Source Code
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[SW Expert Academy] 1974. 스도쿠 검증Source Code/Coding 2022. 7. 16. 01:48
solution.py def check(graph): for i in range(9): # 행 ck = [] for j in range(9): if ck: if graph[i][j] in ck: return '0' ck.append(graph[i][j]) for i in range(9): # 열 ck = [] for j in range(9): if ck: if graph[j][i] in ck: return '0' ck.append(graph[j][i]) for i in range(0,9,3): # 3*3 for j in range(0,9,3): ck = [] for n in range(3): for k in range(3): if ck: if graph[i+n][j+k] in ck: return '0' ..
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[SW Expert Academy] 1961. 숫자 배열 회전Source Code/Coding 2022. 7. 16. 01:19
solution.py # 테스트 케이스 T 입력 T = int(input()) for test_case in range(1, T+1): N = int(input()) arr = [list(map(int, input().split())) for _ in range(N)] # 90도, 180,도 270도 회전한 행렬 초기화 arr1 = [[0 for _ in range(N)] for _ in range(N)] arr2 = [[0 for _ in range(N)] for _ in range(N)] arr3 = [[0 for _ in range(N)] for _ in range(N)] # 90 for i in range(N): for j in range(N): arr1[i][j] = arr[N-1-j][i] #..
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[SW Expert Academy] 2001. 파리 퇴치Source Code/Coding 2022. 7. 13. 17:11
solution.py T = int(input()) # 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다. for test_case in range(1, T + 1): N,M = map(int, input().split()) fly = [list(map(int, input().split())) for _ in range(N)] result = 0 for i in range(N-M+1): for j in range(N-M+1): # 파리채 크기만큼 돌기 tmp = 0 for l in range(i, i+M): for k in range(j, j+M): tmp += fly[l][k] if tmp > result: result = tmp print("#{} {}".format(test_case, resu..
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[SW Expert Academy] 1979. 어디에 단어가 들어갈 수 있을까Source Code/Coding 2022. 7. 13. 00:33
solution.py T = int(input()) # 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다. for test_case in range(1, T + 1): N, K = map( int, input().split()) puzzle = [list(map(int, input().split())) for _ in range(N)] result = 0 # 행을 기준으로 cnt for i in range(N): row = 0 for j in range(N): if puzzle[i][j]==0: if row == K: result += 1 row = 0 else: row += 1 if row == K: result +=1 # 열을 기준으로 cnt for j in range(N): col = 0 fo..