第二次上机作业

1.输入一个年份,判断是不是闰年(能被4整除但不能被100整除,或者能被400整除)

package number2;

import java.util.Scanner;

public class zcc {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner input =new Scanner(System.in);
        System.out.println("请输入一个年份:");
        int i=input.nextInt();
        if(i%4==0&&i%100!=0||i%400==0){
            System.out.println(i+"年是闰年");
        }else{
            System.out.println(i+"年不是闰年");
        }
    }

}

 

 

 2.输入一个4位会员卡号,如果百位数字是3的倍数,就输出是幸运会员,否则就输出不是

package number2;

import java.util.Scanner;

public class zcc {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner input = new Scanner(System.in);
        System.out.println("请输入会员号:");
        int a= input.nextInt();
        if(a/100%10%3==0){
            System.out.println("会员");
        }else{
            System.out.println("非会员");
        }
    }

}

 

 

 3.已知函数,输入x的值,输出对应的y的值

x + 3 ( x > 0 )
y = 0 ( x = 0 )
x*2 –1 ( x < 0 )
package number2;

import java.util.Scanner;

public class zcc {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println("请输入x的值:");
        Scanner input = new Scanner(System.in);
        int x=input.nextInt();
        if(x>0){
            System.out.println("y="+(x+3));
        }else if(x==0){
            System.out.println("y="+0);
        }else if(x<0){
            System.out.println("y="+(x*2-1));
        }

}
}

 

 

 4.输入三个数,判断能否构成三角形(任意两边之和大于第三边)

package number2;

import java.util.Scanner;

public class zcc {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println("请输入三角形的三条边:");
        Scanner input = new Scanner(System.in);
        int x=input.nextInt();
        int y=input.nextInt();
        int z=input.nextInt();
        if(x+y>z&&x+z>y&&y+z>x){
            System.out.println("可以构成三角形");
            }
        else{
            System.out.println("不能构成三角形");
        }
        
}
}

 

posted @ 2020-03-24 23:39  洽多  阅读(135)  评论(0编辑  收藏  举报