結果
問題 | No.2578 Jewelry Store |
ユーザー | chineristAC |
提出日時 | 2023-12-06 00:59:03 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 2,767 ms / 3,500 ms |
コード長 | 4,314 bytes |
コンパイル時間 | 490 ms |
コンパイル使用メモリ | 81,792 KB |
実行使用メモリ | 159,512 KB |
最終ジャッジ日時 | 2024-09-27 00:50:11 |
合計ジャッジ時間 | 21,382 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 67 ms
64,896 KB |
testcase_01 | AC | 67 ms
65,152 KB |
testcase_02 | AC | 141 ms
78,720 KB |
testcase_03 | AC | 122 ms
78,720 KB |
testcase_04 | AC | 186 ms
78,892 KB |
testcase_05 | AC | 117 ms
78,720 KB |
testcase_06 | AC | 134 ms
79,000 KB |
testcase_07 | AC | 185 ms
78,720 KB |
testcase_08 | AC | 120 ms
78,592 KB |
testcase_09 | AC | 183 ms
79,004 KB |
testcase_10 | AC | 247 ms
79,048 KB |
testcase_11 | AC | 130 ms
78,744 KB |
testcase_12 | AC | 124 ms
78,976 KB |
testcase_13 | AC | 133 ms
78,592 KB |
testcase_14 | AC | 118 ms
78,592 KB |
testcase_15 | AC | 119 ms
78,848 KB |
testcase_16 | AC | 248 ms
78,976 KB |
testcase_17 | AC | 182 ms
79,224 KB |
testcase_18 | AC | 119 ms
79,028 KB |
testcase_19 | AC | 151 ms
78,948 KB |
testcase_20 | AC | 185 ms
78,976 KB |
testcase_21 | AC | 186 ms
78,720 KB |
testcase_22 | AC | 120 ms
78,464 KB |
testcase_23 | AC | 131 ms
78,464 KB |
testcase_24 | AC | 103 ms
78,464 KB |
testcase_25 | AC | 121 ms
78,444 KB |
testcase_26 | AC | 118 ms
78,700 KB |
testcase_27 | AC | 301 ms
79,324 KB |
testcase_28 | AC | 503 ms
79,332 KB |
testcase_29 | AC | 487 ms
78,976 KB |
testcase_30 | AC | 613 ms
78,848 KB |
testcase_31 | AC | 297 ms
78,720 KB |
testcase_32 | AC | 269 ms
127,488 KB |
testcase_33 | AC | 291 ms
108,048 KB |
testcase_34 | AC | 348 ms
122,232 KB |
testcase_35 | AC | 160 ms
93,824 KB |
testcase_36 | AC | 307 ms
124,704 KB |
testcase_37 | AC | 233 ms
79,160 KB |
testcase_38 | AC | 245 ms
80,000 KB |
testcase_39 | AC | 244 ms
84,076 KB |
testcase_40 | AC | 241 ms
80,188 KB |
testcase_41 | AC | 239 ms
79,360 KB |
testcase_42 | AC | 252 ms
78,848 KB |
testcase_43 | AC | 298 ms
79,104 KB |
testcase_44 | AC | 193 ms
79,016 KB |
testcase_45 | AC | 253 ms
78,976 KB |
testcase_46 | AC | 225 ms
81,280 KB |
testcase_47 | AC | 1,275 ms
80,060 KB |
testcase_48 | AC | 462 ms
159,512 KB |
testcase_49 | AC | 2,767 ms
111,020 KB |
testcase_50 | AC | 2,708 ms
114,028 KB |
testcase_51 | AC | 233 ms
79,360 KB |
testcase_52 | AC | 516 ms
79,104 KB |
testcase_53 | AC | 141 ms
78,592 KB |
ソースコード
def isPrimeMR(n): if n==1: return 0 d = n - 1 d = d // (d & -d) L = [2, 3, 5, 7, 11, 13, 17] if n in L: return 1 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): from math import gcd 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 += 1 + i % 2 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 import sys from itertools import permutations from heapq import heappop,heappush from collections import deque import random import bisect input = lambda :sys.stdin.readline().rstrip() mi = lambda :map(int,input().split()) li = lambda :list(mi()) K = 17 popcount = [0] * (1<<K) for i in range(K): for S in range(1<<K): if S >> i & 1: popcount[S] ^= 1 mod = 998244353 T,M = mi() pf = primeFactor(M) """ M_div = [1] for p in pf: e = pf[p] nxt_M_div = [] for a in M_div: t = 1 for i in range(e+1): nxt_M_div.append(a * t) t *= p M_div = nxt_M_div M_div.sort() """ M_prime = [pow(p,pf[p]) for p in pf] M_n = len(M_prime) def solve1(N,B,C,D,A): dic = {} w = B for a in A: if M % a: w = (C * w + D) % mod continue if a not in dic: dic[a] = 1 dic[a] *= w + 1 dic[a] %= mod w = (C * w + D) % mod for p in pf: for d in M_div: if d not in dic: continue if M % (d*p): continue if d*p not in dic: dic[d*p] = 1 dic[d*p] *= dic[d] dic[d*p] %= mod for d in dic: dic[d] = (dic[d] - 1 ) % mod for p in pf: for d in M_div[::-1]: if d % p: continue if d // p not in dic: continue dic[d] -= dic[d//p] dic[d] %= mod if M not in dic: return 0 return dic[M] def solve2(N,B,C,D,A): dp = [1] * (1<<M_n) w = B for a in A: if M % a: w = (C * w + D) % mod continue S = 0 for i in range(M_n): if a % M_prime[i] == 0: S ^= 1<<i dp[S] *= 1 + w dp[S] %= mod w = (C * w + D) % mod for i in range(M_n): dp[2**i] *= -1 dp[2**i] %= mod for i in range(M_n): t = 1<<i for S in range(1<<M_n): if S & t: dp[S] *= dp[S^t] dp[S] %= mod res = sum(dp) % mod if M_n & 1: res = (mod-res) % mod return res for _ in range(T): N,B,C,D = mi() A = li() if M!=1: print(solve2(N,B,C,D,A)) else: print((solve2(N,B,C,D,A)-1) % mod) #print(solve1(N,B,C,D,A))