🐱(other): 2019 力扣杯全国秋季编程大赛

Author: JalanJiang

docs/_sidebar.md

@@ -37,4 +37,5 @@
     * [其他](algorithm/other/)
   * 3.3 竞赛
     * [周赛](weekly/)
-    * [双周赛](biweekly/)
+    * [双周赛](biweekly/)
+    * [其他](other/)

docs/other/README.md

@@ -0,0 +1,98 @@
+## 2019 力扣杯全国秋季编程大赛
+
+[点击前往](https://leetcode-cn.com/contest/season/2019-fall/)
+
+### 1. 猜数字
+
+[原题链接](https://leetcode-cn.com/contest/season/2019-fall/problems/guess-numbers/)
+
+#### 思路
+
+```python
+class Solution:
+    def game(self, guess: List[int], answer: List[int]) -> int:
+        res = 0
+        for i in range(3):
+            if guess[i] == answer[i]:
+                res += 1
+        return res
+```
+
+### 2. 分式化简
+
+[原题链接](https://leetcode-cn.com/contest/season/2019-fall/problems/deep-dark-fraction/)
+
+#### 思路
+
+是一个不断“取反、相加”的循环。
+
+```python
+class Solution:
+    def fraction(self, cont: List[int]) -> List[int]:
+        length = len(cont)
+        res = [cont[length - 1], 1]
+        for i in range(length - 2, -1, -1):
+            # 取反
+            cur = cont[i]
+            left = cur * res[0]
+            res = [res[1] + left, res[0]]
+        
+        return res
+```
+
+### 3. 机器人大冒险
+
+[原题链接](https://leetcode-cn.com/contest/season/2019-fall/problems/programmable-robot/)
+
+#### 思路
+
+先记录一波数据：
+
+1. 把执行一次命令能走到的点都记录下来，记为 `path`
+2. 把命令中 `R` 的次数记作 `px`，`U` 的次数记作 `py`
+
+那么拿到一个点 `(x, y)` 时，求出需要循环执行命令的次数：
+
+```
+r = min(x // px, y // py)
+```
+
+循环 `r` 次命令后剩余的步数为：`(x - r * px, y - r * py)`。若 `(x - r * px, y - r * py)` 在 `path` 中，则说明可以走到点 `(x, y)`。
+
+```python
+class Solution:
+    def robot(self, command: str, obstacles: List[List[int]], x: int, y: int) -> bool:
+        path = set()
+        path.add((0, 0))
+        
+        px = 0
+        py = 0
+        for c in command:
+            if c == "R":
+                px += 1
+            else:
+                py += 1
+            path.add((px, py))
+            
+        def can_arrive(x, y):
+            x_round = x // px
+            y_round = y // py
+            r = min(x_round, y_round)
+            lx = x - r * px
+            ly = y - r * py
+            if (lx, ly) in path:
+                return True
+            else:
+                return False
+        
+        if not can_arrive(x, y):
+            return False
+        
+        for ob in obstacles:
+            if ob[0] > x or ob[1] > y:
+                continue
+            if can_arrive(ob[0], ob[1]):
+                return False
+        
+        return True
+```