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[토르] 8주차 과제 제출 #253
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[토르] 8주차 과제 제출 #253
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,57 @@ | ||
from collections import deque | ||
import sys | ||
import copy | ||
from collections import defaultdict | ||
import heapq | ||
|
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print("SLURPYS OUTPUT") | ||
N = int(input()) | ||
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def is_slump(string): | ||
if len(string) >= 1 and not (string[0] == "D" or string[0] == "E"): | ||
return False | ||
for i in range(1, len(string)): | ||
if len(string) > 2 and string[1] != "F": | ||
return False | ||
if len(string) >= 1 and i == len(string) - 1: | ||
return string[-1] == "G" | ||
if string[i] == "F": | ||
continue | ||
else: | ||
return is_slump(string[i:]) | ||
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def is_slimp(string): | ||
if len(string) >= 1 and string[0] != "A": | ||
return False | ||
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if len(string) < 2: | ||
return False | ||
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if len(string) == 2: | ||
return string[1] == "H" | ||
else: | ||
if len(string) >= 2 and string[0] == "A" and string[1] == "B" and is_slimp(string[2:-1]) and string[-1] == "C": | ||
return True | ||
elif len(string) >= 2 and string[0] == "A" and is_slump(string[1:-1]) and string[-1] == "C": | ||
return True | ||
else: | ||
return False | ||
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def is_slurpy(string): | ||
for i in range(len(string)): | ||
if is_slimp(string[:i]): | ||
if is_slump(string[i:]): | ||
return True | ||
return False | ||
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for _ in range(N): | ||
sentence = input() | ||
result = is_slurpy(sentence) | ||
if result: | ||
print("YES") | ||
else: | ||
print("NO") | ||
print("END OF OUTPUT") |
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,14 @@ | ||
from collections import deque | ||
import sys | ||
import copy | ||
from collections import defaultdict | ||
import heapq | ||
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input = sys.stdin.readline | ||
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A, B, C = map(int, input().split()) | ||
a_numbers = sorted(list(map(int, input().split()))) | ||
b_numbers = sorted(list(map(int, input().split()))) | ||
c_numbers = sorted(list(map(int, input().split()))) | ||
|
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answer = sys.maxsize | ||
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정렬한 다음에, 진짜로 모든 부분에 대해 전부 브루트포스하는 게 정해입니다 ㅎㅎ...
단, O(N^3) 풀이는 안됩니다.
최대를 가지는 사람, 최소를 가지는 사람을 고정시켜서 총 3P2가지 경우로 시도해보세요!