CF 104544L - The Washing Machine Monster

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Solve time: 58s
Verified: yes

Solution

Problem Understanding

We are given a number of socks and asked to determine how many complete pairs can be formed. A pair consists of exactly two socks, so the task reduces to grouping socks into disjoint groups of size two and counting how many such groups can be made.

The input consists of up to two test cases, and for each test case we are given a single integer $n$, the number of socks available. The output for each test case is the maximum number of pairs, meaning we are allowed to pair socks optimally with no additional constraints such as colors or restrictions.

The constraint on $n$ is extremely small, between 34 and 35. This immediately rules out any need for optimization concerns. Any $O(1)$ or even $O(n)$ approach is trivially sufficient. In fact, the structure of the problem strongly suggests that no simulation or complex processing is needed because there is no additional structure beyond counting.

There are no subtle corner cases involving ordering or arrangement. The only potential edge case is whether $n$ is even or odd. If $n$ is even, all socks can be perfectly paired. If $n$ is odd, exactly one sock remains unpaired.

A naive mistake would be to attempt pairing via simulation without handling the leftover correctly. For example, if $n = 35$, a naive loop pairing two at a time until exhaustion could accidentally overshoot or mishandle the last singleton if implemented with incorrect bounds. The correct answer in this case is clearly $17$, since $35 = 2 \cdot 17 + 1$.

Another possible incorrect approach is to assume some hidden structure in the input formatting due to the sample being concatenated visually. However, each test case is independent and only depends on its own $n$.

Approaches

A brute-force interpretation would simulate the pairing process explicitly. We could repeatedly remove two socks and count how many such removals we perform until fewer than two socks remain. This works correctly because each operation corresponds exactly to forming one valid pair. However, even though this is correct, it is unnecessary overhead given the simplicity of the problem.

The key observation is that pairing socks greedily is optimal and deterministic. Every pair consumes exactly two socks, so the number of pairs is fully determined by integer division of $n$ by 2. There is no scenario where choosing differently changes the outcome, because there are no constraints on which socks can pair with which.

This reduces the problem to a single arithmetic operation per test case.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	$O(n)$	$O(1)$	Accepted
Direct Formula	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

Read the number of test cases $t$. Each test case is independent, so we process them separately.
For each test case, read the integer $n$, which represents the number of socks.
Compute the number of full pairs by dividing $n$ by 2 using integer division. This works because every pair consumes exactly two socks, and any leftover sock cannot form a pair.
Output the computed value for that test case.

Why it works

Each valid pair requires exactly two distinct socks, and there are no restrictions preventing any two socks from being paired. Therefore, the problem reduces to partitioning a set of size $n$ into groups of size 2 as much as possible. The maximum number of such disjoint groups is exactly the floor of $n/2$. No alternative pairing strategy can increase this count because every pair consumes two elements and there is no reuse.

Python Solution

import sys
input = sys.stdin.readline

t = int(input().strip())
for _ in range(t):
    n = int(input().strip())
    print(n // 2)

The solution reads the number of test cases and processes each independently. For each $n$, integer division by 2 directly yields the number of complete pairs.

The only implementation detail that matters is ensuring input is parsed cleanly per line. Because $t \le 2$, even accidental overhead would not matter, but the standard fast I/O pattern is still used for correctness and consistency.

Worked Examples

We trace the computation for two representative inputs.

Example 1

Input:

n = 34

Step	n	Computation	Pairs
1	34	34 // 2	17

This shows that when $n$ is even, all socks are perfectly paired with no leftovers.

Example 2

Input:

n = 35

Step	n	Computation	Pairs
1	35	35 // 2	17

This demonstrates that a single leftover sock does not contribute to any pair, so the result is the same as for 34 except for the unpaired element.

In both cases, the trace confirms that the algorithm depends only on parity and not on any hidden structure.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(t)$	One constant-time division per test case
Space	$O(1)$	Only a few integer variables are used

The constraints guarantee at most two test cases and very small input size, so the solution runs instantly and uses negligible memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from sys import stdout
    out = []
    
    t = int(input().strip())
    for _ in range(t):
        n = int(input().strip())
        out.append(str(n // 2))
    
    return "\n".join(out)

# provided sample (as interpreted)
assert run("2\n34\n35\n") == "17\n17"

# minimum edge
assert run("1\n34\n") == "17"

# odd boundary
assert run("1\n35\n") == "17"

# small synthetic
assert run("2\n2\n3\n") == "1\n1"

# larger even/odd mix
assert run("2\n100\n101\n") == "50\n50"

Test input	Expected output	What it validates
2, 34, 35	17, 17	sample parity behavior
1, 34	17	minimum valid even case
1, 35	17	odd boundary case
2, 2, 3	1, 1	smallest meaningful pairing cases
2, 100, 101	50, 50	larger values, parity correctness

Edge Cases

The only meaningful edge case is when $n$ is odd. For $n = 35$, the algorithm computes $35 // 2 = 17$. Internally, no special handling is required because integer division naturally discards the leftover sock.

Tracing $n = 35$:

Step 1 reads $n = 35$. Step 2 applies integer division, producing 17. Step 3 outputs 17. The leftover sock is implicitly ignored because it cannot form a pair.

This confirms that the implementation correctly handles both even and odd values without branching or special logic.