HDU 2133---某年某月某日是星期几？

最新推荐文章于 2021-11-01 15:33:49 发布

原创最新推荐文章于 2021-11-01 15:33:49 发布 · 313 阅读

1 ·

CC 4.0 BY-SA版权

本文介绍两种将日期转换为星期的方法，包括自定义公式法和基姆拉尔森计算公式法，并提供C语言实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述

Today is Saturday, 17th Nov,2007. Now, if i tell you a date, can you tell me what day it is ?

Input

There are multiply cases.
One line is one case.
There are three integers, year(0<year<10000), month(0<=month<13), day(0<=day<32).

Output

Output one line.
if the date is illegal, you should output "illegal". Or, you should output what day it is.

Sample Input

2007 11 17

Sample Output

Saturday

solve 1：

公式：W = month+365*(year-1)+r+day

W是公元1年1月1日到所给日期所积天数

365*（year-1)+r是公元1年1月1日到所给年1月1日所积天数

month是所给年1月1日到所给日期所积天数

day是所给月1日到所给日期所积天数

代码

#include<stdio.h>

int run(int year)
{
	return (year%4==0&&year%100!=0||year%400==0);
}
int b[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};



int main()
{
	int year,month,day;
	
	while(~scanf("%d%d%d",&year,&month,&day))
	{
		 if(run(year))
		 {
		 	b[2]=29;
		 }
		 else
		 {
		 	b[2]=28;
		 }
		 if(day>b[month]||year<1||year>10000||day<1||month<1||month>12)
		 {
		 	printf("illegal\n");
		 	continue;
		 }
			int W=0,r=0;
			for(int i=1;i<year;i++)
			{
				if(run(i))
				{
					r++;//标记有几个闰年
				}
			}
			for(int j=1;j<month;j++)
			{
				W=W+b[j];
			}
		
			W = W+365*(year-1)+r+day; //r为多的天数，有几个闰年多几天
			int p=W%7;//最终的W代表从公元1年到题给的日期所积的天数，取余7得星期数
			switch(p)
				{
					case 0:
						printf("Sunday\n");
						break;
					case 1:
						printf("Monday\n");
						break;
					case 2:
						printf("Tuesday\n");
						break;
					case 3:
						printf("Wednesday\n");
						break;
					case 4:
						printf("Thursday\n");
						break;
					case 5:
						printf("Friday\n");
						break;
					case 6:
						printf("Saturday\n");
						break;
					
				}
	
	}
	return 0; 
}

solve 2:

基姆拉尔森计算公式:

w=day+month*2 + 3*（month + 1) / 5 + year + year/4 - year/100 + year/400 + 1 ;

代码：

#include<stdio.h>

int run(int n)
{
	return  (n%4==0&&n%100!=0)||(n%400==0);
}

int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int main()
{
	int d,m,y,w;
	while(~scanf("%d%d%d",&y,&m,&d))
	{
		if(run(y))
		{
			a[2]=29;
		}
		else
		{
			a[2]=28;
		}
		
		if(d<1||d>a[m]||m<1||m>12||y<1||y>10000)
		{
			printf("illegal\n");
		}
		else
		{
			if(m==1|m==2)
			{
				m=m+12;
				y--;
			}
			w=(d+m*2+3*(m+1)/5+y+y/4-y/100+y/400+1)%7;
			switch(w)
			{
				case 0:
						printf("Sunday\n");
						break;
					case 1:
						printf("Monday\n");
						break;
					case 2:
						printf("Tuesday\n");
						break;
					case 3:
						printf("Wednesday\n");
						break;
					case 4:
						printf("Thursday\n");
						break;
					case 5:
						printf("Friday\n");
						break;
					case 6:
						printf("Saturday\n");
						break;

			}
			
		}
	}
	return 0;
 }