2172. 最大公约数

题意

给定一个数组, 每次操作可以选择数组中任意两个相邻的元素 x , y x, yx,y 并将其中的一个元素替换为 gcd ⁡ ( x , y ) \operatorname{gcd}(x, y)gcd(x,y), 其中 gcd ⁡ ( x , y ) \operatorname{gcd}(x, y)gcd(x,y) 表示 x xx 和 y yy 的最大公约数。请问最少需要多少次操作才能让整个数组只含 1 。
思路

流程图如下

其中两个互质元素的最短距离，可以这么求：
- 首先，目的就是为了让数组中变出一个1，而题目中只让相邻gcd，所以两个互质的不能直接gcd
- 所以我们要进行区间gcd，找到最短长度的区间，使得gcd=1
  - 如abcde，a与e互质，其实相当于a或e中无一个质因子相同，所以也就是a和b进行gcd，然后b肯定会带上a的因子，否则b就是1，然后一直到e
- 可以采用二分或双指针，找最短长度区间
- 可以采用st表或线段树，求区间gcd

代码

'''
Author: NEFU AB-IN
Date: 2023-05-24 12:47:50
FilePath: \LanQiao\2172\2172.py
LastEditTime: 2023-05-24 14:52:22
'''
# 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.x < a.x# Final
N = int(1e5 + 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 ——————————————————————
a = [0] * N
dp = [[0] * M for _ in range(N)]
Log = [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 += 1Log[1] = 0for i in range(2, N):Log[i] = Log[i >> 1] + 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()init()
cnt = sum(i == 1 for i in a)
if cnt > 0:print(n - cnt)exit(0)if query(1, n) != 1:print(-1)exit(0)ans = INFfor i in range(1, n + 1):l, r = i, nwhile l < r:mid = (l + r) >> 1if query(i, mid) == 1:r = midelse:l = mid + 1if query(i, r) == 1:ans = min(ans, r - i)print(ans + n - 1)