ABC447をPythonで

AtCoder Beginner Contest 447の解答等の速報的まとめ

A問題

端数切り上げ

A

n, m = map(int, input().split())
print("Yes" if (n + 1) // 2 >= m else "No")

B問題

最大の個数を数える

B

s = input()

count = dict()
for s_i in s:
    if s_i not in count:
        count[s_i] = 0
    count[s_i] += 1

count_max = max(count.values())
ng = set()
for key, val in count.items():
    if count_max == val:
        ng.add(key)

ans = list()
for s_i in s:
    if s_i not in ng:
        ans.append(s_i)

print(*ans, sep="")

C問題

$A$を取った文字列が一致しているときのみ対象
各$A$以外の文字間にある$A$の個数の差が答え

C

def delete_a(S):
    res = list()
    a_list = [0]
    for s_i in S:
        if s_i != "A":
            res.append(s_i)
            a_list.append(0)
        else:
            a_list[-1] += 1

    return res, a_list


s = input()
t = input()

del_s, s_a = delete_a(s)
del_t, t_a = delete_a(t)

if del_s != del_t:
    exit(print(-1))

if len(s_a) < len(t_a):s_a.append(0)
if len(s_a) > len(t_a):t_a.append(0)

ans = 0
for s_i, t_i in zip(s_a, t_a):
    ans += abs(s_i - t_i)

print(ans)

D問題

左から順に$B$に到達したとき、

• $B$の左にまだカウントしていない$A$
• $B$の右にまだカウントしていない$C$

があるとき答えに1を足す

D

s = input()

a_count = 0
c_ind = 0
ans = 0
for i, s_i in enumerate(s):
    if s_i == "A":
        a_count += 1
    elif s_i == "B":
        c_ind = max(c_ind, i)
        while c_ind < len(s) and s[c_ind] != "C":
            c_ind += 1
        if c_ind < len(s) and s[c_ind] == "C" and a_count > 0:
            ans += 1
            a_count -= 1
            c_ind += 1

print(ans)