import math

def solve_n_in_one(problem_id):
    """A dispatcher for multiple elementary math and reasoning problems."""
    solutions = {
        1: lambda: print('I love Luogu!'),
        2: lambda: print(6, 4),  # A + Uim, remaining for B
        3: lambda: (print(3), print(12), print(2)),  # each, given out, returned
        4: lambda: print(f"{500 / 3:.6f}"),
        5: lambda: print((260 + 220) // (12 + 20)),
        6: lambda: print(f"{math.hypot(6, 9):.4f}"),
        7: lambda: (print(110), print(90), print(0)),
        8: lambda: print_geometry(),
        9: lambda: print(calculate_peaches()),
        10: lambda: print(solve_evaluation_machines()),
        11: lambda: print(f"{100 / (8 - 5):.4f}"),
        12: lambda: print_letters(),
        13: lambda: print(cube_edge()),
        14: lambda: print(optimal_pricing())
    }
    problem = solutions.get(problem_id)
    if problem:
        problem()

def print_geometry():
    pi = 3.141593
    r = 5
    circumference = f"{2 * pi * r:.4f}"
    area = f"{pi * r * r:.4f}"
    volume = f"{(4 / 3) * pi * r ** 3:.3f}"
    for value in (circumference, area, volume):
        print(value)

def calculate_peaches():
    remaining = 1
    for _ in range(3):
        remaining = (remaining + 1) * 2
    return remaining

def solve_evaluation_machines():
    # Using the classic "cows eating grass" variant equation:
    # Initial queue + rate * time = machines * time
    # 8 * 30 = I + 30v, 10 * 6 = I + 6v
    rate = (8 * 30 - 10 * 6) / (30 - 6)  # growth rate
    initial = 10 * 6 - rate * 6
    needed = int((initial + rate * 10) / 10)
    return needed

def print_letters():
    m_position = ord('M') - ord('A') + 1
    eighteenth = chr(ord('A') + 17)
    print(m_position, eighteenth, sep='\n')

def cube_edge():
    pi = 3.141593
    radii = [4, 10]
    combined_volume = sum((4 / 3) * pi * r ** 3 for r in radii)
    edge = combined_volume ** (1 / 3)
    return int(edge)

def optimal_pricing():
    price = 110
    enrollees = 10
    candidates = []
    while price > 0:
        revenue = price * enrollees
        if revenue >= 3500:
            # If exact target possible, record the price; stop if revenue goes below 3500 later?
            # The original logic collected ALL prices that yield exactly 3500.
            if revenue == 3500:
                candidates.append(price)
        else:
            break
        price -= 1
        enrollees += 1
    return min(candidates) if candidates else round(3500 / enrollees_estimate())

# Note on Problem 14: original problem expects two answers; take the smaller one after rounding if needed.
# Extended logic ensures minimal price that meets or exceeds 3500, but adjust per statement.

# Problem 10 (machines) can also be derived from linear equations.

Branching Structure with Aples

A situation involving consuming apples over time requires careful condition checks. Consider m apples and a per-apple eating time t. The goal is to determine how many comlpete apples remain after a given duration s.

def remaining_apples(m, t, s):
    """
    Return the count of whole apples left after s time units,
    assuming m starting apples and t time units to finish one apple.
    Several edge cases exist: t = 0, s exceeding total time, partial bites, etc.
    """
    if t == 0:
        # Instant consumption – all apples gone immediately
        return 0

    total_time_needed = m * t
    if s >= total_time_needed:
        return 0

    consumed_completely = s // t
    # If a partial apple is being eaten at the exact moment, it is not counted as remaining.
    partial = 1 if s % t != 0 else 0
    eaten = consumed_completely + partial

    remaining = m - eaten
    return max(0, remaining)

# Example usage:
# m, t, s = 10, 2, 5  -> 7 apples left

This logic avoids floating-point inaccuracies by using integer division and modulo. When t is zero, all apples vanish instantly. If the elapsed time s covers the total eating duration exactly, no apples remain.