takeU

n = int(input()) check = [0] * 26 for _ in range(n): s = input().split() i, ans = 0, False for i in range(len(s)): if not check[ord(s[i][0].lower()) - 97]: check[ord(s[i][0].lower()) - 97] = 1 s[i] = '[' + s[i][0] + ']' + s[i][1:] ans = True break if ans: print(' '.join(s)) continue s = ' '.join(s) for i, c in enumerate(s): if c != ' ' and not check[ord(c.lower()) - 97]: check[ord(c.lower()) - 9..

import sys input = sys.stdin.readline from heapq import heappush, heappop n, k = map(int, sys.stdin.readline().split()) jewelry = sorted([list(map(int, input().split())) for _ in range(n)]) bag = sorted([int(input()) for _ in range(k)]) ans, temp = 0, [] for b in bag: while jewelry and b >= jewelry[0][0]: heappush(temp, -jewelry[0][1]) heappop(jewelry) if temp: ans += heappop(temp) elif not jewe..

from bisect import bisect_left, bisect_right t = int(input()) n = int(input()) li_a = list(map(int, input().split())) m = int(input()) li_b = list(map(int, input().split())) def sum_li(li): res = [] for i in range(len(li)): acc = 0 for j in range(i, len(li)): acc += li[j] res.append(acc) return res s_a, s_b = sorted(sum_li(li_a)), sorted(sum_li(li_b)) ans = 0 for s in s_a: left = bisect_left(s_b..

import sys input = sys.stdin.readline n = int(input()) mat = [[] for _ in range(4)] for _ in range(n): li = list(map(int, input().split())) for i in range(4): mat[i].append(li[i]) d, ans = {}, 0 for i in mat[0]: for j in mat[1]: d[i + j] = d.get(i + j, 0) + 1 for i in mat[2]: for j in mat[3]: if -(i + j) in d: ans += d[-(i + j)] print(ans)

t = int(input()) def extended_gcd(a, b): x0, x1, y0, y1 = 1, 0, 0, 1 while b != 0: n, a, b = a // b, b, a % b x0, x1 = x1, x0 - n * x1 y0, y1 = y1, y0 - n * y1 return x0, x1, y0, y1 for _ in range(t): k, c = map(int, input().split()) x0, x1, y0, y1 = extended_gcd(c, k) if c == 1: print('IMPOSSIBLE' if k >= 1e9 else k + 1) elif x1 != -k and y1 != -c: print('IMPOSSIBLE') else: ans = x0 if x0 > 0 e..

from heapq import heappush, heappop import sys input = sys.stdin.readline n = int(input()) min_h, max_h = [], [] d = {} for _ in range(n): p, l = map(int, input().split()) heappush(min_h, [l, p]) heappush(max_h, [-l, -p]) d[p] = True M = int(input()) for _ in range(M): command = input().split() if command[0] == 'recommend': if command[1] == '1': while not d[-max_h[0][1]]: heappop(max_h) print(-m..

from heapq import heappush, heappop import sys input = sys.stdin.readline n = int(input()) q = sorted([list(map(int, input().split())) for _ in range(n)]) h = [] heappush(h, q[0][1] ) for i in range(1, n): if q[i][0] < h[0]: heappush(h, q[i][1]) else: heappop(h) heappush(h, q[i][1]) print(len(h)) 힙, 그리디

from collections import deque import sys input = sys.stdin.readline def topological_sort(): dp = [0] * (n + 1) q = deque() for i in range(1, n + 1): if indegree[i] == 0: q.append(i) dp[i] = cost[i] while q: cur = q.popleft() for g in graph[cur]: indegree[g] -= 1 dp[g] = max(dp[g], dp[cur] + cost[g]) if indegree[g] == 0: q.append(g) if indegree[last] == 0: print(dp[last]) return t = int(input()) ..

파이썬 그래프, 트리 알고리즘 정리 알고리즘 선택 (최단거리) 가중치가 없는 경우 - BFS 가중치가 있는 경우 음의 가중치가 있을 때 - 벨만-포드 음의 가중치가 없을 때 - 다익스트라 전체 쌍 최단 경로 - 플로이드-워셜 ( n ≤ 100 ) 1. 벨만-포드 (Bellman-Ford Algorithm) 시간 복잡도: O(V * E) 모든 간선을 확인하며 최단 거리를 찾기 때문에 느리다. 음수 간선이 있어도 최단 거리를 찾을 수 있다. import sys input = sys.stdin.readline INF = int(1e9) n, m = map(int, input().split()) edges = [] dist = [INF] * (n + 1) for _ in range(m): u, v, w = ma..

li = list(map(int, input().split()))[:-1] n = len(li) dp = [[[400001] * 5 for _ in range(5)] for _ in range(n + 1)] dp[-1][0][0] = 0 def check(c, n): if c == n: return 1 elif c == 0: return 2 elif n in [c + 1, c - 1, c - 3, c + 3]: return 3 else: return 4 for i in range(n): for j in range(5): for k in range(5): dp[i][li[i]][j] = min(dp[i][li[i]][j], dp[i - 1][k][j] + check(k, li[i])) for j in ra..