第十六届蓝桥杯大赛软件赛省赛 Python 大学 B 组 满分题解

题面链接Htlang/2025lqb_python_b

个人觉得今年这套题整体比往年要简单许多，但是G题想简单了出大问题，预估5+0+10+10+15+12+0+8=60，道阻且长，再接再厉

代码仅供学习参考，满分为赛后洛谷中的测评，蓝桥杯官方测评待完成

A: 攻击次数

答案：103？181？题目没说明白每回合是不是只能使用一个英雄？

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve():xl = 2025cnt = 0for i in range(1, 2025):cnt += 1xl -= 5if i % 2 == 1:xl -= 15else:xl -= 2if i % 3 == 1:xl -= 2elif i % 3 == 2:xl -= 10elif i % 3 == 0:xl -= 7if xl <= 0:breakprint(i, xl)print(cnt)T = 1
for i in range(T):solve()

B: 最长字符串

答案：afplcu

# # 考场错误解
# def I():
#     return input()# def II():
#     return int(input())# def MII():
#     return map(int, I().split())# def LMII():
#     return list(MII())# def solve():
#     with open(r'words.txt', 'r') as file:
#         data = file.readlines()#     dic = {}
#     for i in range(len(data)):
#         x = data[i].strip()
#         data[i] = list(x)
#         dic[x] = 0
#         data[i].sort()
#         dic[x] = data[i]#     can = set()
#     for i in range(100):
#         for x in data:
#             if len(x) == i + 1:
#                 if len(x) == 1:
#                     can.add(tuple(x))
#                 else:
#                     if tuple(x[:len(x) - 1]) in can:
#                         can.add(tuple(x))
#     cnt = 0
#     for x in can:
#         if len(x) > cnt:
#             cnt = len(x)
#     res = []
#     for x in can:
#         if len(x) == cnt:
#             res.append(x)
#     # print(dic)
#     print(res)
#     for x in dic:
#         # print(tuple(dic[x]))
#         if tuple(dic[x]) == res[0]:
#             print(x)# T = 1
# for i in range(T):
#     solve()# 正解
def is_beautiful_words(words):"""计算所有优美单词，并返回一个字典：key: 单词长度value: 集合，每个元素为 (word, sorted(word)) 表示已经确认的优美单词"""# 按长度升序排序（长度相同时字典序排序）words_sorted = sorted(words, key=lambda w: (len(w), w))# 用于存储每个长度的优美单词beautiful_by_length = {}for word in words_sorted:l = len(word)if l == 1:# 长度为1的单词自动是优美字符串（只要在单词本中）beautiful_by_length.setdefault(1, set()).add( (word, word) )  # 此处 sorted(word) == wordelse:# 先看是否存在长度为 l-1 的优美单词if (l - 1) not in beautiful_by_length:continue# 取当前单词的前 l-1 个字符，并计算其排序结果prefix = word[:-1]sorted_prefix = ''.join(sorted(prefix))# 检查是否存在一个长度为 l-1 的优美单词，其排序后的字符与 prefix 一致found = Falsefor bw, bw_sorted in beautiful_by_length[l - 1]:if bw_sorted == sorted_prefix:found = Truebreakif found:beautiful_by_length.setdefault(l, set()).add( (word, ''.join(sorted(word))) )return beautiful_by_lengthdef find_longest_beautiful_word(filename):with open(filename, "r", encoding="utf-8") as f:# 每一行一个单词，去除空白符words = [line.strip() for line in f if line.strip()]beautiful_by_length = is_beautiful_words(words)if not beautiful_by_length:return ""# 找到存在的最大长度max_len = max(beautiful_by_length.keys())# 在最大长度中，找出字典序最小的那个单词candidates = [word for word, _ in beautiful_by_length[max_len]]result = min(candidates) if candidates else ""return resultif __name__ == "__main__":# 假设文件名为 words.txtresult = find_longest_beautiful_word("words.txt")print("最长的优美字符串为:", result)

C: LQ 图形

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve():w, h, v = MII()for i in range(h):print("Q" * w)for i in range(w):print("Q" * (w + v))T = 1
for i in range(T):solve()

D: 最多次数

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve():s = I()check = {'lqb', 'lbq', 'qlb', 'qbl', 'blq', 'bql'}res = 0i = 0# print(len(s))while i < len(s) - 2:# print(i, s[i:i + 3])if s[i:i + 3] in check:res += 1i += 2i += 1print(res)T = 1
for i in range(T):solve()

E: A · B Problem

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve():L = II()res = 0lst = [0] * (L + 5)for i in range(1, L + 1):for j in range(1, L + 1):if i * j >= L:breaklst[i * j] += 1pre = lst.copy()for i in range(1, L):pre[i] += pre[i - 1]for i in range(1, L):cnt1 = icnt2 = L - ires += lst[cnt1] * pre[cnt2]print(res)T = 1
for i in range(T):solve()

赛时对拍

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve1(n):res = 0for i in range(1, n + 1):for j in range(1, n + 1):for k in range(1, n + 1):for l in range(1, n + 1):if i * k + j * l <= n:res += 1# print((i, k), (j, l))print(res)def solve2(L):res = 0lst = [0] * (L + 5)for i in range(1, L + 1):for j in range(1, L + 1):if i * j >= L:breaklst[i * j] += 1pre = lst.copy()for i in range(1, L):pre[i] += pre[i - 1]for i in range(1, L):cnt1 = icnt2 = L - ires += lst[cnt1] * pre[cnt2]print(res)T = 1
for i in range(T):L = II()solve1(L)solve2(L)

F: 园艺

赛时代码，可能超时

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve():n = II()data = LMII()res = 1cnt = 1for i in range(1, n):if data[i] > data[i - 1]:cnt += 1else:res = max(res, cnt)cnt = 1res = max(res, cnt)if res == 1 or res == n:print(res)returnfor jg in range(2, n):  # 间隔if res > (n - 1) // jg + 1:breakfor st in range(n - jg):cnt = 1for idx in range(st + jg, n, jg):if data[idx] > data[idx - jg]:cnt += 1else:res = max(res, cnt)cnt = 1res = max(res, cnt)print(res)T = 1
for i in range(T):solve()

赛后优化及对拍

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve1(n, data):res = 1cnt = 1for i in range(1, n):if data[i] > data[i - 1]:cnt += 1else:res = max(res, cnt)cnt = 1res = max(res, cnt)if res == 1 or res == n:print(res)returnfor jg in range(2, n):  # 间隔if res > (n - 1) // jg + 1:breakfor st in range(n - jg):cnt = 1for idx in range(st + jg, n, jg):if data[idx] > data[idx - jg]:cnt += 1else:res = max(res, cnt)cnt = 1res = max(res, cnt)print(res)def solve2(n, data):dp = [[1 for _ in range(n + 1)] for _ in range(n + 1)]for i in range(0, n):for j in range(0, i):if data[i] > data[j]:dp[i][i - j] = dp[j][i - j] + 1print(max([max(d) for d in dp]))T = 1
for i in range(T):n = II()# data = LMII()import randomdata = [random.randint(1, 2 ** 20) for _ in range(n)]import timet1 = time.time()solve1(n, data)t2 = time.time()solve2(n, data)t3 = time.time()print(t2 - t1, t3 - t2)

G: 书架还原

# # 考场错误解
# def I():
#     return input()# def II():
#     return int(input())# def MII():
#     return map(int, I().split())# def LMII():
#     return list(MII())# def solve():
#     n = II()
#     data = [0] + LMII()
#     res = 0
#     for i in range(1, n + 1):
#         if i != data[i]:
#             if i == data[data[i]]:
#                 res += 1
#                 x = data[data[i]]
#                 data[data[i]] = data[i]
#                 data[i] = x
#     print(res, data)
#     cnt = 0
#     for i in range(1, n + 1):
#         if i != data[i]:
#             cnt += 1
#     if cnt:
#         print(res + cnt - 1)
#     else:
#         print(res)# T = 1
# for i in range(T):
#     solve()# 正解
n = int(input())
a = list(map(int, input().split()))
visited = [False] * n
ans = 0for i in range(n):if not visited[i] and a[i] != i + 1:count = 0j = iwhile not visited[j]:visited[j] = Truecount += 1j = a[j] - 1ans += count - 1print(ans)

H: 异或和

赛时代码，超时

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve1():n = II()data = LMII()res = 0for i in range(n - 1):for j in range(i + 1, n):res += (data[i] ^ data[j]) * (j - i)print(res)T = 1
for i in range(T):solve1()

赛时对拍，超时

def I():return input()def II():return int(input())def MII():return map(int, I().split())def LMII():return list(MII())def solve1(n, data):res = 0for i in range(n - 1):for j in range(i + 1, n):# print(data[i],data[j],data[i] ^ data[j])res += (data[i] ^ data[j]) * (j - i)print(res)def solve2(n, data):res = 0for j_i in range(1, n):cnt = 0for i in range(j_i):for j in range(i + j_i, n, j_i):cnt += data[j - j_i] ^ data[j]res += j_i * cntprint(res)T = 1
for i in range(T):n = II()# data = LMII()import randomdata = [random.randint(1, 2 ** 20) for _ in range(n)]import timet1 = time.time()solve1(n, data)t2 = time.time()solve2(n, data)t3 = time.time()print(t2 - t1, t3 - t2)

赛后补题

n = int(input())
a = list(map(int, input().split()))
ans = 0for k in range(31):  # 考虑31位足以覆盖正整数的情况cnt1 = 0  # 记录当前位为1的个数sum1 = 0  # 记录当前位为1的位置之和for j in range(1, n + 1):b = (a[j - 1] >> k) & 1  # 取出第k位if b:cnt0 = j - 1 - cnt1sum0 = (j - 1) * j // 2 - sum1ans += (j * cnt0 - sum0) * (1 << k)cnt1 += 1sum1 += jelse:ans += (j * cnt1 - sum1) * (1 << k)print(ans)