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‎DivideAndConquer/1629.py‎

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‎DivideAndConquer/1780.py‎

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import sys
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input = sys.stdin.read
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def count_paper(x, y, n):
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if n == 1:
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counts[grid[x][y]] += 1
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return
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# 현재 종이가 모두 같은 수로 이루어져 있는지 확인
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init_val = grid[x][y]
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all_same = True
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for i in range(x, x + n):
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if not all_same:
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break
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for j in range(y, y + n):
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if grid[i][j] != init_val:
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all_same = False
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break
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# 모두 같은 수로 이루어져 있다면 해당 숫자의 카운트를 증가
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if all_same:
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counts[init_val] += 1
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else:
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# 같은 수로만 이루어져 있지 않다면 9개로 분할하여 각각 재귀적으로 처리
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new_size = n // 3
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for i in range(3):
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for j in range(3):
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count_paper(x + i * new_size, y + j * new_size, new_size)
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# 입력 받기
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data = sys.stdin.read().split()
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N = int(data[0])
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grid = [list(map(int, data[i * N + 1: (i + 1) * N + 1])) for i in range(N)]
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# 각 숫자 카운트를 저장할 딕셔너리
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counts = {-1: 0, 0: 0, 1: 0}
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# 전체 종이에 대해 함수 호출
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count_paper(0, 0, N)
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# 결과 출력
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print(counts[-1])
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print(counts[0])
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print(counts[1])

‎DivideAndConquer/1992.py‎

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import sys
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input = sys.stdin.read
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def compress(x, y, n):
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if n == 1:
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return str(grid[x][y])
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# 첫 번째 원소를 기준으로 모든 원소가 같은지 확인
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initial = grid[x][y]
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for i in range(x, x + n):
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for j in range(y, y + n):
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if grid[i][j] != initial:
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# 사분면으로 나누기
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next_n = n // 2
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top_left = compress(x, y, next_n)
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top_right = compress(x, y + next_n, next_n)
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bottom_left = compress(x + next_n, y, next_n)
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bottom_right = compress(x + next_n, y + next_n, next_n)
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return f"({top_left}{top_right}{bottom_left}{bottom_right})"
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# 모든 원소가 같은 경우
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return str(initial)
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# 입력 처리
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data = input().strip().split()
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N = int(data[0])
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grid = [list(map(int, list(data[i + 1]))) for i in range(N)]
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# 압축 결과 출력
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result = compress(0, 0, N)
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print(result)

‎DivideAndConquer/2447.py‎

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import sys
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sys.setrecursionlimit(10**6) #10^6으로 최대 재귀 깊이 설정
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input = sys.stdin.readline
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def star(l):
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if l == 1:
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return['*']
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Stars = star(l//3)
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L = []
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for s in Stars:
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L.append(s*3) # N/3 패턴으로 '둘러싼' -> 둘러싸려면 *3
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for s in Stars:
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L.append(s+' '*(l//3)+s) # N/3의 공백
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for s in Stars:
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L.append(s*3) # N/3 패턴으로 '둘러싼' -> 둘러싸려면 *3
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return L
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n = int(input().strip())
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print('\n'.join(star(n)))

‎DivideAndConquer/2630.py‎

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