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For the calculation of the sum of integers $A, B$ in $N$-decimal, we divide the $(i-1):k$-digit part into $(i-1):j$-digit and $(j-1):k$-digit parts and consider whether the carry-over occurs for each. $(i \gt j \gt k)$

First, the integers $A, B$ are divided into $(i-1):j$ digits in $N$-decimal, and the values $A_{i-1:j}, B_{i-1:j}$ are added to determine whether carry-forward occurs for the digits above the interval, which are classified into three states as follows.

• $A_{i-1:j}+B_{i-1:j} \lt N^{i-j}-1$ ; None
• $A_{i-1:j}+B_{i-1:j} = N^{i-j}-1$ ; Propagate
• $A_{i-1:j}+B_{i-1:j} \gt N^{i-j}-1$ ; Generate

Example for a decimal number:

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use std::collections::HashMap;

fn greet_map(id: usize, name: String) -> HashMap<usize, String> {
    let mut map = HashMap::new();
    let message = format!("Hello, {}!", name);
    map.insert(id, message);
    map
}

fn main() {

mizar

import base64, io, struct, zlib
from typing import BinaryIO, List, Tuple, Union


COLOR_TYPE_V = 0
COLOR_TYPE_VA = 4
COLOR_TYPE_RGB = 2
COLOR_TYPE_RGBA = 6
PixelV = Tuple[int]
PixelVA = Tuple[int, int]

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// -*- coding:utf-8-unix -*-

//! エラトステネスの篩 & 生成した素数の試し割りによる素因数分解
//!
//! ## 素因数分解の実行例
//! 
//! [`main`] を参照
//! 
//! ## ベンチマーク実行例
//!

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mizar

Legal Policy

No	TYPE	Count
01	UP	72
02	UP_LEFT	64
03	UP_RIGHT	64
04	LEFT	72
05	RIGHT	72
06	DOWN	72

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*       text=auto
*.bat   text eol=crlf
*.cmd   text eol=crlf
*.sh    text eol=lf
*.onnx  -text
*.zip   -text

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*       text=auto
*.bat   text eol=crlf
*.cmd   text eol=crlf
*.sh    text eol=lf

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USIエンジン状態遷移図（抄）

graph LR;
flowstart((起動)) -->|usi| usioption[/option/] --> usiok[/usiok/] --> usimode((USIモード)) -->|isready| isready[初期化処理] --> readyok[/readyok/] --> ready((初期化済み))
--> |usinewgame| ingame((対局中));
usimode -->|setoption| setoption[オプション設定] --> usimode;
ingame -->|go| search((探索)) -->|"(探索完了)"| bestmove[/bestmove/] --> ingame;
ingame -->|position| setposition[局面設定] --> ingame;
search -->|stop| bestmove;

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---

👇

$$ \operatorname{I}x(a,b)=\dots, d{2n}=\dots $$

👆

$$ \operatorname{I}_x(a,b)=\dots, d_{2n}=\dots $$