結果
| 問題 | No.243 出席番号(2) |
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-31 17:25:17 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
MLE
|
| 実行時間 | - |
| コード長 | 1,158 bytes |
| 記録 | |
| コンパイル時間 | 240 ms |
| コンパイル使用メモリ | 82,844 KB |
| 実行使用メモリ | 77,944 KB |
| 最終ジャッジ日時 | 2025-03-31 17:25:34 |
| 合計ジャッジ時間 | 3,964 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 5 MLE * 25 |
ソースコード
mod = 10**9 + 7
N = int(input())
A = [int(input()) for _ in range(N)]
from collections import defaultdict
# Count the occurrences of each disliked number
cnt = defaultdict(int)
for a in A:
cnt[a] += 1
# List of unique disliked numbers that appear at least once
vals = list(cnt.keys())
# Precompute factorial modulo mod
fact = [1] * (N + 1)
for i in range(1, N + 1):
fact[i] = fact[i - 1] * i % mod
# Initialize DP: dp[k] is the number of ways to choose k distinct values
dp = [0] * (N + 1)
dp[0] = 1
for v in vals:
c = cnt[v]
# Update dp in reverse to avoid overwriting values we still need to process
for j in range(N, -1, -1):
if dp[j] and j < N:
dp[j + 1] = (dp[j + 1] + dp[j] * c) % mod
# Calculate the result using inclusion-exclusion principle
result = 0
for k in range(0, N + 1):
if dp[k] == 0:
continue
remaining = N - k
if remaining < 0:
continue
# Current term: (-1)^k * dp[k] * (remaining)!
term = dp[k] * fact[remaining] % mod
if k % 2 == 1:
term = (mod - term) % mod # Convert to positive mod value
result = (result + term) % mod
print(result)
lam6er