.

라인/2.py

@@ -0,0 +1,32 @@
+import sys
+input = sys.stdin.readline
+
+K , n_slang = map(int, input().split())
+slangs = [input().strip() for _ in range(n_slang)]
+chat = input().strip()
+
+def match(slang, word):
+    n = len(slang)
+    m = len(word)
+    dy = [[False for _ in range(m+1)] for _ in range(n+1)]
+    dy[0][0] = True
+
+    for i in range(1, n+1):
+        for j in range(1, m+1):
+            if slang[i-1] == word[j-1]:
+                dy[i][j] |= dy[i-1][j-1]
+            if word[j-1] == '.':
+                for k in range(1, K+1):
+                    if i - k >= 0:
+                        dy[i][j] |= dy[i-k][j-1]
+
+    return dy[n][m]
+
+ans = []
+for word in chat.split():
+    if any(match(slang, word) for slang in slangs):
+        ans.append('#' * len(word))
+    else:
+        ans.append(word)
+
+print(*ans)

라인/3.py

@@ -1,7 +1,10 @@
+# 시간이 10억분 주어지지만 방문하는 격자는 최대 30*30
+
+
 from collections import deque
 
 # n=3; m=2; fires=[[1,1]]; ices=[[3,3]]
-n=5; m=3; fires=[[5,5], [1,3], [5,2]]; ices=[[1,5],[3,2]]
+n=5; m=1000000000; fires=[[5,5], [1,3], [5,2]]; ices=[[1,5],[3,2]]
 
 
 dx = [0, 0, -1, 1, 1, 1, -1, -1] # (0~3) 상하좌우 (4~7) 대각 4방향
@@ -10,11 +13,11 @@
 def fire(y, x, mt, t, nn): # 좌표, 주어진 배열, 시간, nn = 가로길이 = 세로길이
     visited = []
     queue = deque([[[y, x], t+1]])
-    mt[y][x] -=1 #
+    mt[y][x] -=1 # 정가운데 0분일때 고려 x
 
     while queue:
         n = queue.popleft()
-        if n[0] not in visited and n[1] > 0: 
+        if n[0] not in visited and n[1] > 0: # 방문하지 않고 시간이 0분 이상 남았을때
             y1 = n[0][0]
             x1 = n[0][1]
             visited.append(n[0])

라인/4.py

@@ -1,5 +1,7 @@
-# 못풀었음ㅜㅠ
 # 조건에 맞는 이동방식을 일일이 계산하여 bfs로 풀음
+dy = [0, 0, 1, -1]
+dx = [1, -1, 0, 0]
+
 
 from collections import deque
 
@@ -25,7 +27,6 @@ def BFS(y, x, wall, n1, m, visited):
                 if visited[y1][x1] > c:
                     visited[y1][x1] = c
 
-
             # case1 오른쪽 2칸
             if y1-1 >= 0 and x1+2 < n1:
                 f = True
@@ -43,10 +44,10 @@ def BFS(y, x, wall, n1, m, visited):
             if y1-1 >= 0 and x1-2 >= 0:
                 f = True
                 for i in range(3):
-                    if wall[y-1][x-i] != '.':
+                    if wall[y1-1][x1-i] != '.':
                         f = False
                         break
-                if wall[y][x-1] != '.':
+                if wall[y1][x1-1] != '.':
                     f = False
                     break
                 if f == True:
@@ -69,10 +70,10 @@ def BFS(y, x, wall, n1, m, visited):
             if y1-1 >= 0 and x1-3 >= 0:
                 f = True
                 for i in range(4):
-                    if wall[y-1][x-i] != '.':
+                    if wall[y1-1][x1-i] != '.':
                         f = False
                         break
-                if wall[y][x-1] != '.' or wall[y][x-2] != '.':
+                if wall[y1][x1-1] != '.' or wall[y1][x1-2] != '.':
                     f = False
                     break
                 if f == True:
@@ -92,6 +93,13 @@ def BFS(y, x, wall, n1, m, visited):
             if y1-1 >= 0 and x1-1>=0:
                 if wall[y1-1][x1] == '.' and wall[y1][x1-1] == '.':
                     queue.append([[y1-1, x1-1],c+1])
+
+            # 상하좌우
+            for i in range(4):
+                yy = y1+dy[i]
+                xx = x1+dx[i]
+                if 0<= yy < m and 0<=xx<n1:
+                    queue.append([[yy, xx], c+1])
 
 
 def solution(wall):

라인/5.py

@@ -0,0 +1,64 @@
+import sys
+input = sys.stdin.readline
+n, T = map(int, input().split())
+
+def gcd(p, q): # 최대공약수 (뉴클리드 호제법)
+    if q==0:
+        return p
+    return gcd(q, p%q)
+
+def get_divisors(x): # 모든 약수 찾기
+    res = []
+    for i in range(1, x+1):
+        if i * i > x: break
+        if x % i == 0:
+            res.append(i)
+            res.append(x // i)
+
+    return res
+
+def solve(A, B, M):
+    # (A, B) -> M일 연체료
+    K = M // A
+    return B * (K+1)
+
+q = [list(map(int, input().split())) for _ in range(n)]
+q.sort(key = lambda x: x[0])
+
+money_gcd = 0
+for i in range(n):
+    money_gcd = gcd(money_gcd, q[i][1])
+
+candidates = get_divisors(money_gcd)
+min_ans, max_ans = sys.maxsize, 0
+for cand_money in candidates:
+    L, R = 1, 1000000
+    for time, cost in q:
+        K = cost//cand_money -1
+        # upper
+        if K > 0:
+            R = min(R, time//K)
+
+        # lower
+        L = max(L, time//(K+1))
+
+    # L < A <= R
+    if L >= R:
+        continue
+
+    max_ans = max(max_ans, solve(L+1, cand_money, T))
+    min_ans = min(min_ans, solve(R, cand_money, T))
+
+if max_ans == 0:
+    print(-1)
+else:
+    print(min_ans, max_ans)
+
+'''
+5 27
+4 1000
+6 1000
+21 3000
+16 2000
+26 3000
+'''