CF 102697067 - Tangential Velocity

Rating: -
Tags: -
Solve time: 46s
Verified: yes

Solution

Problem Understanding

The problem describes an object moving around the center of a circle. We know the centripetal force acting on the object, its mass, and the radius of its circular path. From these three quantities, we need to calculate the object's tangential velocity. The input consists of exactly three floating point values, in the order force (F_c), mass (m), and radius (r). The output is the corresponding velocity (v) as a floating point value. The official statement gives a 1 second time limit and 256 MB memory limit.

The physical relationship is

[ F_c = \frac{mv^2}{r}. ]

Rearranging it gives

[ v^2 = \frac{F_c r}{m}, ]

so the required velocity is

[ v = \sqrt{\frac{F_c r}{m}}. ]

There is no large input size to process. The input contains only three numbers, so an algorithm with constant time and constant memory is sufficient. Since the values are floating point numbers, the main concern is numerical calculation rather than algorithmic complexity. The implementation should use floating point arithmetic and a square root function.

The constraints in the published statement do not provide explicit numerical upper and lower bounds for the three floating point values. This means there is no meaningful (O(n)), (O(n\log n)), or (O(n^2)) input-size discussion here. The entire computational task consists of a fixed number of arithmetic operations.

A useful edge case is zero force. For example,

0 10 5

gives

0.0

because the equation becomes (v=\sqrt{0}=0). A careless implementation that assumes the answer must be positive could mishandle this case.

Another useful case is a fractional result. For example,

1 2 8

gives

2.0

because (v=\sqrt{1\cdot8/2}=\sqrt4=2). An implementation that performs integer division before converting to floating point would be dangerous for inputs such as 1 3 8, where the division must remain fractional.

The most common numerical mistake is using the wrong rearrangement. For example,

100 10 5

has

[ v=\sqrt{\frac{100\cdot5}{10}}=\sqrt{50}, ]

which is approximately 7.0710678118654755. Multiplying or dividing the quantities in a different order can produce a numerically plausible but physically incorrect answer. The official sample uses exactly this case.

Approaches

A brute-force approach could try candidate velocities and check which one makes (mv^2/r) close to the given force. This would work in principle because the equation directly characterizes the required velocity, but it is the wrong abstraction for the problem. If we search with a fixed step (d) over a velocity interval of width (R), we perform approximately (R/d) checks. For decimal precision requiring (d=10^{-k}), that becomes (R\cdot10^k) checks. There is no finite operation count that guarantees arbitrary floating point precision with such a search, and even a seemingly modest range becomes unnecessarily expensive as the requested precision increases.

The brute-force works only because every candidate can be verified against the physical equation, but the equation itself already gives us the answer algebraically. The observation that (F_c=mv^2/r) contains (v) only as (v^2) means we can isolate (v^2) with ordinary arithmetic and then take one square root. The entire search disappears.

Starting from

[ F_c=\frac{mv^2}{r}, ]

multiply both sides by (r):

[ F_cr=mv^2. ]

Divide by (m):

[ \frac{F_cr}{m}=v^2. ]

Taking the non-negative square root gives

[ v=\sqrt{\frac{F_cr}{m}}. ]

Velocity here is treated as the magnitude of the tangential velocity, so we use the non-negative root.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	(O(R/d)) candidate checks	(O(1))	Unnecessary and precision-dependent
Optimal	(O(1))	(O(1))	Accepted

Algorithm Walkthrough

Read the three floating point values (F_c), (m), and (r). They represent the centripetal force, object's mass, and radius of its circular path.
Rearrange the centripetal-force equation to isolate the squared velocity:

[ v^2=\frac{F_c r}{m}. ]

This is the key algebraic step. There is no iterative process because the unknown can be solved directly. 3. Take the square root of the resulting value:

[ v=\sqrt{\frac{F_c r}{m}}. ]

We choose the non-negative root because the requested quantity is the magnitude of the tangential velocity. 4. Print the resulting floating point value. Python's standard floating point formatting is sufficient for this problem, and for the official sample it produces 7.0710678118654755.

The invariant is simple: after computing force * radius / mass, the variable represents exactly the mathematical quantity (v^2). Applying sqrt consequently produces the required non-negative velocity, so there is no approximation loop or intermediate state that can introduce an algorithmic error.

Python Solution

import sys
import math

input = sys.stdin.readline

def solve():
    force, mass, radius = map(float, input().split())
    velocity = math.sqrt(force * radius / mass)
    print(velocity)

if __name__ == "__main__":
    solve()

The first line inside solve reads all three values as float. This is necessary because the statement explicitly allows floating point input.

The expression force * radius / mass follows directly from the rearranged formula. Multiplication is performed before division, so the code computes (F_cr/m) without introducing an integer-division issue.

math.sqrt then computes the non-negative square root. Using math.sqrt makes the intended operation explicit and avoids implementing an unnecessary numerical method.

There are no loops, arrays, or auxiliary data structures. The implementation therefore uses constant memory and performs only a constant number of arithmetic operations.

For valid physical input, the mass and radius are positive, so the division is defined. The zero-force case is naturally handled because sqrt(0.0) is 0.0.

Worked Examples

For the official sample, the input is 100 10 5. The relevant calculation proceeds as follows.

force	mass	radius	(F_cr/m)	velocity
100	10	5	50	7.0710678118654755

The intermediate value 50 is (v^2). Taking its square root gives (\sqrt{50}), which matches the official output.

For a second example, consider

18 2 8

The calculation is

[ v=\sqrt{\frac{18\cdot8}{2}} =\sqrt{72} \approx8.485281374238571. ]

force	mass	radius	(F_cr/m)	velocity
18	2	8	72	8.485281374238571

This example demonstrates that the algorithm does not require the squared velocity to be a perfect square. Floating point arithmetic handles the irrational result directly.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(1))	Only three input conversions and a fixed number of arithmetic operations are performed.
Space	(O(1))	Only three input values and one result are stored.

The published limits are 1 second and 256 MB, while the input itself contains only three floating point values. The solution is far below both limits because its running time and memory usage do not depend on an input size.

Test Cases

import sys
import io
import math

def solve():
    force, mass, radius = map(float, input().split())
    velocity = math.sqrt(force * radius / mass)
    print(velocity)

def run(inp: str) -> str:
    global input
    old_stdin = sys.stdin
    old_input = input

    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        input = old_input

# The helper above needs captured stdout.
# Use this self-contained version for the actual assertions.

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

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

    try:
        force, mass, radius = map(float, sys.stdin.readline().split())
        velocity = math.sqrt(force * radius / mass)
        print(velocity)
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample
assert run("100 10 5\n") == "7.0710678118654755\n", "sample 1"

# Zero force
assert run("0 10 5\n") == "0.0\n", "zero force"

# Exact integer result
assert run("8 2 2\n") == "2.8284271247461903\n", "non-square result"

# Decimal input
assert run("1.5 2.5 10\n") == "2.449489742783178\n", "decimal values"

# Equal values
assert run("9 9 9\n") == "3.0\n", "all equal values"

The custom cases cover the zero-force boundary, a result that is not an integer, fractional input values, and the symmetric all-equal case.

Test input	Expected output	What it validates
`0 10 5`	`0.0`	Zero-force boundary
`8 2 2`	`2.8284271247461903`	Non-perfect-square result
`1.5 2.5 10`	`2.449489742783178`	Floating point input and arithmetic
`9 9 9`	`3.0`	All values equal

Edge Cases

For zero force,

0 10 5

the algorithm computes

[ \frac{0\cdot5}{10}=0, ]

then takes sqrt(0), producing 0.0. No special branch is needed because the formula already handles the boundary correctly.

For fractional arithmetic,

1.5 2.5 10

the intermediate value is

[ \frac{1.5\cdot10}{2.5}=6, ]

so the output is

2.449489742783178

The values must be read as floating point numbers from the beginning. Treating them as integers would either reject the input or lose information.

For a result that is not an integer,

18 2 8

the intermediate value is (72), giving

8.485281374238571

A method that assumes the answer must be an integer would fail here. The direct square root calculation has no such restriction.

For the official sample,

100 10 5

the algorithm computes (100\cdot5/10=50), followed by (\sqrt{50}). The result is 7.0710678118654755, exactly the value shown by the problem's sample.