CF 104941A - Ancient Math

CF 104941A - Ancient Math

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Solution

Problem Understanding

We are given the radius of a circle, and we are asked to construct a square that has exactly the same area as that circle. The task is not geometric construction in the classical sense, but a direct numerical computation: we must output the side length of such a square.

The area of a circle with radius $r$ is $\pi r^2$. If a square has side length $s$, its area is $s^2$. Equating these gives $s^2 = \pi r^2$, so the quantity we need is $s = r\sqrt{\pi}$.

The input size is extremely small, with $r \le 1000$. This means performance is not a concern at all, and any constant-time arithmetic computation is sufficient. The only real requirement is numerical precision, since the answer involves a square root and multiplication by $\pi$, and the judge allows a small relative or absolute error up to $10^{-4}$.

There are no structural edge cases in terms of algorithmic behavior, but there is one practical pitfall: using insufficient floating-point precision or an imprecise value of $\pi$. A naive implementation that uses an overly rounded constant for $\pi$ can still pass, but it risks failing if the approximation is too rough, especially for larger values of $r$ like 1000 where the absolute error scales up.

Another subtle issue is integer division mistakes in languages where $r$ might accidentally be treated as integer arithmetic throughout an expression like r * sqrt(pi) if written incorrectly. However, in Python this is naturally safe.

Approaches

A brute-force interpretation would try to reconstruct the geometry explicitly, perhaps by simulating circle area or searching for a square side whose area matches $\pi r^2$. Such an approach would be unnecessarily complicated and would introduce numerical iteration or root-finding methods. For example, one could binary search for $s$ such that $s^2 \approx \pi r^2$. This would work, but it is overkill: each evaluation is constant time, and convergence is fast, but it adds complexity without any benefit.

The key observation is that the relationship between the circle and square is algebraic and collapses immediately into a closed form. Once we equate areas, no approximation process is needed beyond evaluating a square root. This reduces the problem from a numerical search into a single expression evaluation.

The brute-force idea works because the function $f(s) = s^2 - \pi r^2$ is monotonic in $s$, so root-finding would converge. However, it fails in the sense of being unnecessary and more error-prone than direct computation. The closed form eliminates iteration entirely.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (binary search root)	$O(\log \frac{1}{\epsilon})$	$O(1)$	Accepted but overcomplicated
Optimal (closed form)	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

1. Read the integer radius $r$. This is the only input and fully determines the target area of the circle.
2. Compute the side length using the derived equality $s = r \cdot \sqrt{\pi}$. This step directly translates the geometric condition into arithmetic form, avoiding any approximation loops or iterative refinement.
3. Print the resulting value with sufficient precision. Since the allowed error is $10^{-4}$, standard floating-point formatting with around 10 decimal places is more than enough to guarantee acceptance.

Why it works

The correctness comes from a direct equivalence of areas. Both shapes are fully described by a single parameter, and their areas are deterministic functions of that parameter. Setting $\pi r^2 = s^2$ leaves no ambiguity in $s$, and the positive root is the only meaningful geometric solution. Because we never approximate through iterative refinement, there is no accumulation of error beyond standard floating-point representation error, which stays well within tolerance for the given constraint range.

Python Solution

import sys
import math

input = sys.stdin.readline

r = int(input().strip())
ans = r * math.sqrt(math.pi)
print(ans)

The solution relies entirely on Python’s built-in floating-point arithmetic and math.sqrt, which is accurate enough for the required tolerance. The constant $\pi$ is also provided by math.pi, ensuring high precision compared to manually defined approximations.

The multiplication order is straightforward and safe because both operands are floating-point compatible. There are no integer division issues or overflow concerns given the small constraint.

Worked Examples

Example 1

Input:

1

We compute $s = 1 \cdot \sqrt{\pi}$.

Step	r	sqrt(pi)	s
Read input	1	-	-
Compute	1	1.7724538509	1.7724538509

The output matches the expected square side length for a unit circle, confirming that the formula directly captures the geometry.

Example 2

Input:

42

We compute $s = 42 \cdot \sqrt{\pi}$.

Step	r	sqrt(pi)	s
Read input	42	-	-
Compute	42	1.7724538509	74.4430617380

This demonstrates linear scaling with radius: doubling $r$ doubles the resulting square side length, consistent with the algebraic relationship.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$	Only a constant number of arithmetic operations are performed regardless of input size
Space	$O(1)$	No additional data structures are used

Given the constraint $r \le 1000$, the computation is trivially fast, and floating-point evaluation dominates all runtime considerations, which is negligible in Python.

Test Cases

import sys, io, math

def solve():
    import sys, math
    r = int(sys.stdin.readline())
    print(r * math.sqrt(math.pi))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)
    from io import StringIO
    old_stdout = sys.stdout
    sys.stdout = StringIO()
    solve()
    out = sys.stdout.getvalue()
    sys.stdin = old_stdin
    sys.stdout = old_stdout
    return out.strip()

# provided samples
assert abs(float(run("1")) - 1.7724538509) < 1e-6
assert abs(float(run("42")) - 74.4430617380) < 1e-6

# custom cases
assert abs(float(run("0")) - 0.0) < 1e-6
assert abs(float(run("1000")) - 1000 * math.sqrt(math.pi)) < 1e-6
assert abs(float(run("2")) - 2 * math.sqrt(math.pi)) < 1e-6
assert abs(float(run("10")) - 10 * math.sqrt(math.pi)) < 1e-6

Test input	Expected output	What it validates
0	0	boundary case of zero radius
1000	1000·√π	maximum constraint scaling
2	2·√π	small non-trivial scaling
10	10·√π	intermediate precision stability

Edge Cases

For the smallest radius, such as $r = 1$, the computation reduces to $s = \sqrt{\pi}$. The algorithm simply evaluates the expression and prints it, with no special handling required. Floating-point precision is more than sufficient to keep error within tolerance.

For the largest radius $r = 1000$, the result is about $1772.45$. Even here, the multiplication by $\sqrt{\pi}$ does not introduce instability, since both operands are well within safe floating-point range. The algorithm performs a single multiplication and square root, so no intermediate rounding accumulates across steps.