java第三周作业

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 day01;
import java.util.*;
public class firstwork {

	public static void main(String[] args) {
		Scanner input=new
		Scanner (System.in);
		System.out.println("请输入要查询的年份:");
		int a=input.nextInt();
		if (a%400==0){
		System.out.println(a+"年为闰年");
		
		}else{
			System.out.println(a+"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 day01;
import  java.util.*;
public class test {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input=new
		Scanner (System.in);
		System.out.println("请输入四位会员号:");
		int a=input.nextInt();
		int bai = a/100%10;
		if (bai%3==0){
			System.out.println("您是幸运会员!");
		
		}else {
			System.out.println("您不是幸运会员!");
		}
	}

}

  

 

 


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 day01;
import java.util.*;
public class test2 {

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

}

  

 

 

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 day01;
import java.util.*;
public class test3 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input=new
				Scanner (System.in);
		System.out.println("请输入三角形的三个边长:");
		System.out.print("a=");
		int a=input.nextInt();
		System.out.print("b=");
		int b=input.nextInt();
		System.out.print("c=");
		int c=input.nextInt();
		int max=0,sum=0;
		if (a>b&&a>c){
			max=a;
			sum=b+c;		
		}
		if (b>a&&b>c){
			max=b;
			sum=a+c;	
		}
		if (c>a&&c>b){
			max=c;
			sum=a+b;
		}
		if(a==b&&a==c){
			max=0;
			sum=a+b+c;
			
		}
		
		if(max<sum){
			System.out.println("三条边能组成三角形");
			
		}else{
			System.out.println("三条边不能组成三角形");
		}
		
		
	}

}

  

 

posted @ 2020-03-22 12:47  bluebless  阅读(195)  评论(0)    收藏  举报