• 문제
  • 4963번: 섬의 개수
• 해설

import sys

sys.setrecursionlimit(99999)

def get_data():
    width, height = [int(x) for x in sys.stdin.readline().split()]
    if height == 0:
        sys.exit()
    mapping = []
    for x in range(height):
        mapping.append([int(z) for z in sys.stdin.readline().split()])
    return width, height, mapping

def chk_in_map(x, y, mapping):
    return 0 <= x < len(mapping) and 0 <= y < len(mapping[0])

dx_list = [-1, 0, 1]
dy_list = [-1, 0, 1]

def dfs(x, y, chk, mapping, ans):
    chk[x][y] = True
    ans += 1
    for dx in dx_list:
        for dy in dy_list:
            if dx == 0 and dy == 0:
                continue
            cand_x = x + dx
            cand_y = y + dy
            if chk_in_map(cand_x, cand_y, mapping) and mapping[cand_x][cand_y] == 1 and not chk[cand_x][cand_y]:
                dfs(cand_x, cand_y, chk, mapping, ans)
    return ans

for _ in range(999999999):
    ans = 0
    width, height, mapping = get_data()
    chk = [[False for z in range(width)] for q in range(height)]
    for x in range(height):
        for y in range(width):
            if mapping[x][y] and not chk[x][y]:
                ans = dfs(x, y, chk, mapping, ans)
    print(ans)

• 포인트
  • dfs

• 문제
  • 2667번: 단지번호붙이기
• 해설

import sys

sys.setrecursionlimit(99999)

size = int(sys.stdin.readline().strip())
mapping = [[0] * size for x in range(size)]

for i, x in enumerate(sys.stdin.readlines()):
    for j, y in enumerate(x.strip()):
        mapping[i][j] = int(y)

went_or_not = [[False for _ in range(size)] for _ in range(size)]

def is_in_map(x, y):
    return 0 <= x < size and 0 <= y < size

def dfs(x, y, res):
    went_or_not[x][y] = True
    candidates = [(x - 1, y), (x, y + 1), (x + 1, y), (x, y - 1)]
    for nx, ny in candidates:
        if is_in_map(nx, ny):
            if mapping[nx][ny] and not went_or_not[nx][ny]:
                res.append(1)
                dfs(nx, ny, res)
    return res

final_res = []
for x in range(size):
    for y in range(size):
        if mapping[x][y] != 0 and went_or_not[x][y] == False:
            res_ = dfs(x, y, [1])
            if len(res_) >= 1:
                final_res.append(len(res_))
final_res = sorted(final_res)
print(len(final_res))
for x in final_res:
    print(x)

• 포인트
  • dfs로 풀 생각
  • 각 단지안에 집 개수도 세야하니까 안가본(went_or_not이 False) 그리고 집(mapping에서 1)에서 시작하면서 list를 인자로 받아 늘리기

• 문제
  • 2609번: 최대공약수와 최소공배수
• 해설

import sys

m, n = list(map(int, sys.stdin.readline().split()))
if n > m:
    m, n = n, m

def get_gcd(a, b):
    if b == 0:
        return a
    else:
        return get_gcd(b, a % b)

gcd = get_gcd(m, n)
print(get_gcd(m, n))
print(int((m * n) / gcd))

• 포인트
  • 유클리드 호제법

• 문제
  • 2309번: 일곱 난쟁이
• 해설

import sys

nine_heights = [int(x) for x in sys.stdin.readlines()]
nine_sum = sum(nine_heights)
is_done = False

for left_idx in range(8):
    for right_idx in range(left_idx + 1, 9, 1):
        if (nine_sum - nine_heights[left_idx]) - nine_heights[right_idx] == 100:
            ans = sorted([x for ind, x in enumerate(nine_heights) if ind != left_idx and ind != right_idx])
            for k in ans:
                print(k)
            is_done = True
            break
    if is_done:
        break

• 포인트
  • 왼쪽에서 모든 경우 다 헤아리기

• 문제
  • 2178번: 미로 탐색
• 해설

import sys
from collections import deque

sys.setrecursionlimit(10 ** 7)

N, M = [int(x) for x in sys.stdin.readline().strip().split()]
maze = [list(map(int, x.strip())) for x in sys.stdin.readlines()]

queue = deque()
dx_dy = [(-1, 0), (1, 0), (0, 1), (0, -1)]
went_or_not = [[False for y in range(M)] for x in range(N)]

def can_go(x, y):
    if 0 <= x < N and 0 <= y < M and maze[x][y] and not went_or_not[x][y]:
        return True
    else:
        return False

def is_destination(x, y):
    return x == (N - 1) and y == (M - 1)

ans = [N * M]

def bfs(x, y, cnt):
    went_or_not[x][y] = True
    cnt += 1
    for diff in dx_dy:
        ax, ay = [sum(z) for z in zip((x, y), diff)]
        if can_go(ax, ay):
            went_or_not[ax][ay] = True
            if is_destination(ax, ay):
                ans[0] = ans[0] if ans[0] < cnt else cnt
            else:
                queue.appendleft((ax, ay, cnt))
    if len(queue):
        ax, ay, cnt = queue.pop()
        bfs(ax, ay, cnt)

bfs(0, 0, 1)
print(ans[0])

• 포인트
  • bfs문제

• 문제
  • 2170번: 선 긋기
• 해결

import sys

N = int(sys.stdin.readline())
nlist = [list(map(int, x.strip().split())) for x in sys.stdin.readlines()]
nlist.sort()

cumulative_length = 0
cumulative_right_point = -sys.maxsize

for current_left, current_right in nlist:
    if current_left >= cumulative_right_point:
        cumulative_length += (current_right - current_left)
    else:
        if current_right > cumulative_right_point:
            cumulative_length += (current_right - cumulative_right_point)
    cumulative_right_point = max(current_right, cumulative_right_point)
print(cumulative_length)

• 포인트
  • input을 정렬하고(O(NlogN)) 좌측부터 우측으로 O(N)만 함
  • 이를 sweep line 기법이라 함

• 문제
  • 1978번: 소수 찾기
• 해설

import sys

candidate_count = int(sys.stdin.readline())

def is_prime(n):
    if n == 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        else:
            i += 6
    return True

candidates = [int(x) for x in sys.stdin.readline().split()]
prime_count = 0
for candidate in candidates:
    if is_prime(candidate):
        prime_count += 1
print(prime_count)

• 포인트
  • while 문 안에 소수찾는 로직이 중요
    • 제곱이 n보다 작은 자연수이면서 6으로 나눴을 때 나머지가 1, 5인 것만 확인

• 문제
  • 1934번: 최소공배수
• 해설

import sys

test_case_count = int(sys.stdin.readline())

def get_gcd(less, larger):
    if larger == 0:
        return less
    else:
        return get_gcd(larger, less % larger)

for i in range(test_case_count):
    a, b = [int(x) for x in sys.stdin.readline().split()]
    if b > a:
        a, b = b, a
    gcd = get_gcd(a, b)
    print(int((a * b) / gcd))

• 포인트
  • 쉬움

• 문제
  • 1929번: 소수 구하기
• 해설

import sys

minimum, maximum = [int(x) for x in sys.stdin.readline().split()]


def prime_candidate_generator_less_than(n):
    # 5, 7, 11, 13, 17, 19, ...
    candidate = 5
    increment = 2
    increment_changer = {2: 4, 4: 2}
    while candidate <= n:
        yield candidate
        candidate += increment
        increment = increment_changer[increment]


def get_primes_by_sieve_less_than(n):
    if n == 1:
        return []
    elif n == 2:
        return [2]
    elif n == 3:
        return [2, 3]
    else:
        prime_list = [2, 3]
        sieve_list = [True] * (n + 1)
        for candidate in prime_candidate_generator_less_than(n):
            if sieve_list[candidate]:
                prime_list.append(candidate)
                for i in range(2, 1 + (n // candidate)):
                    sieve_list[candidate * i] = False
        return prime_list


res = get_primes_by_sieve_less_than(maximum)
for i in res:
    if i >= minimum:
        print(i)

• 포인트
  • 에라토스테네스 체(sieve) 사용

1759 암호 만들기

• 문제
  • 1759번: 암호 만들기
• 해설

  • import sys
    
    target_length, C = [int(x) for x in sys.stdin.readline().split()]
    
    available_alphabets = sorted(sys.stdin.readline().split())
    
    
    def is_available(test_password):
        total_jaum = set('aeiou')
        if len(total_jaum & set(test_password)) >= 1 and len(set(test_password) - total_jaum) >= 2:
            return True
        else:
            return False
    
    
    def recursive_finder(test_password):
        # 불가능한 경우
        if len(test_password) > target_length:
            return None
    
        # 정답인 경우
        if len(test_password) == target_length and is_available(test_password):
            print(test_password)
    
        # 계속 호출인 경우
        if len(test_password) == 0:
            for i in available_alphabets:
                recursive_finder(test_password + i)
    
        if 0 < len(test_password) < target_length:
            for i in available_alphabets[1 + available_alphabets.index(test_password[-1]):]:
                recursive_finder(test_password + i)
    
    
    recursive_finder('')

• 포인트
  • Bruteforce
  • Recursive function