CF 102697018 - Machines

The problem asks us to compute the mechanical advantage of several levers. For each test case, we are given two positive lengths, A and B, representing the two sides of the lever. The mechanical advantage is the ratio of these lengths, so the required value is A / B.

codeforcescompetitive-programming

CF 102697018 - Machines

Rating: -
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Solve time: 4m 15s
Verified: yes

Solution

Problem Understanding

The problem asks us to compute the mechanical advantage of several levers. For each test case, we are given two positive lengths, A and B, representing the two sides of the lever. The mechanical advantage is the ratio of these lengths, so the required value is A / B. The first input integer tells us how many lever descriptions follow, and each of the next n lines contains one pair A B. We print one real number for every pair. The official examples confirm the direct ratio, such as 10 / 5 = 2.0 and 2 / 3 = 0.6666666666666666.

The published statement does not provide explicit numerical bounds for n, A, or B, but the time limit is only 1 second and the memory limit is 256 MB. Fortunately, each test case requires only one arithmetic operation, so even a very large number of cases can be handled in linear time. There is no reason to consider quadratic or exponential algorithms here. The main implementation concern is producing the division as a floating-point value rather than accidentally performing integer division.

A simple edge case is equal sides. For input 1 followed by 5 5, the answer is 1.0. An implementation using integer division would still happen to produce 1, which is numerically correct but does not follow the floating-point output demonstrated by the problem.

A more revealing case is a smaller numerator, such as input 1 followed by 2 3. The correct output is 0.6666666666666666. A careless integer-division implementation would produce 0, losing the mechanical advantage completely.

The order of the two sides also matters. For input 1 followed by 3 2, the answer is 1.5, not 0.6666666666666666. Treating the ratio as an unordered pair or reversing the operands changes the physical meaning of the two sides.

Approaches

The most direct approach is to process every lever independently. Read A and B, calculate A / B, and immediately print the result. This is already the optimal asymptotic solution because every input pair must be read and every pair has a distinct answer that must be computed.

There is no meaningful faster algorithm hiding behind the problem. A hypothetical brute-force approach could spend additional work trying to derive the ratio through repeated subtraction or by searching for a decimal representation. Such an approach is correct in principle, but it performs unnecessary operations. For example, computing 10 / 5 through repeated subtraction takes several steps instead of one division, and for large values the number of operations grows with the magnitude of the numbers. If the input contains n pairs with large values, this can become far more expensive than the required n divisions.

The key observation is that the formula itself already gives the answer. There are no interactions between different levers, no optimization over possible configurations, and no state that needs to be carried from one test case to another. Once A and B are known, the answer is completely determined by A / B.

Thus the optimal solution simply performs one floating-point division per test case. Its running time is linear in the number of input pairs and its additional memory usage is constant.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n · max(A, B)) in a repeated-subtraction implementation O(1) Too slow for large values
Direct Division O(n) O(1) Accepted

Algorithm Walkthrough

  1. Read the number of lever configurations, n. This tells us exactly how many pairs must be processed.
  2. Repeat n times and read the two positive lengths A and B. Each pair is independent, so there is no need to store previous pairs.
  3. Compute A / B using Python's / operator. Python's / always performs floating-point division, which is exactly what the output requires.
  4. Print the resulting value. Python's standard floating-point representation matches the representation shown in the examples for these ordinary integer inputs.

Why it works

For every lever, the mechanical advantage is defined by the ratio of the two side lengths, with A as the numerator and B as the denominator. The algorithm computes exactly that ratio once for each input pair and prints it. Since each test case is independent, processing them one at a time cannot affect any other answer. Thus every printed value is the required mechanical advantage.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())

    for _ in range(n):
        a, b = map(int, input().split())
        print(a / b)

if __name__ == "__main__":
    solve()

The first line reads the number of test cases. The loop then executes exactly n times, so there is no risk of accidentally processing an extra line or skipping a pair.

The expression a / b is deliberately used instead of a // b. The former produces floating-point division, while the latter discards the fractional part. This distinction is essential for cases such as 2 3, whose answer is approximately 0.6666666666666666.

The denominator is guaranteed to be positive by the problem statement, so division by zero does not need special handling. Python integers also avoid overflow while reading the input, and the actual division is handled by Python's floating-point arithmetic.

The input is processed with sys.stdin.readline, as required for fast competitive-programming I/O. No array is necessary because each answer can be printed immediately after its corresponding pair is read.

Worked Examples

For the first sample, the algorithm processes four independent lever configurations.

Step A B A / B Output
1 10 5 2.0 2.0
2 2 3 0.6666666666666666 0.6666666666666666
3 7 5 1.4 1.4
4 18 7 2.5714285714285716 2.5714285714285716

This demonstrates that each pair is handled independently and that fractional answers are preserved rather than truncated. The sample itself gives these four outputs.

For a second example, consider two equal-sided and reversed-sided levers.

Step A B A / B Output
1 5 5 1.0 1.0
2 3 2 1.5 1.5
3 2 3 0.6666666666666666 0.6666666666666666

The trace shows why operand order must be preserved. Swapping A and B changes the answer unless the two sides have the same length.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each of the n lever configurations requires one division and one output operation.
Space O(1) Only the current pair of side lengths is stored.

The solution performs constant work per input pair, so its running time grows linearly with the number of configurations. With a 1 second limit, this is the appropriate complexity, and the constant memory usage is comfortably below the 256 MB limit specified by the problem.

Test Cases

import sys
import io

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

    for _ in range(n):
        a, b = map(int, input().split())
        print(a / b)

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

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

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

# Provided sample
assert run(
    "4\n"
    "10 5\n"
    "2 3\n"
    "7 5\n"
    "18 7\n"
) == (
    "2.0\n"
    "0.6666666666666666\n"
    "1.4\n"
    "2.5714285714285716\n"
), "sample 1"

# Minimum-size style case
assert run("1\n1 1\n") == "1.0\n", "minimum-size / equal sides"

# All values equal
assert run(
    "4\n"
    "5 5\n"
    "100 100\n"
    "999 999\n"
    "1000000 1000000\n"
) == (
    "1.0\n"
    "1.0\n"
    "1.0\n"
    "1.0\n"
), "all equal values"

# Fraction smaller than one
assert run(
    "3\n"
    "1 2\n"
    "2 3\n"
    "7 10\n"
) == (
    "0.5\n"
    "0.6666666666666666\n"
    "0.7\n"
), "fractional answers"

# Boundary-oriented large values and reversed operands
assert run(
    "4\n"
    "1000000000 1\n"
    "1 1000000000\n"
    "999999999 1000000000\n"
    "1000000000 999999999\n"
) == (
    "1000000000.0\n"
    "1e-09\n"
    "0.999999999\n"
    "1.000000001\n"
), "large values and operand order"

# Large number of test cases
large_input = "100000\n" + "7 5\n" * 100000
assert run(large_input).count("1.4\n") == 100000, "large n"
Test input Expected output What it validates
1 / 1 1.0 Minimum-size case and equal sides
Several x / x pairs 1.0 for every pair All-equal values
1 / 2, 2 / 3, 7 / 10 Fractional values Correct floating-point division
Very large and reversed pairs Values above and below 1 Large values and operand order
100000 copies of 7 5 100000 copies of 1.4 Linear processing and input handling

Edge Cases

Equal side lengths are handled directly because the division gives exactly 1.0. For the input 1 followed by 5 5, the algorithm reads A = 5 and B = 5, computes 5 / 5, and prints 1.0. No special case is necessary.

A fractional result is also handled without additional logic. For the input 1 followed by 2 3, the algorithm evaluates 2 / 3, producing 0.6666666666666666 with Python's standard floating-point representation. An integer-division implementation would incorrectly produce 0.

Operand order is preserved. For the input 1 followed by 3 2, the algorithm computes 3 / 2 = 1.5. Reversing the pair would instead produce 2 / 3, so reading the two values into the correct numerator and denominator is necessary.

Large values require no special overflow handling in Python. For 1 followed by 1000000000 1, the algorithm computes 1000000000 / 1 and prints 1000000000.0. The same constant-work approach applies regardless of the magnitude of the positive input values.

Finally, a large number of independent test cases does not require storing the entire input structure. Each pair is read, divided, and printed immediately. The memory consumption therefore remains constant even when the number of configurations is large.