#894469 (PyPy3) No.2474 Empty Quartz

• 問題: No.2474 Empty Quartz
• ユーザー: 👑 SPD_9X2
• 提出日時: 2023-07-22 03:39:56
• 言語: PyPy3 (7.3.15)
• 結果: AC
• 実行時間: 167 ms / 2,000 ms
• コード長: 1,525 bytes
• コンパイル時間: 316 ms
• コンパイル使用メモリ: 82,528 KB
• 実行使用メモリ: 92,292 KB
• 最終ジャッジ日時: 2024-06-11 22:48:47
• 合計ジャッジ時間: 2,256 ms
• ジャッジサーバーID （参考情報）: judge5 / judge3

ファイルパターン	結果
sample	AC * 2
other	AC * 5

ソースコード

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"""
Empty Quartz (7) 想定解
"""
import sys
from sys import stdin
mod = 998244353
from collections import deque
def modfac(n, MOD):
 
    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
    return factorials, invs
def modnCr(n,r):
    if n < 0 or r < 0 or n < r:
        return 0
    return fac[n] * inv[n-r] * inv[r] % mod
fac,inv = modfac(200000,mod)
T = int(stdin.readline())
assert 1 <= T <= 10**5
ANS = []
for loop in range(T):
    N,K = map(int,stdin.readline().split())
    
    assert 1 <= N <= 10**5
    assert 0 <= K <= N*(N+1)//2
    # x^2 - (N+1)x + K = 0 を解く
    in_root = (N+1)**2 - 4 * K
    if in_root < 0:
        ANS.append(0)
        continue
    # 10^6 > root の時、pythonでこの処理で大丈夫なことはチェック済(下部)
    root = int(in_root**0.5)
    if root**2 != in_root:
        ANS.append(0)
        continue
    topa = (N+1) + root
    topb = (N+1) - root
    ans = 0
    if topa % 2 == 0 and 1 <= topa // 2 <= N+1:
        ans += modnCr(N,topa//2-1)
    if topb % 2 == 0 and topa != topb and 1 <= topb <= N+1:
        ans += modnCr(N,topb//2-1)
    ANS.append(ans % mod)
print (*ANS,sep="\n")
"""    
# 精度チェック用
for i in range(1,10**6):
    
    x = i**2
    
    y = int(x**0.5)
    
    if y**2 != x:
        print (x,y)
"""
 
 
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