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A 排列字母

我的题解

签到题

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import os
import sys

s = input()
arr = [0]*100

for i in range(len(s)):
  arr[ord(s[i]) - ord('A')] +=1


for i in range(0,26):
  #print(arr[i],end=" ")
  if arr[i] != 0:
    for _ in range(arr[i]):
      print(chr(i+ord('A')),end='')


正确题解

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string = 'WHERETHEREISAWILLTHEREISAWAY'
print(''.join(sorted(string)))

B 寻找整数

正确题解

参考题解0寻找整数 - 蓝桥云课

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L=[]
for i in range(187,10**12,187):
    if i%44==33 and i%45==29 and i%46==15 and i%47==5 and i%48==41 and i%49==46:
    #else:
        L.append(i)
    if len(L)>=2:
        break
arr = [0,0,1,2,1,4,5,4,1,2,9,0,5,10,
11,14,9,0,11,18,9,11,11,15,17,9,
23,20,25,16,29,27,25,11,17,4,29,22,
37,23,9,1,11,11,33,29,15,5,41,46] 
    

print(L)
L=[5458460249, 12590206409]
step=L[1]-L[0]
flag=0
for i in range(L[0],10**17,step):
    flag = False
    for j in range(2,50):
        if i%j!=arr[j]:
            flag =True
            continue
    if flag == False: 
        print(i)
        break

C 纸张尺寸

我的题解

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x = 1189
y = 841

arr1 = [1189,841,594,420,297,210,148,105,74,52,37,26,18]
arr2 = [841,594,420,297,210,148,105,74,52,37,26,18,13]


def cal(a,b,n):
    if n>=20:
        exit(0)
    print(f'{a} {b}')
    cal(b,a//2,n+1)
  
s = input()
pos = int(s[1])
# print(pos)
print(arr1[pos])
print(arr2[pos])


##1189 841
##841 594
##594 420
##420 297
##297 210
##210 148
##148 105
##105 74
##74 52
##52 37
##37 26
##26 18
##18 13
##13 9
##9 6
##6 4
##4 3
##3 2
##2 1
##1 1

D 数位排序

我的题解

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import os
import sys
from functools import cmp_to_key

class Node:
    def __init__(self,v,pos):
        self.v = v
        self.pos = pos
    def __repr__(self):
        return repr((self.v,self,pos))
        
n = int(input())
m = int(input())

li = []


def ini(x):
    sum = 0
    while x//10 !=0:
        sum+=x%10
        x//=10
    return sum+x
def cmp(x,y):
    if x.pos == y.pos:
        return x.v - y.v
    return x.pos - y.pos



for i in range(1,n):
    li.append(Node(i,ini(i)))


li.sort(key = cmp_to_key(cmp))

##for i in li:
##    print(f'{i.v} {i.pos}')
print(li[m-1].v)



正确题解

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import os
import sys

n=int(input())
m=int(input())
s=list(range(1,n+1))
s.sort(key=lambda x:sum(int(i) for i in str(x)))
print(s[m-1])

E 蜂巢

其实这题不是很难，要对自己有信心

正确题解

  • 将六边形转化为直角坐标系，然后进行坐标计算

  • a到b点，考虑斜着走和先斜着走，在横着走的操作z

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import os
import sys

# 请在此输入您的代码
d1,p1,q1,d2,p2,q2=map(int,input().split())

xway = [-1,-0.5,0.5,1,0.5,-0.5]
yway = [0,1,1,0,-1,-1]

xa = xway[d1]*p1 + xway[(d1+2)%6]*q1
ya = yway[d1]*p1 + yway[(d1+2)%6]*q1

xb = xway[d2]*p2 + xway[(d2+2)%6]*q2
yb = yway[d2]*p2 + yway[(d2+2)%6]*q2

disx = abs(xa-xb)
disy = abs(ya-yb)


if disx < disy/2:    #斜着走就能到
    print(int(disy))
else:
    disx -= disy*0.5
    print(int(disy+disx))

F 消除游戏

TODO

我的题解

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print("EMPTY")

正确题解

0全排列的价值 - 蓝桥云课

G 全排列

我的题解

想着dfs枚举推公式，然后推不出来

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st = [False] * 10
n = 3
ans =[]

def judge():
    cnt = 0
    for i in range (1,len(path)):
        for j in range(0,i):
           if path[i] > path[j]:
               cnt += 1
    res.append(cnt) 
def dfs(u):
    if u==n:
       # print(path)
        judge()
        return
    for v in range(1,n+1):
       if st[v] == False:
           st[v] = True
           path.append(v)
           dfs(u+1)
           path.pop()
           st[v] = False
            
for i in range(3,6):
    res = []
    n = i
    path = []
    dfs(0)
    ans.append(sum(res))
    


print(ans)

正确解法

H

线段树，跳过

J 最优清零方案

我的题解

通过枚举每个n//k个区间，减去最小值，加上剩余得单独处理的元素。

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import sys
sys.setrecursionlimit(50000)
input = lambda: sys.stdin.readline().strip()

n, k = map(int,input().split())
arr = list(map(int,input().split()))

cnt = n//k
res = 0
# print(cnt)
for i in range(cnt):
    tmp = i * k
    #print(tmp)
    cntt = 0
    minnum = 100000000
    for j in range(tmp,tmp+k):
        cntt += arr[j]
        minnum = min(arr[j],minnum)
##    print(f'{cntt} {minnum}')
       
    res +=  cntt-minnum*(k-1)

#print(cnt*k)
for j in range(cnt*k,n):
    res += arr[j]




print(res)

正确题解

  • 果然还是不够贪啊

  • 每次操作选择连续的 k 个元素，减去它们的最小值 min_val，操作次数加 min_val。

  • 剩余不足 k 的元素逐个减（操作次数加剩余元素的和）。
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import sys

n,k = map(int,input().split())

nums = list(map(int,input().split()))

cnt = 0

i = 0
while i < n-k+1:
    x = min(nums[i:i+k])
    b = nums[i:i+k].index(x)	#最小值下标，减去后为0
    if x:
        for  a in range(i,i+k):
            nums[a] -= x
        cnt+=x
    i+=b+1					#最小值下标为0，+1
print(cnt+sum(nums))