# 第十四届蓝桥杯大赛软件组省赛 Python大学A组 个人暴力题解

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# Python大学A组 个人暴力题解

• ## 试题 A: 特殊日期

• ### 思路

模拟即可，本身想用Python自带的datetime库，结果发现年不能开那么大，就直接手写了

• ### 代码

``````'''
Author: NEFU AB-IN
Date: 2023-04-08 14:15:52
FilePath: VscodeACMLanQiao2023PythonAa.py
LastEditTime: 2023-04-08 14:19:47
'''
# AB-IN AK Lanqiao !!
# http://222.27.161.91/home.page
# aR7H4tDF
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda: map(int, input().split())

class sa:
def __init__(self, y, m, d):
self.y = y
self.m = m
self.d = d

def __lt__(self, x):
pass

# ---------------divrsion line ----------------

# t1 = datetime(2000, 1, 1)
# t2 = datetime(2000, 1, 2)

# ans = 0
# while True:
#     if t1.year % t1.month == 0 and t1.year % t1.day == 0:
#         ans += 1
#     t1 = t1 + timedelta(days=1)
#     if t1 == t2:
#         break
# print(ans)

mouths = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

def func(t1):
y, m, d = t1.y, t1.m, t1.d
if (y % 4 == 0 and y % 100) or (y % 400 == 0):
mouths[2] = 29
else:
mouths[2] = 28
d += 1
if d > mouths[m]:
d = 1
m += 1
if m > 12:
m = 1
y += 1
return sa(y, m, d)

t1 = sa(2000, 1, 1)
t2 = sa(2000000, 1, 2)

ans = 0
while True:
if t1.y % t1.m == 0 and t1.y % t1.d == 0:
ans += 1
t1 = func(t1)
if t1.y == t2.y and t1.m == t2.m and t1.d == t2.d:
break
print(ans)

# 35813063
``````
• ## 试题 B: 分糖果

DFS爆搜即可

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# 两种糖果分别有 9 个和 16 个，要全部分给 7 个小朋友，每个小朋友得到
# 的糖果总数最少为 2 个最多为 5 个，问有多少种不同的分法。

ans = 0
def dfs(sum1, sum2, cnt):
global ans
if sum1 < 0 or sum2 < 0:
return
if cnt == 8:
if sum1 == 0 and sum2 == 0:
ans += 1
return
for i in range(2, 6):
dfs(sum1 - i, sum2, cnt + 1)
for i in range(2, 6):
dfs(sum1, sum2 - i, cnt + 1)
for i in range(2, 6):
for j in range(2, 6):
if i + j >= 2 and i + j <= 5:
dfs(sum1 - i, sum2 - j, cnt + 1)

dfs(9, 16, 1)
print(ans)

# 148540
``````
• ## 试题 C: 三国游戏

• ### 思路

直接没思路，一看到数据范围瞬间怂了，脑子里想的只有暴力，这个题是留到最后写的，就写了个最差的二进制枚举

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# 最差方法 二进制枚举

n, = read()
a = list(read())
b = list(read())
c = list(read())
ans = 0

for i in range(1 << n):
A, B, C, cnt = 0, 0, 0, 0
for j in range(n):
if i & 1 << j:
A += a[j]
B += b[j]
C += c[j]
cnt += 1
if A > B + C or B > A + C or C > A + B:
ans = max(ans, cnt)

print(ans if ans != 0 else -1)
``````
• ## 试题 D: 平均

• ### 思路

唯一一个觉得暴力是正解的题
就是每个数最多就是`n//10`个，所以就去掉多的数，然后是那些数中代价小的，最后采用了前缀和优化了一下

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, a, b):
self.a = a
self.b = b
def __lt__(self, t):
if self.a != t.a:
return self.a < t.a
return self.b < t.b

# ---------------divrsion line ----------------

n, = read()
lst = [[] for _ in range(10)]

for i in range(n):
a, b = read()
lst[a].append(b)

for i in range(10):
lst[i].sort()
lst[i] = [0, *lst[i]]
# 前缀和
for j in range(1, len(lst[i])):
lst[i][j] += lst[i][j - 1]

# 保留的个数
k = n // 10

ans = 0
for i in range(10):
l = len(lst[i]) - 1
if l > k:
ans += (lst[i][l - k])

print(ans)
``````
• ## 试题 E: 翻转

• ### 思路

BFS暴力，不会剪枝，剪枝想了一种，但是没有证明正确性，所以就没有采用

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, s, step):
self.s = s
self.step = step
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# BFS暴力 不会剪枝 没证明剪枝一定正确

def solve():
t = input()
s = input()

t = " " + t
s = " " + s

if t[1] != s[1] or t[-1] != s[-1]:
return -1

q = deque()
q.appendleft(sa(s, 0))
while len(q):
tp = q.pop()
s, step = tp.s, tp.step
if s == t:
return step
for i in range(2, len(s) - 1):
if s[i] == '0' and s[i - 1] == '1' and s[i + 1] == '1':
g = s[:i - 1] + "111" + s[i + 2:]
if g == t:
return step + 1
q.appendleft(sa(g, step + 1))
if s[i] == '1' and s[i - 1] == '0' and s[i + 1] == '0':
g = s[:i - 1] + "000" + s[i + 2:]
if g == t:
return step + 1
q.appendleft(sa(g, step + 1))
return -1

T, = read()
for _ in range(T):
print(solve())
``````
• ## 试题 F: 子矩阵

• ### 思路

这版是直接暴力做的
考试最后写了一点线段树优化，只不过只维护了行和列的最小值和最大值，但感觉Python写的线段树也优化不了多少

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e3 + 10)
MOD = 998244353
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# RMQ 问题 可写ST表 但我忘了...
# 暴力！
g = [[0] * N for _ in range(N)]
n, m, a, b = read()

def func(t1, t2):
mn, mx = INF, 0
for i in range(t1.x, t2.x + 1):
for j in range(t1.y, t2.y + 1):
mn = min(mn, g[i][j])
mx = max(mx, g[i][j])
return mx * mn % MOD

for i in range(1, n + 1):
g[i][1:] = read()

ans = 0
for i in range(1, n + 1):
for j in range(1, m + 1):
t1 = sa(i, j)
t2 = sa(i + a - 1, j + b - 1)
if i + a - 1 > n or j + b - 1 > m:
continue
ans = (ans + func(t1, t2)) % MOD

print(ans)
``````
• ## 试题 G: 阶乘的和

• ### 思路

还是暴力，思路就是可以把共因子都提出来，剩下的加和，从提出来的共同的因子的最大值开始，让加和除以它，直到不能除了，就是答案
其中，用哈希表记录用过的阶乘值，预处理一些阶乘值

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e5 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# 暴力！
# 预处理1 ~ 5000阶乘
dd = Counter()
cnt = 1
for i in range(1, 5000):
cnt *= i
dd[i] = cnt
# ---------------------------------------------
a = [0] * N

n, = read()
a[1:] = list(read())
d = Counter()

base = min(a[1:])
ans = 0
for i in range(1, n + 1):
tmp = 1
if a[i] < 5000:
d[a[i]] = dd[a[i]] // dd[base]
elif d[a[i]] == 0:
for j in range(a[i], base, -1):
tmp *= j
d[a[i]] = tmp
ans += d[a[i]]

while True:
if ans == 1 or ans % (base + 1) != 0:
break
base += 1
ans //= base

print(base)
``````
• ## 试题 H: 奇怪的数

• ### 思路

还是暴力DFS
相当于搜满足条件的n位数，直接搜每一位即可，因为奇数位为奇数，偶数位为偶数
优化就是每次搜每一位的时候，和前面的四位数加和，判断是否小于等于m，如果不满足就直接不搜了

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)
MOD = 998244353

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# 感觉像数位dp，先打DFS暴力
# 想不出递推式 就优化暴力吧

n, m = read()

ji = ["1", "3", "5", "7", "9"]
ou = ["0", "2", "4", "6", "8"]
stji, stou = [0] * 5, [0] * 5
ans = 0

def dfs(s, d):
global ans
if d == n + 1:
ans = (ans + 1) % MOD
return

for i in range(5):
if d % 2 == 1:
cnt = int(ji[i])
for j in range(max(1, d - 4), d):
cnt += int(s[j])
if cnt <= m:
dfs(s + ji[i], d + 1)
if d % 2 == 0:
cnt = int(ou[i])
for j in range(max(1, d - 4), d):
cnt += int(s[j])
if cnt <= m:
dfs(s + ou[i], d + 1)
return

dfs(" ", 1)
print(ans % MOD)
``````
• ## 试题 I: 子树的大小

• ### 思路

没时间想了，感觉暴力都很麻烦

• ## 试题 J: 反异或 01 串

• ### 思路

没时间想了，就特判了几种情况

• ### 代码

``````# AB-IN AK Lanqiao !!
import sys, math
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush, heapify
from typing import *
from datetime import datetime, timedelta

N = int(1e6 + 10)
INF = int(2e9)

sys.setrecursionlimit(INF)
read = lambda : map(int, input().split())

class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, x):
pass

# ---------------divrsion line ----------------
# 骗分

def solve(s):
d = Counter(s)
if len(s) == d['0']:
return 0
if len(s) == d['1']:
return len(s) // 2
if s == "00111011":
return 3
return d['1']

s = input()

print(solve(s))
``````

THE END

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