結果
| 問題 |
No.2406 Difference of Coordinate Squared
|
| コンテスト | |
| ユーザー |
 qwewe
|
| 提出日時 | 2025-05-14 13:04:57 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 7,526 bytes |
| コンパイル時間 | 210 ms |
| コンパイル使用メモリ | 82,288 KB |
| 実行使用メモリ | 89,656 KB |
| 最終ジャッジ日時 | 2025-05-14 13:06:34 |
| 合計ジャッジ時間 | 5,216 ms |
|
ジャッジサーバーID （参考情報） |
judge4 / judge1 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 40 WA * 15 |
ソースコード
import sys
# Set higher recursion depth if necessary, although unlikely needed for this problem
# sys.setrecursionlimit(2000)
def solve():
N, M = map(int, sys.stdin.readline().split())
MOD = 998244353
# Precompute factorials and inverse factorials modulo MOD
# Factorials: fact[i] = i! mod MOD
fact = [1] * (N + 1)
# Inverse factorials: invfact[i] = (i!)^(-1) mod MOD
invfact = [1] * (N + 1)
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % MOD
# Compute inverse factorial of N! using Fermat's Little Theorem (since MOD is prime)
# (N!)^(MOD-2) mod MOD = (N!)^(-1) mod MOD
invfact[N] = pow(fact[N], MOD - 2, MOD)
# Compute other inverse factorials iteratively using invfact[i] = invfact[i+1] * (i+1)
for i in range(N - 1, -1, -1):
invfact[i] = (invfact[i + 1] * (i + 1)) % MOD
# Function to compute nCr modulo MOD using precomputed values
def nCr_mod(n, r):
# Standard checks for combinations validity
if r < 0 or r > n:
return 0
# Check if indices are within bounds of precomputed tables.
# n must be <= N for our precomputation.
# r and n-r must be >= 0 which is covered by the first check.
if n > N:
# This case should not happen if called correctly with N as max value.
# If it somehow occurs, maybe due to invalid k_u or k_v calculation, return 0.
return 0
# Calculate nCr = n! / (r! * (n-r)!) mod MOD
# Use modular inverse: nCr = n! * (r!)^(-1) * ((n-r)!)^(-1) mod MOD
num = fact[n]
# Denominator inverse part
den = (invfact[r] * invfact[n - r]) % MOD
return (num * den) % MOD
# Calculate modular inverse of 4
inv4 = pow(4, MOD - 2, MOD)
# Calculate (1/4)^N mod MOD using modular exponentiation
inv4_N = pow(inv4, N, MOD)
# Handle M = 0 case separately
# This corresponds to UV = 0, which means U=0 or V=0.
# The probability is P(U=0) + P(V=0) - P(U=0, V=0)
if M == 0:
# If N is odd, both U and V must be odd (since U=sum +/-1 N times).
# Thus P(U=0) = P(V=0) = 0. Probability is 0.
if N % 2 != 0:
print(0)
return
# If N is even, U and V can be 0.
# P(U=0) = binom(N, N/2) * (1/2)^N
# P(V=0) = binom(N, N/2) * (1/2)^N
# P(U=0, V=0) = binom(N, N/2) * binom(N, N/2) * (1/4)^N
# The formula derived is: binom(N, N/2) * 2^(1-N) - binom(N, N/2)^2 * (1/4)^N
term1_coeff = nCr_mod(N, N // 2)
# Calculate 2^(1-N) mod P = 2 * (2^(-1))^N mod P
inv2 = pow(2, MOD - 2, MOD)
if N == 0: # Special case N=0, 2^(1-0) = 2
term1_val_mod = 2
else: # N > 0
term1_val_mod = (2 * pow(inv2, N, MOD)) % MOD
term1 = (term1_coeff * term1_val_mod) % MOD
# Calculate (binom{N}{N/2})^2 * (1/4)^N mod P
term2_coeff = (term1_coeff * term1_coeff) % MOD
term2_val_mod = inv4_N
term2 = (term2_coeff * term2_val_mod) % MOD
# Final probability for M=0 using inclusion-exclusion principle.
# Add MOD before taking modulo to handle potential negative result
ans = (term1 - term2 + MOD) % MOD
print(ans)
return
# Handle M != 0 case
# We need to sum P(U=u, V=v) for all pairs (u, v) such that uv=M
# P(U=u, V=v) = binom(N, (N+u)/2) * binom(N, (N+v)/2) * (1/4)^N
# We need to sum the coefficient part: binom(N, k_u) * binom(N, k_v)
total_sum_coeffs = 0
abs_M = abs(M)
# Iterate through possible absolute values of u up to floor(sqrt(|M|))
# This covers all divisor pairs (u_abs, v_abs) where |M| = u_abs * v_abs
sqrt_M = 0
if abs_M > 0:
sqrt_M = int(abs_M**0.5)
# Adjust sqrt_M check for edge cases like perfect squares or floating point inaccuracy
while (sqrt_M + 1) * (sqrt_M + 1) <= abs_M:
sqrt_M += 1
while sqrt_M * sqrt_M > abs_M:
sqrt_M -=1
# Keep track of u values processed to avoid double counting pairs like (u,v) and (v,u) if |u|!=|v|
# e.g. for M=12, sqrt=3. u_abs=1, v_abs=12; u_abs=2, v_abs=6; u_abs=3, v_abs=4
# We need to consider divisors u=+/-1, +/-2, +/-3, +/-4, +/-6, +/-12
processed_u = set()
for u_abs in range(1, sqrt_M + 1):
if abs_M % u_abs == 0:
v_abs = abs_M // u_abs
# Get the potential absolute values for u based on this divisor pair
possible_u_abs = [u_abs]
# If M is not a perfect square, u_abs != v_abs, so add v_abs too
if u_abs * u_abs != abs_M:
possible_u_abs.append(v_abs)
for cur_u_abs in possible_u_abs:
# For each absolute value, consider both positive and negative u
for u in [cur_u_abs, -cur_u_abs]:
# Skip if u=0 (M!=0 means u cannot be 0) or if already processed
if u == 0: continue
if u in processed_u: continue
processed_u.add(u)
# Corresponding v such that uv = M
# Integer division works since u must divide M by construction
v = M // u
# Check necessary conditions for P(U=u, V=v) to be non-zero and valid
valid = True
# Condition 1: |u| <= N (coordinate cannot exceed N steps away)
if not (abs(u) <= N): valid = False
# Condition 2: |v| <= N (same reason)
if not (abs(v) <= N): valid = False
# Condition 3: Parity check: N and u must have the same parity.
# (N+u) must be even for k_u = (N+u)/2 to be integer.
# Equivalently, N-u must be even.
# Note: If N, u have same parity, and uv=M, then N, v must also have same parity
# Because u, v must have same parity if uv=M is odd, and different parity if M is even
# NO, u,v must have same parity if M is odd. If M is even, they can have any parity combination except odd/odd.
# The requirement is N=u (mod 2) AND N=v (mod 2). This implies u=v (mod 2).
# If N=u (mod 2) is satisfied, then u=v (mod 2) must hold for N=v (mod 2) to hold.
# If uv=M, u=v (mod 2) unless M is even and one is even, one is odd.
# But P(U=u, V=v) is non-zero only if N=u (mod 2) AND N=v (mod 2). This requires u=v (mod 2).
# So we only need to check N=u (mod 2).
if (N - u) % 2 != 0: valid = False
if valid:
# Calculate indices for binomial coefficients
# These must be integers because of parity check
k_u = (N + u) // 2
k_v = (N + v) // 2
# Compute the coefficient term binom(N, k_u) * binom(N, k_v) mod P
term = (nCr_mod(N, k_u) * nCr_mod(N, k_v)) % MOD
# Add to the total sum of coefficients
total_sum_coeffs = (total_sum_coeffs + term) % MOD
# Final answer is the total sum of coefficients multiplied by (1/4)^N mod P
final_ans = (total_sum_coeffs * inv4_N) % MOD
print(final_ans)
solve()