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[해커랭크(Hackerrank)/30 Days of Code/파이썬3(python3)] Day 6: Let's Review 본문
[해커랭크(Hackerrank)/30 Days of Code/파이썬3(python3)] Day 6: Let's Review
규도자 (gyudoza) 2019. 3. 21. 20:57Objective
Today we're expanding our knowledge of Strings and combining it with what we've already learned about loops. Check out the Tutorial tab for learning materials and an instructional video!
Task
Given a string, , of length that is indexed from to , print its even-indexed and odd-indexed characters as space-separated strings on a single line (see the Sample below for more detail).
Note: is considered to be an even index.
Input Format
The first line contains an integer, (the number of test cases).
Each line of the subsequent lines contain a String, .
Constraints
Output Format
For each String (where ), print 's even-indexed characters, followed by a space, followed by 's odd-indexed characters.
Sample Input
2
Hacker
Rank
Sample Output
Hce akr
Rn ak
Explanation
Test Case 0:
The even indices are , , and , and the odd indices are , , and . We then print a single line of space-separated strings; the first string contains the ordered characters from 's even indices (), and the second string contains the ordered characters from 's odd indices ().
Test Case 1:
The even indices are and , and the odd indices are and . We then print a single line of space-separated strings; the first string contains the ordered characters from 's even indices (), and the second string contains the ordered characters from 's odd indices ().
풀이
# Enter your code here. Read input from STDIN. Print output to STDOUT
test_case_count = int(input())
test_case_input = []
for test_case_index in range(test_case_count):
current_word = input()
odd_string = ''
even_string = ''
for string_index in range(len(current_word)):
if not string_index % 2 == 0:
even_string = even_string + current_word[string_index]
else:
odd_string = odd_string + current_word[string_index]
print(odd_string + ' ' + even_string)
설명
0은 짝수로 치고 홀수번째 글자와 짝수번째 글자를 따로 분류하여 출력하는 문제이다.