A排列字母

s='WHERETHEREISAWILLTHEREISAWAY'
s=list(s)
s.sort()
print(''.join(s))
#AAAEEEEEEHHHIIILLRRRSSTTWWWY

B寻找整数

答案：2022040920220409

思路：中国剩余定理应该能做，不过比赛的时候记不清咋写的了，就暴力一波。注意，一个个累加是不行的，自己用python算一下pow(10,17)就知道了。我当时选了最后五个数，因为这几个比较大，先找到能同时满足这五个的数的规律：

tmp.py

i=1
while True:
    flag=True
    if i%49!=46:
        flag=False
    if i%48!=41:
        flag=False
    if i%47!=5:
        flag=False
    if i%46!=15:
        flag=False
    if i%45!=29:
        flag=False
    if flag:
        print(i)
    i+=1

跑出来前几个就够了：4772009, 42909689, 81047369, 119185049,157322729, 195460409, 233598089……不用说肯定是等差数列，再稍微验证一下y=38137680*x+4772009

然后枚举那个x的值，再遍历前面的是否满足即可，这样就能跑出来：

B.py

mod = [(2, 1), (14, 11), (26, 23), (38, 37),
       (3, 2), (15, 14), (27, 20), (39, 23),
       (4, 1), (16, 9), (28, 25), (40, 9),
       (5, 4), (17, 0), (29, 16), (41, 1),
       (6, 5), (18, 11), (30, 29), (42, 11),
       (7, 4), (19, 18), (31, 27), (43, 11),
       (8, 1), (20, 9), (32, 25), (44, 33),
       (9, 2), (21, 11), (33, 11), (45, 29),
       (10, 9), (22, 11), (34, 17), (46, 15),
       (11, 0), (23, 15), (35, 4), (47, 5),
       (12, 5), (24, 17), (36, 29), (48, 41),
       (13, 10), (25, 9), (37, 22)]
i = 0
cha = 38137680
while True:
    flag = True
    num = cha*i+4772009
    for x, y in mod:
        if num % x != y:
            flag = False
            break
    if flag:
        print(num)
        break
    i += 1
# 2022040920220409

C纸张尺寸

没啥好说的，长边整除2即可

name = input()
A0x, A0y = 1189, 841
if name == 'A0':
    print(A0x)
    print(A0y)
else:
    cnt = int(name[1])
    for i in range(cnt):
        A0x, A0y = A0y, A0x//2
    print(A0x)
    print(A0y)

D数位排序

只要会用cmp_to_key就不难

from functools import cmp_to_key


def cmp(x, y):
    hx = getHe(x)
    hy = getHe(y)
    if hx == hy:
        return x-y
    return hx-hy


def getHe(n):
    res = 0
    while n > 0:
        res += n % 10
        n //= 10
    return res


n = int(input())
m = int(input())
num = [i for i in range(1, n+1)]
num.sort(key=cmp_to_key(cmp))
print(num[m-1])

E蜂巢

没思路

F消除游戏

不确定

s = list(input())
pre = s
f = False
for i in range(pow(2, 64)):
    flag = [False]*len(s)
    for j in range(1, len(s)-1):
        if s[j] == s[j-1] and s[j] != s[j+1]:
            flag[j] = True
            flag[j+1] = True
        if s[j] != s[j-1] and s[j] == s[j+1]:
            flag[j-1] = True
            flag[j] = True
    for j in range(len(flag)):
        if flag[j]:
            s[j] = ''
    s = list(''.join(s))
    if len(s) == 0:
        f = True
        break
    if pre == s:
        break
    pre = s


if f:
    print('EMPTY')
else:
    print(''.join(s))

G全排列的价值

写了一点，就不放出来丢人了

H技能升级

求最大值那边应该可以用大顶堆优化一下，时间复杂度可以降到nlogn，比赛时忘了优化，呜呜呜

from math import ceil

n, m = map(int, input().split())
a = [0]
b = [0]
dp = [0]*(m+1)
for i in range(n):
    x, y = map(int, input().split())
    a.append(x)
    b.append(y)

for j in range(1, m+1):
    index = 0
    for k in range(1, n+1):
        if a[k] > a[index]:
            index = k
    dp[j] = dp[j-1]+a[index]
    a[index] -= b[index]
    a[index] = max(a[index], 0)

print(dp[m])

I最长不下降子序列

写了一点，就不放出来丢人了

J最优清零方案