基于有限内置函数的递变资产购买量计算逻辑开发需求

Got it, let's break this down step by step. You need to implement two versions of a Buy() function—one for assets that increase in price by a fixed percentage per unit purchased, and another for assets that decrease in price by a fixed percentage per unit. Since you're working with a simple language with limited built-in math functions, we'll cover both an iterative approach (super compatible, no fancy math needed) and a formula-based approach (if your language has basic log/power functions).

First, let's define some shared variables we'll reference in both implementations:

• base_price: The cost of the first asset unit
• rate: The percentage change per unit (use increase_rate for price hikes, e.g., 0.05 for 5%; use discount_rate for price drops, e.g., 0.1 for 10%)

Scenario: #

Each subsequent asset costs (1 + increase_rate) times the previous one. First unit = base_price, second = base_price*(1+increase_rate), third = base_price*(1+increase_rate)^2, etc. We need to find the maximum number of units we can buy without exceeding the given budget.

Iterative Implementation (Best for Limited Math Functions) #

This approach just accumulates costs until adding the next unit would go over budget—no need for logs or powers. Perfect for simple languages:

function Buy(budget):
    # Edge case: can't even afford the first unit
    if budget < base_price:
        return 0
    
    count = 0
    total_spent = 0
    current_unit_price = base_price
    increase_rate = 0.05  # Replace with your fixed percentage (e.g., 5% = 0.05)
    
    while total_spent + current_unit_price <= budget:
        total_spent += current_unit_price
        count += 1
        # Update price for next unit
        current_unit_price = current_unit_price * (1 + increase_rate)
    
    return count

Formula-Based Implementation (If Your Language Supports Logs/Powers) #

We use the sum of a geometric series to calculate the maximum n directly. The sum of n increasing units is:
Total Cost = base_price * [(1 + increase_rate)^n - 1] / increase_rate

Solving for n gives us:
n ≤ log( (budget * increase_rate / base_price) + 1 ) / log(1 + increase_rate)

We still need to verify for floating-point errors:

function Buy(budget):
    if budget < base_price:
        return 0
    
    P0 = base_price
    r = 0.05  # Your fixed increase rate
    if r == 0:
        # No price change—just divide budget by base price
        return budget // P0  # Use integer division if available
    
    numerator = (budget * r) / P0 + 1
    if numerator <= 1:
        return 0
    
    # Calculate n using logarithms
    n = log(numerator) / log(1 + r)
    count = floor(n)  # Round down to integer
    
    # Verify to fix floating-point inaccuracies
    calculated_total = P0 * ((1 + r)**count - 1) / r
    if calculated_total > budget:
        count -= 1
    elif calculated_total + P0*(1 + r)**count <= budget:
        count += 1
    
    return count

Scenario: #

Each subsequent asset costs (1 - discount_rate) times the previous one. First unit = base_price, second = base_price*(1-discount_rate), third = base_price*(1-discount_rate)^2, etc. Again, find the maximum units for the budget.

Iterative Implementation (Best for Limited Math Functions) #

Similar to the increasing case, but we reduce the price each time. Note the edge case where a 100% discount makes all subsequent units free:

function Buy(budget):
    if budget < base_price:
        return 0
    
    count = 0
    total_spent = 0
    current_unit_price = base_price
    discount_rate = 0.1  # Replace with your fixed discount (e.g., 10% = 0.1)
    
    while total_spent + current_unit_price <= budget:
        total_spent += current_unit_price
        count += 1
        
        # Update price for next unit
        current_unit_price = current_unit_price * (1 - discount_rate)
        
        # Edge case: 100% discount means all remaining units are free
        if discount_rate == 1.0:
            # You can buy as many as you want (since they cost nothing)
            # Return a large integer if your language doesn't support "infinite"
            return count + 999999
    
    return count

Formula-Based Implementation (If Your Language Supports Logs/Powers) #

The sum of n decreasing units is:
Total Cost = base_price * [1 - (1 - discount_rate)^n] / discount_rate

If the budget is large enough to cover the infinite sum (since prices approach 0), you can buy effectively unlimited units. Otherwise, solve for n:
n ≤ log(1 - (budget * discount_rate / base_price)) / log(1 - discount_rate)

Here's the code:

function Buy(budget):
    P0 = base_price
    d = 0.1  # Your fixed discount rate
    if budget < P0:
        return 0
    
    if d == 0:
        return budget // P0
    
    # Check if budget covers the infinite sum (prices approach 0)
    infinite_total = P0 / d
    if budget >= infinite_total:
        return 999999  # Replace with max integer for your language
    
    right_side = 1 - (budget * d) / P0
    if right_side <= 0:
        return 999999
    
    n = log(right_side) / log(1 - d)
    count = floor(n)
    
    # Verify for floating-point errors
    calculated_total = P0 * (1 - (1 - d)**count) / d
    if calculated_total > budget:
        count -= 1
    elif calculated_total + P0*(1 - d)**count <= budget:
        count += 1
    
    return count

Key Notes for Simple Languages: #

• If your language doesn't have floor(), use integer division (e.g., count = int(n) then adjust if needed)
• The iterative method is almost always safer for limited environments—it avoids relying on advanced math functions and floating-point precision issues
• Make sure to test edge cases: budget < base price, 0% rate (fixed price), 100% discount (if applicable)

内容的提问来源于stack exchange，提问作者Blue Blue