interface A {

    public abstract long fact(int m);

    public abstract long intPower(int m, int n);

    public abstract boolean findFactor(int m, int n);
}

public class B implements A {//implement <A>是什么意思？

    int m;
    int n;

    public long fact(int m) {//阶乘
        this.m = m;
        int i;
        long x = 1;
        for (i = 1; i <= m; i++) {
            x *= i;
        }
        System.out.println(m + "的阶乘是：" + x);//放外面
        return x;
    }

    public long intPower(int m, int n) {
        this.m = m;
        this.n = n;
        long y = 0;
        if(n==0){//结束条件
            y = 1;
        }
        else if (n > 0){
            y = m * intPower(m, n - 1);//调用方法，递归
        }
        System.out.println(m + "的"+n+"次方是：" + y);
        return y;
    }

    public boolean findFactor(int m, int n) {//感觉逻辑有些问题。
        boolean flag = false;
        this.m = m;
        this.n = n;
        if (m <= n && n % m == 0) {
            System.out.println("m是n的因子");
            flag = true;
        }
        if (n < m && m % n == 0) {
            System.out.println("n是m的因子");
            flag= true;
        }
        return flag;
    }

    public B(int m, int n) {
        this.m = m;
        this.n = n;
    }

    public static void main(String[] args) {
        B b = new B(2, 4);
        b.fact(2);
        b.intPower(2,4);
    }
}

------解决方案--------------------
interface A
{
	/**
	 * 是接口，为什么还用静态方法呢？
	 * @param m
	 * @return
	 */

	public  long fact(int m);

	public  long intPower(int m, int n);

	public  boolean findFactor(int m, int n);

}

public class B implements A
{

	int m;
	int n;

	public long fact(int m)
	{
		this.m = m;
		int i;
		long x = 1;
		for (i = 1; i <= m; i++)
		{
			x *= i;
			
		}
		System.out.println(m + "的阶乘是：" + x);