CF 102697063 - Schwarzschild Radius

The task is a direct application of the Schwarzschild radius formula. We are given the mass (M) of an object, and we must calculate the radius of its event horizon using [ rs=frac{2GM}{c^2}.

codeforcescompetitive-programming

CF 102697063 - Schwarzschild Radius

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Solution

Problem Understanding

The task is a direct application of the Schwarzschild radius formula. We are given the mass (M) of an object, and we must calculate the radius of its event horizon using

[ r_s=\frac{2GM}{c^2}. ]

For this problem, the constants are deliberately different from their real-world values. We must use (G=6.67\times10^{-1}=0.667) and (c=3\times10^4=30000). The input contains one floating-point mass (M), with (0\le M\le10^9), and the output is the corresponding floating-point radius.

The constraint on (M) does not create any algorithmic difficulty because the answer is obtained from a constant number of arithmetic operations. Even at (M=10^9), we never need to iterate over the value of (M), build an array, or perform any search. A constant-time solution is comfortably inside a one-second limit and uses constant memory.

The main edge case is zero mass. For input 0, the formula gives a radius of exactly zero, so the correct output is 0.0. An implementation that assumes the mass is positive or divides by the mass could fail here, even though the formula itself has no division by (M).

0

The correct output is

0.0

Another common mistake is using the real physical constants instead of the simplified constants specified by the problem. For example,

1000000

must produce

0.0014822222222222222

because the calculation uses (G=0.667) and (c=30000). Using the real gravitational constant and the real speed of light produces a completely different scale and is not the requested computation.

A final boundary case is the maximum allowed mass:

1000000000

The answer is approximately

1.4822222222222222

There is no integer overflow concern in Python, and the intermediate values are tiny compared with the range supported by Python floating-point arithmetic.

Approaches

A brute-force interpretation would try to obtain the radius numerically instead of evaluating the given equation directly. For example, if the mass were an integer, one could compute a quantity proportional to (M) by repeatedly adding the same constant once per unit of mass. At the maximum mass this would require (10^9) iterations, which is far beyond what a one-second competitive-programming solution can afford. The approach is also unnecessary because the problem already gives an explicit closed-form expression.

The brute-force method works only because addition can eventually reproduce multiplication. It fails when (M) becomes large because its running time grows with the numerical value of the input rather than with the size of the input representation.

The key observation is that (G) and (c) are fixed constants. We can combine them into a single constant factor:

[ r_s=M\cdot\frac{2(0.667)}{30000^2}. ]

Once this factor is known, the entire problem is one floating-point multiplication. There is no hidden search, simulation, iteration over possible radii, or numerical method involved. The input has only one value, so the optimal solution performs a constant number of arithmetic operations and immediately prints the result.

Approach Time Complexity Space Complexity Verdict
Brute Force O(M) in a repeated-addition implementation O(1) Too slow for (M=10^9)
Optimal O(1) O(1) Accepted

Algorithm Walkthrough

  1. Read the single floating-point value (M). Reading it as a floating-point number is necessary because the input is explicitly allowed to contain a decimal value.
  2. Set the gravitational constant to (G=0.667) and the speed of light to (c=30000). These are the values required by this problem, rather than their physical values.
  3. Compute (2GM/c^2). The square of the speed of light is (30000^2=900000000), so the calculation is just a few floating-point operations.
  4. Print the resulting value. The problem accepts floating-point output, and the sample itself contains many decimal digits, so there is no reason to round the result manually.

The invariant behind the calculation is simple: after the arithmetic expression is evaluated, the stored value is exactly the mathematical formula specified by the problem, subject only to ordinary floating-point representation. Since every quantity except (M) is a fixed constant and there are no iterative decisions, there is no state that can drift away from the intended result.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    m = float(input())
    G = 0.667
    c = 30000.0

    radius = 2.0 * G * m / (c * c)
    print(radius)

if __name__ == "__main__":
    solve()

The first line imports sys, and input is defined using sys.stdin.readline as requested. Only one input value exists, so the program reads exactly one line and converts it to float.

The constants are written explicitly as 0.667 and 30000.0. Using 0.667 rather than a more precise value for (G) is essential because the statement deliberately changes the physical constant. Likewise, using 30000 for (c) is part of the problem definition.

The expression c * c computes (c^2). Writing the formula directly as 2.0 * G * m / (c * c) closely mirrors the mathematical definition and avoids any opportunity for accidentally using (c) instead of (c^2).

There are no loops, arrays, recursion, or special boundary checks. Zero is handled naturally because multiplying the mass by zero produces a zero radius. Python's floating-point representation is also more than sufficient for the scale of the values involved.

Worked Examples

For the provided sample, the input mass is (1{,}000{,}000). The calculation proceeds as follows.

Step (M) (G) (c) Radius
Read input 1000000 0.667 30000 not computed
Compute numerator 1000000 0.667 30000 (2\cdot0.667\cdot1000000=1334000)
Compute denominator 1000000 0.667 30000 (30000^2=900000000)
Divide 1000000 0.667 30000 (1334000/900000000=0.0014822222222222222)

The output is therefore 0.0014822222222222222. This demonstrates that the only state needed by the algorithm is the input mass and the constants.

For a second example, take the maximum allowed mass.

1000000000

The trace is

Step (M) Numerator (2GM) Denominator (c^2) Radius
Read input 1000000000 not computed not computed not computed
Compute numerator 1000000000 1334000000 not computed not computed
Compute denominator 1000000000 1334000000 900000000 not computed
Divide 1000000000 1334000000 900000000 1.4822222222222222

This trace confirms that the maximum input does not change the algorithm at all. The number of operations remains constant regardless of the magnitude of (M).

Complexity Analysis

Measure Complexity Explanation
Time O(1) One input value and a constant number of arithmetic operations are processed.
Space O(1) Only the mass, constants, and result are stored.

The maximum mass is (10^9), but the algorithm never loops up to (M). Its running time is independent of the numerical value of the mass, so it easily fits the one-second time limit. The memory usage consists of a few floating-point variables and is negligible compared with the 256 MB limit.

Test Cases

import sys
import io

def calculate(m):
    G = 0.667
    c = 30000.0
    return 2.0 * G * m / (c * c)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    m = float(sys.stdin.readline())
    print(calculate(m))

    out = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout

    return out

# Provided sample
assert run("1000000\n") == "0.0014822222222222222\n", "sample 1"

# Minimum-size input
assert run("0\n") == "0.0\n", "zero mass"

# Smallest positive boundary
assert run("1\n") == "1.4822222222222223e-09\n", "minimum positive mass"

# Maximum allowed mass
assert run("1000000000\n") == "1.4822222222222222\n", "maximum mass"

# Decimal input and repeated equal values
assert run("12.5\n") == "1.8527777777777778e-08\n", "decimal mass"

assert abs(calculate(500000000) - 0.7411111111111111) < 1e-15, "half maximum"
Test input Expected output What it validates
1000000 0.0014822222222222222 Provided sample and the required constants
0 0.0 Minimum boundary and zero handling
1 1.4822222222222223e-09 Small positive value and floating-point output
1000000000 1.4822222222222222 Maximum allowed mass
12.5 1.8527777777777778e-08 Decimal input rather than only integers
500000000 approximately 0.7411111111111111 Linear scaling of the result

Edge Cases

The zero-mass case is handled directly by the formula. With input

0

the algorithm reads (M=0), computes (2\cdot0.667\cdot0/30000^2=0), and prints 0.0. There is no division by (M), so no special branch is required.

The minimum positive mass also exercises the scale of the floating-point result. With

1

the numerator is (1.334), while the denominator is (900000000), giving approximately 1.4822222222222223e-09. A solution that performs integer division would lose the fractional result, which is why the computation must use floating-point arithmetic.

The maximum mass

1000000000

produces

1.4822222222222222

The algorithm performs exactly the same operations as it does for any smaller input. There is no overflow issue in Python, and the intermediate numerator is only (1.334\times10^9).

The decimal-input case

12.5

confirms that the input must not be parsed with int(). The correct calculation is approximately 1.8527777777777778e-08. Parsing with float() preserves the fractional part and lets the same formula handle both integer-looking and genuinely decimal masses.

Finally, using the physical values of (G) and (c) is a correctness error rather than a precision improvement. The problem explicitly defines (G=0.667) and (c=30000), so every test must be evaluated with those constants.