chore: add daily leetcode post 1828统计一个圆中点的数目_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/1828统计一个圆中点的数目_translated.md

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+---
+title: 1828. Statistics the number of a circle mid -point One question daily
+date: '2024.01.01 0:00'
+tags:
+  - - Python
+  - - answer
+  - - math
+abbrlink: 3277549c
+---
+# topic：
+
+
+[1828. Statistics the number of a circle mid -point](https://leetcode.cn/problems/queries-on-number-of-points-inside-a-circle/description/)
+
+# Thought：
+
+Today's question is very simple，I wonder if it is against yesterday's question bidding。
+Ask`queries`There are a few in the circle`points`Points in an array。
+To be honest, I was scared，I think this question is the question of the picture again。不过仔细读题了后发现只是一道简单的math题，
+We can use European -style distance to solve（Time complexity isOn^2，I thought I had a better solution，At first glance everyone is like this（））
+
+The formula of the European -style distance is as follows：
+
+$\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$
+
+We look at the code for specific operations：
+
+# Code
+
+```python
+class Solution:
+    def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
+        # Seek to solve whether the European -style distance from the center is less than r
+        ans = [0] * len(queries)
+        flag = 0
+        for x, y, r in queries:
+            for i, j in points:
+                if ((x - i) ** 2 + (y - j) ** 2) ** (1 / 2) <= r:
+                    ans[flag] += 1
+            flag += 1
+        return ans
+```
+
+Take a look at someone who can't understandpythonCode：
+
+
+```python
+class Solution:
+    def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
+        points = sorted(points)
+
+        res = [0 for _ in range(len(queries))]
+
+        for i, (u, v, r) in enumerate(queries):
+            left, right = u - r, u + r
+
+            idx1 = bisect_left(points, [left, -inf])
+            idx2 = bisect_right(points, [right, inf])
+
+            for x, y in points[idx1: idx2 + 1]:
+                if (v - r <= y <= v + r and 
+                    (x - u) * (x - u) + (y - v) * (y - v) <= r * r):
+  
+                    res[i] += 1
+
+        return res
+```