zoukankan      html  css  js  c++  java
  • 3.26上机练习

    1.输出x=1,如果是5,输出x=5,如果是 10,输出 x=10,除了以上几个值,都输出x=none。(知识点:if条件语句)

    package text1;
    
    import java.util.Scanner;
    
    public class test14 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            //输入变量值x=1输出x=1,x=5--x=5,x=10--x=10,其他输出x=none 
            @SuppressWarnings("resource")
            Scanner sc=new Scanner(System.in);
            int x=sc.nextInt();
            if(x==1||x==5||x==10) {
                System.out.println("x="+x);
            }else  {
                System.out.println("x=none");
            }
        }
    
    }

    2.用switch结构实现第1题

    package text1;
    
    import java.util.Scanner;
    
    public class test14 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
          
            @SuppressWarnings("resource")
            Scanner sc=new Scanner(System.in);
            int x=sc.nextInt();
            switch(x) {
            case 1:System.out.println("x=1");break;
            case 5:System.out.println("x=5");break;
            case 10:System.out.println("x=10");break;
            default :System.out.println("x=none");break;
            }
        }
    
    }

    3.判断一个数字是否能被5和6同时整除(打印能被5和6整除),或只能被5整除(打印能被5整 除),或只能被6整除,(打印能被6整除),不能被5或6整除,(打印不能被5或6整除)

    package text1;
    
    import java.util.Scanner;
    
    public class test14 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
           
            @SuppressWarnings("resource")
            Scanner sc=new Scanner(System.in);
            int x=sc.nextInt();
            if(x%5==0&&x%6==0) {
                System.out.println("能被5和6整除");
            }else if(x%5==0&&x%6!=0) {
                System.out.println("能被5整除");
            }else if(x%6==0&&x%5!=0) {
                System.out.println("能被6整除");
            }else if(x%5!=0&&x%6!=0) {
                System.out.println("不能被5或6整除");
            }
        }
    
    }

    4.输入一个0~100的分数,如果不是0~100之间,打印分数无效,根据分数等级打印 A(90-100),B(80-89),C,D,E(知识点:条件语句if elseif)

    package text1;
    
    import java.util.Scanner;
    
    public class test14 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            //A(90-100),B(80-89),C,D,E
            @SuppressWarnings("resource")
            Scanner sc=new Scanner(System.in);
            int x=sc.nextInt();
            if(x>=0&&x<=100) {
                if(x<=100&&x>=90) {
                    System.out.println("A");
                }else if(x<90&&x>=80) {
                    System.out.println("B");
                }else if(x<80&&x>=70) {
                    System.out.println("C");
                }else if(x<70&&x>=60) {
                    System.out.println("D");
                }else if(x<60&&x>=0) {
                    System.out.println("E");
                }
            }else {
                System.out.println("此分数无效");
            }
        }
    
    }

    5.输入三个整数x,y,z,请把这三个数由小到大输出(知识点:条件语句)

    package text1;
    
    import java.util.Scanner;
    
    public class test14 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            //A(90-100),B(80-89),C,D,E
            @SuppressWarnings("resource")
            Scanner sc=new Scanner(System.in);
            int a=sc.nextInt();
            int b=sc.nextInt();
            int c=sc.nextInt();
            if(a>b&&b>c) {
                System.out.println(c+"<"+b+"<"+a);
            }else if(a>b&&b<c) {
                System.out.println(b+"<"+c+"<"+a);
            }else if(b>a&&a>c) {
                System.out.println(c+"<"+a+"<"+b);
            }else if(b>c&&c>a) {
                System.out.println(a+"<"+c+"<"+b);
            }else if(c>a&&a>b) {
                System.out.println(b+"<"+a+"<"+c);
            }else if(c>b&&b>a) {
                System.out.println(a+"<"+b+"<"+c);
            }
        }
    
    }
  • 相关阅读:
    css优先级及匹配原理
    [BZOJ1208]宠物收养所
    [BZOJ1588]营业额统计
    [AtCoder2558]Many Moves
    [POJ3416]Crossing
    [POJ3378]Crazy Thairs
    高精度大整数模板
    py3实现维吉尼亚加解密
    Codeforces 617E: XOR and Favorite Number(莫队算法)
    hdu 1695: GCD 【莫比乌斯反演】
  • 原文地址:https://www.cnblogs.com/F-dl/p/12573030.html
Copyright © 2011-2022 走看看