題目1043：Day of Week 九度OJ

題目1043：Day of Week

時間限制：1 秒

記憶體限制：32 兆

特殊判題：否

送出：9211

解決：3167

題目描述：

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.

輸入：

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.

輸出：

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.

樣例輸入：

9 October 2001
14 October 2001
      

樣例輸出：

Tuesday
Sunday      

提示：

Month and Week name in Input/Output:

January, February, March, April, May, June, July, August, September, October, November, December

Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

來源：

2008年上海交通大學計算機研究所學生機試真題

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib> 
#define ISLEAP(x) (x%4==0&&x%100!=0)||x%400==0 ? 1:0
using namespace std;

char week[7][15]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};
char month[13][15]={"","January","February","March","April","May","June","July","August","September","October","November","December"};
int months[13][2]={
0,0,
31,31,
28,29,
31,31,
30,30,
31,31,
30,30,
31,31,
31,31,
30,30,
31,31,
30,30,
31,31
};

struct Date{
	int y,m,d;
	void nextday(){
		d++;
		if(d>months[m][ISLEAP(y)]){
			m++;
			d=1;
			if(m>12){
				y++;
				m=1;
			} 
		}	
	}
};
int buf[3001][13][33];
int main(){
	Date tmp;
	tmp.y=0;
	tmp.m=1;
	tmp.d=1;
	int cnt=0;
	while(tmp.y!=3001){//其實我們并不知道0年1月1日是星期幾 
		buf[tmp.y][tmp.m][tmp.d]=cnt;
		tmp.nextday();
		cnt++;
	}
	int day,year,mon,all,dayofweek;	
	char s[15];
	while(scanf("%d%s%d",&day,s,&year)!=EOF){
		for(int i=1;i<13;i++){
			if(strcmp(s,month[i])==0){
				mon=i;break;
			}
		}
		all=buf[year][mon][day]-buf[2001][10][14];//兩天的天數間隔，可能為負  
		if(all>=0){
			dayofweek=abs(all)%7;			
		}else{
			if(abs(all)%7==0){
				dayofweek=0;	
			}else{
				dayofweek=7-abs(all)%7;
			}	
		}
		cout<<week[dayofweek]<<endl;
	}
	return 0;
}