CF 1672A - Log Chopping
We are asked to determine the winner in a simple two-player game with logs of integer lengths. There are $n$ logs, and two players take turns splitting a single log into two positive integer pieces. The first player unable to make a move loses.
Rating: 800
Tags: games, implementation, math
Solve time: 1m 28s
Verified: yes
Solution
Problem Understanding
We are asked to determine the winner in a simple two-player game with logs of integer lengths. There are $n$ logs, and two players take turns splitting a single log into two positive integer pieces. The first player unable to make a move loses. Each test case provides the number of logs and their lengths, and the output must indicate which player wins assuming both play optimally.
The input constraints are small: at most 50 logs per test case, and log lengths up to 50. There can be up to 100 test cases, but the total number of logs across all test cases is still manageable. These bounds suggest that we do not need highly optimized data structures or algorithms; a linear scan per test case is sufficient.
The key edge cases arise when all logs are length 1. In that situation, the first player cannot make any move and immediately loses. For example, if the input is 1\n1, the output should be maomao90 because errorgorn has no available moves. Another subtle case is when the sum of logs is even or odd: because each split increases the total number of logs by one, the parity of the total number of splits can determine which player will run out of moves first.
Approaches
A brute-force approach would simulate the game turn by turn. We could pick the largest log, split it, alternate turns, and continue until no log can be split. While this is conceptually correct, it would require tracking the entire set of logs at every step. In the worst case, each split doubles the number of logs, producing an exponential blowup in the number of possible sequences. This is clearly inefficient even for the small input sizes.
The insight that simplifies the problem is to notice that a log of length $1$ cannot be split, while any log of length greater than $1$ can always be split. Each split increases the total number of logs by exactly one. Therefore, the game can be analyzed purely by the total number of splits available. Let $k$ be the number of logs of length greater than $1$. Each of these logs will require exactly one split to become two smaller logs, eventually reducing everything to 1-length logs. The total number of possible splits is therefore equal to the number of logs with length greater than $1$. Errorgorn moves first, so if this number is odd, he will make the last split and win; if it is even, maomao90 will win. This reduces the problem to counting the number of logs with length greater than 1 and checking the parity, which is linear in $n$.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force Simulation | O(2^n) | O(n) | Too slow |
| Count and Parity | O(n) | O(1) | Accepted |
Algorithm Walkthrough
- Read the number of test cases $t$.
- For each test case, read the number of logs $n$ and the list of log lengths $a_1, a_2, \dots, a_n$.
- Count the number of logs whose length is greater than $1$. This is the total number of splits that can happen in the game.
- If this count is odd, errorgorn, who moves first, will make the last split. Print
"errorgorn". - If the count is even, maomao90 will make the last split. Print
"maomao90".
Why it works: Each log of length greater than 1 contributes exactly one available move. The players alternate moves, and the parity of the number of moves determines who will have the last turn. This simple invariant fully determines the winner.
Python Solution
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
logs = list(map(int, input().split()))
moves = sum(1 for x in logs if x > 1)
if moves % 2 == 1:
print("errorgorn")
else:
print("maomao90")
The solution reads input efficiently using sys.stdin.readline and processes each test case separately. Counting logs greater than 1 is done with a generator expression, which avoids creating an intermediate list. The modulo operation determines parity and directly decides the winner. No additional data structures are needed.
Worked Examples
Sample 1:
| Step | Logs | Splits >1 | Parity | Winner |
|---|---|---|---|---|
| Input | [2,4,2,1] | 3 | odd | errorgorn |
Here, there are three logs longer than 1. Errorgorn moves first and will also make the last move. Sample 2:
| Step | Logs | Splits >1 | Parity | Winner |
|---|---|---|---|---|
| Input | [1] | 0 | even | maomao90 |
No log can be split, so errorgorn loses immediately.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n) per test case | We scan the list of logs once to count logs >1. |
| Space | O(n) | We store the list of logs for each test case. |
Given $n \le 50$ and $t \le 100$, the total work is small and fits easily within 1 second.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
out = io.StringIO()
sys.stdout = out
t = int(input())
for _ in range(t):
n = int(input())
logs = list(map(int, input().split()))
moves = sum(1 for x in logs if x > 1)
if moves % 2 == 1:
print("errorgorn")
else:
print("maomao90")
return out.getvalue().strip()
# Provided samples
assert run("2\n4\n2 4 2 1\n1\n1") == "errorgorn\nmaomao90", "sample 1 and 2"
# Custom cases
assert run("1\n3\n1 1 1") == "maomao90", "all logs length 1"
assert run("1\n5\n2 2 2 2 2") == "errorgorn", "all logs length >1 odd count"
assert run("1\n4\n2 2 2 2") == "maomao90", "all logs length >1 even count"
assert run("1\n1\n50") == "errorgorn", "single long log"
| Test input | Expected output | What it validates |
|---|---|---|
| 3 logs of length 1 | maomao90 | first player cannot move |
| 5 logs of length 2 | errorgorn | odd number of moves |
| 4 logs of length 2 | maomao90 | even number of moves |
| 1 log of length 50 | errorgorn | single long log handled correctly |
Edge Cases
For the all-ones case, such as input 1\n3\n1 1 1, the count of logs >1 is zero. Zero is even, so the algorithm correctly outputs maomao90, reflecting that errorgorn has no moves.
For a single long log, input 1\n1\n50, the count of logs >1 is one, which is odd. The algorithm outputs errorgorn, reflecting that errorgorn will make the only move and win. These edge cases confirm the parity-based logic is robust even at the extremes.