[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
s0n9h2 wants to merge 4 commits into
base:main
Choose a base branch
from
Open

[CR] 10월 2, 3주차 #98

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
4 commits
Select commit Hold shift + click to select a range
e5f25db
[baekjoon-11559] PUYO PUYO
s0n9h2 Oct 7, 2022
5213e25
[programmers] 모음사전
s0n9h2 Oct 7, 2022
15d4582
[baekjoon-9935] 문자열 폭발
s0n9h2 Oct 10, 2022
47add06
[programmers] 모음사전
s0n9h2 Oct 15, 2022
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
94 changes: 94 additions & 0 deletions SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py
View file
Open in desktop
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
global trigger
dx = [1, -1, 0, 0]
dy = [0, 0, 1, -1]

count = 1
puyo = [(i, j)]
q = deque([(i, j)])
visited[i][j] = True
while q:
x, y = q.popleft()
for d in range(4):
nx = x + dx[d]
ny = y + dy[d]
if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
# 상하좌우의 같은 색 뿌요 찾기
if state[x][y] == state[nx][ny]:
visited[nx][ny] = True
q.append((nx, ny))
count += 1
puyo.append((nx, ny))

if count >= 4:
# 뿌요가 한 번이라도 터지면 콤보!
trigger = True
# 뿌요를 터뜨리는 함수 돌리기
for px, py in puyo:
state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
for j in range(6):
q = deque([])
for i in range(11, -1, -1):
# 한 줄에 남아있는 뿌요들만 모으기
if state[i][j] != '.':
q.append(state[i][j])
# 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
for i in range(11, -1, -1):
if q:
state[i][j] = q.popleft()
else:
state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
combo = 0
while trigger == True:
trigger = False
visited = [[False] * 6 for _ in range(12)]
for i in range(11, -1, -1):
for j in range(6):
if visited[i][j] == False and state[i][j] != ".":
# 현상황에서 터뜨리는 함수 돌리기
PuyoPop(i, j)
# print("")
# for i in range(12):
# print("".join(state[i]))
# print("=================")

# 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기
if trigger == True:
combo += 1
PuyoDrop(state)

# for i in range(12):
# print("".join(state[i]))
# print("=================")

print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.
29 changes: 29 additions & 0 deletions SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py
View file
Open in desktop
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
global num, answer
# print(tmp_word, num)
if tmp_word == word:
return num
if len(tmp_word) == 5:
return

for w in word_list:
num += 1
tmp_word.append(w)
answer = backtracking(word, tmp_word)
if answer:
return answer
tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
tmp_word = []
backtracking(list(word), tmp_word)

return answer
20 changes: 20 additions & 0 deletions SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py
View file
Open in desktop
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
string = list(input().rstrip())
bomb_str = list(input().rstrip())
# 임시 스택에 넣어 비교하기
tmp = []
for i in range(len(string)):
tmp.append(string[i])
if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
if tmp[-len(bomb_str):] == bomb_str:

@blcklambblcklambOct 13, 2022

Copy link
Copy Markdown
Contributor

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,
송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.

iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:

=> 548ms

iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:

=> 580ms

# for i in range(len(bomb_str)):
# tmp.pop()
del tmp[-len(bomb_str):]

if tmp:
print("".join(tmp))
else:
print("FRULA")

@blcklambblcklambOct 13, 2022

Copy link
Copy Markdown
Contributor

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i'); if (__m === '*' || __re.test(location.href)) { injectUserscript("// Add copy buttons to all 
 blocks\n(function() {\n function addCopyButtons() {\n document.querySelectorAll('pre code').forEach(function(codeBlock) {\n if (codeBlock.parentElement.hasAttribute('data-copy-added')) return;\n codeBlock.parentElement.setAttribute('data-copy-added', 'true');\n \n var btn = document.createElement('button');\n btn.textContent = 'Copy';\n btn.style.cssText = 'position:absolute;top:4px;right:4px;padding:2px 8px;font-size:11px;background:#4ecdc4;border:none;border-radius:4px;color:#1a1a2e;cursor:pointer;opacity:0.7;transition:opacity 0.2s;';\n btn.onmouseover = function() { this.style.opacity = '1'; };\n btn.onmouseout = function() { this.style.opacity = '0.7'; };\n btn.onclick = function() {\n navigator.clipboard.writeText(codeBlock.textContent).then(function() {\n btn.textContent = 'Copied!';\n setTimeout(function() { btn.textContent = 'Copy'; }, 1500);\n });\n };\n codeBlock.parentElement.style.position = 'relative';\n codeBlock.parentElement.appendChild(btn);\n });\n }\n \n addCopyButtons();\n \n // Re-run on dynamic content\n var observer = new MutationObserver(addCopyButtons);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Add Copy Buttons to Code Blocks");
}
} catch(__e) { console.warn('[Userscript:Add Copy Buttons to Code Blocks]', __e); }
})();
(function(){
try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Force GitHub README to respect dark mode\n(function() {\n var style = document.createElement('style');\n style.textContent = '\n .markdown-body {\n color-scheme: dark light;\n }\n .markdown-body pre { background: #161b22 !important; }\n .markdown-body code { background: rgba(110, 118, 129, 0.4) !important; }\n .markdown-body table th, .markdown-body table td { border-color: #30363d !important; }\n .markdown-body img { background: #0d1117; }\n .markdown-body blockquote { border-left-color: #8b949e; }\n .markdown-body hr { border-color: #30363d; }\n ';\n document.head.appendChild(style);\n})();", "GitHub Dark Mode README Fix");
}
} catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Highlight search terms from Google/DuckDuckGo/Bing referrer\n(function() {\n var ref = document.referrer;\n var terms = [];\n \n if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) {\n var url = new URL(ref);\n var q = url.searchParams.get('q') || url.searchParams.get('p');\n if (q) {\n terms = q.split(/\\s+/).filter(function(t) { return t.length > 2; });\n }\n }\n \n if (terms.length === 0) return;\n \n var style = document.createElement('style');\n style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }';\n document.head.appendChild(style);\n \n function highlight(node) {\n if (node.nodeType === 3) { // text node\n var text = node.textContent;\n var found = false;\n terms.forEach(function(term) {\n var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\\]\\\\]/g, '\\\\') + ')', 'gi');\n if (regex.test(text)) {\n found = true;\n var frag = document.createDocumentFragment();\n var parts = text.split(regex);\n parts.forEach(function(part, i) {\n if (i % 2 === 0) {\n frag.appendChild(document.createTextNode(part));\n } else {\n var span = document.createElement('span');\n span.className = 'userscript-highlight';\n span.textContent = part;\n frag.appendChild(span);\n }\n });\n node.parentNode.replaceChild(frag, node);\n }\n });\n } else if (node.nodeType === 1 && node.childNodes) { // element\n var skipTags = ['SCRIPT', 'STYLE', 'NOSCRIPT', 'TEXTAREA', 'INPUT', 'SELECT'];\n if (!skipTags.includes(node.tagName)) {\n Array.from(node.childNodes).forEach(highlight);\n }\n }\n }\n \n highlight(document.body);\n \n // Re-highlight on dynamic content\n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1 || node.nodeType === 3) highlight(node);\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Highlight Search Terms");
}
} catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Strip utm_, fbclid, gclid, etc. from all links on page\n(function() {\n var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content',\n 'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid',\n 'ref', 'ref_src', 'source', 'medium', 'campaign'];\n \n function cleanUrl(url) {\n try {\n var u = new URL(url, window.location.origin);\n var changed = false;\n trackingParams.forEach(function(p) {\n if (u.searchParams.has(p)) {\n u.searchParams.delete(p);\n changed = true;\n }\n });\n return changed ? u.toString() : url;\n } catch (e) {\n return url;\n }\n }\n \n function cleanLinks() {\n document.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n \n cleanLinks();\n \n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1) {\n if (node.tagName === 'A') cleanLinks();\n node.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Remove Tracking Parameters from Links");
}
} catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); }
})();
(function(){
try {
var __m = "youtube.com";
var __re = new RegExp('^' + "youtube\\.com" + '

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Auto-enable theater mode on YouTube\n(function() {\n function tryTheater() {\n var btn = document.querySelector('button[aria-label=\"Theater mode\"], ytd-player #player button[title=\"Theater mode\"]');\n if (btn && !btn.classList.contains('activated')) {\n btn.click();\n }\n }\n \n // Try immediately\n tryTheater();\n \n // Try after navigation (SPA)\n var lastUrl = location.href;\n setInterval(function() {\n if (location.href !== lastUrl) {\n lastUrl = location.href;\n setTimeout(tryTheater, 500);\n }\n }, 1000);\n \n // Also try on player load\n var observer = new MutationObserver(tryTheater);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "YouTube Theater Mode Default");
}
} catch(__e) { console.warn('[Userscript:YouTube Theater Mode Default]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Remove or un-stick sticky/fixed headers that block content\n(function() {\n function unstick() {\n document.querySelectorAll('header, nav, [role=\"banner\"], .header, .navbar, .sticky, .fixed-top, [style*=\"position: fixed\"], [style*=\"position:sticky\"]').forEach(function(el) {\n if (el.style.position === 'fixed' || el.style.position === 'sticky' || \n getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') {\n el.style.position = 'static';\n el.style.top = 'auto';\n el.style.zIndex = 'auto';\n }\n });\n }\n \n unstick();\n \n var observer = new MutationObserver(unstick);\n observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] });\n})();", "Kill Sticky Headers");
}
} catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Universal Dark Mode - works on any site\n(function() {\n var enabled = true;\n \n function applyDarkMode() {\n if (!enabled) return;\n \n // Create style element if it doesn't exist\n var style = document.getElementById('universal-dark-mode-style');\n if (!style) {\n style = document.createElement('style');\n style.id = 'universal-dark-mode-style';\n document.head.appendChild(style);\n }\n \n // Dark mode CSS - inverts colors but preserves images/video\n style.textContent = '\n /* Invert everything except media */\n html {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #1a1a2e !important;\n }\n \n /* Restore images, videos, iframes, canvas */\n img, video, iframe, canvas, svg, picture, [style*=\"background-image\"] {\n filter: invert(1) hue-rotate(180deg) !important;\n }\n \n /* Preserve specific elements that should not be inverted */\n .no-dark-mode, .no-dark-mode *,\n [data-theme=\"light\"], [data-theme=\"light\"],\n .ace_editor, .ace_editor *,\n .CodeMirror, .CodeMirror *,\n .monaco-editor, .monaco-editor *,\n .markdown-body pre, .markdown-body pre *,\n .highlight, .highlight *,\n pre code, pre code * {\n filter: none !important;\n }\n \n /* Fix common UI elements */\n .modal, .popup, .dropdown-menu, .tooltip, .popover {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #2d2d44 !important;\n border-color: #444 !important;\n }\n \n /* Scrollbars */\n ::-webkit-scrollbar { background: #1a1a2e !important; }\n ::-webkit-scrollbar-thumb { background: #444 !important; }\n ::-webkit-scrollbar-thumb:hover { background: #555 !important; }\n \n /* Selection */\n ::selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ';\n }\n \n function removeDarkMode() {\n var style = document.getElementById('universal-dark-mode-style');\n if (style) style.remove();\n }\n \n // Toggle with Alt+Shift+D\n document.addEventListener('keydown', function(e) {\n if (e.altKey && e.shiftKey && e.key === 'D') {\n e.preventDefault();\n enabled = !enabled;\n if (enabled) {\n applyDarkMode();\n console.log('[Universal Dark Mode] Enabled');\n } else {\n removeDarkMode();\n console.log('[Universal Dark Mode] Disabled');\n }\n }\n });\n \n // Apply on load\n applyDarkMode();\n \n // Re-apply on dynamic content\n var observer = new MutationObserver(function(mutations) {\n if (enabled && !document.getElementById('universal-dark-mode-style')) {\n applyDarkMode();\n }\n });\n observer.observe(document.head, { childList: true });\n \n console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle');\n})();", "Universal Dark Mode");
}
} catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); }
})();
})();

Skip to content
[CR] 10월 2, 3주차#98

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

 Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and
privacy statement. We’ll occasionally send you account related emails.

Already on GitHub?
Sign in
to your account

 Open

s0n9h2
wants to merge
4
commits into

base:main
Choose a base branch

from

 Open


[CR] 10월 2, 3주차
#98
Show file tree
Hide file tree
 Changes from all commits
Commits

Show all changes


4 commits


Select commit
Hold shift + click to select a range
e5f25db

[baekjoon-11559] PUYO PUYO


s0n9h2 Oct 7, 2022
5213e25

[programmers] 모음사전


s0n9h2 Oct 7, 2022
15d4582

[baekjoon-9935] 문자열 폭발


s0n9h2 Oct 10, 2022
47add06

[programmers] 모음사전


s0n9h2 Oct 15, 2022

File filter

Filter by extension
Filter by extension
Conversations

Failed to load comments. 
Loading
Jump to

Jump to file


Failed to load files. 
Loading
 Diff view

 Diff view


94 changes: 94 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/11559_PUYOPUYO.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,94 @@
from collections import deque
from sys import stdin
input = stdin.readline

# 4개 이상의 뿌요가 모이면 터진다!
# 여러 그룹이 있으면 동시에 터지고 이후 상황에서 터지면 콤보가 쌓인다!
def PuyoPop(i, j):
 global trigger
 dx = [1, -1, 0, 0]
 dy = [0, 0, 1, -1]

 count = 1
 puyo = [(i, j)]
 q = deque([(i, j)])
 visited[i][j] = True
 while q:
 x, y = q.popleft()
 for d in range(4):
 nx = x + dx[d]
 ny = y + dy[d]
 if 0 <= nx < 12 and 0 <= ny < 6 and visited[nx][ny] == False:
 # 상하좌우의 같은 색 뿌요 찾기
 if state[x][y] == state[nx][ny]:
 visited[nx][ny] = True
 q.append((nx, ny))
 count += 1
 puyo.append((nx, ny))

 if count >= 4:
 # 뿌요가 한 번이라도 터지면 콤보!
 trigger = True
 # 뿌요를 터뜨리는 함수 돌리기
 for px, py in puyo:
 state[px][py] = "."


# 중력에 의해 떨어지는 뿌요!
def PuyoDrop(state):
 for j in range(6):
 q = deque([])
 for i in range(11, -1, -1):
 # 한 줄에 남아있는 뿌요들만 모으기
 if state[i][j] != '.':
 q.append(state[i][j])
 # 바닥부터 차례로 채워주고 남는 공간은 .으로 채우기
 for i in range(11, -1, -1):
 if q:
 state[i][j] = q.popleft()
 else:
 state[i][j] = "."


state = [list(input().strip()) for _ in range(12)]
trigger = True

if __name__ == "__main__":
 combo = 0
 while trigger == True:
 trigger = False
 visited = [[False] * 6 for _ in range(12)]
 for i in range(11, -1, -1):
 for j in range(6):
 if visited[i][j] == False and state[i][j] != ".":
 # 현상황에서 터뜨리는 함수 돌리기
 PuyoPop(i, j)
 # print("")
 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 # 한 번이라도 터졌으면 콤보 올리고 빈 칸 채우기 
 if trigger == True:
 combo += 1
 PuyoDrop(state)

 # for i in range(12):
 # print("".join(state[i]))
 # print("=================")

 print(combo)

# 문제가 된 예제
# ......
# ......
# ......
# ......
# ......
# ......
# .G....
# RR....
# RY....
# RYG...
# RRY...
# RYYGG.

29 changes: 29 additions & 0 deletions
SH_Noh/season3/baekjoon/10/02-1/pro_모음사전.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
# 사전에 A E I O U 만 사용하여 길이 5 이하의 모든 단어가 수록되어 있음
# 사전 특성상 알파벳 순으로 정렬되어 있음

# 백트래킹
def backtracking(word, tmp_word):
 global num, answer
 # print(tmp_word, num)
 if tmp_word == word:
 return num
 if len(tmp_word) == 5:
 return

 for w in word_list:
 num += 1
 tmp_word.append(w)
 answer = backtracking(word, tmp_word)
 if answer:
 return answer
 tmp_word.pop()

word_list = ["A", "E", "I", "O", "U"]
num = 0
answer = ''

def solution(word):
 tmp_word = []
 backtracking(list(word), tmp_word)

 return answer

20 changes: 20 additions & 0 deletions
SH_Noh/season3/baekjoon/10/03-2/9935_문자열폭발.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
from sys import stdin
input = stdin.readline

if __name__ == "__main__":
 string = list(input().rstrip())
 bomb_str = list(input().rstrip())
 # 임시 스택에 넣어 비교하기
 tmp = []
 for i in range(len(string)):
 tmp.append(string[i])
 if tmp[-1] == bomb_str[-1] and len(tmp) >= len(bomb_str):
 if tmp[-len(bomb_str):] == bomb_str:
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리할 때 처음에 최대한 조건을 줄여야지 시간이 적게 나올 것 같은데,

송희님께서 하신 부분이 가장 적게 나오네요!(536ms) 흥미로운 지점입니다.
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str) andtmp[-len(bomb_str):] ==bomb_str:
=> 548ms
iftmp[-1] ==bomb_str[-1]:
iflen(tmp) >=len(bomb_str):
iftmp[-len(bomb_str):] ==bomb_str:
=> 580ms
 # for i in range(len(bomb_str)):
 # tmp.pop()
 del tmp[-len(bomb_str):]

 if tmp:
 print("".join(tmp))
 else:
 print("FRULA")
@blcklambblcklambOct 13, 2022

Copy link


Copy Markdown

Contributor
Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

조건문 처리 외에 나머지는 저와 거의 동일합니다!