jxtxzzw edited 7 年，3 月前
题目的意思是给定一天
然后输出这是星期几
这道题的测试数据有误
下详
用Java做竟然是Wrong Answer
用ZYG的模板是Accept
可见模板是错的
下面先给出AC的代码
注意因为测试数据有误
所以这段AC的代码实际上是错误的
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
string dayOfWeek[] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};
int DateToInt (int m, int d, int y){
return
1461 * (y + 4800 + (m - 14) / 12) / 4 +
367 * (m - 2 - (m - 14) / 12 * 12) / 12 -
3 * ((y + 4900 + (m - 14) / 12) / 100) / 4 +
d - 32075;
}
void IntToDate (int jd, int &m, int &d, int &y){
int x, n, i, j;
x = jd + 68569;
n = 4 * x / 146097;
x -= (146097 * n + 3) / 4;
i = (4000 * (x + 1)) / 1461001;
x -= 1461 * i / 4 - 31;
j = 80 * x / 2447;
d = x - 2447 * j / 80;
x = j / 11;
m = j + 2 - 12 * x;
y = 100 * (n - 49) + i + x;
}
string IntToDay (int jd){
return dayOfWeek[jd % 7];
}
int main()
{
int m,d,y;
scanf("%d-%d-%d",&y,&m,&d);
cout<<IntToDay(DateToInt(m,d,y))<<endl;
return 0;
}
下面给出正确的代码
注意由于测试数据有误
所以这段代码运行的结果是WA
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
final String namesOfWeekdays[] = { "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday",
"Saturday" };
Scanner in = new Scanner(System.in);
Calendar calendarInstance = Calendar.getInstance();
Date dateInformation = new Date();
try {
dateInformation = new SimpleDateFormat("yyyy-MM-dd").parse(in.nextLine());
} catch (ParseException e) {
e.printStackTrace();
}
calendarInstance.setTime(dateInformation);
int dayOfWeek = calendarInstance.get(Calendar.DAY_OF_WEEK) - 1;
if (dayOfWeek < 0)
dayOfWeek = 0;
System.out.println(namesOfWeekdays[dayOfWeek]);
}
}
原因如下
罗马的某个教皇把1582年10月5日至14日这10天撤销了
（详询Wiki）
1582年10月4日后紧跟着就是15日
所以1582-10-15是星期五
它前面的1天是1582-10-4是星期四
注意到中间的10天是不存在的
如果按照一般的情况
10月15日是星期五
10月4日比15日相差了11天
所以是星期一
按照模板做下来确实
1582-10-15是星期五
1582-10-04是星期一
然而Java的Calendar在这里做了特殊的处理
1582-10-15是星期五
1582-10-04是星期四
同样的
select to_date(‘1582-10-4’,’yyyy-mm-dd’)+1 from dual
答案是1582-10-15
由此可见
模板仅适用于1582-10-15及以后的日历
但是题目说到年份包含前导0
那么就应该分类讨论
分10-04之前的和10-15之后的两个公式
附ZYG的模板
/*
// Routines for performing computations on dates. In these routines,
// months are exprsesed as integers from 1 to 12, days are expressed
// as integers from 1 to 31, and years are expressed as 4-digit
// integers.
string dayOfWeek[] = {"Mo", "Tu", "We", "Th", "Fr", "Sa", "Su"};
// converts Gregorian date to integer (Julian day number)
int DateToInt (int m, int d, int y){
return
1461 * (y + 4800 + (m - 14) / 12) / 4 +
367 * (m - 2 - (m - 14) / 12 * 12) / 12 -
3 * ((y + 4900 + (m - 14) / 12) / 100) / 4 +
d - 32075;
}
// converts integer (Julian day number) to Gregorian date: month/day/year
void IntToDate (int jd, int &m, int &d, int &y){
int x, n, i, j;
x = jd + 68569;
n = 4 * x / 146097;
x -= (146097 * n + 3) / 4;
i = (4000 * (x + 1)) / 1461001;
x -= 1461 * i / 4 - 31;
j = 80 * x / 2447;
d = x - 2447 * j / 80;
x = j / 11;
m = j + 2 - 12 * x;
y = 100 * (n - 49) + i + x;
}
// converts integer (Julian day number) to day of week
string IntToDay (int jd){
return dayOfWeek[jd % 7];
}
*/