update: 添加问题“1780.判断一个数字是否可以表示成三的幂的和”的代码(并更新其题解) (#1081)

326: AC.rust+java+go+py+cpp (#1078)

cpp - AC,100.00%,43.53%
py - AC,100.00%,59.18%
go - AC,100.00%,90.00%
java - AC,100.00%,75.56%
rust - AC,100.00%,-%  # 是的，没看错

Signed-off-by: LetMeFly666 <[email protected]>

Author: LetMeFly666

.commitmsg

@@ -1,8 +1,7 @@
 326: AC.rust+java+go+py+cpp (#1078)
 
-rust - AC,51.11%,100.00%
-java - AC,89.67%,5.37%
-go - AC,20.00%,41.05%
-py - AC,46.89%,5.52%
-cpp - AC,25.79%,19.01%
-今天(相比于之前)倒着写的
+cpp - AC,100.00%,43.53%
+py - AC,100.00%,59.18%
+go - AC,100.00%,90.00%
+java - AC,100.00%,75.56%
+rust - AC,100.00%,-%  # 是的，没看错

Codes/1780-check-if-number-is-a-sum-of-powers-of-three_20250814.cpp

@@ -0,0 +1,22 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:41:02
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    bool checkPowersOfThree(int n) {
+        while (n) {
+            if (n % 3 == 2) {
+                return false;
+            }
+            n /= 3;
+        }
+        return true;
+    }
+};

Codes/1780-check-if-number-is-a-sum-of-powers-of-three_20250814.go

@@ -0,0 +1,16 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:43:30
+ */
+package main
+
+func checkPowersOfThree(n int) bool {
+    for ; n > 0; n /= 3 {
+        if n % 3 == 2 {
+            return false
+        }
+    }
+    return true
+}

Codes/1780-check-if-number-is-a-sum-of-powers-of-three_20250814.java

@@ -0,0 +1,17 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:44:38
+ */
+class Solution {
+    public boolean checkPowersOfThree(int n) {
+        while (n > 0) {
+            if (n % 3 == 2) {
+                return false;
+            }
+            n /= 3;
+        }
+        return true;
+    }
+}

Codes/1780-check-if-number-is-a-sum-of-powers-of-three_20250814.py

@@ -0,0 +1,13 @@
+'''
+Author: LetMeFly
+Date: 2025-08-14 10:28:59
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-08-14 18:43:00
+'''
+class Solution:
+    def checkPowersOfThree(self, n: int) -> bool:
+        while n:
+            if n % 3 == 2:
+                return False
+            n //= 3
+        return True

Codes/1780-check-if-number-is-a-sum-of-powers-of-three_20250814.rs

@@ -0,0 +1,17 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:45:31
+ */
+impl Solution {
+    pub fn check_powers_of_three(mut n: i32) -> bool {
+        while n > 0 {
+            if n % 3 == 2 {
+                return false;
+            }
+            n /= 3;
+        }
+        return true;
+    }
+}

Codes/lib.rs

@@ -5,4 +5,4 @@
  * @LastEditTime: 2025-08-11 22:07:08
  */
 pub struct Solution;
-include!("0326-power-of-three_20250813.rs");  // 这个fileName是会被脚本替换掉的
+include!("1780-check-if-number-is-a-sum-of-powers-of-three_20250814.rs");  // 这个fileName是会被脚本替换掉的

Solutions/LeetCode 1780.判断一个数字是否可以表示成三的幂的和.md

@@ -134,18 +134,109 @@ public:
 #### C++
 
 ```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:41:02
+ */
 class Solution {
 public:
     bool checkPowersOfThree(int n) {
         while (n) {
-            if (n % 3 == 2)
+            if (n % 3 == 2) {
                 return false;
+            }
             n /= 3;
         }
         return true;
     }
 };
 ```
 
-> 同步发文于CSDN，原创不易，转载请附上[原文链接](https://blog.letmefly.xyz/2022/12/09/LeetCode%201780.%E5%88%A4%E6%96%AD%E4%B8%80%E4%B8%AA%E6%95%B0%E5%AD%97%E6%98%AF%E5%90%A6%E5%8F%AF%E4%BB%A5%E8%A1%A8%E7%A4%BA%E6%88%90%E4%B8%89%E7%9A%84%E5%B9%82%E7%9A%84%E5%92%8C/)哦~
-> Tisfy：[https://letmefly.blog.csdn.net/article/details/128248159](https://letmefly.blog.csdn.net/article/details/128248159)
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-08-14 10:28:59
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-08-14 18:43:00
+'''
+class Solution:
+    def checkPowersOfThree(self, n: int) -> bool:
+        while n:
+            if n % 3 == 2:
+                return False
+            n //= 3
+        return True
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:44:38
+ */
+class Solution {
+    public boolean checkPowersOfThree(int n) {
+        while (n > 0) {
+            if (n % 3 == 2) {
+                return false;
+            }
+            n /= 3;
+        }
+        return true;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:43:30
+ */
+package main
+
+func checkPowersOfThree(n int) bool {
+    for ; n > 0; n /= 3 {
+        if n % 3 == 2 {
+            return false
+        }
+    }
+    return true
+}
+```
+
+#### Rust
+
+```rust
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-14 10:28:59
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-14 18:45:31
+ */
+impl Solution {
+    pub fn check_powers_of_three(mut n: i32) -> bool {
+        while n > 0 {
+            if n % 3 == 2 {
+                return false;
+            }
+            n /= 3;
+        }
+        return true;
+    }
+}
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/128248159)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2022/12/09/LeetCode%201780.%E5%88%A4%E6%96%AD%E4%B8%80%E4%B8%AA%E6%95%B0%E5%AD%97%E6%98%AF%E5%90%A6%E5%8F%AF%E4%BB%A5%E8%A1%A8%E7%A4%BA%E6%88%90%E4%B8%89%E7%9A%84%E5%B9%82%E7%9A%84%E5%92%8C)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

Solutions/Other-English-LearningNotes-SomeWords.md

@@ -1358,6 +1358,8 @@ categories: [自用]
 |provocation|n. 挑衅，刺激，激怒|
 |dignity|n. 尊重，高贵，庄严，端庄，尊严|
 |commonsense|n. 常识，直觉判断力<br/>adj. 常识的，具有常识的|
+|||
+|solemn|adj. 严肃的，庄严的，正式的，郑重的|
 
 + 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
 + 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)

newSolution.py

@@ -2,7 +2,7 @@
 Author: LetMeFly
 Date: 2022-07-03 11:21:14
 LastEditors: LetMeFly.xyz
-LastEditTime: 2025-08-12 22:11:48
+LastEditTime: 2025-08-14 18:47:41
 Command: python newSolution.py 102. 二叉树的层序遍历
 What's more: 当前仅支持数字开头的题目
 What's more: 代码结构写的很混乱 - 想单文件实现所有操作
@@ -225,7 +225,7 @@ def refreshPublistTime(solution: str) -> str:
     with open("README.md", "r", encoding="utf-8") as f:
         readme = f.read()
 else:
-    input(f'手动更新完{solutionName}了吗？: ')
+    input(f'手动更新完 {solutionName} 了吗？: ')
     # TODO: 有空时候可以写个手动更新题解中代码的函数 - 最简单的办法是代码追加到(最后一种方法的)后面然后手动删
 
 def readmeNewLine(readme: str) -> str: