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No.681 Fractal Gravity Glue

レベル : / AI推定難易度 : 1650 / 実行時間制限 : 1ケース 2.000秒 / メモリ制限 : 512 MB / 標準ジャッジ問題
タグ : (AC するまで非表示) / 解いたユーザー数 33
作問者 : はむこ / テスター : 夕叢霧香(ゆうむらきりか)
お気に入りにしたユーザー ProblemId : 1853 / 自分の提出
問題文最終更新日: 2026-06-26 06:07:15
This statement was translated by AI (claude-opus-5-5, effort medium) and may contain errors or differ in nuance from the original. The original statement takes precedence. Back to the original
yukicoder contest 189 (順位表) の他の問題:

Problem Statement

Hamuko is going to try Gravity Glue.
Gravity Glue is an art form in which stones of various sizes are stacked high without using any adhesive.

Since Hamuko is captivated by fractals, she has decided to stack stones so that their sizes form a fractal.
The Gravity Glue $G_{b, d}$ is a way of stacking stones uniquely determined by its beauty $b>0$ and difficulty $d>0$.

The Gravity Glue $G_{1, d}$ of beauty $1$ is a stack of $d$ stones of size $1$ ($d$ is any positive integer).
The Gravity Glue $G_{i+1, d}$ of beauty $i+1$ uses $d$ stones of size $i+1$, stacked in the sandwich form
($G_{i, d}$, stone of size $i+1$, $G_{i, d}$, stone of size $i+1$, ..., stone of size $i+1$, $G_{i, d}$).

Now, Hamuko has finished building $G_{b, d}$ up to the $n$-th level from the bottom.
Find the total size of the stones needed to complete the rest, modulo $10^9+7$.

Input

n
b d

$0 \le n \le 10^9$, integer
$1 \le b \le 10^9$, integer
$1 \le d \le 10^9$, integer

It is guaranteed that $n$ does not exceed the height of $G_{b, d}$.

Output

Print a newline at the end.

Sample

Sample 1
Input
0
3 1
Output
11

The target Gravity Glue is, from the bottom, 1213121.
Since $n=0$, all the stones still need to be stacked.

Sample 2
Input
4
3 1
Output
4

The target Gravity Glue is, from the bottom, 1213121.
Since $n=4$, 1213 has already been stacked. The remaining part is 121, so the total size is 4.

Sample 3
Input
7
3 1
Output
0

All the stones have already been stacked.

Sample 4
Input
4
3 2
Output
31

The target Gravity Glue is, from the bottom, 11211211311211211311211211.
Since $n=4$, 1121 has already been stacked. The total of the remaining part is 31.

Sample 5
Input
114514
1919114 5141919
Output
666638169

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