CF 102697045 - Planet Omicron Persei VIII

The problem gives three physical quantities describing an object falling on Planet Omicron Persei VIII. The first value is the gravitational force (Fg) acting on a person, the second is that person's mass (M), and the third is the distance (D) through which the object falls.

codeforcescompetitive-programming

CF 102697045 - Planet Omicron Persei VIII

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Solve time: 49s
Verified: yes

Solution

Problem Understanding

The problem gives three physical quantities describing an object falling on Planet Omicron Persei VIII. The first value is the gravitational force (F_g) acting on a person, the second is that person's mass (M), and the third is the distance (D) through which the object falls. The object starts from rest, so its initial velocity is zero. We need to compute the fall time (T), then print it together with the distance, with both numerical values shown to one decimal place. The original problem specifies a 1 second time limit and 256 MB memory limit.

The first equation gives the gravitational acceleration directly:

[ F_g = MG ]

so

[ G = \frac{F_g}{M}. ]

The second equation describes the object's displacement:

[ D = V_iT + \frac12GT^2. ]

Because the object is released from rest, (V_i=0). The equation becomes

[ D = \frac12GT^2. ]

Solving for positive time gives

[ T = \sqrt{\frac{2D}{G}}. ]

Substituting (G=F_g/M) gives the most convenient form for the implementation:

[ T = \sqrt{\frac{2DM}{F_g}}. ]

There are no explicit numeric bounds for (F_g), (M), or (D) in the published statement. That means there is no meaningful input-size complexity issue here. The algorithm performs a fixed number of arithmetic operations and one square root, so it easily fits the one-second limit.

A subtle edge case is that the initial velocity must be treated as zero. For example, with input

500
70
10

the correct output is

It will take the object 1.7 seconds to fall 10.0 meters

because the object starts from rest. A careless implementation that keeps an unspecified (V_i) as a nonzero value would solve a different physical situation. This is also the sample supplied by the problem.

Another common mistake is formatting only the calculated time. For input

100
50
5

we have (G=2) and (T=\sqrt{5}\approx2.236), so the required output is

It will take the object 2.2 seconds to fall 5.0 meters

The distance itself must also be printed with one digit after the decimal point. Printing 5 instead of 5.0 does not match the required format.

A third edge case is a distance that is already very small. For example,

10
2
0.1

gives (G=5) and (T=\sqrt{0.04}=0.2), so the output is

It will take the object 0.2 seconds to fall 0.1 meters

A careless implementation that truncates instead of rounding can fail on values that lie between tenths.

Approaches

A brute-force way to think about the problem is to simulate possible times in increments of one tenth of a second. For each candidate (t), we could calculate the distance

[ D(t)=\frac12Gt^2 ]

and stop when the candidate reaches the required distance. This is correct if the candidates are checked in increasing order and the result is rounded appropriately. If the true answer is (T), however, this requires roughly (10T) checks. More precisely, checking (0.0,0.1,\ldots,\lceil10T\rceil/10) requires (\lceil10T\rceil+1) candidate evaluations. Since the statement provides no upper bound on (T), there is no finite worst-case operation count for this simulation. Even if a bound existed, numerical searching would be unnecessary for a problem whose equation can be solved exactly.

The brute-force approach works because the distance is a monotonic function of time, but it fails because it repeatedly evaluates an equation whose unknown can be isolated algebraically. The key observation is that the object starts from rest, so the velocity term disappears completely. Once (V_i=0), the remaining equation contains only (T^2), allowing us to solve for (T) directly.

First calculate the gravitational acceleration from the force and mass. Then substitute that acceleration into the falling-distance equation and isolate the positive square root. This reduces the entire problem to constant-time arithmetic.

Approach Time Complexity Space Complexity Verdict
Brute Force (O(T)) candidate checks (O(1)) Too slow in principle
Optimal (O(1)) (O(1)) Accepted

Algorithm Walkthrough

  1. Read (F_g), (M), and (D). These are the gravitational force, the mass used to determine the planet's gravitational acceleration, and the falling distance.
  2. Compute the gravitational acceleration as

[ G=\frac{F_g}{M}. ]

This follows directly from (F_g=MG), so there is no need to know the planet's gravity beforehand. 3. Set the initial velocity to zero because the object is released from rest. The general equation (D=V_iT+\frac12GT^2) therefore reduces to

[ D=\frac12GT^2. ] 4. Rearrange the equation to obtain

[ T^2=\frac{2D}{G}. ]

Time cannot be negative, so we take the positive square root. 5. Substitute the expression for (G) and calculate

[ T=\sqrt{\frac{2DM}{F_g}}. ]

This avoids an unnecessary intermediate variable and makes the implementation shorter. 6. Print the required sentence using one decimal place for both (T) and (D). Python's floating-point formatting with .1f performs the required decimal rounding.

The key invariant is that every calculated value of (T) is obtained from the exact physical equation after applying the condition (V_i=0). The computed acceleration satisfies (F_g=MG), and the resulting time satisfies (D=\frac12GT^2). Since the physical time is nonnegative, choosing the positive square root gives the unique valid solution.

Python Solution

import sys
import math

input = sys.stdin.readline

def solve():
    fg = float(input())
    mass = float(input())
    distance = float(input())

    time = math.sqrt(2.0 * distance * mass / fg)

    print(
        f"It will take the object {time:.1f} seconds "
        f"to fall {distance:.1f} meters"
    )

if __name__ == "__main__":
    solve()

The first three reads correspond directly to the three physical quantities in the input. They are parsed as floating-point numbers because the statement does not restrict the values to integers.

The expression 2.0 * distance * mass / fg is exactly the rearranged formula (2DM/F_g). Using math.sqrt then produces the positive solution for (T).

The output uses {time:.1f} and {distance:.1f}. The first formatting operation rounds the calculated fall time to one decimal place, while the second guarantees that the supplied distance appears in the required format even when the input was an integer such as 10.

There is no loop, recursion, array, or other data structure. The calculation uses constant memory and performs only a fixed amount of work.

Worked Examples

For the first example, the input is

500
70
10

The algorithm proceeds as follows.

(F_g) (M) (D) (G=F_g/M) (T=\sqrt{2DM/F_g}) Printed (T)
500 70 10 7.142857... 1.673320... 1.7

The gravitational acceleration is approximately (7.1429\text{ m/s}^2). Substituting it into the falling-distance equation gives a time of approximately (1.6733) seconds, which rounds to (1.7). The distance is printed as 10.0.

The resulting output is

It will take the object 1.7 seconds to fall 10.0 meters

This example exercises the main formula and also confirms that both numerical fields need one decimal place.

For a second example, consider

100
50
5

The calculation is

(F_g) (M) (D) (G=F_g/M) (2DM/F_g) (T) Printed (T)
100 50 5 2 5 2.236067... 2.2

Here the gravitational acceleration is exactly (2\text{ m/s}^2). The resulting time is (\sqrt5), approximately (2.2361) seconds, so one-decimal formatting produces 2.2.

The output is

It will take the object 2.2 seconds to fall 5.0 meters

This trace demonstrates that the program must round the mathematical result rather than truncate it to an integer.

Complexity Analysis

Measure Complexity Explanation
Time (O(1)) A fixed number of arithmetic operations and one square root are performed.
Space (O(1)) Only three input values and a few scalar intermediate values are stored.

The published limits are 1 second and 256 MB, while the problem contains only three input values and requires a constant-size calculation. The solution is far below both limits.

Test Cases

import sys
import io
import math

def solve():
    input = sys.stdin.readline

    fg = float(input())
    mass = float(input())
    distance = float(input())

    time = math.sqrt(2.0 * distance * mass / fg)

    print(
        f"It will take the object {time:.1f} seconds "
        f"to fall {distance:.1f} meters"
    )

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

    try:
        sys.stdin = io.StringIO(inp)
        sys.stdout = io.StringIO()
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample
assert run("500\n70\n10\n") == (
    "It will take the object 1.7 seconds to fall 10.0 meters\n"
), "sample 1"

# Minimum-style positive values
assert run("1\n1\n1\n") == (
    "It will take the object 1.4 seconds to fall 1.0 meters\n"
), "basic positive values"

# All quantities chosen so that the time is exactly 0.2
assert run("10\n2\n0.1\n") == (
    "It will take the object 0.2 seconds to fall 0.1 meters\n"
), "small distance"

# Exact integer time
assert run("100\n50\n8\n") == (
    "It will take the object 2.8 seconds to fall 8.0 meters\n"
), "non-integer rounded time"

# Large values, useful for checking floating-point arithmetic
assert run("1000000000000\n1000000000000\n1000000000000\n") == (
    "It will take the object 1414213.6 seconds to fall 1000000000000.0 meters\n"
), "large values"
Test input Expected output What it validates
500 / 70 / 10 1.7 seconds, 10.0 meters Provided sample and required formatting
1 / 1 / 1 1.4 seconds, 1.0 meters Small positive values
10 / 2 / 0.1 0.2 seconds, 0.1 meters Small distance and decimal input
100 / 50 / 8 2.8 seconds, 8.0 meters Correct square-root calculation and rounding
10^12 / 10^12 / 10^12 1414213.6 seconds, 1000000000000.0 meters Large values and floating-point handling

The published statement does not provide a formal numeric maximum for the three inputs, so the final test uses large values rather than claiming a particular official maximum.

Edge Cases

The first edge case is the initial velocity. With

500
70
10

the object is explicitly released from rest, so (V_i=0). The algorithm removes the entire (V_iT) term before solving the equation. Keeping that term with an arbitrary nonzero velocity would produce a physically different answer.

The second edge case concerns output formatting. For

100
50
5

the computed time is approximately (2.2360679). The required result is 2.2, not 2 and not 2.3. The :.1f format specifier performs the rounding while also guaranteeing exactly one digit after the decimal point. The distance is formatted separately, producing 5.0.

The third edge case is a fractional distance:

10
2
0.1

Here

[ T=\sqrt{\frac{2(0.1)(2)}{10}} =\sqrt{0.04} =0.2. ]

The program prints

It will take the object 0.2 seconds to fall 0.1 meters

This catches implementations that assume the distance is an integer.

Finally, large numerical values do not change the algorithm. For

1000000000000
1000000000000
1000000000000

the mass and gravitational force cancel in the formula, leaving (T=\sqrt{2\cdot10^{12}}), approximately (1414213.562). The program prints 1414213.6. The calculation still uses only constant memory and constant time, regardless of the magnitude of the input values.