Enter a date to determine which day of the year the date is, the days of the week, the day of the two days, and the date used for the string output

Before writing a blog post (http://blog.csdn.net/shiwazone/article/details/45053739) is implemented with basic functions, this time using the class design method, that is, the object-oriented approach to rewrite, And added a date to convert the implementation into a string. The program here can also solve the problem of programming Zhu Ji Nanxiong exercise 3.4.

#include "calendar.h" int main () {time t;t.initialtime (); T.show (); T.strshow (); Time T1;t1.initialtime (); T1. Strshow (); Time T2;t2.initialtime (); T2. Strshow (); cout << "The days between T1 and T2 are:" << t1.countdays (T2) <<endl;cout << "The Days be Tween T2 and T1 is: "<< t2.countdays (T1) <<endl;return 0;}

#ifndef canlendar_h#define canlendaa_h#include<iostream> #include <string>using namespace std;string ntos (int t);//number converted to character class Time{private:int year;int month;int day;unsigned int weekday;int weekdaycount () const;// Calculates the day of the week on which the day of the int daycount () const;//calculates the day of the year, the number of days int daysyearcount () const;//calculates the day of the day is the number of days bool Isleapyear () const;// Determine if the year is not a leap years bool Check () const;//Check if the time format is correct void Numtostr (string &sy, String &sm, String &sd, String &sw) const;//numeric date converted to character date public:void initialtime ();//input initialization time void Show () const;//display time information void Strshow () const;// Use the expression of a string to display the date int countdays (const time & T2) const;//calculates the number of days of the two-day period};int Time::weekdaycount () const//calculates the day of the week {return Daycount ()% 7;} int time::D aycount () const//calculates the day of the day {int days = 0;days = (year-1) * 365 + (YEAR-1)/4-(YEAR-1)/100+ (year-1 )/+ + daysyearcount (); return days;} int time::D aysyearcount () const//calculates the day of the year, {int days = 0;int mtemp = month-1;while (mtemp > 0) {switch (mtemp) {case (1 ): Case (3): Case (5): Case (7): Case (8): CASE (+): Case (N): Days + = 31; Break;case (4): Case (6): Case (9): Case (one): Days + = 30; Break;case (2):d ays + = 28; Break;default:break;} --mtemp;} if (Isleapyear ()) ++days;//if it is a leap year, plus a day return days + day;//Returns the calculated number plus the number of days in the month}bool time::isleapyear () const//Determine if the years are leap Year% 4 = = 0 && Yearly% = 0) Return true;//is a multiple of four and is not a multiple of 100, is a leap years if (year% = = 0) return true;else return FA LSE;} BOOL Time::check () const//Check if the time format is correct {if (year <= 0 | | (Month <= 0 | | month>12) | |  Day <= 0) return false;else{if (month = = 1 | |  Month = = 3 | |  Month = = 5 | |  Month = = 7| |  Month = = 8 | |  Month = = 10 | |  month = =) && Day > Return false;else{if (month = = 4 | |  Month = = 6 | |  Month = = 9 | |  month = = one) && Day > Return false;else{if (month = = 2) {if (Isleapyear ()) {if (Day >) return false; else return true;} Else{if (Day >) return false; else return true;}}}}} void Time::numtostr (String &sy, String &sm, String &sd, string &sw) const//The numeric date is converted to the character date {sy = Ntos (year/1000) + ntos (year/100-10 * (year/1000)) + Ntos (year/10-10 * (year/100)) + Ntos (ye Ar%), SD = Ntos (DAY/10) + ntos (day%), switch (month) {case (1): SM = "January"; Break;case (2): SM = "February"; BR Eak;case (3): SM = "March"; Break;case (4): SM = "April"; Break;case (5): SM = "may"; Break;case (6): SM = "June"; Break;case (7): SM = "July"; Break;case (8): SM = "August"; Break;case (9): SM = "September"; Break;case (Ten): SM = "October"; Break;case (one): SM = "November"; Break;case (): SM = "December"; Break;default:break;} Switch (Weekday) {case (1): SW = "Monday"; Break;case (2): SW = "Tuesday"; Break;case (3): SW = "Wednesday"; Break;case (4) : SW = "Thursday"; Break;case (5): SW = "Friday"; Break;case (6): SW = "Saturday"; Break;case (7): SW = "Sunday"; Break;default:break;}} void Time::initialtime ()//input initialization time {cout << "Enter the Times (year,month,day): \ n"; Cin >> Year;cin.get (); CIN >> Month;cin.get (); Cin >> Day;cin.get (); if (!check ()) {cout << "Try again:\n"; Initialtime ();} Elseweekday = Weekdaycount ();} void Time::show () const//display time information {cout << year: "<< years <<" \ T "; cout <<" Month: "<< Month & lt;< "\ t"; cout << "Day:" << day << "\ t"; cout << "Weekday:" << Weekday << Endl;cou T << "This is a"; if (Isleapyear ()) cout << "Leap"; else cout << "Nonleap"; cout << "year.\n"; cout << "Today is the" << Daysyearcount () << "Da Ys of the year.\n ";}  void Time::strshow () const//uses the expression of the string to display the date {string St, sy, SM, SD, SW;NUMTOSTR (sy, SM, SD, SW); st = sy + "/" + SM + "/" + SD + "," + "Today is" + SW + ".";  String::iterator it = St.begin (); for (; It! = St.end (); ++it) cout << *it;cout << endl;cout << "This is a "; if (Isleapyear ()) cout <<" Leap "; else cout << "Nonleap"; cout << "year.\n"; cout << "Today is the" << Daysyearcount () << "Da Ys of the YeaR.\n ";} the int time::countdays (const time & T2) const//calculates the number of days in the two-day period {int t = daycount ()-T2. Daycount (); if (T < 0) Return-t;return T;} string ntos (int t)//number converted to character {switch (t) {case (0): return "0", Case (1): Return "1", Case (2): Return "2", Case (3): Return "3"  Case (4): Return "4", Case (5): Return "5", Case (6): Return "6", Case (7): Return "7", Case (8): Return "8", Case (9): return "9";d Efault:break;}} #endif

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Enter a date to determine which day of the year the date is, the days of the week, the day of the two days, and the date used for the string output