pta-2024年秋面向对象程序设计实验一-java

2024-09-07 08:44

本文主要是介绍pta-2024年秋面向对象程序设计实验一-java,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

文章申明:作者也为初学者,解答仅供参考,不一定是最优解;

一:7-1 sdut-sel-2 汽车超速罚款(选择结构)

答案:

import java.util.Scanner;

        public class Main {

public static void main(String[] arg){

        Scanner sc=new Scanner(System.in);

        int a=sc.nextInt();

        int b=sc.nextInt();

        int c=b-a;

        if(c>=1&&c<=20) {

        System.out.println("You are speeding and your fine is 100.");

                }         

        else if(c>=21&&c<=30){

        System.out.println("You are speeding and your fine is 270.");

                }

        else if(c>=31)System.out.println("You are speeding and your fine is 500.");

        else System.out.println("Congratulations, you are within the speed limit!");

}

}

 

二: 7-2 Java中二进制位运算

答案:

import java.util.Scanner;

public class Main{

        public static void main(String[] args){

        Scanner sc=new Scanner(System.in);

        int a=sc.nextInt();

        String c=sc.next();

        int b=sc.nextInt();

        switch(c){

        case "&":

        System.out.println(a+" & "+b+" = "+(a&b));

        break;

        case "|":

        System.out.println(a+" | "+b+" = "+(a|b));

        break;

        case "^":

        System.out.println(a+" ^ "+b+" = "+(a^b));

        break;

        default:

        System.out.println("ERROR");

        break;

                }

        }

}

 

三:7-3 sdut-最大公约数和最小公倍数 

 答案:

import java.util.Scanner;

public class Main{

        public static void main(String[] args){

                Scanner sc=new Scanner(System.in);

                  while(sc.hasNext()){

                        int a=sc.nextInt();

                        int b=sc.nextInt();

                        System.out.print(ans(a,b));

                        System.out.print(" "+a*b/ans(a,b));

                        System.out.println();

                                                        }

                                        }

public static int ans( int a, int b){

                if(b>a){

                        int tem=a;

                        a=b;

                        b=tem;

                        }

                        int c=a%b;

                        while(c!=0){

                        a=b;

                        b=c;

                        c=a%b;

                                        }

                        return b;

                                                }

}

 

四:7-4 判断回文 

答案: 

import java.util.Scanner;

public class Main {

        static int ans(int a) {//判断位数

        int count = 0;

        while (a != 0) {

        count++;

        a = a / 10;

        }

        return count;

        }

                        public static void main(String[] args) {

                        Scanner sc = new Scanner(System.in);

                                int a = sc.nextInt();

                                int ret = ans(a);

                                 System.out.println(ret);

                                int sum=0;

                                int tem=a;

                        while (a != 0) {

                                int c = a % 10;

                                int f=ret;

                                        while(f--!=1){

                                                c*=10;

                                                                }

                                        sum+=c;

                                        ret--;

                                        a=a/10;

                                                                        }

                if (tem==sum) System.out.println("Y");

                else System.out.println("N");

        }

}

7-5 字符串操作

 

答案:

import java.util.Scanner;

public class Main {

        public static void main(String[] args) {

                Scanner sc = new Scanner(System.in);

                String input=sc.nextLine();

                StringBuilder nodigit=new StringBuilder();//创建可变字符串类

                for(char a:input.toCharArray()){//调用string中的方法,将input字符串转变为字符数组

                        if(!Character.isDigit(a)){//调用isdigit方法判断是否为数字

                                nodigit.append(a);//可变字符串末尾插入该字符

                        }

                }

                String ans=nodigit.reverse().toString();//对可变字符串翻转并转换为string类型

                System.out.println(ans);

                }

}

 

六:7-6 N个数的排序与查 

 

答案:

import java.util.Scanner;

public class Main {

        static void maopao(int a[]){//写一个冒泡排序吧

                for(int i=0;i<a.length-1;i++){

                        for(int j=0;j<a.length-i-1;j++){

                                if(a[j]>a[j+1]){

                                        int temp=a[j];

                                        a[j]=a[j+1];

                                        a[j+1]=temp;

                                        }

                                }

                }

}

        public static void main(String[] args) {

                Scanner sc = new Scanner(System.in);

                int c= sc.nextInt();

                int a[]= new int[c];

                        for(int i=0;i<c;i++){

                        a[i]=sc.nextInt();

                        }

                maopao(a);

                int x=sc.nextInt();

                        for (int i = 0; i < c; i++) {

                                if (a[i]==x) {System.out.println(i+1);

                                return;}

                                }

                System.out.println(-1);

        }

}

 

7-7 二进制的前导的零 

 

答案:

 

import java.util.Scanner;

public class Main {

        public static void main(String[] args) {

                Scanner sc = new Scanner(System.in);

                int a = sc.nextInt();

                int count=0;

                        while(a!=0){

                                a>>>=1;//无符号右移直至全部32位均为0

                                count++;

                        }

                System.out.println(32-count);//

        }

}

解法二:这个好理解一些

import java.util.Scanner;

public class Main {

        public static void main(String[] args) {

                Scanner sc = new Scanner(System.in);

                int a = sc.nextInt();

                int count=0;

                        for (int i = 31; i >= 0; i--) {

                                if ((a & (1 << i)) == 0) {//直到第一位为1停止

                                count++;

                                } else {

                                        break; // 找到第一个1,停止循环

                        }

                }

        System.out.println(count);

        }

}

 

解法三:直接调用 Integer.numberOfLeadingZeros(a)方法,可以直接计算a前导的0的个数

7-8 古时年龄称谓知多少? 

 

答案:

import java.util.Scanner;

public class Main{

public static void main(String[] arg){

Scanner sc=new Scanner(System.in);

int age=sc.nextInt();

if(age>=0&&age<=9)

System.out.println("垂髫之年");

else if(age<=19)System.out.println("志学之年");

else if(age<=29)System.out.println("弱冠之年");

else if(age<=39)System.out.println("而立之年");

else if(age<=49)System.out.println("不惑之年");

else if(age<=59)System.out.println("知命之年");

else if(age<=69)System.out.println("花甲之年");

else if(age<=79)System.out.println("古稀之年");

else if(age<=89)System.out.println("杖朝之年");

else if(age<=99)System.out.println("耄耋之年");

}

}

 

7-9 期末编程作业计分规则 

  

答案:

import java.util.Scanner;

public class Main {

        public static void main(String[] args) {

                Scanner sc = new Scanner(System.in);

                int num=sc.nextInt();//学生总数

                int max=sc.nextInt();//最高分

                int arr[]=new int[num+1];

                for(int i=1;i<=num;i++){

                arr[i]=sc.nextInt();

                }

                int a=sc.nextInt();//第一档百分比

                int b=sc.nextInt();//第二档百分比

                int c=sc.nextInt();//第三档百分比

                int sb1=sc.nextInt();//减分序列1

                int sb2=sc.nextInt()+sb1;//减分序列2

                int sb3=sc.nextInt()+sb2;//减分序列3

                int sb4=sc.nextInt()+sb3;//减分序列4

                double k1=Math.ceil(num*a/100.0);//第一档人数//Math.ceil可以向上取整

                double k2=Math.ceil(num*b/100.0)+k1;//第二档人数

                double k3 =Math.ceil(num*c/100.0)+k2;//第三档人数

                for(int i=1;i<=num;i++){

                if(i<= k1)arr[i]=(int)(arr[i]/(max/100.0))-sb1;

                else if (i<=k2)arr[i]=(int)(arr[i]/(max/100.0))-sb2;

                else if (i<=k3)arr[i]=(int)(arr[i]/(max/100.0))-sb3;

                else arr[i]=(int)(arr[i]/(max/100.0))-sb4;

        }

                for (int i =1; i <=num; i++) {

                System.out.print(arr[i]+" ");

                }

        }

}

 

7-10 sdut-array2-3 二维方阵变变变 

 

答案:写一个翻转函数,不同角度对应不同翻转次数

 

import java.util.Scanner;

public class Main {

        static int n;//行列数

        static int[][] reverse(int arr[][]){                //翻转函数

        int f[][]=new int[n][n];//创建新数组用于反转操作,不用自身进行反转的原因是会产生覆盖问题

        for(int i=0;i<f.length;i++){

                for(int j=0;j<f[i].length;j++){

                f[i][j]=arr[n-1-j][i];

                }

        }

        return f;

    }

                public static void main(String[] args) {

                        Scanner sc = new Scanner(System.in);

                        n = sc.nextInt();

                        int k=sc.nextInt();        //翻转次数

                        if(k==90)k=1;

                        else if(k==180)k=2;

                        else if(k==-90)k=3;

                        int arr[][]=new int[n][n];

                        for (int i = 0; i < n; i++) {

                                for (int j = 0; j < n; j++) {

                                arr[i][j]=sc.nextInt();

                                }

                        }

                        for(int i=0;i<k;i++){        //翻转k次,每次翻转90°,-90°就是顺时针翻转3次

                        arr= reverse(arr);

                        }

                        for(int i=0;i<n;i++){

                                for(int j=0;j<n;j++){

                                        if(j!=n-1){        //本题严格控制行末空格所以不要多打印空格了

                                        System.out.print(arr[i][j]+" ");

                                        }

                                        else System.out.print(arr[i][j]);

                                }

                                if(i!=n-1){        //避免多打印一个换行

                                System.out.println();

                         }

                    }

            }

}

7-11 直角三角形 

 

答案: 

import java.util.Scanner;

public class Main {

        public static void main(String[] args) {

                Scanner sc = new Scanner(System.in);

                double a = sc.nextDouble();

                double b = sc.nextDouble();

                double c = sc.nextDouble();

                double max=a>b?(a>c?a:c):(b>c?b:c);//求三边最大的值

                if (a*a+b*b==c*c||c*c+b*b==a*a||a*a+c*c==b*b){        //如果能构成直角三角形

                        if(c==max) System.out.println(a*b/2);

                                else if(b==max) System.out.println(a*c/2);

                                        else System.out.println(b*c/2);

                    }

                else System.out.println(0.0);

        }

}

这篇关于pta-2024年秋面向对象程序设计实验一-java的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/1144607

相关文章

Spring Cloud LoadBalancer 负载均衡详解

《SpringCloudLoadBalancer负载均衡详解》本文介绍了如何在SpringCloud中使用SpringCloudLoadBalancer实现客户端负载均衡,并详细讲解了轮询策略和... 目录1. 在 idea 上运行多个服务2. 问题引入3. 负载均衡4. Spring Cloud Load

Springboot中分析SQL性能的两种方式详解

《Springboot中分析SQL性能的两种方式详解》文章介绍了SQL性能分析的两种方式:MyBatis-Plus性能分析插件和p6spy框架,MyBatis-Plus插件配置简单,适用于开发和测试环... 目录SQL性能分析的两种方式:功能介绍实现方式:实现步骤:SQL性能分析的两种方式:功能介绍记录

在 Spring Boot 中使用 @Autowired和 @Bean注解的示例详解

《在SpringBoot中使用@Autowired和@Bean注解的示例详解》本文通过一个示例演示了如何在SpringBoot中使用@Autowired和@Bean注解进行依赖注入和Bean... 目录在 Spring Boot 中使用 @Autowired 和 @Bean 注解示例背景1. 定义 Stud

如何通过海康威视设备网络SDK进行Java二次开发摄像头车牌识别详解

《如何通过海康威视设备网络SDK进行Java二次开发摄像头车牌识别详解》:本文主要介绍如何通过海康威视设备网络SDK进行Java二次开发摄像头车牌识别的相关资料,描述了如何使用海康威视设备网络SD... 目录前言开发流程问题和解决方案dll库加载不到的问题老旧版本sdk不兼容的问题关键实现流程总结前言作为

SpringBoot中使用 ThreadLocal 进行多线程上下文管理及注意事项小结

《SpringBoot中使用ThreadLocal进行多线程上下文管理及注意事项小结》本文详细介绍了ThreadLocal的原理、使用场景和示例代码,并在SpringBoot中使用ThreadLo... 目录前言技术积累1.什么是 ThreadLocal2. ThreadLocal 的原理2.1 线程隔离2

springboot将lib和jar分离的操作方法

《springboot将lib和jar分离的操作方法》本文介绍了如何通过优化pom.xml配置来减小SpringBoot项目的jar包大小,主要通过使用spring-boot-maven-plugin... 遇到一个问题,就是每次maven package或者maven install后target中的ja

Java中八大包装类举例详解(通俗易懂)

《Java中八大包装类举例详解(通俗易懂)》:本文主要介绍Java中的包装类,包括它们的作用、特点、用途以及如何进行装箱和拆箱,包装类还提供了许多实用方法,如转换、获取基本类型值、比较和类型检测,... 目录一、包装类(Wrapper Class)1、简要介绍2、包装类特点3、包装类用途二、装箱和拆箱1、装

如何利用Java获取当天的开始和结束时间

《如何利用Java获取当天的开始和结束时间》:本文主要介绍如何使用Java8的LocalDate和LocalDateTime类获取指定日期的开始和结束时间,展示了如何通过这些类进行日期和时间的处... 目录前言1. Java日期时间API概述2. 获取当天的开始和结束时间代码解析运行结果3. 总结前言在J

Java深度学习库DJL实现Python的NumPy方式

《Java深度学习库DJL实现Python的NumPy方式》本文介绍了DJL库的背景和基本功能,包括NDArray的创建、数学运算、数据获取和设置等,同时,还展示了如何使用NDArray进行数据预处理... 目录1 NDArray 的背景介绍1.1 架构2 JavaDJL使用2.1 安装DJL2.2 基本操

最长公共子序列问题的深度分析与Java实现方式

《最长公共子序列问题的深度分析与Java实现方式》本文详细介绍了最长公共子序列(LCS)问题,包括其概念、暴力解法、动态规划解法,并提供了Java代码实现,暴力解法虽然简单,但在大数据处理中效率较低,... 目录最长公共子序列问题概述问题理解与示例分析暴力解法思路与示例代码动态规划解法DP 表的构建与意义动