CF 102697089 - The Model

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Tags: -
Solve time: 45s
Verified: yes

Solution

Problem Understanding

The problem models the process of winding a model airplane's rubber band. One complete turn of the crank winder produces a fixed number of winds in the rubber band. The first input value, (n), is the total number of winds we want, while the second value, (m), tells us how many winds are produced by one turn of the winder. We need to determine how many complete turns of the winder are required to reach exactly (n) winds. The statement guarantees that this number is an integer.

Mathematically, if every turn produces (m) winds and we make (w) turns, the rubber band receives (w \cdot m) winds. We need

[ w \cdot m = n, ]

so the required number of turns is simply

[ w = \frac{n}{m}. ]

The official limits give a one-second time limit and 256 MB of memory, but the input values themselves are only described as positive integers with the divisibility condition. Since the entire task reduces to one arithmetic operation, even extremely large integer values would not create an algorithmic problem. The only practical consideration in Python is that the input should be parsed as integers rather than floating-point values, because the answer is guaranteed to be an exact integer.

There are two small edge cases worth checking. First, the answer can be exactly one. For example,

64
64

requires exactly one turn, so the output is

An implementation that accidentally assumes at least two turns would fail here.

Second, the required number of turns can be much larger than one. For example,

100
1

requires 100 turns, so the output is

A solution that reverses the division and computes (m/n) would produce a value smaller than one instead of the required number of turns. The sample 2250 15 similarly gives 150 turns, confirming that the desired quantity is total winds divided by winds produced per turn.

Approaches

A direct simulation would start with zero winds and repeatedly turn the winder, adding (m) winds after every turn, until the total reaches (n). Because the input guarantees that (n) is divisible by (m), this process eventually stops exactly at (n), so it is correct. If (n/m=k), however, the simulation performs exactly (k) iterations. In the worst case, with (m=1), it performs (n) iterations. Since the statement does not give a small upper bound on (n), this is unnecessary work and could become too slow for a sufficiently large input.

The key observation is that every turn contributes exactly the same amount, so there is no changing state to simulate. If (k) turns produce (n) winds, then (k m=n). Solving this equation directly gives (k=n/m). The brute-force method works because it repeatedly performs the same addition, while the optimal method replaces all those identical additions with one division.

The two approaches are therefore:

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n / m), worst case O(n)	O(1)	Too slow for unnecessarily large inputs
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read (n), the total number of winds required.
Read (m), the number of winds produced by one complete turn of the winder.
Compute n // m. The divisibility guarantee means there is no remainder, so integer division gives the exact number of turns rather than an approximation.
Print the result.

Why it works

Suppose the winder is turned (w) times. Each turn contributes exactly (m) winds, so the total number of winds is (w m). We need this total to equal (n), giving (w m=n). Since (m>0), dividing both sides by (m) gives (w=n/m). The input guarantees that this quotient is an integer, so the algorithm prints exactly the required number of turns.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
m = int(input())

print(n // m)

The first two lines read the two positive integers directly from standard input. Using int keeps the calculation exact, which is preferable to floating-point division because the required answer is an integer.

The expression n // m performs integer division. Here it is exactly the mathematical quotient because the problem guarantees that the number of winds requested is divisible by the number of winds produced by one turn. Using / would create a floating-point value such as 150.0, which the original checker may also accept according to the statement, but integer division expresses the required result more precisely.

There are no loops, arrays, recursion, or auxiliary data structures. The order of the inputs also matters: n is the total desired number of winds, while m is the number of winds generated by one turn, so the division must be n // m, not m // n.

Worked Examples

Sample 1

Input:

2250
15

The algorithm starts with n = 2250 and m = 15.

Variable	Value
`n`	2250
`m`	15
`n // m`	150
Output	150

The result means that 150 turns of the winder produce (150 \times 15 = 2250) winds. This is exactly the required number.

Sample 2

Input:

64
4

The algorithm reads n = 64 and m = 4, then divides the total number of required winds by the number generated per turn.

Variable	Value
`n`	64
`m`	4
`n // m`	16
Output	16

Sixteen turns produce (16 \times 4 = 64) winds, so the answer is exactly 16.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only two integers are read and one division is performed
Space	O(1)	Only the two input integers and the result are stored

The solution easily fits the one-second time limit and 256 MB memory limit because its running time and auxiliary memory do not depend on the magnitude of the input values.

Test Cases

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

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

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

        input = sys.stdin.readline

        n = int(input())
        m = int(input())
        print(n // m)

        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# provided samples
assert run("2250\n15\n") == "150\n", "sample 1"
assert run("64\n4\n") == "16\n", "sample 2"

# custom cases
assert run("1\n1\n") == "1\n", "minimum-size values"
assert run("100\n1\n") == "100\n", "one wind per turn"
assert run("1000000000\n1000\n") == "1000000\n", "large values"
assert run("72\n8\n") == "9\n", "exact division boundary"

Test input	Expected output	What it validates
`1 / 1`	`1`	Smallest positive values and the one-turn case
`100 / 1`	`100`	Large quotient and the worst case for a repeated-addition approach
`1000000000 / 1000`	`1000000`	Large integer values without floating-point arithmetic
`72 / 8`	`9`	Exact divisibility and correct direction of division

Edge Cases

For the one-turn case, consider:

64
64

The algorithm computes 64 // 64 = 1 and prints 1. This works because the entire required winding can be completed by a single turn. A simulation also works, but there is no reason to perform any iteration when the quotient is already available directly.

For the one-wind-per-turn case, consider:

100
1

The algorithm computes 100 // 1 = 100 and prints 100. A repeated-addition solution would need 100 iterations here, while the optimal solution performs one arithmetic operation. This is also the worst possible case for the simulation when (m) is allowed to be as small as 1.

For a large-value case, consider:

1000000000
1000

The algorithm computes

[ 1000000000 / 1000 = 1000000 ]

and prints:

The calculation remains exact because Python integers have arbitrary precision, and no floating-point conversion is involved.

Finally, the order of the inputs can be checked with:

72
8

The correct result is 9, because each turn contributes 8 winds and nine turns give (9 \times 8=72). Reversing the operands would be logically incorrect, so this case catches implementations that confuse the desired total with the per-turn rate.