CF 102697064 - Gas Cost

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Solve time: 50s
Verified: yes

Solution

Problem Understanding

The task is to calculate the fuel expense of one car trip. The input gives three floating-point values: the car's fuel efficiency in miles per gallon, the distance driven in miles, and the price of one gallon of gasoline.

If the car travels distance miles and gets mpg miles from each gallon, the amount of gasoline required is

[ \frac{\text{distance}}{\text{mpg}} ]

gallons. Multiplying that quantity by the price of one gallon gives the total trip cost:

[ \text{cost} = \frac{\text{distance}}{\text{mpg}} \times \text{price per gallon}. ]

The required output is this resulting dollar amount as a floating-point value. The official statement specifies one test case containing exactly three floating-point values.

There are no explicit numerical bounds for the three floating-point values in the published statement. That makes an asymptotic algorithm based on the magnitude of the inputs unnecessary. The entire problem consists of reading three numbers and performing two arithmetic operations, so the running time is constant regardless of their values.

A common edge case is a trip of zero distance. For example, with input 20 0 5, the car consumes zero gallons, so the correct output is 0.0. A careless implementation that multiplies the distance by the fuel efficiency instead of dividing by it would produce the wrong result.

Another useful boundary case is a fuel price of zero. For example, 20 100 0 requires five gallons, but every gallon is free, so the answer is 0.0. The formula handles this naturally.

A third case is when the fuel efficiency is not an integer. For input 7.5 30 4, the car needs 30 / 7.5 = 4 gallons, giving an answer of 16.0. An implementation that reads the values as integers would fail before it even reaches the calculation.

The published sample is 10 100 3.5. The car needs 100 / 10 = 10 gallons, and ten gallons at $3.50 each cost $35.00, so the output is 35.0.

Approaches

A literal brute-force approach would simulate the trip one unit of distance at a time and accumulate the corresponding fuel cost. If the distance were an integer D, this would require D iterations, each representing another mile driven. Its worst-case operation count would consequently be proportional to D, or O(D). That is already unnecessary for this problem, and the actual distance is a floating-point value, so there is not even a natural discrete unit over which to iterate.

The brute-force idea works only because the total cost can be viewed as the sum of the cost of every small piece of the trip. The key observation is that every mile has the same fuel cost, so there is no state that changes during the trip. We can calculate the number of gallons consumed directly instead of simulating consumption. Dividing the total distance by the miles-per-gallon value gives the exact fuel requirement in one operation, and multiplying by the gallon price gives the answer.

The entire problem can thus be reduced to the expression

[ \frac{distance \times price}{mpg}. ]

There is no need for arrays, loops, dynamic programming, graph algorithms, or numerical search.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(D) for an integer distance D	O(1)	Unnecessary and unsuitable for arbitrary floating-point distance
Direct Formula	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the three floating-point values as mpg, distance, and price.

Floating-point input is required because both fuel efficiency and gasoline price can contain fractional values. 2. Divide the distance by the fuel efficiency to calculate the number of gallons consumed.

A car that travels mpg miles per gallon needs distance / mpg gallons to travel the requested distance. 3. Multiply the number of gallons by the price per gallon.

This converts the fuel quantity into the total monetary cost of the trip. 4. Print the resulting floating-point value.

Python's standard floating-point output is sufficient for this problem and produces the sample output as 35.0.

Why it works

The fuel efficiency says that every gallon moves the car exactly mpg miles. Reversing that relationship, traveling one mile consumes 1 / mpg gallons, so traveling distance miles consumes distance / mpg gallons. Since every gallon costs the same price, multiplying the consumed gallons by that price gives exactly the total amount paid for the trip. The algorithm performs precisely these two conversions, so its result is the required gas cost.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    mpg, distance, price = map(float, input().split())

    gallons = distance / mpg
    cost = gallons * price

    print(cost)

if __name__ == "__main__":
    solve()

The first line imports sys, and input is assigned to sys.stdin.readline as requested for fast standard input. Although input size is tiny here, this is the usual competitive-programming setup.

The three values are parsed with float because the statement explicitly allows floating-point values. The variable gallons represents the physical quantity of fuel consumed, which makes the calculation easier to read and mirrors the mathematical derivation.

The division must happen before applying the price. distance / mpg has units of gallons, and multiplying that by dollars per gallon leaves dollars. Writing the calculation in this order also avoids confusing fuel efficiency with fuel consumption.

Python's float uses double-precision floating-point arithmetic. There is no integer overflow concern because the calculation never uses integer-sized accumulators, and only three values are processed.

Worked Examples

For the official sample, the input is:

10 100 3.5

The algorithm proceeds as follows.

mpg	distance	price	gallons	cost
10.0	100.0	3.5	10.0	35.0

The car travels 10 miles per gallon, so 100 miles requires 10 gallons. At $3.50 per gallon, the final cost is $35.00. This confirms that the formula matches the intended interpretation of the three input values.

A second example can exercise fractional fuel efficiency:

7.5 30 4

mpg	distance	price	gallons	cost
7.5	30.0	4.0	4.0	16.0

The car covers 7.5 miles per gallon, so 30 miles require exactly four gallons. Four gallons at four dollars each cost 16 dollars. The example demonstrates why all three values should be parsed as floating-point numbers.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only three values are read and a constant number of arithmetic operations are performed.
Space	O(1)	Only a few scalar floating-point variables are stored.

The published limits are one second and 256 MB. The solution uses constant time and constant memory, so the resource usage is negligible compared with those limits.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    input = sys.stdin.readline
    mpg, distance, price = map(float, input().split())

    gallons = distance / mpg
    cost = gallons * price

    print(cost)

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("10 100 3.5\n") == "35.0\n", "sample 1"

# minimum-style zero-distance case
assert run("1 0 10\n") == "0.0\n", "zero distance"

# all values are equal
assert run("5 5 5\n") == "5.0\n", "all equal values"

# fractional miles per gallon
assert run("7.5 30 4\n") == "16.0\n", "fractional mpg"

# zero fuel price
assert run("20 100 0\n") == "0.0\n", "free fuel"

The zero-distance case checks that no fuel is required when the trip has no length. The all-equal case checks that the formula is not accidentally using the inputs in the wrong order. The fractional-efficiency case verifies floating-point parsing and arithmetic. The zero-price case confirms that the final multiplication correctly eliminates the cost even though fuel is consumed.

Test input	Expected output	What it validates
`10 100 3.5`	`35.0`	Official sample and basic formula
`1 0 10`	`0.0`	Zero-distance boundary
`5 5 5`	`5.0`	All-equal values and operand order
`7.5 30 4`	`16.0`	Fractional fuel efficiency
`20 100 0`	`0.0`	Zero fuel price

Edge Cases

For zero distance, the exact input is 1 0 10. The algorithm calculates 0 / 1 = 0 gallons and then 0 * 10 = 0, producing 0.0. No special conditional is necessary because the formula already captures the physical situation correctly.

For a zero fuel price, the input 20 100 0 requires 100 / 20 = 5 gallons. The final multiplication is 5 * 0 = 0, so the output is 0.0. This confirms that the algorithm separates fuel consumption from fuel price correctly.

For fractional fuel efficiency, consider 7.5 30 4. The division gives 30 / 7.5 = 4 gallons, and the final multiplication gives 4 * 4 = 16. Reading the input with float is what allows 7.5 to be represented correctly.

For the official sample 10 100 3.5, the calculation is 100 / 10 = 10 gallons followed by 10 * 3.5 = 35. The printed result is 35.0, matching the official output.

The main implementation pitfall is reversing the first operation. Miles per gallon tells us how many miles one gallon can cover, so fuel consumed is distance divided by miles per gallon. Multiplying distance by miles per gallon would have the wrong units and would produce an incorrect answer.