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清溪镇网站建设,小企业网站建设有什么用,企业的管理系统有哪些,一般使用的分辨率的显示密度是多少dpi?写在前面包含本次模拟赛的10道题题解能过样例，应该可以AC若有错误，欢迎评论区指出本次题目除了最后两题有些难度，其余题目较为简单，我只将代码和结果给出，如果不能理解欢迎私信我，我会解答滴。start 2022…

写在前面

包含本次模拟赛的10道题题解
能过样例，应该可以AC
若有错误，欢迎评论区指出
本次题目除了最后两题有些难度，其余题目较为简单，我只将代码和结果给出，如果不能理解欢迎私信我，我会解答滴。

start = 2022 def check(num) : s = str(hex(num))[2:] for i in range(len(s)) : if ord(s[i]) < ord('a') or ord(s[i]) > ord('f') : return False return True while 1 : start += 1 if check(start) : break print(start)

#没代码，直接草稿纸手推就行了

import datetime start = datetime.date(1900,1,1) 
end = datetime.date(9999,12,31) 
diff = datetime.timedelta(1) def check(num) : year, month, day = num.split('-')if sum(list(map(int,year))) == sum(list(map(int,month))) + sum(list(map(int,day))) : return True return False ans = 0 
while start <= end : if check(str(start)) : ans += 1 if start == end : break start += diff print(ans)


s = '99,22,51,63,72,61,20,88,40,21,63,30,11,18,99,12,93,16,7,53,64,9,28,84,34,96,52,82,51,77'lst = list(map(int,s.split(',')))ans = 0 for i in range(29) : for j in range(i + 1, 30) : if lst[i] * lst[j] >= 2022 : ans += 1 print(ans)


s = '110010000011111110101001001001101010111011011011101001111110\
010000000001010001101100000010010110001111100010101100011110\
001011101000100011111111111010000010010101010111001000010100\
101100001101011101101011011001000110111111010000000110110000\
010101100100010000111000100111100110001110111101010011001011\
010011011010011110111101111001001001010111110001101000100011\
101001011000110100001101011000000110110110100100110111101011\
101111000000101000111001100010110000100110001001000101011001\
001110111010001011110000001111100001010101001110011010101110\
001010101000110001011111001010111111100110000011011111101010\
011111100011001110100101001011110011000101011000100111001011\
011010001101011110011011111010111110010100101000110111010110\
001110000111100100101110001011101010001100010111110111011011\
111100001000001100010110101100111001001111100100110000001101\
001110010000000111011110000011000010101000111000000110101101\
100100011101011111001101001010011111110010111101000010000111\
110010100110101100001101111101010011000110101100000110001010\
110101101100001110000100010001001010100010110100100001000011\
100100000100001101010101001101000101101000000101111110001010\
101101011010101000111110110000110100000010011111111100110010\
101111000100000100011000010001011111001010010001010110001010\
001010001110101010000100010011101001010101101101010111100101\
001111110000101100010111111100000100101010000001011101100001\
101011110010000010010110000100001010011111100011011000110010\
011110010100011101100101111101000001011100001011010001110011\
000101000101000010010010110111000010101111001101100110011100\
100011100110011111000110011001111100001110110111001001000111\
111011000110001000110111011001011110010010010110101000011111\
011110011110110110011011001011010000100100101010110000010011\
010011110011100101010101111010001001001111101111101110011101'def check(x, y) : res = 0 queue = [(x, y)] while queue : nx, ny = queue.pop(0) res += 1 st[nx][ny] = False for i, j in [(1,0), (-1,0), (0,1), (0,-1)] : px = nx + i ; py = ny + j if 0 <= px < 30 and 0 <= py < 60 and Map[px][py] == '1' and st[px][py] : queue.append((px,py)) return res Map = [[0] * 60 for i in range(30)]for i in range(30) :for j in range(60) :Map[i][j] = s[i*30+j]st = [[True] * 60 for i in range(30)] ans = 0 
for i in range(30) : for j in range(60) : if Map[i][j] == '1' and st[i][j] : ans = max(ans, check(i, j)) print(ans)


x = int(input())
n = int(input()) x = x + (n % 7)print(x if x <= 7 else x - 7)

from math import sqrt, piW,H,n,r = map(int,input().split()) def get_dist(x1, y1, x2, y2) : dx = x1 - x2 dy = y1 - y2 return sqrt(dx * dx + dy * dy) def check(x, y) : for i, j in loc : if get_dist(x, y, i, j) <= r : return True return False loc = [] 
for i in range(n) : x,y = map(int,input().split()) loc.append((x,y))ans = 0 
for i in range(W + 1) : for j in range(H + 1) : if check(i, j) : ans += 1 print(ans)


N,M = map(int,input().split()) 
T = int(input())Map = [[True] * (M + 1) for i in range(N + 1)] 
for _ in range(T) : r1,c1,r2,c2 = map(int,input().split()) for i in range(r1, r2 + 1) : for j in range(c1, c2 + 1) : Map[i][j] = False ans = 0 
for i in range(1, N + 1) : ans += Map[i][1:].count(True)print(ans)

这道题是经典的滑雪道记忆化搜索的题目，我写过一篇很详细的题解，请自行参考。

蓝桥杯/洛谷：最长滑雪道/滑雪 + 扩展题目（记忆化搜索 Python）_蓝桥杯滑雪_正在黑化的KS的博客-CSDN博客


N,M = map(int,input().split()) 
Map = [] for i in range(N) : Map.append(list(map(int,input().split()))) dist = [[-1] * M for i in range(N)] def find(x, y) : if dist[x][y] != -1 : return dist[x][y]dist[x][y] = 0 for i,j in [(1,0),(0,1),(-1,0),(0,-1)] : px = x + i ; py = y + j if 0 <= px < N and 0 <= py < M and Map[px][py] < Map[x][y] :  dist[x][y] = max(dist[x][y], find(px, py)) dist[x][y] += 1 return dist[x][y] ans = 0 
for i in range(N) : for j in range(M) : ans = max(ans, find(i, j)) print(ans)

重点题目来了！（敲黑板

有几位小伙伴私信问我这道题目该怎么解，我先说明我是用线段树解决的，但我认为应该还有别的解法，只是我目前还没有想到，如果有与我解法不同的方法可以放在评论区，大家一起讨论哈。

题目大意就是说，对于每一个点i而言，它的前 k 个点到后 k 个点这段 2k + 1 的区间中，最小数是多少。（区间边界是 0 ～ n - 1）

首先先说下超时做法，我们很容易想到，可以依次枚举每个点i，对每个点i再枚举它的

2k + 1 这个区间，维护一个最小值。这样的做法最坏情况下的时间复杂度是 $O(n^{2})$ ，对于这道题而言，n最大有10^6，因此肯定会超时。

正解如下，其实本质上这道题目是一道线段树的简化版，因为只涉及区间查询，不涉及区间修改，所以建树之后每次按要求查询区间就可以了。线段树的理论有些复杂，这里就不再给大家详细讲了，网上有很多篇关于线段树优秀的讲解。大家可以自行翻阅，如果已经理解了线段树还是不能看懂我的代码的话，可以来私信我。

Python3代码：


N = 10 ** 6 + 10 
INF = float('inf')n = int(input()) 
w = list(map(int,input().split())) class Node : def __init__(self, l, r, v) : self.l = l  self.r = r  self.v = v tr = [None] * (4 * N) def pushup(u) : tr[u].v = min(tr[u << 1].v, tr[u << 1 | 1].v) def build(u, l, r) :if l == r : tr[u] = Node(l, r, w[r]) else : tr[u] = {l, r, INF}  mid = l + r >> 1 build(u << 1, l, mid) ; build(u << 1 | 1, mid + 1, r) pushup(u) def query(u, l, r) :if l <= tr[u].l and r >= tr[u].r : return tr[u].v mid = tr[u].l + tr[u].r >> 1 res = INF if l <= mid : res = min(res, query(u << 1, l, r))if r > mid : res = min(res, query(u << 1 | 1, l, r)) return res build(1, 0, n - 1) k = int(input()) 
for i in range(n) : print(query(1, max(0, i - k), min(n - 1, i + k)), end = ' ')

Tip：考虑到很多小伙伴是C++组选手，我这里也贴一份C++的代码，以供参考。

Python写线段树真的有点反人类..

C++代码：

#include <iostream>
#include <algorithm>
#include <cstring>using namespace std ;typedef long long LL ; const int N = 1e6 + 10 ; 
int n, k ; 
int w[N] ; struct Node 
{int l, r, mn ; 
}tr[4 * N] ; void pushup(int u) 
{tr[u].mn = min(tr[u << 1].mn, tr[u << 1 | 1].mn) ; 
}void build(int u, int l, int r) 
{if (l == r) tr[u] = {l, r, w[r]} ; else {tr[u] = {l, r} ; int mid = l + r >> 1 ; build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r) ; pushup(u) ; } 
}int query(int u, int l, int r) 
{if (l <= tr[u].l && r >= tr[u].r) return tr[u].mn ; int mid = tr[u].l + tr[u].r >> 1 ;int res = 0x3f3f3f3f ;  if (l <= mid) res = min(res, query(u << 1, l, r)) ; if (r > mid) res = min(res, query(u << 1 | 1, l, r)) ; return res ;
}int main() 
{ios::sync_with_stdio(false) ; cin >> n ; for (int i = 0 ; i < n ; i ++ ) cin >> w[i] ;     build(1, 0, n - 1) ; cin >> k ; for (int i = 0 ; i < n ; i ++ ) cout << query(1, max(0, i - k), min(n - 1, i + k)) << " " ; return 0 ; 
}