如何仅使用加减运算求解自然数的平方根？技术实现咨询

Got it, let's break this down! Since we can only use addition and subtraction to find the square root of a natural number, we can lean on a clever mathematical property to make this work—no fancy built-in functions like sqrt() allowed.

Here's the key trick: the sum of the first n odd integers equals n². For example:

• 1 = 1²
• 1 + 3 = 4 = 2²
• 1 + 3 + 5 = 9 = 3²
• 1 + 3 + 5 + 7 = 16 = 4²

So to find the integer square root of a natural number x, we just keep subtracting consecutive odd numbers starting from 1 until we can't subtract anymore without going negative. The number of successful subtractions gives us the integer part of the square root.

Here's the C function implementation:

int integer_sqrt(int x) {
    // Guard clause: input is a natural number, so x < 0 is invalid
    if (x < 0) return -1;
    
    int root = 0;
    int current_odd = 1;
    
    while (x >= current_odd) {
        x -= current_odd;
        root++;
        current_odd += 2; // Move to the next odd number
    }
    
    return root;
}

How it works: #

• For x = 16: We subtract 1, 3, 5, 7 (four times) until x becomes 0. Return 4 (exact square root).
• For x = 17: After subtracting the same four odds, x is left with 1 (can't subtract the next odd, 9). Return 4 (the integer part of √17).

If you need a decimal approximation instead of just the integer part, we can extend the same logic by scaling the input. For example, to get a result accurate to 2 decimal places:

1. Multiply the input x by 10000 (since 10^4 = (10^2)^2—scaling by 100 for each decimal place, squared to keep the square root relationship).
2. Compute the integer square root of this scaled value.
3. Divide the result by 100 to convert back to a decimal.

Here's the code for this:

#include <stdio.h>

int integer_sqrt(long long x) {
    if (x < 0) return -1;
    
    int root = 0;
    long long current_odd = 1;
    
    while (x >= current_odd) {
        x -= current_odd;
        root++;
        current_odd += 2;
    }
    
    return root;
}

double sqrt_approx(int x) {
    int int_root = integer_sqrt(x);
    // Check if x is a perfect square to avoid unnecessary scaling
    if ((long long)int_root * int_root == x) {
        return (double)int_root;
    }
    
    // Scale x to get 2 decimal places of precision
    long long scaled_x = (long long)x * 10000;
    int scaled_root = integer_sqrt(scaled_x);
    
    return (double)scaled_root / 100.0;
}

// Example usage
int main() {
    printf("√16 = %.2f\n", sqrt_approx(16));  // Output: 4.00
    printf("√17 = %.2f\n", sqrt_approx(17));  // Output: 4.12
    printf("√2 = %.2f\n", sqrt_approx(2));    // Output: 1.41
    return 0;
}

Notes: #

• We switched to long long for the scaled values to avoid integer overflow with larger inputs.
• To get more decimal places, just adjust the scaling factor: for 3 decimal places, multiply by 1000000 (10^6) and divide by 1000.

• x = 0: The function returns 0, which is correct (√0 = 0).
• x = 1: Returns 1, which is exact.
• Large values: Using long long ensures we handle bigger natural numbers without overflow.

内容的提问来源于stack exchange，提问作者ProgrammingEnthusiast