[알고리즘] 22. Drawing Book

Brie’s Drawing teacher asks her class to open their books to a page number. Brie can either start turning pages from the front of the book or from the back of the book. She always turns pages one at a time. When she opens the book, page  is always on the right side:

When she flips page , she sees pages  and . Each page except the last page will always be printed on both sides. The last page may only be printed on the front, given the length of the book. If the book is  pages long, and she wants to turn to page , what is the minimum number of pages she will turn? She can start at the beginning or the end of the book.

Given  and , find and print the minimum number of pages Brie must turn in order to arrive at page .

Function Description

Complete the pageCount function in the editor below. It should return the minimum number of pages Brie must turn.

pageCount has the following parameter(s):

• n: the number of pages in the book
• p: the page number to turn to

Input Format

The first line contains an integer , the number of pages in the book.
The second line contains an integer, , the page that Brie's teacher wants her to turn to.

Constraints

Output Format

Print an integer denoting the minimum number of pages Brie must turn to get to page .

Sample Input 0

6
2

Sample Output 0

1

Explanation 0

If Brie starts turning from page , she only needs to turn  page:

If Brie starts turning from page , she needs to turn  pages:

Because we want to print the minumum number of page turns, we print  as our answer.

Sample Input 1

5
4

Sample Output 1

0

Explanation 1

If Brie starts turning from page , she needs to turn  pages:

If Brie starts turning from page , she doesn't need to turn any pages:

Because we want to print the minimum number of page turns, we print  as our answer.

답 : 

#!/bin/python3

import os

import sys

#

# Complete the pageCount function below.

#

def pageCount(n, p):

right = n-p

count = 0

if right < p:

if n % 2 != 0:

if n-p >1:

count = (n-p)//2

else:

count = (n+1-p)//2

else:

count += p//2

return count

if __name__ == '__main__':

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

n = int(input())

p = int(input())

result = pageCount(n, p)

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

fptr.close()

앞에서부터 시작할지 뒤에서 부터 시작할지를 먼저 정하기 위해 (총 페이지수 - 목표 페이지) 와 (목표페이지)의 크기를 비교해서 나눴다.

총 홀수 페이지면 마지막장이 양면, 짝수면 단면에 존재하기에 n+1로 표형했고,