**11.17 (代数: 完全平方)编写一个Java程序，提示用户输人一个整数m,然后找到最小的整数n,使得 m*n是一个完全平方。 提示:存储所有m的最小因子到一个数组列表，则n是列表中出现奇数次的因子的乘积。例如，考虑m=90的情况，保存因子2,3,3,5到一个数组列表中。列表中2和5出现了奇数次数，因此，n是10。) 这里是一个运行示例: . Enter an integer m: 1500 The smallest number n for m * n to be a perfect square is 15 m×n is 22500 Enter an integer m: 63 The smalle st number n for m*n to be a perfect square is 7 m*n is 441

好的，以下是Java程序的代码：

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();
        input.close();

        ArrayList<Integer> factors = new ArrayList<Integer>();
        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++) {
            if (i == 0 || factors.get(i) != factors.get(i - 1)) {
                if (Collections.frequency(factors, factors.get(i)) % 2 == 1) {
                    n *= factors.get(i);
                }
            }
        }

        System.out.println("The smallest number n for m * n to be a perfect square is " + n);
        System.out.println("m * n is " + m * n);
    }
}

示例输出：

Enter an integer m: 1500
The smallest number n for m * n to be a perfect square is 15
m * n is 22500

Enter an integer m: 63
The smallest number n for m * n to be a perfect square is 7
m * n is 441