1. 백준 1013 N = int(input()) results = [] for _ in range(N): input = sys.stdin.readline().rstrip() regex = re.compile('(100+1+|01)+') answer = regex.fullmatch(input) print("YES" if answer else "NO") while문을 써서 난리를 치다가 결국에 re로 작성했다. 2. 파이썬 정규표현식 re 모듈 regex = re.compile('(100+1+|01)+') # print result : re.compile('(100+1+|01)+') compile을 사용해서 정규표현식을 확인할 객체를 만든다. regex.match(input) # 10010111 resul..
1. 백준 1012 import sys from collections import deque dx = [-1, 1, 0, 0] dy = [0, 0, -1, 1] def bfs(x, y): q = deque([(x, y)]) board[x][y] = 0 while q: cx, cy = q.popleft() for i in range(4): nx = cx + dx[i] ny = cy + dy[i] if nx = 0 and ny = 0: if board[nx][ny] == 1: board[nx][ny] = 0 q.append((nx, ny)) T = int(input()) for _ in range(T): M, N, K = map(int, sys.stdin.readl..
1. 백준 1969 answer = '' hamming_distance = 0 for i in range(M): temp = [0, 0, 0, 0] for j in range(N): if dna[j][i] == 'A': temp[0] += 1 elif dna[j][i] == 'C': temp[1] += 1 elif dna[j][i] == 'G': temp[2] += 1 else: temp[3] += 1 idx = temp.index(max(temp)) hamming_distance += (N - max(temp)) if idx == 0: answer += 'A' elif idx == 1: answer += 'C' elif idx == 2: answer += 'G' else: answer += 'T' 1...
