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打印日期

[编程题]打印日期

热度指数：50535 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

给出年分m和一年中的第n天，算出第n天是几月几号。

输入描述:

输入包括两个整数y(1<=y<=3000)，n(1<=n<=366)。

输出描述:

可能有多组测试数据，对于每组数据，
按 yyyy-mm-dd的格式将输入中对应的日期打印出来。

示例1

输入

输出

算法知识视频讲解

Python

to_day

def IsLeapYear(year):

if (year % 4 == 0 and year % 100 != 0) or (year % 400) == 0:
return 1
else:
return 0

month_day = [
[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31],
[31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
]
while True:
try:
year, num = map(int, input().split())
row = IsLeapYear(year)
month = 0
while num > month_day[row][month]:
num -= month_day[row][month]
month += 1
day = num
print("%d-%02d-%02d" % (year, month + 1, day))
except:
break

发表于 2022-08-01 16:04:55 回复(0)

Python

lx的流星

while True:
    try:
        year,day=map(int,input().strip().split(' '))
        def isrunnian(year):
            if year%4==0 and year%100!=0:
                return True
            elif year%400==0:
                return True
            else:
                return False
        list1=[31,28,31,30,31,30,31,31,30,31,30,31]
        list2=[31,29,31,30,31,30,31,31,30,31,30,31]
        sum1=0
        if isrunnian(year):
            for i in range(1,13):
                sum1=sum(list2[0:i])
                if sum1>=day:
                    month=i
                    date=day-(sum1-list2[i-1])
                    break
        else:
            for i in range(1,13):
                sum1=sum(list1[0:i])
                if sum1>=day:
                    month=i
                    date=day-(sum1-list1[i-1])
                    break
        result=[str(year)]
        if len(str(month))<2:
            result.append('0'+str(month))
        else:
            result.append(str(month))
        if len(str(date))<2:
            result.append('0'+str(date))
        else:
            result.append(str(date))
        print('-'.join(result))

    except:
        break

发表于 2019-08-22 08:56:18 回复(0)

Python

华科平凡

使用python的datetime库，美滋滋。

import datetime

while True:
    try:
        year, day = map(int, input().split())
        first_day = datetime.datetime(year, 1, 1)
        delta = datetime.timedelta(days=day - 1)
        print(datetime.datetime.strftime(first_day + delta, "%Y-%m-%d"))


    except:
        break

发表于 2017-10-04 08:50:08 回复(0)

Python

Daws0n

import sys

months = {0:0,1:31,2:28,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}

def sum_month(num):

sum_days = 0

for n in range(1,num+1):

sum_days += months[n]

return sum_days

for line in sys.stdin:

years,days = line.split(' ')

year = int(years)

days = int(days)

if year%4==0 and year%100 != 0:

months[2] = 29

elif year%4 == 0 and year%100 == 0 and year%400 == 0:

months[2] = 29

else:

months[2] = 28

for mon in range(1,13):

if sum_month(mon) >= days:

month = mon

day = days - sum_month(mon-1)

break

print '%4d-%02d-%02d' % (year,month,day)

发表于 2017-02-19 15:24:03 回复(0)

Python

fhqplzj

import datetime as D

while True:
    try:
        parts = map(int, raw_input().split())
        print (D.datetime(parts[0], 1, 1) + D.timedelta(parts[1] - 1)).strftime('%Y-%m-%d')
    except EOFError:
        break

in java:

import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Scanner;

/**
 * Created by fhqplzj on 17-1-6 at 下午7:52.
 */
public class Main {
    public static void main(String[] args) {
        Calendar calendar = Calendar.getInstance();
        SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd");
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int year = scanner.nextInt();
            int delta = scanner.nextInt();
            calendar.set(year, 0, 0);
            calendar.add(Calendar.DAY_OF_YEAR, delta);
            System.out.println(simpleDateFormat.format(calendar.getTime()));
        }
    }
}

编辑于 2017-01-06 20:03:36 回复(0)

Python

questRush

import datetime as D
try:
    while 1:
        a = map(int, raw_input().split())
        print (D.datetime(a[0] - 1,12,31) + D.timedelta(days=a[1])).strftime('%Y-%m-%d')
except:
    pass

发表于 2016-12-23 12:05:31 回复(0)