結果
| 問題 |
No.1796 木上のクーロン
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-16 16:03:29 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 1,782 bytes |
| コンパイル時間 | 416 ms |
| コンパイル使用メモリ | 82,260 KB |
| 実行使用メモリ | 174,936 KB |
| 最終ジャッジ日時 | 2025-04-16 16:10:00 |
| 合計ジャッジ時間 | 14,303 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 17 TLE * 1 -- * 16 |
ソースコード
import sys
from collections import deque
MOD = 998244353
def main():
input = sys.stdin.read().split()
ptr = 0
N = int(input[ptr])
ptr += 1
Q = list(map(int, input[ptr:ptr+N]))
ptr += N
edges = [[] for _ in range(N+1)]
for _ in range(N-1):
u = int(input[ptr])
v = int(input[ptr+1])
ptr += 2
edges[u].append(v)
edges[v].append(u)
# Precompute factorial and k0 = (N!)^2 mod MOD
fact = [1] * (N+1)
for i in range(1, N+1):
fact[i] = fact[i-1] * i % MOD
k0 = fact[N] * fact[N] % MOD
# Precompute inv_sq[d] = 1/(d^2) mod MOD for d from 1 to N+1
max_d = N
inv = [1] * (max_d + 2)
inv[1] = 1
for i in range(2, max_d + 2):
inv[i] = MOD - MOD // i * inv[MOD % i] % MOD
inv_sq = [1] * (max_d + 2)
for d in range(1, max_d + 2):
inv_sq[d] = inv[d] * inv[d] % MOD
# For each vertex p, compute sum of Q_i / (d(p,i)+1)^2
E = [0] * (N+1) # 1-based
# Since BFS for each node is O(N^2), which is not feasible, this approach will not work for large N.
# The following code is a placeholder to illustrate the intended logic but will not pass due to time constraints.
for p in range(1, N+1):
dist = [-1] * (N+1)
q = deque()
q.append(p)
dist[p] = 0
while q:
u = q.popleft()
for v in edges[u]:
if dist[v] == -1:
dist[v] = dist[u] + 1
q.append(v)
total = 0
for i in range(1, N+1):
d = dist[i]
total = (total + Q[i-1] * inv_sq[d+1]) % MOD
E[p] = total * k0 % MOD
for p in range(1, N+1):
print(E[p])
if __name__ == '__main__':
main()
lam6er