2940. 花坛的最小改变次数

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2940. 花坛的最小改变次数

• 题意

  略

• 思路

  首先需要区间查询gcd，想到st表
  其次思路，固定左端点，二分右端点，找gcd与区间长度相等的右端点，个人是这么理解的：

  • 区间长度 mid - i + 1
  • gcd
  • 区间长度随mid增大而增大，gcd随mid增大而减小或不变
  • 区间长度开始为1，gcd开始大于等于1，所以两者如果无限延伸一定有交点（可能不止一个），所以找到最右边的设为x，那么x往左的，都是gcd大于等于区间长度的，那么把这个区间放进答案数组
    在答案数组里，按右端点排序，如果两个端点可以合并，如果某个区间左端点可以小于前哥区间右端点，说明可以一起改，统计改几次即可

• 代码

  '''
  Author: NEFU AB-IN
  Date: 2023-06-09 18:00:12
  FilePath: \LanQiao\2940\2940.py
  LastEditTime: 2023-06-09 20:09:28
  '''
  # import
  from sys import setrecursionlimit, stdin, stdout, exit
  from collections import Counter, deque
  from heapq import heapify, heappop, heappush, nlargest, nsmallest
  from bisect import bisect_left, bisect_right
  from datetime import datetime, timedelta
  from string import ascii_lowercase, ascii_uppercase
  from math import log, gcd, sqrt, fabs, ceil, floorclass sa:def __init__(self, x, y):self.x = xself.y = ydef __lt__(self, a):return self.y < a.y# Final
  N = int(2e5 + 10)
  M = 20
  INF = int(2e9)# Define
  setrecursionlimit(INF)
  input = lambda: stdin.readline().rstrip("\r\n")  # Remove when Mutiple data
  read = lambda: map(int, input().split())
  LTN = lambda x: ord(x.upper()) - 65  # A -> 0
  NTL = lambda x: ascii_uppercase[x]  # 0 -> A# —————————————————————Division line ——————————————————————
  dp = [[0] * M for _ in range(N)]
  Log = [0] * N
  a = [0] * Ndef init():for j in range(M):i = 1while i + (1 << j) - 1 <= n:if j == 0:dp[i][j] = a[i]else:dp[i][j] = gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1])i += 1for i in range(2, N):Log[i] = Log[i // 2] + 1def query(l, r):k = Log[r - l + 1]return gcd(dp[l][k], dp[r - (1 << k) + 1][k])n, = read()
  a[1:] = read()ans = []
  init()for i in range(1, n + 1):l, r = i, nwhile l < r:mid = l + r + 1 >> 1if query(i, mid) >= mid - i + 1:l = midelse:r = mid - 1if query(i, l) == l - i + 1:ans.append(sa(i, l))cnt = 1
  if len(ans) == 0:print(0)
  else:ans.sort()tmp = ans[0].yfor i in ans:if i.x > tmp:cnt += 1tmp = i.yprint(cnt)