[알고리즘] 43. Library Fine

Your local library needs your help! Given the expected and actual return dates for a library book, create a program that calculates the fine (if any). The fee structure is as follows:

1. If the book is returned on or before the expected return date, no fine will be charged (i.e.: .
2. If the book is returned after the expected return day but still within the same calendar month and year as the expected return date, .
3. If the book is returned after the expected return month but still within the same calendar year as the expected return date, the .
4. If the book is returned after the calendar year in which it was expected, there is a fixed fine of .

Charges are based only on the least precise measure of lateness. For example, whether a book is due January 1, 2017 or December 31, 2017, if it is returned January 1, 2018, that is a year late and the fine would be .

Function Description

Complete the libraryFine function in the editor below. It must return an integer representing the fine due.

libraryFine has the following parameter(s):

• d1, m1, y1: returned date day, month and year
• d2, m2, y2: due date day, month and year

Input Format

The first line contains  space-separated integers, , denoting the respective , , and  on which the book was returned.
The second line contains  space-separated integers, , denoting the respective , , and  on which the book was due to be returned.

Constraints

Output Format

Print a single integer denoting the library fine for the book received as input.

Sample Input

9 6 2015
6 6 2015

Sample Output

45

Explanation

Given the following dates:
Returned: 
Due: 

Because , we know it is less than a year late.
Because , we know it's less than a month late.
Because , we know that it was returned late (but still within the same month and year).

Per the library's fee structure, we know that our fine will be . We then print the result of  as our output

답 :

#!/bin/python3

import math

import os

import random

import re

import sys

# Complete the libraryFine function below.

def libraryFine(d1, m1, y1, d2, m2, y2):

# 1이 반납일 / 2가 대여일

fee = 0

if y1 > y2:

fee += 10000

elif y1 == y2:

if m1 > m2:

fee += (500*(m1-m2))

elif m1 == m2:

if d1 > d2:

fee += (15*(d1-d2))

return fee

if __name__ == '__main__':

fptr = open(os.environ['OUTPUT_PATH'], 'w')

d1M1Y1 = input().split()

d1 = int(d1M1Y1[0])

m1 = int(d1M1Y1[1])

y1 = int(d1M1Y1[2])

d2M2Y2 = input().split()

d2 = int(d2M2Y2[0])

m2 = int(d2M2Y2[1])

y2 = int(d2M2Y2[2])

result = libraryFine(d1, m1, y1, d2, m2, y2)

fptr.write(str(result) + '\n')

fptr.close()

연도/월/일 순으로 판단 반납연도가 대여연도를 넘어버리면 무조건 10000 * 연도간 차이수 하고 계산 끝

연도는 같은데 반납월이 대여월을 넘으면 500 * 월간 차이수 하고 계산 끝

연도 월도 같은데 반납일이 대여일을 넘으면 15 * 일간 차이수 하고 계산 끝