ABC384をPythonで（A～E）

トヨタ自動車プログラミングコンテスト2024#12（AtCoder Beginner Contest 384）の解答等の速報的まとめ

A問題

c1と一致しなかったやつをc2に置き換える

A

n, c1, c2 = input().split()
s = input()

ans = []
for s_i in s:
    ans.append(c1 if c1 == s_i else c2)

print(*ans, sep="")

B問題

問題文の通り

B

n, r = map(int, input().split())

for _ in range(n):
    d, a = map(int, input().split())
    if d == 1 and 1600 <= r <= 2799:
        r += a
    elif d == 2 and 1200 <= r <= 2399:
        r += a

print(r)

C問題

全参加者のスコアを計算してソートする

C

p = list(map(int, input().split()))

data = []
for bit in range(1, 1 << 5):
    name = ""
    score = 0
    for i in range(5):
        if bit >> i & 1:
            name += chr(ord("A") + i)
            score += p[i]
    data.append((-score, name))

data.sort()
for _, name in data:
    print(name)

D問題

根本は尺取り法
複数回Aを跨ぐと尺取りはうまくいかないので、事前に合計でmodを取って端の部分だけ尺取りを行えばいいようにSを小さくする

D

n, s = map(int, input().split())
a = list(map(int, input().split()))

if s >= sum(a):
    s = s % sum(a)

a += a

left = 0
sum_a = 0
for right in range(n * 2):
    while left < right and sum_a > s:
        sum_a -= a[left]
        left += 1
    if sum_a == s:
        exit(print("Yes"))

    sum_a += a[right]

print("No")

E問題

自身に隣接するスライムをheapqで管理して、小さいほうから吸収できるか確かめていく

E

from heapq import heappop, heappush

h, w, x = map(int, input().split())
p, q = map(lambda x:int(x) - 1, input().split())

field = [list(map(int, input().split())) for _ in range(h)]

check = [[False] * w for _ in range(h)]
check[p][q] = True
power = field[p][q]
heap = []
for i, j in [(p, q + 1), (p, q - 1), (p + 1, q), (p - 1, q)]:
    if 0 <= i < h and 0 <= j < w:
        heappush(heap, (field[i][j], i, j))
        check[i][j] = True

while heap and heap[0][0] * x < power:
    p_i, x_i, y_i = heappop(heap)
    power += p_i
    for i, j in [(x_i, y_i + 1), (x_i, y_i - 1), (x_i + 1, y_i), (x_i - 1, y_i)]:
        if 0 <= i < h and 0 <= j < w and not check[i][j]:
            heappush(heap, (field[i][j], i, j))
            check[i][j] = True

print(power)