No.3770 Sum of Pow of Distinct Count
レベル :
/ AI推定難易度 : 2700
/ 実行時間制限 : 1ケース 4.000秒 / メモリ制限
: 1024 MB / 標準ジャッジ問題
タグ : / 解いたユーザー数 5
作問者 :
apricity
/ テスター :
kaichou243
marc2825
タグ : / 解いたユーザー数 5
作問者 :
apricity
/ テスター :
問題文最終更新日: 2026-10-15 04:58:20
English (AI translation)
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yukicoder contest 518
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Problem Statement
You are given positive integers $N,M,K$.
For an integer sequence $A$, let $f(A)$ be the number of distinct values appearing in $A$.
There are $M^N$ possible integer sequences $A$ of length $N$ consisting of integers between $1$ and $M$, inclusive. Let $S_k$ be the sum of $f(A)^k$ over all of them.
For each $k = 1,2,\dots,K$, find $S_k$ modulo $998244353$.
Input
$N\ M\ K$
- All input values are integers.
- $1 \leq N \leq 10^{18}$
- $1 \leq M \leq 10^{18}$
- $1 \leq K \leq 2\times 10^5$
Output
Print $K$ lines. The $k$-th line should contain $S_k$ modulo $998244353$.
Sample
Sample 1
Input
3 2 4
Output
14 26 50 98
There are $2^3 = 8$ possible sequences $A$. Among them, $f(A) = 1$ for the $2$ sequences $(1,1,1)$ and $(2,2,2)$, and $f(A) = 2$ for the other $6$ sequences. Therefore,
- $S_1 = 1^1\times 2 + 2^1\times 6 = 14,$
- $S_2 = 1^2\times 2 + 2^2\times 6 = 26,$
- $S_3 = 1^3\times 2 + 2^3\times 6 = 50,$
- $S_4 = 1^4\times 2 + 2^4\times 6 = 98$
.
Sample 2
Input
5 1 2
Output
1 1
Sample 3
Input
18392867552 14161132551 8
Output
726713581 100998767 286996549 468345593 527607776 620383472 238722459 972963401
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