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);

        }

}

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