> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/math/1344.-angle-between-hands-of-a-clock.md).

# 1344. Angle Between Hands of a Clock

Given two numbers, `hour` and `minutes`. Return the smaller angle (in degrees) formed between the `hour` and the `minute` hand.

**Example 1:**

![](https://assets.leetcode.com/uploads/2019/12/26/sample_1_1673.png)

```
Input: hour = 12, minutes = 30
Output: 165
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2019/12/26/sample_2_1673.png)

```
Input: hour = 3, minutes = 30
Output: 75
```

**Example 3:**

![](https://assets.leetcode.com/uploads/2019/12/26/sample_3_1673.png)

```
Input: hour = 3, minutes = 15
Output: 7.5
```

**Example 4:**

```
Input: hour = 4, minutes = 50
Output: 155
```

**Example 5:**

```
Input: hour = 12, minutes = 0
Output: 0
```

**Constraints:**

* `1 <= hour <= 12`
* `0 <= minutes <= 59`
* Answers within `10^-5` of the actual value will be accepted as correct.

问给定时间在时钟上时针和分针相差的角度（小的那个）是多少。时针把360°分成了12份格子，分针分成了60份，且时钟会随着当前分钟数在相关格子内移动对应长度。

```python
class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        abd = abs(hour * 30 + minutes / 60 * 30 - minutes * 6)
        return min(360 - abd, abd)
```


---

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