java - 可以计算 2 个区间之间的数字的最大平方根次数

Question

我写了一个简单的程序来计算一个数字可以计算平方根的最大次数，输入是从num1到num2的区间例如:如果输入是 (1,20)，则答案为 2，因为 16 的平方根是 4 ，而 4 的平方根是 2 。

 int max = 0;
    for (int i = num1; i <= num2; i++) {
        boolean loop = true;
        int count = 0;
        int current = i;
        if (i == 1) {
            count++;
        } else {
            while (loop) {
                double squareRoot = Math.sqrt(current);
                if (isCurrentNumberPerfectSquare(squareRoot)) {
                    count++;
                    current = (int) squareRoot;
                } else {
                    loop = false;
                }
            }
        }

        if (count > max) {
            max = count;
        }
    }
    return max;


static boolean isCurrentNumberPerfectSquare(double number) {
    return ((number - floor(number)) == 0);
}

我得到了答案，但想知道是否可以使用某种数学方法来改进？有什么建议吗？

Answer 1

为了避免更多困惑，我在此给​​出对这个主题的最终答案。前面提到的两种方法的组合。

“Parameswar”正在寻找的是由最低底形成的最大完美正方形。

Step 1 -
To get that calculate the largest possible perfect square based on your num2 value.
If it is outside your range, you have no perfect square within.

Step 2 -
If it is within your range, you have to check all perfect square formed by a lower base value with a higher number of times.

Step 3 -
If you find one that is within your range, replace your result with the new result and proceed to check lower values. (go back to Step 2)

Step 4 -
Once the value you check is <= 2 you have already found the answer.

这里有一些示例实现:

    static class Result {
        int base;
        int times;
    }

    static boolean isCurrentNumberPerfectSquare(double number) {
        return ((number - Math.floor(number)) == 0);
    }

    private static int perfectSquare(int base, int times) {

        int value = base;
        for (int i = times; i > 0; i--) {
            value = (int) Math.pow(base, 2);
        }
        return value;
    }

    private static Result calculatePerfectSquare(int perfectSquare) {

        Result result = new Result();
        result.base = (int) Math.sqrt(perfectSquare);
        result.times = 1;

        while (result.base > 2 && isCurrentNumberPerfectSquare(Math.sqrt(result.base))) {
            result.base = (int) Math.sqrt(result.base);
            result.times += 1;
        }

        System.out.println(perfectSquare + " -> " + result.base + " ^ " + result.times);
        return result;
    }

    static int maxPerfectSquares(int num1, int num2) {

        int largestPerfectSqr = (int) Math.pow(Math.floor(Math.sqrt(num2)), 2);
        if (largestPerfectSqr < num1) {
            return 0;
        }

        Result result = calculatePerfectSquare(largestPerfectSqr);

        int currentValue = result.base;

        while (currentValue > 2) {

            // check lower based values
            currentValue--;

            int newValue = perfectSquare(currentValue, result.times + 1);
            if (newValue >= num1 && newValue < num2) {

                result = calculatePerfectSquare(newValue);
                currentValue = result.base;
            }
        }

        return result.times;
    }