結果

問題 No.2125 Inverse Sum
ユーザー mkawa2
提出日時 2025-03-11 14:29:34
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,634 bytes
コンパイル時間 298 ms
コンパイル使用メモリ 82,344 KB
実行使用メモリ 82,000 KB
最終ジャッジ日時 2025-03-11 14:29:38
合計ジャッジ時間 3,016 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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ファイルパターン 結果
sample AC * 3
other AC * 22 WA * 8
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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
# sys.setrecursionlimit(200005)
# sys.set_int_max_str_digits(200005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
# inf = -1-(-1 << 31)
inf = -1-(-1 << 62)
# md = 10**9+7
md = 998244353
from math import gcd
def prime_factorization(a):
    pp, ee = [], []
    if a & 1 == 0:
        pp += [2]
        ee += [0]
        while a & 1 == 0:
            a >>= 1
            ee[-1] += 1
    p = 3
    while p**2 <= a:
        if a%p == 0:
            pp += [p]
            ee += [0]
            while a%p == 0:
                a //= p
                ee[-1] += 1
        p += 2
    if a > 1:
        pp += [a]
        ee += [1]
    return pp, ee
P,Q = LI()
g=gcd(P,Q)
P//=g
Q//=g
pp, ee = prime_factorization(Q)
ff = [1]
for p, e in zip(pp, ee):
    a = p
    nf = []
    for _ in range(e*2):
        for f in ff:
            nf.append(f*a)
        a *= p
    ff += nf
# print(ff)
ans = []
for f in ff:
    g = Q*Q//f
    if (f+Q)%P or (g+Q)%P: continue
    n = (f+Q)//P
    m = (g+Q)//P
    ans.append((n, m))
print(len(ans))
for n, m in ans: print(n, m)
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