POJ 2080 Calendar

最新推荐文章于 2018-11-07 22:59:01 发布

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最新推荐文章于 2018-11-07 22:59:01 发布

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本文链接：https://blog.youkuaiyun.com/b_r_e_a_d/article/details/76734615

Calendar

Description

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

Sample Output

2004-09-26 Sunday
2004-10-06 Wednesday
2004-10-16 Saturday
2004-10-17 Sunday

这道题错得很惨，一直WA。每次遇到这种题都有点不稳的，一直有细节问题，最后全部删掉用另一种方式写了一遍，过了，但罚时也是真的惨。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
string week[9]={"Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday","Friday"};
int day_y(int x)
{
	if((x%4==0&&x%100!=0)||x%400==0)
		return 366;
	return 365;
}
int day_m(int y,int x)
{
	if(x==1||x==3||x==5||x==7||x==8||x==10||x==12)
		return 31;
	if(((y%4==0&&y%100!=0)||y%400==0)&&x==2)
		return 29;
	else if(x==2)
		return 28;
	return 30;
}
int main()
{
	int n;
	int year,month,day;
	while(cin>>n)
	{
		if(n==-1)
			break;
		year=2000;
		month=1;
		day=n%7;
		while(n>=day_y(year))
		{
			n-=day_y(year);
			year++;
		}
		while(n>=day_m(year,month))
		{
			n-=day_m(year,month);
			month++;
		}
		printf("%d-%02d-%02d ",year,month,n+1);
		cout<<week[day]<<endl;
	}
	return 0;
}