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最新推荐文章于 2023-03-21 12:32:38 发布

最新推荐文章于 2023-03-21 12:32:38 发布 · 149 阅读

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这篇深情的诗篇表达了作者与其爱人之间紧密相连的情感纽带。无论身在何处，心中的爱意始终伴随，彼此的命运交织在一起，共同经历生活的点滴。

I carry your heart with me , I carry it in my heart , I am never without it .

Anywhere I go , you go , my dear .

And whatever is done by only me , is you doing , my darling.

I fear no fate.. ...

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【文章翻译+笔记】Towards the Next Generation of Recommender Systems:A Survey of the State-of-the-Art and Pos

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Write a C++ program that defines a class DateV2 that (1) Contains all the members in the class DateV1; Programming for Engineers C++ (2) Has two constructors as follows: One takes three parameters, int y, int m, int n; The other is the default constructor that takes no parameter (3) Has additional public member functions as follows: string getWeekDay(); // return the week day, for example, Sunday if day is 0, etc bool Leap(); // return if the year is leap int differFrom(DateV2& oneDate); // return the difference in days between the calling object // and the oneDate object void printDate(); // print the year, the month in English, the day, and the week day Test class DateV2 in the main function as follows: (1) Declare and set the objects today and tomorrow as in Problem 2. (2) Declare and initialize (by a constructor) an object to represent your OWN birthday. (3) Use the member function printDate to print today, tomorrow, and your birthday. (4) Output the weekday of today, tomorrow, and your own birthday. (5) Output how many days has passed since your birth (the difference between your birthday and today). Hint: i) We can use another string array to store the English name for week days (Sunday, Monday, through Saturday) ii) We know that it is Monday on Year 1, Month 1, and Day 1 iii) A good idea is to first design a function to compute the number of days that has passed since Year 1, Month 1, and Day 1, and then to use this function to compute the week day for a give date and to compute the difference between two dates. You can store the number of days for each of the 12 months in an integer array, which helps in counting the days.

05-22

```c++ #include <iostream> #include <string> using namespace std; class DateV1 { public: int year, month, day; DateV1() { year = 2021; month = 10; day = 1; } DateV1(int y, int m, int d) { year = y; month = m; day = d; } }; class DateV2 : public DateV1 { public: DateV2() : DateV1() {} DateV2(int y, int m, int d) : DateV1(y, m, d) {} string getWeekDay() { string weekdays[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}; int days = countDays(); return weekdays[days % 7]; } bool Leap() { if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) return true; return false; } int differFrom(DateV2& oneDate) { int days = countDays() - oneDate.countDays(); return days < 0 ? -days : days; } void printDate() { string months[] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"}; cout << year << "-" << months[month - 1] << "-" << day << endl; } private: int countDays() { int days = 0; for (int i = 1; i < year; i++) { if (Leap()) days += 366; else days += 365; } int monthDays[] = {31,28,31,30,31,30,31,31,30,31,30,31}; if (Leap()) monthDays[1] = 29; for (int i = 1; i < month; i++) { days += monthDays[i - 1]; } days += day - 1; return days; } }; int main() { DateV2 today(2021, 10, 1); DateV2 tomorrow(2021, 10, 2); DateV2 myBirthday(2000, 10, 1); cout << "Today's date: "; today.printDate(); cout << "Tomorrow's date: "; tomorrow.printDate(); cout << "My birthday: "; myBirthday.printDate(); cout << endl << "Weekday of today: " << today.getWeekDay() << endl; cout << "Weekday of tomorrow: " << tomorrow.getWeekDay() << endl; cout << "Weekday of my birthday: " << myBirthday.getWeekDay() << endl; cout << endl << "Days since my birthday: " << today.differFrom(myBirthday) << endl; return 0; } ```