无小数支持编程语言中缩放算法的适配实现问询

Alright, let's break down how to adapt your two real-number formulas to a language that only supports integers but allows precision control. The core approach here is fixed-point arithmetic: we'll scale all values by a fixed precision factor (let's call it SCALE) to represent decimals as integers, then rewrite every operation in the formulas to use integer arithmetic only.

Step 1: Define Your Precision Factor (SCALE) #

First, pick a SCALE value based on your required precision and value ranges. For example:

• Use SCALE = 1000 for 3 decimal places of precision.
• Use SCALE = 10000 for 4 decimal places.

All input variables (c, s, a, T, E, b) will be converted to integers by multiplying them by SCALE (e.g., 0.1 becomes 100 if SCALE=1000). We'll denote these scaled integers as c_scaled, s_scaled, etc.

Step 2: Adapt the First Formula #

Original formula:
$$\frac{\sqrt{c^2 + 2sa + s^2T2 + 2scT }- c}{s}-T$$

First, simplify the radicand (the part under the square root) to reduce computation and avoid overflow:
$$c^2 + 2sa + s^2T2 + 2scT = (c + sT)^2 + 2sa$$

Integer Arithmetic Rewrite #

We'll translate this step-by-step to integer operations (all variables are scaled integers):

1. Calculate the term inside the squared parentheses:

   term1 = c_scaled * SCALE + s_scaled * T_scaled
   

   (This represents (c + s*T) * SCALE² to keep the math aligned for integer operations.)

2. Compute the full radicand (scaled to SCALE^4 to avoid decimals):

   radicand = term1 * term1 + 2 * s_scaled * a_scaled * SCALE * SCALE
   

3. Compute the integer square root of radicand. You'll need an integer sqrt function (use built-in if available, or implement Newton-Raphson/binary search for large numbers):

   sqrt_radicand = integer_sqrt(radicand)
   

   (This gives us sqrt(radicand) = sqrt((c+sT)² + 2sa) * SCALE².)

4. Calculate the final scaled result:

   numerator = sqrt_radicand - c_scaled * SCALE - s_scaled * T_scaled
   result_scaled = numerator / s_scaled
   

   (If you need rounded division instead of truncation, use (numerator + s_scaled / 2) / s_scaled.)

Step 3: Adapt the Second Formula #

Original formula:
$$E -\frac{1}{2}sT^2 + sTb + \frac{1}{2}sb^2-cT + cb$$

Integer Arithmetic Rewrite #

To eliminate the 1/2 terms, multiply all terms by 2 first, then divide by 2 at the end (avoids floating-point operations):

1. Compute all terms scaled to SCALE^3 (to maintain precision through multiplications):

   term_E = 2 * E_scaled * SCALE * SCALE
   term_sT2 = -s_scaled * T_scaled * T_scaled
   term_sTb = 2 * s_scaled * T_scaled * b_scaled
   term_sb2 = s_scaled * b_scaled * b_scaled
   term_cT = -2 * c_scaled * T_scaled * SCALE
   term_cb = 2 * c_scaled * b_scaled * SCALE
   

2. Sum all terms, then divide by 2 * SCALE * SCALE to get the final scaled result:

   total = term_E + term_sT2 + term_sTb + term_sb2 + term_cT + term_cb
   result_scaled = total / (2 * SCALE * SCALE)
   

   (Again, use (total + (2 * SCALE * SCALE) / 2) / (2 * SCALE * SCALE) for rounded division.)

Critical Implementation Notes #

• Avoid Overflow: Use the largest integer type available (e.g., 64-bit integers like int64_t in C/C++, long long in Java) to handle large intermediate values from multiplications.
• Validate Radicand: Before computing the square root, check that radicand >= 0 to avoid undefined behavior.
• Precision Adjustments: If your language supports it, use rounding instead of truncation for divisions to minimize precision loss.
• Test with Known Values: Validate your implementation using test cases where you know the real-number result (e.g., input c=1, s=1, a=1, T=1 and compare the scaled result to the expected real value multiplied by SCALE).

内容的提问来源于stack exchange，提问作者PloxCake