結果

問題 No.2218 Multiple LIS
コンテスト
ユーザー タコイモ
提出日時 2023-02-17 21:51:57
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 447 ms / 3,000 ms
コード長 3,961 bytes
コンパイル時間 259 ms
コンパイル使用メモリ 82,560 KB
実行使用メモリ 90,784 KB
最終ジャッジ日時 2024-07-19 13:05:22
合計ジャッジ時間 7,981 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 39
権限があれば一括ダウンロードができます

ソースコード

diff # 

import sys
#sys.setrecursionlimit(500000)
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def TI(): return tuple(map(int,sys.stdin.readline().rstrip().split()))
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip())
#for i, pi in enumerate(p):
from collections import defaultdict,deque
import bisect
import itertools
dic = defaultdict(int)
d = deque()
YN = ['No','Yes']
def gcd(a, b):
    while b: a, b = b, a % b
    return a
def isPrimeMR(n):
    d = n - 1
    d = d // (d & -d)
    L = [2, 7, 61] if n < 1<<32 else [2, 3, 5, 7, 11, 13, 17] if n < 1<<48 else [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
    for a in L:
        t = d
        y = pow(a, t, n)
        if y == 1: continue
        while y != n - 1:
            y = y * y % n
            if y == 1 or t == n - 1: return 0
            t <<= 1
    return 1
def findFactorRho(n):
    m = 1 << n.bit_length() // 8
    for c in range(1, 99):
        f = lambda x: (x * x + c) % n
        y, r, q, g = 2, 1, 1, 1
        while g == 1:
            x = y
            for i in range(r):
                y = f(y)
            k = 0
            while k < r and g == 1:
                ys = y
                for i in range(min(m, r - k)):
                    y = f(y)
                    q = q * abs(x - y) % n
                g = gcd(q, n)
                k += m
            r <<= 1
        if g == n:
            g = 1
            while g == 1:
                ys = f(ys)
                g = gcd(abs(x - ys), n)
        if g < n:
            if isPrimeMR(g): return g
            elif isPrimeMR(n // g): return n // g
            return findFactorRho(g)
def primeFactor(n):
    i = 2
    ret = {}
    rhoFlg = 0
    while i * i <= n:
        k = 0
        while n % i == 0:
            n //= i
            k += 1
        if k: ret[i] = k
        i += i % 2 + (3 if i % 3 == 1 else 1)
        if i == 101 and n >= 2 ** 20:
            while n > 1:
                if isPrimeMR(n):
                    ret[n], n = 1, 1
                else:
                    rhoFlg = 1
                    j = findFactorRho(n)
                    k = 0
                    while n % j == 0:
                        n //= j
                        k += 1
                    ret[j] = k

    if n > 1: ret[n] = 1
    if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
    return ret
def divisors(N): #約数全列挙
    pf = primeFactor(N)
    ret = [1]
    for p in pf:
        ret_prev = ret
        ret = []
        for i in range(pf[p]+1):
            for r in ret_prev:
                ret.append(r * (p ** i))
    return sorted(ret)
def divisors_pf(pf):
    ret = [1]
    for p in pf:
        ret_prev = ret
        ret = []
        for i in range(pf[p]+1):
            for r in ret_prev:
                ret.append(r * (p ** i))
    return sorted(ret)
def isPrime(n):
    if n in {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}: return 1
    if n <= 100: return 0
    for i in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]:
        if n % i == 0: return 0
    return isPrimeMR(n)
def findPrime(n):
    if n <= 2: return 2
    i = n | 1
    while 1:
        if isPrime(i): return i
        i += 2
def findNttFriendlyPrime(n, k, m=1):
    a = (n >> k) + 1
    i = (a << k) + 1
    while 1:
        if (i - 1) % m == 0:
            if isPrime(i):
                g = primitiveRoot(i)
                ig = pow(g, i - 2, i)
                return (i, g, ig) # p, g, invg
        i += 1 << k
N = I()
A = LI()
dp = defaultdict(int)
ans = 1

for i in range(N):
  ai = A[i]
  pf = divisors(ai)
  
  for kk in range(len(pf)-1,-1,-1):
    j = pf[kk]  
   
    if dp[j]:
      k = dp[j]+1
      ans = max(ans,k)
      dp[ai] = max(k,dp[ai])
      
  dp[ai] = max(1,dp[ai])
#print(dp)
print(ans)
0