結果

問題 No.1200 お菓子配り-3
コンテスト
ユーザー neterukun
提出日時 2020-08-28 22:39:59
言語 PyPy3
(7.3.15)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 3,030 bytes
コンパイル時間 416 ms
コンパイル使用メモリ 82,428 KB
実行使用メモリ 80,208 KB
最終ジャッジ日時 2024-11-14 16:08:24
合計ジャッジ時間 6,461 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 29 WA * 2
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
input = sys.stdin.buffer.readline


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]
    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 += 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

def make_divisors(n):
    def solve(li, ind, val):
        if ind == len(li):
            res.append(val)
            return
        for v in li[ind]:
            solve(li, ind + 1, val * v)

    factors = primeFactor(n)
    li = []
    for v in factors:
        tmp = [1]
        val = v
        for _ in range(factors[val]):
            tmp.append(val)
            val *= v
        li.append(tmp)
    res = []
    solve(li, 0, 1)
    return res
    

s = int(input())
memo = {}
for _ in range(s):
    x, y = map(int, input().split())
    
    if x < y:
        x, y = y, x
    
    sum_ = x + y
    diff = x - y
    
    cnt = 0

    if diff == 0:
        cnt += x - 1
        # b * (a + 1) = x
        cnt += (len(make_divisors(x)) - 1)
        print(cnt)
        continue

    for div in make_divisors(diff):
        a = div + 1

        # b + c = sum_ // (a + 1)
        # b - c = diff // (a - 1)
        if sum_ % (a + 1) != 0:
            continue

        t1, t2 = sum_ // (a + 1), diff // (a - 1)
        if (t1 + t2) % 2 == 1:
            continue
        b, c = (t1 + t2) // 2, (t1 - t2) // 2

        if b > 0 and c > 0:
            cnt += 1

    print(cnt)
0