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Several bug fixes.

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krahets
2023-09-24 20:38:21 +08:00
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commit ff8e7ceec5
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docs/chapter_computational_complexity/iteration_and_recursion.md
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@ -429,7 +429,7 @@
以上述的求和函数为例，设问题 $f(n) = 1 + 2 + \dots + n$ 。
- **迭代**：在循环中模拟求和过程，从 $1$ 遍历到 $n$ ，每轮执行求和操作，即可求得 $f(n)$ 。
- **递归**：将问题分解为子问题 $f(n) = n + f(n-1)$ ，不断（递归地）分解下去，直至基本情况 $f(0) = 0$ 时终止。
- **递归**：将问题分解为子问题 $f(n) = n + f(n-1)$ ，不断（递归地）分解下去，直至基本情况 $f(1) = 1$ 时终止。
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