CF 1812E - Not a Geometry Problem

Rating: -
Tags: *special, constructive algorithms, geometry, math
Solve time: 1m 4s
Verified: yes

Solution

Problem Understanding

We are given three integers that should be interpreted as coordinates in a three-dimensional space. The task is to compute a single real number derived from these three values, and the sample output reveals what this quantity represents: it is the Euclidean length of the vector from the origin to the point $(x, y, z)$.

Geometrically, this is the straight-line distance from $(0, 0, 0)$ to $(x, y, z)$. In algebraic terms, this distance is determined entirely by the Pythagorean theorem extended into three dimensions.

The constraints are extremely small, with each coordinate bounded between -1000 and 1000. This immediately tells us that any arithmetic involving squares will stay comfortably within 32-bit integer limits. The computation is constant-time, so there is no concern about performance.

The main subtlety in this kind of problem is not computational complexity but numerical precision. A naive implementation using integer arithmetic is safe up to the squaring step, but the final square root must be computed in floating point. If implemented incorrectly, especially by mixing integer division or rounding too early, the result can drift.

There are no real structural edge cases beyond sign handling. Squaring removes sign information, so inputs like $(1000, -1000, 0)$ behave identically to $(1000, 1000, 0)$. The only important boundary case is when all inputs are zero, where the output should be exactly zero.

A common mistake is attempting to approximate the value manually or using integer square root logic, which would lose the required precision.

Approaches

The brute-force perspective is to think in terms of geometry: measure the distance between two points in space by decomposing it into orthogonal components. One might imagine simulating movement along axes and accumulating distance, but since the axes are perpendicular, the correct aggregation rule is fixed: each coordinate contributes independently through squaring.

A naive but still correct approach is to explicitly compute the squared distance by summing the squares of each coordinate and then taking a square root. This already is the optimal mathematical formulation. Any attempt to decompose further, such as iterative geometric construction or simulation, is unnecessary overhead and would only complicate the solution.

The key observation is that the Euclidean norm in three dimensions is defined directly as:

$$\sqrt{x^2 + y^2 + z^2}$$

There is no hidden structure beyond this identity. The problem is essentially testing whether one recognizes this as a direct application of the distance formula in 3D space.

Thus the solution reduces to computing a sum of squares followed by a square root.

Approach	Time Complexity	Space Complexity	Verdict
Direct geometric computation	O(1)	O(1)	Accepted
Any alternative construction	O(1)	O(1)	Accepted but unnecessary

Algorithm Walkthrough

Read the three integers $x, y, z$. These represent a point in 3D space relative to the origin.
Compute the squared contributions of each coordinate independently. This isolates each axis component before combining them.
Sum the squared values to obtain the squared Euclidean distance from the origin.
Compute the square root of this sum to recover the actual geometric distance.
Output the result as a floating-point number with default precision, which is sufficient given the very large allowed error tolerance.

Why it works

The Euclidean distance formula is derived from repeatedly applying the Pythagorean theorem. First, the distance in the $xy$-plane is $\sqrt{x^2 + y^2}$. Extending this into the third dimension treats that planar distance as one leg of a right triangle with height $z$, giving $\sqrt{(x^2 + y^2) + z^2}$. Since squaring eliminates direction and preserves magnitude, the order of accumulation does not affect correctness.

Python Solution

import sys
import math

input = sys.stdin.readline

def solve():
    x, y, z = map(int, input().split())
    ans = math.sqrt(x*x + y*y + z*z)
    print(ans)

if __name__ == "__main__":
    solve()

The solution reads the input in constant time and immediately applies the Euclidean norm formula. Using math.sqrt ensures correct floating-point behavior. Multiplication is used instead of exponentiation for efficiency, although both are equivalent here.

There are no loops or branches, so the implementation is minimal and avoids any risk of off-by-one or accumulation errors.

Worked Examples

We trace the computation for two representative inputs.

Example 1: (1, 1, 1)

Step	x	y	z	x² + y² + z²	Result
Initial	1	1	1	-	-
Squaring	1	1	1	3	-
Square root	-	-	-	3	1.7320508075688772

This confirms the standard unit diagonal in a cube, matching the expected value.

Example 2: (0, 0, 0)

Step	x	y	z	x² + y² + z²	Result
Initial	0	0	0	-	-
Squaring	0	0	0	0	-
Square root	-	-	-	0	0.0

This confirms that the origin maps to zero distance.

In both cases, the computation behaves consistently with geometric expectations, reinforcing that no special-case handling is required.

Complexity Analysis

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

Given the extremely small input size and constant-time operations, the solution is trivially within limits.

Test Cases

import sys, io
import math

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    x, y, z = map(int, sys.stdin.readline().split())
    return str(math.sqrt(x*x + y*y + z*z))

# provided sample
assert abs(float(run("1 1 1\n")) - 1.7320508075688772) < 1e-9

# origin
assert abs(float(run("0 0 0\n")) - 0.0) < 1e-9

# axis-aligned
assert abs(float(run("3 0 0\n")) - 3.0) < 1e-9

# mixed signs
assert abs(float(run("-3 4 12\n")) - 13.0) < 1e-9

# symmetric point
assert abs(float(run("1000 1000 1000\n")) - math.sqrt(3000000)) < 1e-6

Test input	Expected output	What it validates
1 1 1	√3	basic correctness
0 0 0	0	origin edge case
3 0 0	3	axis alignment
-3 4 12	13	classic Pythagorean triple in 3D
1000 1000 1000	large float	upper bound stability

Edge Cases

For the input 0 0 0, the algorithm computes $0^2 + 0^2 + 0^2 = 0$, and the square root is exactly 0. The implementation handles this naturally without any special branching, since math.sqrt(0) returns 0.

For negative coordinates such as -3 4 12, squaring ensures the sign is eliminated before aggregation, so the computation proceeds as $9 + 16 + 144 = 169$, and the square root yields 13. This confirms that direction does not affect magnitude, and no conditional handling of sign is necessary.

For large equal inputs like 1000 1000 1000, the squared sum becomes 3,000,000, which remains safely within floating-point precision for Python's double-precision arithmetic. The algorithm does not lose accuracy within the required tolerance, and the output remains stable.