結果
問題 |
No.3118 Increment or Multiply
|
ユーザー |
|
提出日時 | 2025-04-20 22:59:27 |
言語 | Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1) |
結果 |
AC
|
実行時間 | 687 ms / 2,000 ms |
コード長 | 1,890 bytes |
コンパイル時間 | 390 ms |
コンパイル使用メモリ | 12,544 KB |
実行使用メモリ | 12,536 KB |
最終ジャッジ日時 | 2025-04-20 22:59:38 |
合計ジャッジ時間 | 10,643 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 35 |
ソースコード
import sys def main(): data = sys.stdin.read().split() it = iter(data) T = int(next(it)) mod = 998244353 i2 = pow(2, mod-2, mod) out_lines = [] for _ in range(T): N = int(next(it)) A = int(next(it)) # Case A == 1: sum of first N-1 pairs if A == 1: n_mod = N % mod ret = n_mod * (n_mod - 1) % mod * i2 % mod out_lines.append(str(ret)) continue # Determine maximum exponent e such that A^e <= N b = 1 e = 0 while b <= N // A: b *= A e += 1 size = e + 10 x = [0] * size power = [0] * size res = [0] * size # x[i] = floor(N / A^i) x[0] = N for i in range(1, size): x[i] = x[i-1] // A # reduce modulo for later arithmetic # power[i] = (A^i) % mod power[0] = 1 a_mod = A % mod for i in range(1, size): power[i] = power[i-1] * a_mod % mod # res[i] = sum_{j < i} (floor(N/A^j) mod A) for i in range(1, size): res[i] = (res[i-1] + (x[i-1] % A)) % mod for i in range(size): x[i] %= mod ans = 0 for i in range(e + 1): if i != e: d = (power[e-i] - x[i+1] - 1) % mod if d > 0: ans = (ans + d * (d + 1) % mod * i2) % mod ans = (ans + d * (x[i] - power[e-i]) % mod) % mod ans = (ans + d * (i + res[i]) % mod) % mod # For range A^{e-i} <= z <= floor(N/A^i) t = (x[i] - power[e-i]) % mod ans = (ans + t * (t + 1) % mod * i2) % mod ans = (ans + (i + res[i]) * (t + 1) % mod) % mod out_lines.append(str(ans)) sys.stdout.write("\n".join(out_lines)) if __name__ == '__main__': main()