第十四届蓝桥杯大赛软件赛省赛pythonB(含代码)

A.2023

暴力即可

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")# INF = 0x3f3f3f3f3f3f# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())a = 12345678
b = 98765432
cnt_all = 0
ans = 0for num in range(a, b + 1):cnt_all += 1tmp = [int(i) for i in str(num)]flag_2_0 = Falseflag_0 = Falseflag_2_1 = Falseflag_3 = Falsefor i in range(len(tmp)):if flag_2_0 == False and tmp[i] == 2:flag_2_0 = Trueif flag_2_0 == True and flag_0 == False and tmp[i] == 0:flag_0 = Trueif flag_2_0 == True and flag_0 == True and flag_2_1 == False and tmp[i] == 2:flag_2_1 = Trueif flag_2_0 == True and flag_0 == True and flag_2_1 == True and flag_3 == False and tmp[i] == 3:flag_3 = Trueans += 1break
print(cnt_all - ans)
# 85959030

B.硬币兑换

注意，只有新币可以兑换，最后结果不论硬币新旧。
故应当将1 + 2022， 2 + 2021， … ，1011 + 1012。

a = [i for i in range(2024)]
a = a + [0] * 2024
left = 1
right = 2022
while left < right:a[left + right] += a[left]a[right] -= a[left]a[left] = 0left += 1right -= 1print(max(a))
# 513589

C.松散子序列

import sysinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())s = input()
n = len(s)
f = [0] * n
if s == '':print(0)
elif n == 1:print(ord(s[0]) - ord('a') + 1)
else:f[0] = ord(s[0]) - ord('a') + 1f[1] = max(f[0], ord(s[1]) - ord('a') + 1)for i in range(2, n):f[i] = max(f[i - 1], f[i - 2] + ord(s[i]) - ord('a') + 1)print(max(f))

D.管道

直接遍历每一个阀门，定义数组 v i s i vis_i visi 为第 i i i 个阀门最早感应到水流的时间。

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")INF = int(1e12)# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())n, len = read()
a = []
for _ in range(n):l, s = read()a.append([l, s])vis = [INF] * (len + 1)for i in range(n):l, s = a[i]vis[l] = min(vis[l], s)for i in range(n):l, s = a[i]t = s + 1left = l - 1right = l + 1while left >= 1 and t < vis[left]:vis[left] = tt += 1left -= 1t = s + 1while right <= len and t < vis[right]:vis[right] = tt += 1right += 1print(max(vis[1::]))

E.保险箱

写的贪心，感觉不对。

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")# INF = 0x3f3f3f3f3f3f# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())n = I()
x = input()
y = input()
prev = 0
ans = 0
for i in range(n-1, -1, -1):tmp_x = int(x[i]) + prevtmp_y = int(y[i])if tmp_x > tmp_y:if (tmp_x - tmp_y) - (tmp_y + 10 - tmp_x) > 1:ans += (tmp_y + 10 - tmp_x)prev = 1else:ans += tmp_x - tmp_yprev = 0elif tmp_x == tmp_y:prev = 0else:if (tmp_y - tmp_x) - (tmp_x + 10 - tmp_y) > 1:ans += (tmp_x + 10 - tmp_y)prev = -1else:ans += tmp_y - tmp_xprev = 0print(ans)

F.树上选点

暴力枚举第一个点，再找第二个点。

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")# INF = 0x3f3f3f3f3f3f# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())def dfs_deep(u, depth):deep[u] = depthfor v in g[u]:dfs_deep(v, depth + 1)def dfs(u, fa, x):if fa == x:vis[u] = Truefor v in g[u]:if v == x:vis[u] = Truedfs(v, u, x)n = I()
g = [[] for _ in range(n + 1)]f = list(read())for i in range(n - 1):g[f[i]].append(i + 2)
value = list(read())
value = [0] + value
deep = [0] * (n + 1)
dfs_deep(1, 0)
ans = 0for x in range(1, n + 1):vis = [False] * (n + 1)vis[0] = Truefor i in range(1, n + 1):if deep[i] == deep[x]:vis[i] = Truedfs(1, 0, x)tmp = []maxm1 = 0for i in range(1, n + 1):if not vis[i] and i != x:if value[i] > maxm1:maxm1 = value[i]ans = max(ans, value[x] + maxm1)print(ans)

G.T字消除

没做。

H.独一无二

开个cnt数组，每次dis[u]+w == dis[v]的时候+1

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")INF = 0x3f3f3f3f3f3f# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())class Edge:v = 0w = 0ne = -1def add(a, b, c):global idxe[idx].v = be[idx].w = ce[idx].ne = h[a]h[a] = idxidx += 1def dijkstra(s):vis = [False] * (n + 1)dis = [INF] * (n + 1)cnt = [0] * (n + 1)dis[0] = dis[s] = 0vis[0] = Trueq = []heapq.heappush(q, [0, s])while q:_, u = heapq.heappop(q)if vis[u]:continuevis[u] = Truei = h[u]while i != -1:v = e[i].vw = e[i].wi = e[i].neif dis[u] + w < dis[v]:dis[v] = dis[u] + wif not vis[v]:heapq.heappush(q, [dis[v], v])if dis[u] + w == dis[v]:cnt[v] += 1cnt[s] = 1for i in range(1, n + 1):print(cnt[i] - 1)n, m = read()
h = [-1] * (n + 1)
idx = 0
e = [Edge() for _ in range(m)]
for _ in range(m):u, v, c = read()add(u, v, c)
dijkstra(1)

I.异或和

a维护以x为根的子树异或和, prev维护每个结点的点权。
每次change只需要向上更改贡献即可。
正解应该是 dfs序+树状数组/线段树维护

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")# INF = 0x3f3f3f3f3f3f# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())def dfs(u):for v in g[u]:a[u] ^= dfs(v)return a[u]def change(u, x, y):if u == x:a[x] = a[x] ^ prev[x] ^ y# 注意要把点权也修改了，比赛的时候忘记了。。。。prev[x] = yreturn prev[x] ^ yfor v in g[u]:a[u] ^= change(v, x, y)return 0n, m = read()
a = list(read())
a = [0] + a
prev = copy.deepcopy(a)
g = [[] for _ in range(n + 1)]
for i in range(n - 1):u, v = read()g[u].append(v)dfs(1)for _ in range(m):op = list(read())if op[0] == 1:x, y = op[1], op[2]change(1, x, y)if op[0] == 2:x = op[1]print(a[x])

J.混乱的数组

手算 x ≤ 10 x\le10 x≤10

import bisect
import sys
import copy
from collections import deque, defaultdict
import heapq
from itertools import accumulate, permutations, combinations
import mathinput = lambda: sys.stdin.readline().rstrip("\r\n")
printf = lambda d: sys.stdout.write(str(d) + "\n")# INF = 0x3f3f3f3f3f3f# sys.setrecursionlimit(100000)def read():line = sys.stdin.readline().strip()while not line:line = sys.stdin.readline().strip()return map(int, line.split())def I():return int(input())x = I()
if x == 1:print(2)print(2, 1)
if x == 2:print(3)print(2, 1, 1)
if x == 3:print(3)print(3, 2, 1)
if x == 4:print(4)print(2, 2, 1, 1)
if x == 5:print(4)print(3, 2, 1, 1)
if x == 6:print(4)print(4, 3, 2, 1)
if x == 7:print(5)print(3, 2, 2, 2, 1)
if x == 8:print(5)print(3, 3, 2, 1, 1)
if x == 9:print(5)print(4, 3, 2, 1, 1)
if x == 10:print(5)print(5, 4, 3, 2, 1)

注意

非官方题解，只是我自己的做法，如果有错误，或者有更好的做法都欢迎指正。

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