[toc]

A 穿越时空之门

特别要注意python整除是//

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cnt = 0
for i in range(1,2025):
    B = i
    cntB = 0
    while B:
        cntB += B%2
        B//=2
    BB = i
    cntBB = 0
    while BB:
        cntBB += BB %4
        BB//=4
    if cntB == cntBB:
        print(i)
        cnt+=1
print(cnt)

B 数字串个数

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mod = 10**9 + 7

x = (9**10000 - 2*8**10000 + 7**10000 )% mod
print(x)

C 连连看

我的题解

对一半，其余超时超时，暴力枚举对角线

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from collections import deque
n , m = map(int , input().split())

arr = []

for i in range(n):
    arr.append(list(map(int , input().split())))

xway = [-1,-1,1,1]
yway = [-1,1,-1,1]

    
cnt = 0
for i in range(n):
    for j in range(m):
       for way in range(4):
            x = i + xway[way]
            y = j + yway[way]
            while 0<=x<n and 0<=y<m:
                if arr[x][y] == arr[i][j]:
                    cnt+=1
                x+= xway[way]
                y+= yway[way]
            
print(cnt)

正确解法

类似于八皇后问题，取主对角线为i-j+1005，副对角线为i+j。

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import sys
n , m = map(int , input().split())

arr = []
zhu = [[0]*2050 for _ in range(2025)]
fu = [[0]*2050 for _ in range(2025)]


for i in range(n):
    arr.append(list(map(int,input().split())))
ans = 0    
for i in range(n):
    for j in range(m):
        ans += zhu[i-j+1005][arr[i][j]] + fu[i+j][arr[i][j]]
        zhu[i-j+1005][arr[i][j]]+= 1
        fu[i+j][arr[i][j]]+= 1
        
print(ans*2)

D 神奇的闹钟

使用datetime库解决

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from datetime import datetime, timedelta


st = datetime(1970,1,1) 
#print(st)

n = int(input())

for i in range(n):
    ymd, hms, m = input().split()
    st = datetime(1970,1,1) 
    #print(st)
    m = int(m)
    
    tt = datetime.fromisoformat(ymd+" "+hms)
    #print(tt)
    d = timedelta(minutes=m)
    #print(d)
    while st<tt:
        st+=d
    print(st-d)

E 蓝桥村的真相

暴力枚举然后找规律

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def check(ll):
    cnt = 0
    for i in range(len(ll)):
        tmp = ll[i]
        leng = len(ll[i])
        flag = False
        for j in range(leng*100):
            pos = j%leng
            if tmp[pos] == 1:
                if tmp[(pos+1)%leng] == 1 and tmp[(pos+2)%leng] == 1:
                    flag = True
                    break
                if tmp[(pos+1)%leng] ==0 and tmp[(pos+2)%leng] == 0:
                    flag = True
                    break
            if tmp[pos] == 0:
                if tmp[(pos+1)%leng] == 1 and tmp[(pos+2)%leng] == 0:
                    flag = True
                    break
                if tmp[(pos+1)%leng] ==0 and tmp[(pos+2)%leng] == 1:
                    flag = True
                    break
        if flag:
            continue
        
        for j in range(leng):
            if tmp[j] == 0:
                cnt += 1
        #print(cnt)
    return cnt
       
def dfs(path):
    if len(path) == n:
        ll.append(path.copy())  # 传入副本而不是地址
        return
    for i in range(0,2):
        path.append(i)
        #print(path)
        dfs(path)
        path.pop()

for i in range(1,10):
    n = i
    ll = []
    dfs([])
    #print(ll)
    print(check(ll))

F 魔法巡游

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import os
import sys

n = int(input())
a = [0] + list(map(int, input().split()))
b = [0] + list(map(int, input().split()))

inf = float('inf')


dp1 = [[-inf]*3 for _ in range(n+1)]
dp2 = [[-inf]*3 for _ in range(n+1)]

# dp1[0] = [0]*3	dp1不能初始化为0，因为小桥不是先手，无法从dp1转移
dp2[0] = [0]*3	


for i in range(1,n+1):
    len = -inf
    x = a[i]
    e = set()
    
    dp1[i] = dp1[i-1].copy()
    for j in str(x):
        if j == '0':
            e.add(0)
            len = max(len,dp2[i-1][0])
        if j == '2':
            e.add(1)
            len = max(len,dp2[i-1][1])
        if j == '4':
            e.add(2)
            len = max(len,dp2[i-1][2])
    for k in e:
        dp1[i][k] = max(dp1[i][k],len+1)
        
    len = -inf
    e = set()
    
    x = b[i]
    dp2[i] = dp2[i-1].copy()
    for j in str(x):
        if j == '0':
            e.add(0)
            len = max(len,dp1[i-1][0])
        if j == '2':
            e.add(1)
            len = max(len,dp1[i-1][1])
        if j == '4':
            e.add(2)
            len = max(len,dp1[i-1][2])
    
    for k in e:
        dp2[i][k] = max(dp2[i][k],len+1)

if n == 1:
    print(1)
else:
    print(max(max(dp1[-1]),max(dp2[-1])))

H纯职业小组

参考题解

有思路但是写不出来

第十五届蓝桥杯PythonB组H题纯职业小组题解 - AcWing

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T = int(input())
for _ in range(T):
    n, k = map(int, input().split())
    h = {}
    for i in range(n):
        a, b = map(int, input().split())
        if a not in h:
            h[a] = 0
        h[a] += b

    b = []
    for x in h:
        b.append(h[x])

    n = len(b)

    cnt = 0
    res = 0
    for i in range(n):		# 去除为2的小组
        cnt += b[i] // 3
        u = min(b[i], 2)
        b[i] -= u
        res += u

    if cnt < k:		#无法构成
        print(-1)
        continue
	
    c1, c2, c3 = 0, 0, 0	#记录能组成1,2,3小组的个数
    for x in b:
        c3 += x // 3		#记录c3小组个数
        x %= 3				
        if x == 1:			
            c1 += 1
        elif x == 2:
            c2 += 1

     # k为1时，再取一个即可
    k -= 1
    res += 1

    # 优先取3的组， 
    v = min(k, c3)
    k -= v
    c3 -= v
    res += v * 3

    v = min(k, c2)
    k -= v
    c2 -= v
    res += v * 2

    v = min(k, c1)
    k -= v
    c1 -= v
    res += v * 1

    print(res)