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Diff for: ‎README.md

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<a href="https://github.com/guanghechen/algorithm.ts/actions/workflows/ci.yml">
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<img
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alt="CI Workflow"
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src="https://github.com/guanghechen/algorithm.ts/workflows/Build/badge.svg?branch=main"
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src="https://github.com/guanghechen/algorithm.ts/actions/workflows/ci.yml/badge.svg"
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/>
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</a>
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<a href="https://github.com/facebook/jest">

Diff for: ‎packages/binary-index-tree/README-zh.md

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<header>
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<h1 align="center">
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<a href="https://github.com/guanghechen/algorithm.ts/tree/main/packages/binary-index-tree#readme">@algorithm.ts/binary-index-tree</a>
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</h1>
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<div align="center">
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<a href="https://www.npmjs.com/package/@algorithm.ts/binary-index-tree">
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<img
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alt="Npm Version"
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src="https://img.shields.io/npm/v/@algorithm.ts/binary-index-tree.svg"
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/>
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</a>
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<a href="https://www.npmjs.com/package/@algorithm.ts/binary-index-tree">
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<img
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alt="Npm Download"
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src="https://img.shields.io/npm/dm/@algorithm.ts/binary-index-tree.svg"
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/>
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</a>
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<a href="https://www.npmjs.com/package/@algorithm.ts/binary-index-tree">
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<img
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alt="Npm License"
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src="https://img.shields.io/npm/l/@algorithm.ts/binary-index-tree.svg"
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/>
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</a>
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<a href="#install">
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<img
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alt="Module Formats: cjs, esm"
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src="https://img.shields.io/badge/module_formats-cjs%2C%20esm-green.svg"
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/>
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</a>
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<a href="https://github.com/nodejs/node">
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<img
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alt="Node.js Version"
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src="https://img.shields.io/node/v/@algorithm.ts/binary-index-tree"
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/>
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</a>
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<a href="https://github.com/facebook/jest">
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<img
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alt="Tested with Jest"
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src="https://img.shields.io/badge/tested_with-jest-9c465e.svg"
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/>
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</a>
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<a href="https://github.com/prettier/prettier">
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<img
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alt="Code Style: prettier"
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src="https://img.shields.io/badge/code_style-prettier-ff69b4.svg?style=flat-square"
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/>
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</a>
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</div>
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</header>
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<br/>
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**树状数组** 的 Typescript 实现。
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树状数组是一个树形的数组结构，用于高效地维护前缀和。通常有两种操作方式：
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1. 单点更新，区间查询： 修改数列中某个元素的数值，以及求解某个位置开始的前缀和；
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求解任意区间 $[L, R]$ 的和可以拆成求解区间 $[1,R]$ 的和与区间 $[1,L-1]$ 的和
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之差，即转换成求解两个前缀和的问题。
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2. 区间更新，单点查询： 给数列中前 $x$ 个元素的值同时加上某个值，以及求解数列中
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任意位置上的元素当前值。同样地，如果要给任意区间 $[L, R]$ 加上一个共同的值 $x$，
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可以先给 $[1,R]$ 中的元素同时加上 $x$，再给 $[1,L-1]$ 中的元素同时加上 $-x$。
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上述操作全是在 $O(\log N)$ 的均摊复杂度下完成。
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树状数组能解决的问题是线段树的子集，但是其相比于线段树拥有更小的复杂度常数，以及
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更简单的实现且更易理解。
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## Install
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* npm
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```bash
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npm install --save @algorithm.ts/binary-index-tree
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```
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* yarn
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```bash
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yarn add @algorithm.ts/binary-index-tree
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```
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## Usage
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### 单点更新，区间查询
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* Solve numbers:
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```typescript {3}
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import { createBinaryIndexTree1 } from '@algorithm.ts/binary-index-tree'
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const MAX_N = 10
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const bit = createBinaryIndexTree1<number>(MAX_N, 0)
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bit.init(MAX_N)
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// Add 10 on the 2th element.
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bit.add(2, 10)
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// Get the prefix sums.
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bit.query(1) // => 0
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bit.query(2) // => 10
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bit.query(/* any integer between [2, 10] */) // => 10
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// Add 7 on the 4th element.
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bit.add(4, 7)
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// Get the prefix sums.
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bit.query(1) // => 0
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bit.query(2) // => 10
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bit.query(3) // => 10
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bit.query(4) // => 17
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bit.query(/* any integer between [4, 10] */) // => 17
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```
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* Solve bigint:
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```typescript {6}
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import { createBinaryIndexTree1 } from '@algorithm.ts/binary-index-tree'
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const MAX_N = 10
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// Please note that the second parameter is `0n`, which represents the zero
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// element of bigint, and 0 is passed-in in the above example.
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const bit = createBinaryIndexTree1<number>(MAX_N, 0n)
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bit.init(MAX_N)
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// Add 10n on the 2th element.
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bit.add(2, 10n)
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// Get the prefix sums.
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bit.query(1) // => 0n
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bit.query(2) // => 10n
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bit.query(/* any integer between [2, 10] */) // => 10n
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// Add 7n on the 4th element.
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bit.add(4, 7)
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// Get the prefix sums.
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bit.query(1) // => 0n
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bit.query(2) // => 10n
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bit.query(3) // => 10n
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bit.query(4) // => 17n
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bit.query(/* any integer between [4, 10] */) // => 17n
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```
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### 区间更新，单点查询
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* Solve numbers:
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```typescript {3}
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import { createBinaryIndexTree2 } from '@algorithm.ts/binary-index-tree'
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const MAX_N = 10
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const bit = createBinaryIndexTree2<number>(MAX_N, 0)
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bit.init(MAX_N)
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// Add 10 on the first two elements.
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bit.add(2, 10)
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// Get the value of x-st element.
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bit.query(1) // => 10
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bit.query(2) // => 10
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bit.query(/* any integer between [3, 10] */) // => 0
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// Add 7 on the first four elements.
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bit.add(4, 7)
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// Get the value of x-st element.
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bit.query(1) // => 17
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bit.query(2) // => 17
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bit.query(3) // => 17
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bit.query(4) // => 17
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bit.query(/* any integer between [5, 10] */) // => 0
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```
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* Solve bigint:
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```typescript {6}
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import { createBinaryIndexTree2 } from '@algorithm.ts/binary-index-tree'
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const MAX_N = 10
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// Please note that the second parameter is `0n`, which represents the zero
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// element of bigint, and 0 is passed-in in the above example.
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const bit = createBinaryIndexTree2<number>(MAX_N, 0n)
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bit.init(MAX_N)
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// Add 10 on the first two elements.
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bit.add(2, 10n)
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// Get the value of x-st element.
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bit.query(1) // => 10n
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bit.query(2) // => 10n
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bit.query(/* any integer between [3, 10] */) // => 0n
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// Add 7 on the first four elements.
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bit.add(4, 7)
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// Get the value of x-st element.
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bit.query(1) // => 17n
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bit.query(2) // => 17n
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bit.query(3) // => 17n
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bit.query(4) // => 17n
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bit.query(/* any integer between [5, 10] */) // => 0n
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```
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## Related
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[homepage]: https://github.com/guanghechen/algorithm.ts/tree/main/packages/binary-index-tree#readme
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[binary-index-tree]: https://me.guanghechen.com/post/algorithm/shuffle/#heading-binary-index-tree

Diff for: ‎packages/binary-index-tree/README.md

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<br/>
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[中文文档](./README-zh.md)
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A typescript implementation of the **Binary Index Tree**.
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A Binary Index tree is a tree array, used to efficiently maintain the prefix
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sum. Such as support single point update with interval query, or interval
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update with single point query. The amortized complexity of each operation is
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$O(log N)$.
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The Binary Index Tree is a tree-shaped array structure used to efficiently
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maintain the prefix sum. There are usually two modes of operation:
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1. Single point update, interval query. Modify the value of an element in the
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number sequence, and solve the prefix sum at a certain position. Solve the
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sum of any interval $[L, R]$ can be divided into the sum of interval $[1,R]$
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and the sum of interval $[1, L-1]$, then perform a subtraction operation.
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2. Interval update, single-point query. Add a value to the value of the first
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$x$ elements in the sequence, and solve the current value of the element at
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any position in the sequence. Similarly, if you want to add a common value $x$
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to any interval $[L, R]$, you can first add $x$ to all elements in [1,R], and
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then add $-x$ to all elements in [1,L-1].
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The above operations are all done under the amortized complexity of $O(\log N)$.
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The problem that the Binary Index Tree can solve is a subset of the Segment
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Tree. Its advantage is that the complexity constant is smaller, and the

Diff for: ‎packages/binary-index-tree/package.json

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"package.json",
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"CHANGELOG.md",
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"LICENSE",
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"README.md"
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"README.md",
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"README-zh.md"
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],
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"scripts": {
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"build": "cross-env NODE_ENV=production rollup -c ../../rollup.config.js",

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