CF 125A - Measuring Lengths in Baden
In Baden, the conversion rules are different from the real world. One inch equals 3 centimeters, and one foot contains 12 inches. We are given a length measured in centimeters. The task is to express that length as a combination of feet and inches.
CF 125A - Measuring Lengths in Baden
Rating: 1400
Tags: math
Solve time: 1m 27s
Verified: yes
Solution
Problem Understanding
In Baden, the conversion rules are different from the real world. One inch equals 3 centimeters, and one foot contains 12 inches.
We are given a length measured in centimeters. The task is to express that length as a combination of feet and inches. Since the original length may not correspond to an exact integer number of inches, we first need to round it to the nearest whole inch according to the special rounding rule described in the statement. After that, among all equivalent representations, we must maximize the number of feet.
Since one inch is exactly 3 centimeters, converting centimeters to inches means computing the nearest integer to $n / 3$. The statement gives examples such as 1 cm becoming 0 inches and 2 cm becoming 1 inch, which confirms ordinary rounding to the nearest integer.
The constraint is very small, $1 \le n \le 10000$. Even an inefficient solution would easily fit within the limits. The challenge is not performance but correctly handling the rounding rule and then converting inches into feet and remaining inches.
The most common source of mistakes is the rounding step.
Consider input:
1
We have $1/3 = 0.333...$, which rounds to 0 inches. The correct output is:
0 0
A careless implementation that always rounds up fractional values would incorrectly produce 1 inch.
Consider input:
2
We have $2/3 = 0.666...$, which rounds to 1 inch. The correct output is:
0 1
A careless implementation using floor division would produce 0 inches.
Another subtle case occurs when rounding increases the total inch count enough to create an additional foot.
For example:
35
Since $35/3 = 11.666...$, the rounded value is 12 inches. Twelve inches equals one foot, so the correct output is:
1 0
An implementation that converts centimeters directly into feet and inches before rounding would miss this transition.
Approaches
A brute-force solution could try every possible number of feet, convert it back into inches, and check which representation matches the rounded length. Since the input size is tiny, this would still run instantly. For a rounded length of at most about 3333 inches, we would test only a few hundred possibilities.
The real structure of the problem is much simpler. After converting the length into a rounded integer number of inches, the requirement to maximize feet has an obvious interpretation: use as many groups of 12 inches as possible.
The problem naturally splits into two independent parts.
First, compute the nearest integer number of inches. Since one inch equals three centimeters, we need the nearest integer to $n/3$. For positive integers, this can be done entirely with integer arithmetic:
$$\text{inches} = \left\lfloor \frac{n+1}{3} \right\rfloor$$
This formula reproduces the required rounding behavior:
- 1 cm → 0 inches
- 2 cm → 1 inch
- 4 cm → 1 inch
- 5 cm → 2 inches
Once we know the total number of inches, maximizing feet means taking as many complete groups of 12 inches as possible. The number of feet is integer division by 12, and the remaining inches are the remainder modulo 12.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(I) where I is the rounded inch count | O(1) | Accepted |
| Optimal | O(1) | O(1) | Accepted |
Algorithm Walkthrough
- Read the integer $n$, the length in centimeters.
- Convert centimeters to the nearest integer number of inches.
Since one inch equals three centimeters, the exact value is $n/3$. The required rounding can be implemented as:
$$\text{inches} = (n + 1) // 3$$
This works because the fractional parts are always 0, 1/3, or 2/3. 3. Compute the maximum number of feet.
Every foot contains 12 inches, so:
$$\text{feet} = \text{inches} // 12$$ 4. Compute the remaining inches.
$$\text{rem} = \text{inches} \bmod 12$$
5. Output feet and rem.
Why it works
After rounding, the length is represented by an exact integer number of inches. Any representation in feet and inches must satisfy:
$$12 \cdot \text{feet} + \text{inches_remaining} = \text{total_inches}$$
with the remaining inches between 0 and 11 inclusive.
Integer division by 12 produces the largest possible number of complete groups of 12 inches. The remainder is exactly the number of inches left over after taking those groups. Since no representation can contain more complete groups than integer division provides, the resulting number of feet is maximal.
Python Solution
import sys
input = sys.stdin.readline
n = int(input())
inches = (n + 1) // 3
feet = inches // 12
rem = inches % 12
print(feet, rem)
The first step computes the rounded number of inches. Using (n + 1) // 3 avoids floating-point arithmetic entirely and exactly matches the rounding rule from the statement.
The next two lines perform a standard conversion from inches into feet and remaining inches. Integer division counts how many complete feet fit into the total inch count, while modulo gives the leftover inches.
The order matters. We must round the centimeter value to inches first, then convert to feet. Performing the conversion before rounding would give incorrect answers for values such as 35 cm, where rounding changes 11.666... inches into 12 inches and creates an additional foot.
Worked Examples
Example 1
Input:
42
| n | Rounded inches | Feet | Remaining inches |
|---|---|---|---|
| 42 | (42 + 1) // 3 = 14 | 14 // 12 = 1 | 14 % 12 = 2 |
Output:
1 2
This example shows the normal case. After rounding, we have 14 inches. One complete foot uses 12 of them, leaving 2 inches.
Example 2
Input:
35
| n | Rounded inches | Feet | Remaining inches |
|---|---|---|---|
| 35 | (35 + 1) // 3 = 12 | 12 // 12 = 1 | 12 % 12 = 0 |
Output:
1 0
This example demonstrates why rounding must happen first. The exact value is 11.666... inches, which rounds to 12 inches and becomes exactly one foot.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | Only a fixed number of arithmetic operations are performed |
| Space | O(1) | Only a few integer variables are stored |
The algorithm performs constant-time arithmetic regardless of the input value. With $n \le 10000$, it easily fits within both the time and memory limits.
Test Cases
# helper: run solution on input string, return output string
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
n = int(sys.stdin.readline())
inches = (n + 1) // 3
feet = inches // 12
rem = inches % 12
return f"{feet} {rem}"
# provided sample
assert run("42\n") == "1 2", "sample 1"
# custom cases
assert run("1\n") == "0 0", "minimum value"
assert run("2\n") == "0 1", "round up from 2/3 inch"
assert run("35\n") == "1 0", "rounding creates a new foot"
assert run("10000\n") == "277 10", "maximum constraint"
| Test input | Expected output | What it validates |
|---|---|---|
1 |
0 0 |
Smallest input, rounds down |
2 |
0 1 |
Correct handling of 2/3 rounding |
35 |
1 0 |
Rounding before foot conversion |
10000 |
277 10 |
Largest allowed input |
Edge Cases
The first tricky case is:
1
The algorithm computes:
$$(1+1)//3 = 0$$
So the total length is 0 inches. Then:
$$0//12 = 0,\quad 0%12 = 0$$
The output becomes:
0 0
This matches the required rounding rule.
The second tricky case is:
2
The algorithm computes:
$$(2+1)//3 = 1$$
So the rounded length is 1 inch. Then:
$$1//12 = 0,\quad 1%12 = 1$$
The output is:
0 1
This confirms that 2 cm rounds upward to 1 inch.
The third tricky case is:
35
The algorithm computes:
$$(35+1)//3 = 12$$
Then:
$$12//12 = 1,\quad 12%12 = 0$$
The output is:
1 0
This case verifies that rounding is completed before splitting the result into feet and inches. If the order were reversed, the answer would be wrong.