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def Pair_bracket():
    for i in range(L):
        if q[i] in check:
            ni = i - 1
            while True:
                if q[ni] == 0:
                    ni -= 1
                else:
                    break
            a = q[ni]
            b = check[q[i]]
            if a == b:
                q[i] = 0
                q[ni] = 0
            else:
                return 0
    return 1

check = {")" :"(", "]":"[", "}":"{", ">":"<"}
for T in range(10):
    L = int(input())
    q = list(input())

    print(f'#{T+1} {Pair_bracket()}')

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from collections import deque

def BFS():
    visited[0] = 1
    while q:
        a = q.popleft()
        if a == 99:
            return 1
        if visited[a] == 0:
            for i in range(len(arr[a])):
                q.append(arr[a][i])
            visited[a] = 1
    return 0   

for _ in range(10):
    T, N = map(int, input().split())
    arr = [[] for _ in range(100)]
    route = list(map(int, input().split()))
    visited = [0 for _ in range(100)] 

    for i in range(0, N*2, 2):
        arr[route[i]].append(route[i+1])
        
    q = deque()
    for i in range(len(arr[0])):
        q.append(arr[0][i])

    print(f'#{T} {BFS()}')

# 빈 2차원 리스트 만드는 법
# [[] * 100]  은 안됨
# [[] for _ in range(N)] 으로 생성

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import sys
from collections import deque

text = sys.stdin.readline().rstrip()

# 후위 표기법 알고리즘
operator = {"+":1, "-":1, "*":2 , "/":2, "(":0, ")":0}
stack = deque()
result = ""

for i in text:
    if i not in operator:
        result += i
    elif i == "(":
        stack.append(i)
    elif i == ")":
        while stack and stack[-1] != "(":
            result += stack.pop()
        stack.pop()
    else:
        while stack and operator[i] <= operator[stack[-1]]:
            result += stack.pop()
        stack.append(i)

while stack:
    result += stack.pop()
                
print(result)

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for T in range(int(input())):
    N = input()
    result = ["A", "B"]
    turn = 1
    while len(N) > 1:
        N = str(int(N[0]) + int(N[1])) + N[2:]
        turn += 1
    print(f'#{T+1} {result[turn%2]}')
    
# 단순 산수

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for T in range(int(input())):
    N, M = map(int, input().split())
    a = [input() for _ in range(N)]
    result = N * M

    for i in range(1, N - 1):
        for j in range(1, N - i):
            count = 0
            for x in range(N):
                if 0 <= x < i:
                    count += M - a[x].count('W')
                elif i <= x < i + j:
                    count += M - a[x].count('B')
                else:
                    count += M - a[x].count('R')
            if result > count:
                result = count

    print(f'#{T+1} {result}')

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N = int(input())
wine = [0 for _ in range(N)]
total = [0 for _ in range(N)]

for i in range(N):
    wine[i] = int(input())

if N == 1:
    total[0] = wine[0]
else:
    total[0] = wine[0]
    total[1] = wine[0] + wine[1]

for i in range(2, N):
    total[i] = max(total[i-1], total[i-2] + wine[i], total[i-3] + wine[i] + wine[i-1])

print(total[N-1])

# N == 1인 경우를 생각하지 않았을 땐 런타임 오류에 걸렸었다.
# 그리고 그걸 생각하는데 오래 걸렸다.

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import sys

N = int(sys.stdin.readline().rstrip())
arr = list(map(int, sys.stdin.readline().split()))

result = [arr[0]]
for i in range(N-1):
    result.append(max(result[i]+arr[i+1], arr[i+1]))

print(max(result))

# dp 연습하기 좋은 문제

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def w(a,b,c):

    if a <= 0 or b <= 0 or c <= 0:
        return 1

    if a > 20 or b > 20 or c > 20:
        return w(20, 20, 20)

    if dp[a][b][c]:
        return dp[a][b][c]
    
    if a < b and b < c:
        dp[a][b][c] = w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c) 
        return dp[a][b][c]

    dp[a][b][c] = w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1)   
    return dp[a][b][c]

dp = [[[0]*21 for _ in range(21)] for _ in range(21)]

while True:
    a, b, c = map(int, input().split())

    if a == -1 and b == -1 and c == -1:
        break

    print(f'w({a}, {b}, {c}) = {w(a, b, c)}')

# 갑자기 난이도가 높아진듯..
# 3차원 dp까지는 어찌저찌 고안했으나
#     if dp[a][b][c]:
#         return dp[a][b][c]
# 를 어디에 넣어야 할지 고민하는데 
# 시간을 소요.

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for T in range(int(input())):
    S = " " + input()
    L = len(S)
    result = 1
    for i in range(2, L):
        for j in range(L-i):
            if S[j+1:j+i+1] == S[j+i:j:-1]:
                result = i

    print(f'#{T+1} {result}')
    
# 슬라이싱을 잘 고민 하면 쉽게 해결

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for T in range(int(input())):
    q = []
    n = ""
    for N in range(int(input())):
        q.append(input())

    S = 0
    L = N

    while S != L:
        if q[S] < q[L]:
            n += q[S]
            S += 1
        elif q[S] > q[L]:
            n += q[L]
            L -= 1
        else:
            W = int((L-S)//2)
            for i in range(W):
                flag = True
                c = q[S+i+1]
                d = q[L-(i+1)]
                if c == d:
                    continue
                elif c < d:
                    n += q[S]
                    S += 1
                    flag = False
                    break
                elif c > d:
                    n += q[L]
                    L -= 1
                    flag = False
                    break
            if flag:
                n += q[S]
                S += 1
    n += q[L]

    print(f'#{T+1} {n}')

# 조건은 다 쉬웠으나, 검사하는 횟수?에 착각을해
# 생각보다 시간이 많이 걸린 문제