CF 915A - Garden

CF 915A - Garden

Rating: 900
Tags: implementation
Solve time: 11m 44s
Verified: yes

Solution

Problem Understanding

We are asked to water a linear garden of length k using one of n buckets. Each bucket waters a fixed segment of the garden every hour, specifically a continuous stretch of length a_i. Luba must water the entire garden without leaving gaps and cannot water any part twice. Our task is to determine which bucket minimizes the total number of hours required.

The input gives the number of buckets and the garden length on the first line, followed by the lengths that each bucket can water in one hour. The output is a single integer, the minimum number of hours needed to water the garden using a single bucket.

The constraints are small: n and k are both at most 100. That means even a straightforward approach iterating over all buckets is efficient. We can safely check each bucket in turn and perform simple arithmetic to see if it divides the garden evenly. The edge case that could cause a naive implementation to fail is when a bucket’s watering length does not divide the garden length evenly. For example, if k = 6 and the buckets are [5, 2], the bucket with length 5 cannot fully water the garden, while 2 divides 6 exactly, giving 3 hours.

Approaches

The brute-force approach is simple. For each bucket, check if its watering length divides the garden length exactly. If it does, compute the number of hours required as k // a_i. Keep track of the minimum number of hours across all buckets. This is correct because watering can only be done in continuous segments of exactly the bucket’s length, and partial watering of a segment is not allowed.

A naive variation might attempt to simulate placing segments of length a_i along the garden, but that is unnecessary because division handles it directly. Since the maximum value for n and k is 100, iterating over all buckets and performing integer division is extremely fast, giving an operation count of at most 100, which is negligible.

The key insight that simplifies this problem is realizing that the problem reduces to integer division. For each bucket that divides the garden length exactly, the number of hours is k // a_i. We only need to pick the largest such bucket because larger segments reduce the number of hours. The observation that k % a_i == 0 allows us to discard buckets that cannot fully water the garden in integer hours.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Accepted
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of buckets n and garden length k.
2. Read the list of bucket lengths a.
3. Initialize a variable min_hours to a large number to track the minimal hours.
4. Iterate through each bucket length length in a.
5. Check if k % length == 0. If it does not, skip this bucket because it cannot water the garden fully in integer hours.
6. Compute the number of hours as hours = k // length.
7. Update min_hours if hours is smaller than the current min_hours.
8. After checking all buckets, print min_hours.

The invariant here is that we only consider bucket lengths that divide the garden exactly. Since the problem guarantees at least one valid bucket, min_hours will always be updated with a correct value. Because larger segments reduce the number of hours, this approach ensures we find the minimal number of hours.

Python Solution

import sys
input = sys.stdin.readline

n, k = map(int, input().split())
a = list(map(int, input().split()))

min_hours = float('inf')
for length in a:
    if k % length == 0:
        hours = k // length
        if hours < min_hours:
            min_hours = hours

print(min_hours)

The solution reads input efficiently with sys.stdin.readline. We initialize min_hours to infinity to ensure any valid number of hours is smaller. We iterate over all buckets, check divisibility with the garden length, and update the minimum hours. The check k % length == 0 is crucial because it guarantees full coverage of the garden. The final print outputs the minimal hours.

Worked Examples

Sample 1

Input:

3 6
2 3 5

Bucket Length	k % length	Hours (k//length)	min_hours
2	0	3	3
3	0	2	2
5	1	-	2

This demonstrates that only buckets dividing the garden length exactly are considered. The minimal hours is 2.

Sample 2

Input:

2 7
1 3

Bucket Length	k % length	Hours (k//length)	min_hours
1	0	7	7
3	1	-	7

Here only bucket length 1 works, giving 7 hours.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Iterate over each bucket once
Space	O(n)	Store the list of bucket lengths

Given the small constraints n ≤ 100 and k ≤ 100, this algorithm runs instantly within the 1-second time limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    min_hours = float('inf')
    for length in a:
        if k % length == 0:
            hours = k // length
            if hours < min_hours:
                min_hours = hours
    return str(min_hours)

# provided samples
assert run("3 6\n2 3 5\n") == "2", "sample 1"
assert run("1 1\n1\n") == "1", "sample 2"

# custom cases
assert run("2 7\n1 3\n") == "7", "only bucket 1 works"
assert run("5 100\n1 2 4 5 10\n") == "10", "bucket 10 is optimal"
assert run("3 15\n4 5 6\n") == "3", "bucket 5 is best"
assert run("1 100\n100\n") == "1", "single bucket matches exactly"

Test input	Expected output	What it validates
2 7\n1 3	7	Only bucket 1 works
5 100\n1 2 4 5 10	10	Bucket with largest divisor is chosen
3 15\n4 5 6	3	Correct choice among multiple divisors
1 100\n100	1	Single bucket equals garden length

Edge Cases

If the garden length equals the largest bucket, such as k = 100 and bucket = 100, the algorithm correctly identifies 1 hour. If the smallest bucket is the only divisor, like k = 7 and bucket = 1, the algorithm produces 7 hours. The algorithm ignores non-dividing buckets automatically, preventing incorrect partial watering. All edge cases are covered by the divisibility check and the guarantee that at least one bucket works.