CF 104447L - Amazing Teacher
We are given a sequence of independent test cases. Each test case contains a single integer representing a student's original score on a 0 to 10 scale.
Rating: -
Tags: -
Solve time: 36s
Verified: yes
Solution
Problem Understanding
We are given a sequence of independent test cases. Each test case contains a single integer representing a student's original score on a 0 to 10 scale. The task is to transform this score according to a simple rule set that models a teacher who upgrades passing or nonzero submissions to a perfect score, but refuses to do so for completely blank submissions.
For each test case, if the original score is strictly greater than zero, the output becomes 10. If the score is exactly zero, the output remains 0.
The constraints are extremely small. There are at most 11 test cases and each score lies in the range from 0 to 10. This immediately rules out any need for complex data structures or optimization. Any solution that processes each test case in constant time is sufficient.
The only subtle edge case is the boundary between zero and positive values. A naive mistake would be to treat zero as valid for upgrade or to incorrectly assume all inputs should map to 10 regardless of value. For example, input 0 must remain 0, while input 1 must become 10. Another possible mistake is overcomplicating the rule and accidentally introducing off-by-one logic, such as checking n >= 0, which would incorrectly upgrade zero as well.
Approaches
A brute-force interpretation of the problem might attempt to explicitly simulate grading logic with multiple conditional branches or even store a mapping table for all possible values from 0 to 10. This would work correctly because the input domain is tiny, and we could simply predefine outputs for each possible score.
However, such an approach is unnecessary. The structure of the problem reveals that only one condition matters: whether the value is zero or not. The brute-force approach treats each value independently, but we can compress all nonzero cases into a single outcome because they behave identically. This observation reduces the problem to a single conditional check.
The key insight is that the entire transformation is a binary classification: zero maps to zero, everything else maps to ten. Once this is recognized, the solution becomes constant-time per test case.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force Mapping Table | O(1) per case | O(1) | Accepted |
| Optimal Conditional Check | O(1) per case | O(1) | Accepted |
Algorithm Walkthrough
- Read the number of test cases. This determines how many independent transformations must be applied.
- For each test case, read the integer score.
- Check whether the score is equal to zero. This is the only meaningful decision point because all positive values behave identically.
- If the score is zero, output zero. Otherwise, output ten.
Each decision is local to its test case, so no state needs to be carried across iterations.
Why it works
The correctness rests on the fact that the transformation depends only on membership in a two-element partition of the input domain: the singleton set containing zero, and the set of all positive integers in range. Since all nonzero inputs are mapped to the same output, distinguishing among them is unnecessary. The algorithm therefore preserves correctness by applying the exact rule definition without approximation or aggregation beyond what the rule itself allows.
Python Solution
import sys
input = sys.stdin.readline
def solve():
t = int(input())
for _ in range(t):
n = int(input())
if n > 0:
print(10)
else:
print(0)
if __name__ == "__main__":
solve()
The implementation follows the algorithm directly. Input is handled using fast I/O because it is standard practice in competitive programming, even though the constraints here do not require it.
The conditional if n > 0 is the core of the solution. It intentionally avoids writing if n != 0 or if n >= 0 because the latter would incorrectly map zero to ten. The separation of the two outputs is explicit and avoids any hidden arithmetic tricks that could introduce mistakes.
Worked Examples
Example 1
Consider the input:
3
0
5
1
| Test case | n | Condition (n > 0) | Output |
|---|---|---|---|
| 1 | 0 | false | 0 |
| 2 | 5 | true | 10 |
| 3 | 1 | true | 10 |
The first test case confirms that zero is preserved. The remaining cases show that all positive values collapse into the same output.
Example 2
Input:
4
10
2
0
7
| Test case | n | Condition (n > 0) | Output |
|---|---|---|---|
| 1 | 10 | true | 10 |
| 2 | 2 | true | 10 |
| 3 | 0 | false | 0 |
| 4 | 7 | true | 10 |
This trace demonstrates that the upper bound of the input range does not matter. Even the maximum value 10 is treated identically to any other positive value.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(t) | Each test case requires a single comparison and constant-time output |
| Space | O(1) | No auxiliary data structures are used beyond input variables |
Given that t is at most 11, the solution runs in constant time in practice and is trivially within limits.
The memory usage is minimal since no arrays or accumulators are required.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
from contextlib import redirect_stdout
import io as sysio
out = sysio.StringIO()
with redirect_stdout(out):
solve()
return out.getvalue().strip()
def solve():
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
if n > 0:
print(10)
else:
print(0)
# provided samples
assert run("3\n0\n5\n1\n") == "0\n10\n10"
assert run("4\n10\n2\n0\n7\n") == "10\n10\n0\n10"
# custom cases
assert run("1\n0\n") == "0", "zero stays zero"
assert run("1\n1\n") == "10", "smallest positive becomes 10"
assert run("1\n10\n") == "10", "max value still becomes 10"
assert run("5\n0\n0\n0\n0\n0\n") == "0\n0\n0\n0\n0", "all zeros"
| Test input | Expected output | What it validates |
|---|---|---|
| single 0 | 0 | zero boundary |
| single 1 | 10 | minimal positive case |
| single 10 | 10 | upper bound case |
| all zeros | all 0 | repeated edge stability |
Edge Cases
The only meaningful edge case is the value zero itself. For input 0, the algorithm evaluates the condition n > 0 as false and correctly outputs 0. A correct trace is:
| Step | n | Condition (n > 0) | Output |
|---|---|---|---|
| read | 0 | - | - |
| evaluate | 0 | false | 0 |
This confirms that zero is never upgraded.
For a minimal positive value such as 1, the condition evaluates to true, producing 10, and this behavior is consistent across all positive inputs regardless of magnitude.