CF 102697077 - That Hertz!

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Tags: -
Solve time: 58s
Verified: yes

Solution

Problem Understanding

The task is a direct application of the temperature-dependent speed of sound. For every test case, we are given the air temperature in degrees Celsius and the distance, in kilometers, from a speaker to the furthest edge of the crowd. We need to calculate how many seconds sound needs to travel that distance.

The speed of sound is

[ V = 331.3 + 0.606\vartheta ]

where (\vartheta) is the temperature in Celsius and (V) is measured in meters per second. Since the input distance is in kilometers, it must first be converted to meters. The travel time is then simply distance divided by speed. The required answer is printed to two decimal places. The official statement gives a one-second time limit and 256 MB of memory. It does not expose a useful numeric upper bound for the number of test cases on the problem page, but the algorithm uses only a constant number of arithmetic operations per case, so the running time is linear in the number of cases. There is no reason to use an algorithm more complicated than (O(n)).

The first edge case is a negative temperature. For example,

1
-1 1.2

gives a speed of (331.3 - 0.606 = 330.694) m/s. The distance is 1200 meters, so the answer is approximately 3.63. A solution that assumes the temperature must be positive would reject a valid input or calculate the wrong speed.

The second edge case is a fractional distance. For example,

1
7 3.6

has a distance of 3600 meters, not 3.6 meters. The sound speed is (335.542) m/s, giving a travel time of about 10.73 seconds. Forgetting the kilometer-to-meter conversion produces an answer that is smaller by a factor of 1000.

The third edge case is a result that needs rounding. For example,

1
24 1.1

gives a speed of (345.844) m/s and a travel time of approximately 3.18068 seconds, so the required output is 3.18. A program that prints the raw floating-point value does not follow the required output format.

The fourth edge case is zero distance, if it is allowed by the input range. For example,

1
20 0

requires no travel at all, so the answer is 0.00. The formula handles this naturally because the numerator is zero.

Approaches

A literal brute-force solution could simulate the sound travelling toward the crowd, advancing its position repeatedly until the target distance is reached. If the distance is (D) kilometers and the speed is (V) meters per second, such a simulation would require roughly (1000D/V) iterations when advancing by one second, or even more iterations if it uses smaller time increments to obtain the required precision. The exact worst-case count depends on the maximum distance and temperature, and the published problem page does not provide those numeric bounds. More importantly, the simulation throws away the central fact that the speed is constant for a test case, so every iteration repeats work that can be replaced by one division.

The direct mathematical approach is both simpler and faster. The brute-force simulation works because after each time interval we know how far the sound has travelled. Since the speed does not change during a test case, the total travel time is exactly distance divided by speed. The observation that the physical model already gives a constant velocity lets us reduce the entire simulation to a few arithmetic operations.

For each test case, compute the speed from the temperature, convert kilometers to meters, divide the distance by the speed, and print the result with two digits after the decimal point. There is no hidden search, graph traversal, dynamic programming state, or iterative approximation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force simulation	(O(D/V)) iterations per case	(O(1))	Too slow and unnecessary
Direct formula	(O(n))	(O(n)) for stored output, or (O(1)) extra	Accepted

Algorithm Walkthrough

Read the number of test cases. Each following line describes one independent temperature and distance pair.
Read the temperature (\vartheta) and the distance (d) in kilometers. The temperature can be negative, so it should be treated as an ordinary signed number.
Calculate the speed of sound using

[ V = 331.3 + 0.606\vartheta. ]

This is the only physical formula needed by the problem.

Convert the distance from kilometers to meters by multiplying it by 1000. This is necessary because the speed is measured in meters per second.
Calculate the travel time with

[ t = \frac{1000d}{V}. ]

The units now cancel correctly: meters divided by meters per second gives seconds.

Print (t) with exactly two digits after the decimal point. Python's floating-point formatting performs the required rounding for ordinary contest inputs.

Why it works

For every test case, the speed calculated from the temperature is constant during the sound's journey. At constant speed, the fundamental relation is (d = Vt), so rearranging gives (t=d/V). The algorithm uses exactly this relation after converting the input distance into the same unit used by the speed formula. Since every computed value represents the physical travel time for that test case, formatting it to two decimal places produces the required answer.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    ans = []

    for _ in range(n):
        temperature, distance_km = input().split()
        temperature = int(temperature)
        distance_km = float(distance_km)

        speed = 331.3 + 0.606 * temperature
        distance_m = distance_km * 1000.0
        time = distance_m / speed

        ans.append(f"{time:.2f}")

    sys.stdout.write("\n".join(ans))

if __name__ == "__main__":
    solve()

The first conversion reads the temperature as an integer because the statement specifies an integer temperature. The distance is read as a floating-point value because examples such as 3.6 and 1.1 are valid.

The speed calculation follows the given formula directly. Keeping the constants as floating-point values avoids accidental integer arithmetic.

The multiplication by 1000.0 is the unit conversion from kilometers to meters. It must happen before the division because the speed is expressed in meters per second.

The expression f"{time:.2f}" both rounds the value to two decimal places and guarantees that trailing zeroes are printed. For example, 3.1 becomes 3.10, which is required by the output format.

The answers are accumulated and written once at the end. This avoids repeatedly calling print for large numbers of test cases and keeps the I/O overhead small.

Python integers and floating-point values are more than sufficient for the arithmetic involved here. There is no integer overflow issue in Python, and the calculation itself needs only ordinary double-precision floating-point accuracy.

Worked Examples

The first sample is:

For the first test case, the temperature is (7), so the sound speed is (331.3 + 0.606 \times 7 = 335.542) m/s. The distance is (3600) meters, giving a time of approximately (10.73) seconds.

Temperature	Distance km	Speed m/s	Distance m	Time s	Output
7	3.6	335.542	3600	10.73	`10.73`
24	1.1	345.844	1100	3.18	`3.18`
17	4.4	341.602	4400	12.88	`12.88`
-1	1.2	330.694	1200	3.63	`3.63`
8	5.9	336.148	5900	17.55	`17.55`

This trace exercises the main calculation, fractional distances, negative temperature, and two-decimal rounding. It also shows why the unit conversion cannot be skipped. The official sample output is 10.73, 3.18, 12.88, 3.63, and 17.55.

A smaller example focusing on a zero distance is:

2
20 0
0 1

Temperature	Distance km	Speed m/s	Distance m	Time s	Output
20	0	343.42	0	0.00	`0.00`
0	1	331.3	1000	3.0184...	`3.02`

The first row demonstrates that a zero distance needs no special branch. The formula naturally produces zero. The second row checks the unit conversion at exactly one kilometer and also exercises rounding upward from approximately 3.0184 to 3.02.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(n))	Each of the (n) test cases uses a constant number of arithmetic operations.
Space	(O(n))	The implementation stores the formatted answers before writing them.

The one-second limit is easily handled because there is no iteration depending on the physical distance or required precision. Even if the number of test cases is large, the work grows linearly, with only a few arithmetic operations per case. The memory usage can also be reduced to (O(1)) extra space by printing each result immediately, but storing the output is already small and is convenient for fast I/O. The official memory limit is 256 MB.

Test Cases

import sys
import io

def solve_data(inp: str) -> str:
    data = inp.split()
    it = iter(data)

    n = int(next(it))
    out = []

    for _ in range(n):
        temperature = int(next(it))
        distance_km = float(next(it))

        speed = 331.3 + 0.606 * temperature
        distance_m = distance_km * 1000.0
        time = distance_m / speed

        out.append(f"{time:.2f}")

    return "\n".join(out)

# Provided sample
assert solve_data(
    """5
7 3.6
24 1.1
17 4.4
-1 1.2
8 5.9
"""
) == """10.73
3.18
12.88
3.63
17.55""", "sample"

# Minimum-size style case
assert solve_data(
    """1
0 0
"""
) == "0.00", "zero distance"

# All values use the same temperature and distance
assert solve_data(
    """3
20 1
20 1
20 1
"""
) == """2.91
2.91
2.91""", "all equal values"

# Negative temperature and fractional distance
assert solve_data(
    """1
-1 1.2
"""
) == "3.63", "negative temperature"

# Boundary around a one-kilometer distance
assert solve_data(
    """2
0 1
10 1
"""
) == """3.02
2.96""", "unit conversion and temperature effect"

# Large number of test cases
assert solve_data(
    "1000\n" + "\n".join(["20 1"] * 1000) + "\n"
) == "\n".join(["2.91"] * 1000), "many test cases"

Test input	Expected output	What it validates
`1 / 0 0`	`0.00`	Zero distance and required trailing zeroes
Three identical `20 1` cases	Three times `2.91`	Repeated identical inputs and independent test cases
`-1 1.2`	`3.63`	Negative temperature and fractional distance
`0 1`, `10 1`	`3.02`, `2.96`	Kilometer-to-meter conversion and temperature-dependent speed
1000 identical cases	1000 lines of `2.91`	Linear processing and output handling

Edge Cases

For a negative temperature, consider

1
-1 1.2

The algorithm computes the speed as (331.3 + 0.606(-1)=330.694) m/s. It converts 1.2 km to 1200 m, then calculates (1200/330.694), which is approximately 3.63. Nothing in the calculation assumes a nonnegative temperature, so the case is handled directly.

For a fractional distance, consider

1
7 3.6

The speed is (335.542) m/s. The distance becomes (3600) m, and (3600/335.542) is approximately 10.73. The multiplication by 1000 is the critical operation. If a solution divides 3.6 directly by the speed, it mixes kilometers with meters per second and produces a result 1000 times too small.

For zero distance, consider

1
20 0

The speed is (343.42) m/s and the converted distance is zero. The division gives exactly 0.0, which is formatted as 0.00. No special case is needed.

For rounding, consider

1
0 1

The speed is (331.3) m/s and the distance is (1000) m. The time is approximately 3.0184, which must be printed as 3.02. Formatting with .2f handles this automatically.

For repeated test cases, consider

Each case is processed independently. Every speed is (343.42) m/s, every distance is (1000) m, and every result is 2.91. There is no state carried from one test case to the next, so the same input pair correctly produces the same output each time.