Question

定义一个接口，接口中有两个抽象的方法：long factorial(int m);求参数的阶乘。long myPower(int m,int n)

接口中有两个抽象的方法：long factorial(int m);求参数的阶乘。long myPower(int m,int n);求参数m的n次方。boolean findFactorSum(int m,int n);判断参数中较小数是否是较大数的因子。实现该接口

Answer 1

interface Test{
long factorial(int m);
long myPower(int m,int n);
boolean findFactorSun(int m,int n);
}

class Fa implements Test{
public int Fm=1;

public long factorial(int m){
for(int i=1;i<=m;i++){
Fm*=i;
}
return Fm;
}

public long myPower(int m,int n){
if(n==1)
return m;

else
return myPower(m,n-1)*m;

}

public boolean findFactorSun(int m,int n){
return m>=n? m%n==0?true:false:n%m==0?true:false;
}

}

public class First {
public static void main(String[] args) {
Fa f=new Fa();
System.out.println("M的阶乘为："+f.factorial(8));
System.out.println("m的n次方为："+f.myPower(8, 4));
System.out.println("参数中较小数是否是较大数的因子："+f.findFactorSun(4, 8));
}

Answer 2

import java.io.InputStream;
import java.util.Scanner;

public class TestSum
{
public static void main(String[] args)
{
InputStream is = System.in;
Scanner scan = new Scanner(is);
System.out.print("请输入一个整数:");
int num = scan.nextInt();
System.out.println(num+"的阶乘为："+factorial(num));
System.out.println(num+"的2次方为："+myPower(num,2));
System.out.println(num+"与2次,小者是否为大的因子："+findFactorSum(num,2));
}

public static long factorial(int m){
int flag = m >=0 ? 1 : -1;
long result = 1;
for(int i=2;i<=Math.abs(m);i++){
result *= i;
}
return Math.pow(flag,Math.abs(m)) > 0 ? result : 0-result;
}

public static long myPower(int m,int n){
return ((Double)Math.pow(m,n)).longValue();
}

public static boolean findFactorSum(int m,int n){
return m>=n ? m%n == 0 ? true : false : n%m == 0 ? true :false;
}
}

Answer 3

自己到图书馆有许多这方面的介绍