Ex - Fibonacci-Revisited
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Ex - Fibonacci: Revisited
Score : $600$ points
Problem Statement
We define the general term $a_n$ of a sequence $a_0, a_1, a_2, \dots$ by:
$a_n = \begin{cases} 1 & (0 \leq n \lt K) \\ \displaystyle{\sum_{i=1}^K} a_{n-i} & (K \leq n). \\ \end{cases}$
Given an integer $N$, find the sum, modulo $998244353$, of $a_m$ over all non-negative integers $m$ such that $m\text{ AND }N = m$. ($\text{AND}$ denotes the bitwise AND.)
What is bitwise AND?
The bitwise AND of non-negative integers and , , is defined as follows.
・When is written in binary, its s place () is if s places of and are both , and otherwise.
Constraints
- $1 \leq K \leq 5 \times 10^4$
- $0 \leq N \leq 10^{18}$
- $N$ and $K$ are integers.
Input
The input is given from Standard Input in the following format:
Output
Print the answer.
2 6
21
$a_0$ and succeeding terms are $1, 1, 2, 3, 5, 8, 13, 21, \dots$. Four non-negative integers, $0, 2, 4$, and $6$, satisfy $6 \text{ AND } m = m$, so the answer is $1 + 2 + 5 + 13 = 21$.
2 8
35
1 123456789
65536
300 20230429
125461938
42923 999999999558876113
300300300