蓝桥杯python第三期模拟赛 python题解
文章目录
想着在考前模拟测试一下，就写了这个
题目1
print(int('2021ABCD',16))
思路：python一行搞定
答案：539077581
题目2
from math import gcd
def g(a,b):
return a*b//gcd(a,b)
cnt = 0
for i in range(1,2022):
if g(i,2021) == 4042:
cnt += 1
print(cnt) # 答案3
思路：直接暴力搜索，轻松通过
题目3
思路：这道题来说的话，两两配对嘛，直接暴力搜索，然后判断条件，符合+1，轻松通过咯
a = [i*i for i in range(45)]
s = set()
for i in range(len(a)):
for j in range(len(a)):
if a[i] + a[j] not in s and 1 <= a[i] + a[j] <= 2021:
s.add(a[i]+a[j])
# print(s)
print(len(s)) # 答案 624
题目4
思路这道题的话就是最小生成数嘛，我们可以直接套模板，也是很简单的
a = ['0 9 3 0 0 0 0 9',
'9 0 8 1 4 0 0 0',
'3 8 0 9 0 0 0 0',
'0 1 9 0 3 0 0 5',
'0 4 0 3 0 7 0 6',
'0 0 0 0 7 0 5 2',
'0 0 0 0 0 5 0 4',
'9 0 0 5 6 2 4 0']
graph = [[0]*9 for _ in range(9)]
for i in range(8):
a[i] = a[i].split(' ')
for j in range(8):
graph[i+1][j+1] = int(a[i][j])
def find(x):
if father[x] != x:
father[x] = find(father[x])
return father[x]
def union(x,y):
fx,fy = find(x),find(y)
if fx!=fy:
father[fx] = fy
father = [i for i in range(9)]
ans = 0
for x in range(1,8):
w = float('inf')
for i in range(1,9):
for j in range(1,9):
# 首先有边
if graph[i][j] != 0 and find(i) != find(j):
if w > graph[i][j]:
w = graph[i][j]
a = i
b = j
ans += w
union(a,b)
print(a,b,w)
print(ans) # 答案26
题目5
思路，简单点就是暴力搜索呀，有些人用前缀和，我个人觉得填空题没必要
a = ['69859241839387868941',
'17615876963131759284',
'37347348326627483485',
'53671256556167864743',
'16121686927432329479',
'13547413349962773447',
'27979945929848824687',
'53776983346838791379',
'56493421365365717745',
'21924379293872611382',
'93919353216243561277',
'54296144763969257788',
'96233972513794732933',
'81443494533129939975',
'61171882988877593499',
'61216868895721348522',
'55485345959294726896',
'32124963318242554922',
'13593647191934272696',
'56436895944919899246']
ans = 0
for i in range(16):
for j in range(16):
sum = 0
for x in range(i,i+5):
for y in range(j,j+5):
sum += int(a[x][y])
if ans < sum:
ans = sum
print(ans) # 154
题目6
思路：其实这道题也很简单的，最小编号就是我们的a*(p-1) + 1，就是上一页的最大值+1，最大编号是min(t, a * p)
t,a,p = map(int,input().split())
print(a*(p-1)+1,min(t,a*p))
题目7
思路：对于python来说，就是逐一判断了，不符合直接退出，O n的算法
s = input()
for i in range(len(s)-1,0,-1):
if s[i] > s[i-1]:
flag = True
else:
flag = False
break
if not flag:
print('NO')
else:
print('YES')
题目8
思路：灵活运用datetime库，完美嘻嘻
import datetime
s = input()
e = input()
def change(x):
x = x.split(':')
date = datetime.datetime(2022,4,8,hour=int(x[0]),minute=int(x[1]),second=int(x[2]))
return date
s = change(s)
e = change(e)
print(e-s)
题目9
思路，有些人用状压慢慢排序，我们不用，python有key排序，多简单，我们只要指定key就可以了，嘻嘻
n,m,k = map(int,input().split())
import time
a = []
for i in range(n):
a.append(list(map(int,input().split()))+[i])
s = [0]*n
for i in range(m):
a.sort(key=lambda x:x[i])
#a = sorted(a,key=lambda x:x[i])
for j in range(n):
s[a[j][-1]] += j
print(' '.join(map(str,s)))
题目10
思路：压轴题还是难，不过怎么说呢，捞点分吧，我这里借鉴了别人的写法，其实就是高级的LIS
最后思路是动态规划，想懂的可以详细研究代码，明天考试，还要复习，就不详细写了。
# DFS写法，过不了30%
n,k = map(int,input().split())
a = list(map(int,input().split()))
s = input()
ans = 0
# cnt 代表s中下标的个数，idx是从这部分可以深度搜索
# length是长度
def dfs(cnt,idx,length):
global ans
if s[idx] == '1': # 如果遇到1
cnt += 1
length += 1
if idx == n-1 and cnt >= k:
ans = max(ans,length)
for i in range(idx+1,n):
if a[i] > a[idx]:
dfs(cnt,i,length)
for i in range(n):
dfs(0,i,0)
print(ans)
# 邪恶的DP
# dp[i][j]表示取i个数中，至少有j个数在s下标下的长度
n,k = map(int,input().split())
a = list(map(int,input().split()))
s = input()
dp = [[-1]*(k+1) for _ in range(n+1)]
dp[0][0] = 1
if s[0] == '1':
dp[0][1] = 1
for i in range(1,n):
dp[i][0] = 1 # 单独一个i最少可以
for j in range(k): # 去s中的数
# a[i]为下标数
if s[i] == '1' and j > 0:
flag = True
else:
flag = False
for x in range(1,i):
# 从前面寻找最大的序列
if a[x] > a[j]: continue
# a[i] 在s中
if flag:
if dp[x][j-1] == -1:
continue
dp[i][j] = max(dp[i][j],dp[x][j-1] + 1)
else:
if dp[x][j] == -1:
continue
dp[i][j] = max(dp[i][j],dp[x][j] + 1)
print(dp[n-1][k])
