結果

問題 No.2011 Arbitrary Mod (Hidden)
ユーザー chineristAC
提出日時 2022-07-15 22:06:22
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 51 ms / 2,000 ms
コード長 2,317 bytes
コンパイル時間 326 ms
コンパイル使用メモリ 81,920 KB
実行使用メモリ 56,064 KB
最終ジャッジ日時 2024-06-27 18:08:51
合計ジャッジ時間 5,750 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
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ファイルパターン 結果
other AC * 43
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ソースコード

diff #
プレゼンテーションモードにする

import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
def isPrimeMR(n):
    if n==1:
        return 0
    d = n - 1
    d = d // (d & -d)
    L = [2, 3, 5, 7, 11, 13, 17,19,23,29,31,37,41,43,47]
    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
"""
2^n mod phi(M) = 0
かつ
a^2-398 \neq 0 mod M
→ M=2^n+1 prime 
"""
M = (2**16+1)*(2**4+1)*(2+1)*5
n = int(input())
print(M)
print(1)
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