5 8 0

The search starts from 5. The moves generated by adding or subtracting zero return to 5, which has already been visited, so they are ignored. Breadth First Search continues through 6, 7, and 8, producing the correct answer 3.

When a detour is required, Breadth First Search still succeeds because it explores all states at the current distance before moving to longer paths. For the input

1 4 5

the search reaches 0 after one step, then 5 after two steps, and finally 4 after three steps. A greedy algorithm that always reduced the absolute difference to the target would never consider this sequence.

When the current volume is already at the maximum boundary, illegal moves are discarded immediately. For

100 99 10

the transition to 110 is ignored because it is outside the valid range. The remaining legal move to 99 reaches the target in one second, which is the optimal answer.