BLOG-1 目录

• BLOG-1 目录
• （1）前言
• （2）设计与分析
• （3）踩坑心得
• （4）改进建议
• （5）总结

（1）前言

1. 知识点

   • 第一次题目集：基本数据类型的运用、不同变量类型的定义和赋值、操作符的简单运用，以及字符串的初步认识。

   • 第二次题目集：字母-数字转换、串口字符解析、String的格式判断与内容提取。

   • 第三次题目集：定义类解决复杂问题，认识并尝试类的不同聚合的方式。

2. 题量

   • 第一次题目集：一共9道题目。

   • 第二次题目集：一共3道题目。

   • 第三次题目集：一共4道题目。

3. 难度

   • 第一次题目集：难度较小。

   • 第二次题目集：难度适中。

   • 第三次题目集：难度较大。

（2）设计与分析

1. 题目集2的7-2

• 源码:

import java.util.Scanner;
public class Main {
    public static void main(String[] args){
        Scanner input = new Scanner (System.in);
        String a=input.nextLine();
        int n=a.length();
        if(n<11){
            System.out.print ("null data");
            return;
        }
        int t=0;
        for(int i=0;i<n;i++){
            if(a.charAt(i)=='1'){
                t++;
            }
        }
        if(t==n){
            System.out.print ("null data");
            return;
        }
        t=1;
        int k=0;
        for(int i=0;i<n;i++){
            if(a.charAt (i)=='0'&&i+10<n){
                for(int j=i+1;j<=i+8;j++){
                    if(a.charAt (j)=='1'){
                        k++;
                    }
                }
                if(k%2==0) {
                    if (a.charAt (i + 9) == '1' && a.charAt (i + 10) == '1'&&i+10<n) {
                        System.out.println (t + ":" + a.charAt (i + 1) + "" + a.charAt (i + 2) + a.charAt (i + 3) + "" + a.charAt (i + 4) + "" + a.charAt (i + 5) + "" + a.charAt (i + 6) + "" + a.charAt (i + 7) + "" + a.charAt (i + 8));
                        t++;
                        i+=10;
                    } else if (a.charAt (i + 9) == '0' && a.charAt (i + 10) == '1'&&i+10<n) {
                        System.out.println (t + ":" + "parity check error");
                        t++;
                        i+=10;
                    } else if (a.charAt (i + 9) == '1' && a.charAt (i + 10) == '0'&&i+10<n) {
                        System.out.println (t + ":" + "validate error");
                        t++;
                        i+=10;
                    } else if (a.charAt (i + 9) == '0' && a.charAt (i + 10) == '0'&&i+10<n) {
                        System.out.println (t + ":" + "validate error");
                        t++;
                        i+=10;
                    }
                }
                else{
                    if (a.charAt (i + 9) == '0' && a.charAt (i + 10) == '1'&&i+10<n) {
                        System.out.println (t + ":" + a.charAt (i + 1) + "" + a.charAt (i + 2) + a.charAt (i + 3) + "" + a.charAt (i + 4) + "" + a.charAt (i + 5) + "" + a.charAt (i + 6) + "" + a.charAt (i + 7) + "" + a.charAt (i + 8));
                        t++;
                        i+=10;
                    }else if (a.charAt (i + 9) == '1' && a.charAt (i + 10) == '1'&&i+10<n) {
                        System.out.println (t + ":" + "parity check error");
                        t++;
                        i+=10;
                    } else if (a.charAt (i + 9) == '0' && a.charAt (i + 10) == '0'&&i+10<n) {
                        System.out.println (t + ":" + "validate error");
                        t++;
                        i+=10;
                    } else if (a.charAt (i + 9) == '1' && a.charAt (i + 10) == '0'&&i+10<n) {
                        System.out.println (t + ":" + "validate error");
                        t++;
                        i+=10;
                    }
                }
            }
            k=0;
        }
    }
}

• 思路分析:
  ①采用nextLine()获取数据;
  ②采用for循环遍历验证数据合法性;
  ③输出数据对应结果;

• 对源代码的分析:
  ①先用nextLine()读取数据;
  ②然后for循环遍历，将错误数据情况全用if进行判断;
  ③输出数据对应if的结果;

2. 题目集3的7-1

• 源码：

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);

        double a = Double.parseDouble(input.next());
        double b = Double.parseDouble(input.next());
        double c = Double.parseDouble(input.next());

        if(a == 0){
            System.out.println("Wrong Format");
            System.exit(0);
        }

        //create a QuadraticEquation object
        QuadraticEquation equation = new QuadraticEquation(a, b, c);
        //get value of b * b - 4 * a * c
        double discriminant = equation.getDiscriminant();

        System.out.println("a=" + equation.getA() +
                ",b=" + equation.getB() +
                ",c=" + equation.getC()+":");

        if (discriminant < 0) {
            System.out.println("The equation has no roots.");
        }
        else if (discriminant == 0)
        {
            System.out.println("The root is " +
                    String.format("%.2f", equation.getRoot1()));
        }
        else
        {
            System.out.println("The roots are " +
                    String.format("%.2f", equation.getRoot1())
                    + " and " +  String.format("%.2f", equation.getRoot2()));
        }
    }
}

class QuadraticEquation{
    double a,b,c;

    public QuadraticEquation(double a, double b, double c) {
        this.a=a;
        this.b=b;
        this.c=c;
    }

    public double getDiscriminant() {
        return b * b - 4 * a * c;
    }

    public double getRoot1() {
        double x=(-b+Math.sqrt ( b * b - 4 * a * c))/(2*a);
        return x;
    }

    public double getRoot2() {
        double x=(-b-Math.sqrt ( b * b - 4 * a * c))/(2*a);
        return x;
    }

    public double getA() {
        return a;
    }

    public double getB() {
        return b;
    }

    public double getC() {
        return c;
    }
}

• 思路分析:
  ①依题目要求建立QuadraticEquation类求根;
  ②根据求根公式判断并计算一元二次方程的根;
  ③return相应的结果;

• 对源代码的分析：
  ①首先在QuadraticEquation类定义并初始化方程系数的值;
  ②然后分别创建判别根合法性方法及求根方法;
  ③计算方程的根并return给main;
  ④powerDesigner的相应类图:
  

3. 题目集3的7-2

• 源码:

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int year = 0;
        int month = 0;
        int day = 0;

        int choice = input.nextInt();

        if (choice == 1) { // test getNextNDays method
            int m = 0;
            year = Integer.parseInt(input.next());
            month = Integer.parseInt(input.next());
            day = Integer.parseInt(input.next());

            DateUtil date = new DateUtil(year, month, day);

            if (!date.checkInputValidity()) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            m = input.nextInt();

            if (m < 0) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            System.out.print(date.getYear() + "-" + date.getMonth() + "-" + date.getDay() + " next " + m + " days is:");
            System.out.println(date.getNextNDays(m).showDate());
        } else if (choice == 2) { // test getPreviousNDays method
            int n = 0;
            year = Integer.parseInt(input.next());
            month = Integer.parseInt(input.next());
            day = Integer.parseInt(input.next());

            DateUtil date = new DateUtil(year, month, day);

            if (!date.checkInputValidity()) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            n = input.nextInt();

            if (n < 0) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            System.out.print(
                    date.getYear() + "-" + date.getMonth() + "-" + date.getDay() + " previous " + n + " days is:");
            System.out.println(date.getPreviousNDays(n).showDate());
        } else if (choice == 3) {    //test getDaysofDates method
            year = Integer.parseInt(input.next());
            month = Integer.parseInt(input.next());
            day = Integer.parseInt(input.next());

            int anotherYear = Integer.parseInt(input.next());
            int anotherMonth = Integer.parseInt(input.next());
            int anotherDay = Integer.parseInt(input.next());

            DateUtil fromDate = new DateUtil(year, month, day);
            DateUtil toDate = new DateUtil(anotherYear, anotherMonth, anotherDay);

            if (fromDate.checkInputValidity() && toDate.checkInputValidity()) {
                System.out.println("The days between " + fromDate.showDate() +
                        " and " + toDate.showDate() + " are:"
                        + fromDate.getDaysofDates(toDate));
            } else {
                System.out.println("Wrong Format");
                System.exit(0);
            }
        }
        else{
            System.out.println("Wrong Format");
            System.exit(0);
        }
    }
}

class DateUtil{
    int year,month,day;

    public DateUtil(int year, int month, int day) {
        this.year=year;
        this.month=month;
        this.day=day;
    }

    public boolean checkInputValidity(){
        return year >= 1820 && year <= 2020 && month >= 1 && month <= 12 && day >= 1 && day <= 31;
    }//检测输入的年、月、日是否合法

    public boolean isLeapYear(int year){
        return year % 400 == 0 || year % 100 != 0 && year % 4 == 0;
    }//判断year是否为闰年

    public DateUtil getNextNDays(int n){
        if(n==0)
            return new DateUtil(year, month, day);
        int k = 0;
        int[] a = {0,31,28,31,30,31,30,31,31,30,31,30,31};
        for (int i = this.month+1; i <=12; i++) {
            k+=a[i];
        }
        k+=a[this.month]-this.day;
        if(isLeapYear(this.year)&&this.month<=2)
            k++;

        int year=0,month,day;
        if(k>n){
            int dayl = a[this.month];
            if (isLeapYear(this.year)&&this.month==2) {
                dayl++;
            }
            dayl-=this.day;
            if(dayl>=n){
                month=this.month;
                day=n+this.day;
            }
            else{
                n-=dayl;
                month=this.month+1;
                int j=month;
                if (n - a[k] > 0 && j <= 12) {
                    do {
                        n -= a[k];
                        month++;
                        j++;
                    } while (n - a[k] > 0 && j <= 12);
                }
                day=n;
            }
        }
        else{
            n-=k;
            year = this.year+1;
            int y=365;
            if(isLeapYear (year)){
                y++;
            }
            if (n - y > 0) {
                do {
                    n -= y;
                    year++;
                    y = 365;
                    if (isLeapYear (year))
                        y++;
                } while (n - y > 0);
            }
            int j=1;
            if (n - a[j] > 0) {
                do {
                    n -= a[j];
                    j++;
                } while (n - a[j] > 0);
            }
            month=j;
            day=n;
        }
        return new DateUtil(year, month, day);
    }//取得year-month-day的下n天日期

    public DateUtil getPreviousNDays(int n){
        int k = 0;
        int[] a = {0,31,28,31,30,31,30,31,31,30,31,30,31};
        for (int i = this.month+1; i <=12; i++) {
            k+=a[i];
        }
        k+=a[this.month]-this.day;
        if(isLeapYear(this.year)&&this.month<=2)
            k++;

        int year, month, day;
        int r = 365-k;
        if (isLeapYear(this.year)) {
            r++;
        }
        if (r>n) {//本年
            year=this.year;
            int dayl=this.day;//本月剩余的日期
            if (dayl>n) {    //本月
                month = this.month;
                day = dayl-n;
            }
            else{    //其他月
                n-=dayl;
                month = this.month-1;
                if (month==0) {
                    month = 12;
                    year=this.year-1;
                }
                int j = month;
                if (n - a[j] > 0) {
                    do {
                        n -= a[j];
                        month--;
                        j--;
                    } while (n - a[j] > 0);
                }
                day = a[j]-n;
                if (isLeapYear(year)&&month==2) {
                    day++;
                }
            }
        }
        else {
            n-=r;
            year = this.year-1;
            int y = 365;
            if (isLeapYear(year)) {
                y++;
            }
            if (n - y > 0) {
                do {
                    n -= y;
                    year--;
                    y = 365;
                    if (isLeapYear (year))
                        y++;
                } while (n - y > 0);
            }
            int j = 12;
            if (n - a[j] > 0) {
                do {
                    n -= a[j];
                    j--;
                } while (n - a[j] > 0);
            }
            month = j;
            day = a[j]-n;
            if (isLeapYear(year)&&month==2) {
                day++;
            }
        }
        return new DateUtil(year, month, day);
    }//取得year-month-day的前n天日期

    public boolean compareDates(DateUtil date){
        if (date.year<this.year) {
            return false;
        }
        else if (date.year==this.year&&date.month<this.month) {
            return false;
        }
        if (date.year==this.year&&date.month==this.month&&date.day<this.day) {
            return false;
        }
        return true;
    }//比较当前日期与date的大小（先后）

    public boolean equalTwoDates(DateUtil date){
        return year == date.year && month == date.month && day == date.day;
    }//判断两个日期是否相等

    public int getDaysofDates(DateUtil date){
        int[] a;
        a = new int[]{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int days=0;
        if(date.year<year) {
            for (int i = year - 1; i > date.year; i--)
                if (isLeapYear(i)) {
                    days += 366;
                } else {
                    days += 365;
                }
            if (isLeapYear(year)) a[2] = 29;
            for (int i = 1; i < month; i++) days += a[i];
            days += day;
            a[2]=28;
            if (isLeapYear(date.year)) a[2] = 29;
            for (int i = 12; i > date.month; i--) days += a[i];
            days += a[date.month]-date.day;
            return days;
        }
        else if(date.year>year){
            for (int i = date.year - 1; i > year; i--)
                if (isLeapYear(i)) {
                    days += 366;
                } else {
                    days += 365;
                }
            if (isLeapYear(date.year)) a[2] = 29;
            for (int i = 1; i < date.month; i++) days += a[i];
            days += date.day;
            a[2]=28;
            if (isLeapYear(year)) a[2] = 29;
            for (int i = 12; i > month; i--) days += a[i];
            days += a[month]-day;
            return days;
        }
        else{
            if(date.month > month){
                if (isLeapYear(date.year)) a[2] = 29;
                for (int i = month; i < date.month; i++) days += a[i];
                days += date.day;
                days -= day;
                return days;
            }
            else if(month>date.month){
                if (isLeapYear(date.year)) a[2] = 29;
                for (int i = date.month; i < month; i++) days += a[i];
                days += day;
                days -= date.day;
                return days;
            }
            else {
                return Math.abs (date.day-day);
            }
        }
    }//求当前日期与date之间相差的天数

    public String showDate(){
        return this.year + "-" + this.month + "-" + this.day;
    }//以“year-month-day”格式返回日期值

    public int getYear() {
        return year;
    }

    public int getMonth() {
        return month;
    }

    public int getDay() {
        return day;
    }
}

• 思路分析:
  ①根据题目要求建立DateUtil类并构建各个方法;
  ③return相应的结果;

• 对源代码的分析：
  ①首先在DateUtil类定义并初始化year,month,day的值;
  ②然后分别建立求前n天的方法getPreviousNDays,求当前日期与date之间相差的天数的方法getDaysofDates,取得year-month-day的下n天日期的方法getNextNDays;
  ③getPreviousNDays：循环从总前n天的n中减去当前月的天数且对year,month,day实时改变，同时判断是否闰月,当n为0是即为前n天的日期。
  getNextNDays与getPreviousNDays同理。
  getPreviousNDays：将两个日期的day单独相加，然后对其余月份直接加；
  ④计算相应的结果并return给main;
  ⑤powerDesigner的相应类图:
  

4. 题目集3的7-3

• 源码：

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int year = 0;
        int month = 0;
        int day = 0;

        int choice = input.nextInt();

        if (choice == 1) { // test getNextNDays method
            int m = 0;
            year = Integer.parseInt(input.next());
            month = Integer.parseInt(input.next());
            day = Integer.parseInt(input.next());

            DateUtil date = new DateUtil(year, month, day);

            if (!date.checkInputValidity()) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            m = input.nextInt();

            if (m < 0) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            System.out.println(date.getNextNDays(m).showDate());
        } else if (choice == 2) { // test getPreviousNDays method
            int n = 0;
            year = Integer.parseInt(input.next());
            month = Integer.parseInt(input.next());
            day = Integer.parseInt(input.next());

            DateUtil date = new DateUtil(year, month, day);

            if (!date.checkInputValidity()) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            n = input.nextInt();

            if (n < 0) {
                System.out.println("Wrong Format");
                System.exit(0);
            }

            System.out.println(date.getPreviousNDays(n).showDate());
        } else if (choice == 3) { //test getDaysofDates method
            year = Integer.parseInt(input.next());
            month = Integer.parseInt(input.next());
            day = Integer.parseInt(input.next());

            int anotherYear = Integer.parseInt(input.next());
            int anotherMonth = Integer.parseInt(input.next());
            int anotherDay = Integer.parseInt(input.next());

            DateUtil fromDate = new DateUtil(year, month, day);
            DateUtil toDate = new DateUtil(anotherYear, anotherMonth, anotherDay);

            if (fromDate.checkInputValidity() && toDate.checkInputValidity()) {
                System.out.println(fromDate.getDaysofDates(toDate));
            } else {
                System.out.println("Wrong Format");
                System.exit(0);
            }
        }
        else{
            System.out.println("Wrong Format");
            System.exit(0);
        }
    }
}

class DateUtil {
    private final int year,month,day;
    private final int [] Month1 = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    private final int[] Month2 ={0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    public DateUtil(int year, int month, int day) {
        this.year = year;
        this.month = month;
        this.day = day;
    }

    private int getDay(int year, int month) {
        int days;
        days = Month1[month];
        if (isLeapYear(year) && month == 2) {
            days = 29;
        }
        return days;
    }

    public boolean checkInputValidity(){
        if(isLeapYear(year)) {
            return year >= 1900 && year <= 2050 && month >= 1 && month <= 12 && day >= 1 && day <= Month2[month];
        }
        else{
            return year >= 1900 && year <= 2050 && month >= 1 && month <= 12 && day >= 1 && day <= Month1[month];
        }
    }//检测输入的年、月、日是否合法

    public boolean isLeapYear(int year)//判断year是否为闰年
    {
        return (year % 4 == 0 && year % 100 != 0) || year % 400 == 0;
    }


    public DateUtil getNextNDays(int n)//取得year-month-day的下n天日期
    {
        int year = this.year;
        int month = this.month;
        int day = this.day;
        int i = 0;
        if (i < n) {
            do {
                day++;
                int t = getDay (year, month);
                if (day > t) {
                    month++;
                    day = 1;
                    if (month > 12) {
                        year++;
                        month = 1;
                    }
                }
                i++;
            } while (i < n);
            return new DateUtil(year, month, day);
        } else {
            return new DateUtil(year, month, day);
        }
    }

    public DateUtil getPreviousNDays(int n)//取得year-month-day的前n天日期
    {
        int year = this.year;
        int month = this.month;
        int day = this.day;
        int i = 0;
        if (i < n) {
            do {
                day--;
                if (day < 1) {
                    do {
                        month--;
                        if (month < 1) {
                            month = 12;
                            year--;
                        }
                        day += getDay (year, month);
                    } while (day < 1);
                }
                i++;
            } while (i < n);
            return new DateUtil(year, month, day);
        } else {
            return new DateUtil(year, month, day);
        }
    }

    public boolean compareDates(DateUtil date)//比较当前日期与date的大小（先后）
    {
        if (this.year > date.year) return true;
        if (this.year == date.year) {
            if (this.month > date.month) return true;
            if (this.month == date.month) {
                return this.day >= date.day;
            }
        }
        return false;
    }

    public boolean equalTwoDates(DateUtil date)//判断两个日期是否相等
    {
        if (date != null)
            return year == date.year && month == date.month && day == date.day;
        else
            return false;
    }

    public int getDaysofDates(DateUtil date)//求当前日期与date之间相差的天数
    {
        int[] a;
        a = new int[]{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int days=0;
        if(date.year<year) {
            for (int i = year - 1; i > date.year; i--)
                if (isLeapYear(i)) {
                    days += 366;
                } else {
                    days += 365;
                }
            if (isLeapYear(year)) a[2] = 29;
            for (int i = 1; i < month; i++) days += a[i];
            days += day;
            a[2]=28;
            if (isLeapYear(date.year)) a[2] = 29;
            for (int i = 12; i > date.month; i--) days += a[i];
            days += a[date.month]-date.day;
            return days;
        }
        else if(date.year>year){
            for (int i = date.year - 1; i > year; i--)
                if (isLeapYear(i)) {
                    days += 366;
                } else {
                    days += 365;
                }
            if (isLeapYear(date.year)) a[2] = 29;
            for (int i = 1; i < date.month; i++) days += a[i];
            days += date.day;
            a[2]=28;
            if (isLeapYear(year)) a[2] = 29;
            for (int i = 12; i > month; i--) days += a[i];
            days += a[month]-day;
            return days;
        }
        else{
            if(date.month > month){
                if (isLeapYear(date.year)) a[2] = 29;
                for (int i = month; i < date.month; i++) days += a[i];
                days += date.day;
                days -= day;
                return days;
            }
            else if(month>date.month){
                if (isLeapYear(date.year)) a[2] = 29;
                for (int i = date.month; i < month; i++) days += a[i];
                days += day;
                days -= date.day;
                return days;
            }
            else {
                return Math.abs (date.day-day);
            }
        }
    }

    public String showDate()//以“year-month-day”格式返回日期值
    {
        return year + "-" + month + "-" + day;
    }

    public int getYear() {
        return year;
    }

    public int getMonth() {
        return month;
    }

    public int getDay() {
        return day;
    }
}

• 思路分析:
  ①依题目要求建立DateUtil类、Year类、Month类、Day类;
  ②将类间关系调整为题目所诉;

• 对源代码的分析：
  ①在DateUtil类只能调用Day类的方法,Day类只能调用Month类的方法,Month类只能调用Year类的方法;
  ②在算后n天时,day在计算中只能一天一天增加,当day突破对应month的最大天数,则day=1,month++,当month突破12时month=1,year++,同理求出前n天;
  ③计算并return对应结果给main;
  ④powerDesigner的相应类图:
  

（3）踩坑心得

1. 题目集2的7-2：

   • 题目模拟的串口接收处理程序有效数据是8位,写代码时未对每一段串口数据判断,导致小于8位的串口数据也输出了出来。

2. 题目集3的7-1：

   • 参数全为double型,防止运算精度缺失。

3. 题目集3的7-2：

   • 对每个月的最后一天,每年的最后一月,闰年2月的最后一天都要单独讨论。

   • 对整形最大值进行化小后运算,防止爆int。

4. 题目集3的7-3：

   • 对每个月的最后一天,每年的最后一月,闰年2月的最后一天都要单独讨论。

   • 对整形最大值进行化小后运算,防止爆int。

   • 日期越界讨论。

（4）改进建议

我觉得最需要改进的是题目集3的7-3：

本题源码的类没有满足题目要求的类间关系，只是单纯的仿照前一题仿写。代码在上文给出，自我感觉十分糟糕。对此我有以下修改方案：

1. 重新将类拆分，严格按照题目给出的类和方法重新构建；

2. 严格按照题目给出类间关系进行方法调用。

（5）总结

对本阶段三次题目集的综合性总结，学到了什么，哪些地方需要进一步学习及研究，对教师、课程、作业、实验、课上及课下组织方式等方面的改进建议及意见：

1. 学到的知识

   • 基本数据类型的运用、不同变量类型的定义和赋值、操作符的简单运用，以及字符串的初步认识。

   • 字母-数字转换、串口字符解析、String的格式判断与内容提取。

   • 定义类解决复杂问题，认识并学会类的不同聚合方式。

2. 需要进一步学习及研究的地方

   • 我认为应该要重新学习构建类间关系的相关知识，多学习一些java各种语法的基本原理。

   • 熟练掌握构造函数的基本运用

3. 对教师、课程、作业、实验、课上及课下组织方式等方面的改进建议及意见：

   • 老师发送的PTA作业有一些的题目测试点过于刁钻，而样例和题目信息中又未给出相应提示，导致难以发现。

   • 希望PTA作业能够在一些难度较大的题目集后面能够再次发布补题集，让并没有满分的同学能够再次练习测试。