public class PerfectSquare { public static void main(String[] args) { for (int i = 1; ; i++) { int x = i + 100; // 加上 100 后的数 int y = x + 168; // 再加上 168 后的数 // 判断 x 是否为完全平方数，y 是否为完全平方数 if (isPerfectSquare(x) && isPerfectSquare(y)) { System.out.println("符合条件的整数为：" + i); break; } } } // 判断一个数是否为完全平方数 public static boolean isPerfectSquare(int n) { int sqrt = (int) Math.sqrt(n); return sqrt * sqrt == n; } }解释一下这段代码

这段代码的作用是寻找符合一定条件的整数。具体来说，这个程序会从1开始不断枚举整数 i，然后判断 i + 100 和 i + 100 + 168 是否都为完全平方数。如果是的话，就输出符合条件的整数 i，并停止程序。

其中，isPerfectSquare(int n) 方法是用来判断一个数是否为完全平方数的。它通过计算这个数的平方根，然后判断这个平方根的平方是否等于原数，从而得出结论。

相关问题

Java (Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.)

Here is the Java code to implement the above program:

import java.util.ArrayList;
import java.util.Scanner;

public class PerfectSquare {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Enter an integer m: ");
        int m = input.nextInt();

        ArrayList<Integer> factors = new ArrayList<>();
        int factor = 2;
        while (m > 1) {
            if (m % factor == 0) {
                factors.add(factor);
                m /= factor;
            } else {
                factor++;
            }
        }

        int n = 1;
        for (int i = 0; i < factors.size(); i++) {
            int count = 0;
            for (int j = 0; j < factors.size(); j++) {
                if (factors.get(i) == factors.get(j)) {
                    count++;
                }
            }
            if (count % 2 != 0) {
                n *= factors.get(i);
            }
        }

        System.out.println("The smallest integer n such that m * n is a perfect square is " + n);
    }
}

Explanation:

- We first prompt the user to enter an integer m using the Scanner class.
- We then create an ArrayList called factors to store all the smallest factors of m.
- We use a while loop to find all the smallest factors of m and add them to the ArrayList.
- We then use two nested for loops to find the factors that appear an odd number of times in the ArrayList and store their product in the variable n.
- Finally, we print out the value of n as the smallest integer such that m * n is a perfect square.

用java编写程序：输出1~20以内所有的完全平方数c（c^2=a^2=b^2）

可以使用循环语句和条件判断语句来解决这个问题。以下是一个Java程序，可以输出1~20以内所有的完全平方数：

public class PerfectSquare {
    public static void main(String[] args) {
        for (int i = 1; i <= 20; i++) {
            int c = i * i;
            int a = (int) Math.sqrt(c);
            int b = (int) Math.sqrt(c);
            if (a * a == c && b * b == c) {
                System.out.println(c);
            }
        }
    }
}

程序首先使用for循环从1到20遍历所有的数字，然后计算每个数字的平方值c。接着使用Math类中的sqrt()方法计算c的平方根，并将结果赋值给a和b。如果a和b的平方与c相等，则输出c。