问题 B: Day of Week

最新推荐文章于 2022-05-04 00:20:42 发布

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最新推荐文章于 2022-05-04 00:20:42 发布

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分类专栏： codeup

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本文介绍了一个程序，用于将格里高利历（俄历）的特定日期转换为对应的星期几。程序接收年月日输入，考虑到闰年的规则，并输出对应日期的英文星期名称。示例输入包括2012年12月21日和2013年1月5日，输出分别为Friday和Saturday。

题目描述

We now use the Gregorian style of dating in Russia. The leap years are years with number divisible by 4 but not divisible by 100, or divisible by 400.
For example, years 2004, 2180 and 2400 are leap. Years 2004, 2181 and 2300 are not leap.
Your task is to write a program which will compute the day of week corresponding to a given date in the nearest past or in the future using today’s agreement about dating.

输入

There is one single line contains the day number d, month name M and year number y(1000≤y≤3000). The month name is the corresponding English name starting from the capital letter.

输出

Output a single line with the English name of the day of week corresponding to the date, starting from the capital letter. All other letters must be in lower case.

样例输入

21 December 2012
5 January 2013

样例输出

Friday
Saturday

#include <stdio.h>
#include <string.h>
char month[13][20]={
	"",
	"January",
	"February",
	"March",
	"April",
	"May",
	"June",
	"July",
	"August",
	"September",
	"October",
	"November",
	"December"
};
char week[7][20]={
	"Sunday",
	"Monday",
	"Tuesday",
	"Wednesday",
	"Thursday",
	"Friday",
	"Saturday"
};

int year[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},{0,31,29,31,30,31,30,31,31,30,31,30,31}};

int isLeap(int year){
	return ((year%4==0&&year%100!=0)||year%400==0);
}
struct date{
	int t;
	int y;
	int m;
	int d;
};
int main()
{
	struct date d1,base_date={20130106,2013,1,6};
	char m[20];
	while(scanf("%d %s %d", &d1.d, m, &d1.y)!=EOF){
		int diff=0,sign=-1; 
		struct date d2=base_date;
		for(int i=1;i<=12;i++){
			if(!strcmp(m,month[i])) {d1.m=i;break;}
		}
		d1.t= d1.y*10000 + d1.m*100 + d1.d;
		if(d1.t>d2.t){int tmp; tmp=d1.t; d1.t=d2.t; d2.t=tmp; sign=1;}
		if(d1.t==d2.t){printf("%s\n", week[0]);continue;}
		d1.y=d1.t/10000; d1.m=d1.t%10000/100; d1.d=d1.t%100;
		d2.y=d2.t/10000; d2.m=d2.t%10000/100; d2.d=d2.t%100;
		while(d1.t<d2.t){
			diff++;
			int le=isLeap(d1.y);
			if(d1.m==12 && d1.d==31){d1.y++; d1.m=1;d1.d=1;}
			else if(d1.d == year[le][d1.m]){d1.m++; d1.d=1;}
			else d1.d++;
			d1.t= d1.y*10000 +d1.m*100 +d1.d;
		}
		if(sign>0) printf("%s\n", week[diff%7]);
		if(sign<0) printf("%s\n", week[6-(diff-1)%7]);
	};
	return 0;
}