Revisit the English version (#1885)

* Update giscus scroller.

* Refine English docs and landing page

* Sync the headings.

* Update landing pages.

* Update the avatar

* Update Acknowledgements

* Update landing pages.

* Update contributors.

* Update

* Fix the formula formatting.

* Fix the glossary.

* Chapter 6. Hashing

* Remove Chinese chars.

* Fix headings.

* Update giscus themes.

* fallback to default giscus theme to solve 429 many requests error.

* Add borders for callouts.

* docs: sync character encoding translations

* Update landing page media layout and i18n
This commit is contained in:
Yudong Jin
2026-04-10 23:03:03 +08:00
committed by GitHub
parent ae03a167a4
commit b01036b09e
132 changed files with 1702 additions and 1508 deletions
@@ -95,7 +95,7 @@ The two-dimensional $dp$ table has size $(n+1) \times (amt+1)$.
This problem differs from the unbounded knapsack problem in the following two aspects regarding the state transition equation.
- This problem seeks the minimum value, so the operator $\max()$ needs to be changed to $\min()$.
- The optimization target is the number of coins rather than item value, so when a coin is selected, simply execute $+1$.
- The optimization target is the number of coins rather than item value, so when a coin is selected, simply add $1$.
$$
dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
@@ -109,7 +109,7 @@ When there are no coins, **it is impossible to make up any amount $> 0$**, which
### Code Implementation
Most programming languages do not provide a $+ \infty$ variable, and can only use the maximum value of integer type `int` as a substitute. However, this can lead to large number overflow: the $+ 1$ operation in the state transition equation may cause overflow.
Most programming languages do not provide a $+ \infty$ variable, and can only use the maximum value of integer type `int` as a substitute. However, this can lead to integer overflow: the $+ 1$ operation in the state transition equation may cause overflow.
For this reason, we use the number $amt + 1$ to represent invalid solutions, because the maximum number of coins needed to make up $amt$ is at most $amt$. Before returning, check whether $dp[n, amt]$ equals $amt + 1$; if so, return $-1$, indicating that the target amount cannot be made up. The code is as follows:
@@ -172,7 +172,7 @@ The space optimization for the coin change problem is handled in the same way as
[file]{coin_change}-[class]{}-[func]{coin_change_dp_comp}
```
## Coin Change Problem Ii
## Coin Change Problem II
!!! question