Merge branch 'AlgorithmStudy-Allumbus:main' into main

Author: YoonYn9915

_WeeklyChallenges/W03-[IMPLEMENTATION]/Assignment_BOJ_13335_트럭.py

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+'''
+BOJ #13335. 수강과목 (실버1)
+https://www.acmicpc.net/problem/13335
+유형: 구현, 시뮬레이션, 자료구조
+
+출처:https://velog.io/@mimmimmu/12%EC%A3%BC%EC%B0%A8-%EB%B0%B1%EC%A4%80-13335%EB%B2%88-%ED%8A%B8%EB%9F%AD-%ED%8C%8C%EC%9D%B4%EC%8D%AC
+'''
+
+
+from collections import deque
+
+n, w, L = map(int, input().split())
+cars = list(map(int, input().split()))
+
+queue = deque()
+for _ in range(w):
+    queue.append(0)
+
+time = 0
+idx = 0
+while idx < n:
+    time += 1
+    queue.popleft()
+
+    if sum(queue) + cars[idx] <= L:
+        queue.append(cars[idx])
+        idx += 1
+    else:
+        queue.append(0)
+
+while queue:
+    time += 1
+    queue.popleft()
+
+print(time)

_WeeklyChallenges/W03-[IMPLEMENTATION]/README.md

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+## 🚀 12월 1주차 (12/02) 스터디 발제 주제: Implementation With Queue
+> 발제자: 조윤상
+
+### 🗂️ 스터디 자료
+- PDF: [바로가기](https://github.com/AlgorithmStudy-Allumbus/codingtest_algorithm_study/blob/main/_WeeklyChallenges/W03-%5BIMPLEMENTATION%5D/Study_BOJ_1202.pdf)
+![image](https://github.com/user-attachments/assets/493b84c6-6c3a-4faa-85d6-0ccefde1ff78)
+
+
+### 📖 문제
+- [백준 #3190. 뱀](https://www.acmicpc.net/problem/3190): 구현, 시뮬레이션, 큐 / 골드5
+- 정답 코드: [Study_BOJ_3190_뱀.py](https://github.com/AlgorithmStudy-Allumbus/codingtest_algorithm_study/blob/main/_WeeklyChallenges/W03-%5BIMPLEMENTATION%5D/Study_BOJ_3190_%EB%B1%80.py)
+  https://github.com/AlgorithmStudy-Allumbus/codingtest_algorithm_study/blob/aa53c6266f5a4d2a9a37b1f087511ed327e8dc59/_WeeklyChallenges/W03-%5BIMPLEMENTATION%5D/Study_BOJ_3190_%EB%B1%80.py#L8-L51
+
+### 💻 과제
+- [백준 #13335. 트럭](https://www.acmicpc.net/problem/13335):  구현, 시뮬레이션, 큐 / 실버1
+- 정답 코드: [Assignment_BOJ_13335_트럭.py](https://github.com/AlgorithmStudy-Allumbus/codingtest_algorithm_study/blob/main/_WeeklyChallenges/W03-%5BIMPLEMENTATION%5D/Assignment_BOJ_13335_%ED%8A%B8%EB%9F%AD.py)
+  https://github.com/AlgorithmStudy-Allumbus/codingtest_algorithm_study/blob/aa53c6266f5a4d2a9a37b1f087511ed327e8dc59/_WeeklyChallenges/W03-%5BIMPLEMENTATION%5D/Assignment_BOJ_13335_%ED%8A%B8%EB%9F%AD.py#L10-L35

_WeeklyChallenges/W03-[IMPLEMENTATION]/Study_BOJ_1202.pdf

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+'''
+BOJ #3190. 뱀 (골드5)
+https://www.acmicpc.net/problem/3190
+유형: 구현, 시뮬레이션, 자료구조
+'''
+
+
+import sys
+from collections import deque
+input = sys.stdin.readline
+n = int(input())
+k = int(input())
+maps = [[0] * (n+1) for _ in range(n+1)]
+for _ in range(k):#사과의 위치
+    x,y = map(int,input().split())
+    maps[x][y] = 2
+info = {}
+l = int(input())
+for _ in range(l):# 뱀의 방향변환정보 (초, 방향 L:왼쪽 D:오른쪽)
+    sec, direct = input().split()
+    info[int(sec)] = direct
+time = 0
+dx = [1,0,-1,0]
+dy = [0,1,0,-1]
+x, y = 1, 1
+maps[y][x] = 1
+d = 0
+snakes = deque([(1, 1)])
+
+while True:
+    nx, ny = x+dx[d], y+dy[d]
+    # 뱀의 몸통에 닿거나 벽에 부딪히는 경우 종료
+    if nx<=0 or ny<=0 or nx>n or ny>n or (nx,ny) in snakes:
+        break
+    # 사과를 먹지 못하면 꼬리 없애기
+    if maps[ny][nx]!=2:
+        a,b = snakes.popleft()
+        maps[b][a]=0
+    x, y = nx, ny
+    maps[y][x] = 1
+    snakes.append((nx, ny))
+    time+=1
+	
+    # 시간에 해당하는 방향전환 정보가 있을 경우
+    if time in info.keys():
+        if info[time] == "D":
+            d = (d+1)%4
+        else:
+            nd = 3 if d==0 else d-1
+            d = nd
+print(time+1)

_WeeklyChallenges/W03-[IMPLEMENTATION]/Study_BOJ_3190_뱀.py

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+import sys
+input = sys.stdin.readline
+
+N, K = map(int, input().split()) # 물품의 개수, 버틸 수 있는 무게 
+
+items = []
+for _ in range(N):
+    W, V = map(int, input().split()) 
+    items.append((W, V)) # (물건의 개수, 물건의 가치)
+
+def backpack(N, K, items):
+    dp = [0] * (K+1) # 배낭 크기만큼 DP 테이블 초기화 
+    
+    for weight, value in items:
+        for w in range(K, weight-1, -1): # 역순으로 반복
+            dp[w] = max(dp[w], dp[w-weight] + value)
+    
+    return dp[K]
+
+print(backpack(N, K, items))

learntosurf/Dynamic Programming/2024-11-25-[BOJ]-#12865-평범한배낭.py

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+n = int(input())
+
+def fibonacci(n):
+    dp = [0] * (n+1)
+    
+    dp[0] = 0
+    dp[1] = 1
+    
+    for i in range(2, n+1):
+        dp[i] = dp[i-1] + dp[i-2]
+    
+    return dp[n]
+
+print(fibonacci(n))

learntosurf/Dynamic Programming/2024-11-25-[BOJ]-#2748-피보나치수2.py

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+N = int(input())
+
+def operation(N):
+    dp = [0] * (N+1)
+    
+    for i in range(2, N+1): # 1은 연산이 필요하지 않음 
+        dp[i] = dp[i-1] + 1
+        
+        if i%2==0:
+            dp[i] = min(dp[i], dp[i//2]+1)
+        
+        if i%3==0:
+            dp[i] = min(dp[i], dp[i//3]+1)
+
+    return dp[N]
+
+print(operation(N))

learntosurf/Dynamic Programming/2024-11-26-[BOJ]-#1463-1로만들기.py

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+import sys 
+input = sys.stdin.readline
+
+N = int(input())
+stair = [int(input()) for _ in range(N)]
+
+# N에 따른 예외 처리
+if N == 1:
+    print(stair[0])
+    exit()
+
+if N == 2:
+    print(stair[0] + stair[1])
+    exit()
+
+dp = [0] * (N+1)
+dp[0] = stair[0]
+dp[1] = stair[0] + stair[1]
+dp[2] = max(stair[0] + stair[2], stair[1] + stair[2])
+
+for i in range(3, N):
+    dp[i] = max(dp[i-2] + stair[i], dp[i-3] + stair[i-1] + stair[i])
+
+print(dp[N-1])

learntosurf/Dynamic Programming/2024-11-27-[BOJ]-#2579-계단오르기.py

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+import sys 
+input = sys.stdin.readline
+INF = sys.maxsize
+
+N = int(input())
+blocks = input().split()
+
+dp = [INF] * N
+dp[0] = 0 # 1번 블록에서 출발 
+
+for i in range(1, N): # 목표 블록 (1번 블록 이후부터 N번까지)
+    for j in range(i): # 이전 블록들 (0번부터 i-1번까지 탐색)
+        if blocks[j] == 'B' and blocks[i] != 'O':
+            continue
+        elif blocks[j] == 'O' and blocks[i] != 'J':
+            continue
+        elif blocks[j] == 'J' and blocks[i] != 'B':
+            continue
+        dp[i] = min(dp[i], dp[j] + (i-j)**2)
+
+if dp[-1] == INF:
+    print(-1)
+else:
+    print(dp[-1])

learntosurf/Dynamic Programming/2024-11-28-[BOJ]-#12026-BOJ거리.py

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+import sys
+input = sys.stdin.readline
+
+N,S,M = map(int,input().split())
+V = list(map(int,input().split()))
+
+dp = [[0]*(M+1) for _ in range(N+1)]
+dp[0][S] = 1 # 시작 볼륨을 설정 
+
+for i in range(1, N+1): # 각 곡에 대해
+    for j in range(M+1): # 가능한 볼륨에 대해
+        if dp[i-1][j] != 0: # 이전 곡에서 해당 볼륨이 가능하다면
+            if 0 <= j + V[i-1] <= M: # 볼륨을 올릴 수 있다면
+                dp[i][j + V[i-1]] = 1
+            if 0 <= j - V[i-1] <= M: # 볼륨을 낮출 수 있다면
+                 dp[i][j-V[i-1]] = 1
+volume = -1
+for i in range(M, -1, -1): # 최대 볼륨부터 탐색
+    if dp[N][i] == 1: # 가능한 볼륨을 찾으면 
+        volume = i
+        break
+    
+print(volume)