Solution of ZOJ 2420 Calendar

A calendar is a system for measuring time, from hours and minutes, to months and days, and finally to years and centuries. The terms of hour, day, month, year and century are all units of time measurements of a calender system.

According to the Gregorian calendar, which is the civil calendar in use today, years evenly divisible by 4 are leap years, with the exception of centurial years that are not evenly divisible by 400. Therefore, the years 1700, 1800, 1900 and 2100 are not leap years, but 1600, 2000, and 2400 are leap years.

Given the number of days that have elapsed since January 1, 2000 A.D, your mission is to find the date and the day of the week.

 

Input

The input consists of lines each containing a positive integer, which is the number of days that have elapsed since January 1, 2000 A.D. The last line contains an integer -1, which should not be processed. You may assume that the resulting date won't be after the year 9999.

 

Output

For each test case, output one line containing the date and the day of the week in the format of "YYYY-MM-DD DayOfWeek", where "DayOfWeek" must be one of "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" and "Saturday".

 

Sample Input

1730
1740
1750
1751
-1

 

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下面是我的实现代码：

 

#include <iostream> #include <cstdio> using namespace std; string week_name[] = { "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday" }; int _days_of_month[13] = { -1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; bool is_leap_year( int year ); void calc_date( int days ); int days_of_year( int year ); int days_of_month( int year, int month ); int main() { int days_elapsed; while( cin >> days_elapsed ) { if( days_elapsed == -1 ) break; calc_date( days_elapsed ); } return 0; } void calc_date( int days ) { int year = 2000, month = 1, day = 6; day += days++; day %= 7; int days_year = days_of_year( year ); while( days > days_year ) { days -= days_year; days_year = days_of_year( ++year ); } int days_month = days_of_month( year, month ); while( days > days_month ) { days -= days_month; days_month = days_of_month( year, ++month ); } printf( "%d-%02d-%02d %s/n", year, month, days, week_name[day].c_str() ); } bool is_leap_year( int year ) { return ( year % 4 == 0 && year % 100 != 0 ) || ( year % 400 == 0 ); } int days_of_year( int year ) { return is_leap_year(year) ? 366 : 365; } int days_of_month( int year, int month ) { return ( month == 2 && is_leap_year( year ) ) ? 29 : _days_of_month[month]; }