# Problem: Fast Power (LintCode)

> <http://www.lintcode.com/en/problem/fast-power/>

## 思路

* 对时间复杂度有要求 O(logn)，所以要考虑对半分
* 取 mod 的分配率：(a *b) % c = \[(a % c)* (b % c)] % c;

```java
class Solution {
    /*
     * @param a, b, n: 32bit integers
     * @return: An integer
     */
    public int fastPower(int a, int b, int n) {
        if (n == 1) {
            return a % b;
        }
        if (n == 0) {
            return 1 % b;
        }

        long product = fastPower(a, b, n / 2);
        product = product * product % b;
        if (n % 2 == 1) {
            product = (product * a) % b;
        }

        return (int) product;
    }
};
```

## 易错点

1. n == 0 的情况

   ```java
   if (n == 0) {
       return 1 % b;
   }
   ```

   当时错写成 `return b;`
2. 中间的计算结果用 long 存储，最后再转成 int 类型

   ```java
   long product = fastPower(a, b, n / 2);
   ```

   ```java
   return (int) product;
   ```
3. 如果 power n 是奇数的话，记得再多乘一次 a

   ```java
   if (n % 2 == 1) {
        product = (product * a) % b;
   }
   ```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://liuyang89116.gitbook.io/my-leetcode-book/chapter_3_binary_search/pow-and-sqrt/problem_fast_power_lintcode.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.