update: 添加问题“3546.等和矩阵分割I”的代码和题解 (#1464)

LetMeFly666 · LetMeFly666 · commit ede867319540 · 2026-03-25T22:28:57.000+08:00
2906: AC.cpp ($1463) - AC,92.86%,83.33%

中断了一阵子，其实写不了这么久

Signed-off-by: LetMeFly666 &lt;Tisfy@qq.com&gt;
diff --git a/.commitmsg b/.commitmsg
@@ -0,0 +1,3 @@
+2906: AC.cpp ($1463) - AC,92.86%,83.33%
+
+中断了一阵子，其实写不了这么久
diff --git a/Codes/3546-equal-sum-grid-partition-i.cpp b/Codes/3546-equal-sum-grid-partition-i.cpp
@@ -0,0 +1,51 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-03-25 21:42:32
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-03-25 22:05:04
+ */
+#ifdef _DEBUG
+#include "_[1,2]toVector.h"
+#endif
+
+#define CHECK_AND_BREAK(now, sum) \
+    if ((now) == (sum)) {         \
+        return true;              \
+    }                             \
+    if ((now) > (sum)) {          \
+        break;                    \
+    }
+typedef long long ll;
+class Solution {
+public:
+    bool canPartitionGrid(vector<vector<int>>& grid) {
+        ll sum = 0;
+        for (vector<int>& row : grid) {
+            for (int& v : row) {
+                sum += v;
+            }
+        }
+        if (sum % 2) {
+            return false;
+        }
+        sum /= 2;
+
+        ll now = 0;
+        for (vector<int>& row : grid) {
+            for (int& v : row) {
+                now += v;
+            }
+            CHECK_AND_BREAK(now, sum)
+        }
+
+        now = 0;
+        for (int j = 0; j < grid[0].size(); j++) {
+            for (int i = 0; i < grid.size(); i++) {
+                now += grid[i][j];
+            }
+            CHECK_AND_BREAK(now, sum)
+        }
+
+        return false;
+    }
+};
diff --git a/README.md b/README.md
@@ -1126,6 +1126,7 @@
 |3516.找到最近的人|简单|<a href="https://leetcode.cn/problems/find-closest-person/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/04/LeetCode%203516.%E6%89%BE%E5%88%B0%E6%9C%80%E8%BF%91%E7%9A%84%E4%BA%BA/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151184074" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-closest-person/solutions/3772009/letmefly-3516zhao-dao-zui-jin-de-ren-ji-8rlbz/" target="_blank">LeetCode题解</a>|
 |3531.统计被覆盖的建筑|中等|<a href="https://leetcode.cn/problems/count-covered-buildings/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/11/LeetCode%203531.%E7%BB%9F%E8%AE%A1%E8%A2%AB%E8%A6%86%E7%9B%96%E7%9A%84%E5%BB%BA%E7%AD%91/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/155824933" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-covered-buildings/solutions/3854864/letmefly-3531tong-ji-bei-fu-gai-de-jian-9kgng/" target="_blank">LeetCode题解</a>|
 |3541.找到频率最高的元音和辅音|简单|<a href="https://leetcode.cn/problems/find-most-frequent-vowel-and-consonant/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/13/LeetCode%203541.%E6%89%BE%E5%88%B0%E9%A2%91%E7%8E%87%E6%9C%80%E9%AB%98%E7%9A%84%E5%85%83%E9%9F%B3%E5%92%8C%E8%BE%85%E9%9F%B3/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151653999" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-most-frequent-vowel-and-consonant/solutions/3780744/letmefly-3541zhao-dao-pin-lu-zui-gao-de-yv5vs/" target="_blank">LeetCode题解</a>|
+|3546.等和矩阵分割I|中等|<a href="https://leetcode.cn/problems/equal-sum-grid-partition-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/03/25/LeetCode%203546.%E7%AD%89%E5%92%8C%E7%9F%A9%E9%98%B5%E5%88%86%E5%89%B2I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/159476515" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/equal-sum-grid-partition-i/solutions/3937005/letmefly-3546deng-he-ju-zhen-fen-ge-iji-0n0ee/" target="_blank">LeetCode题解</a>|
 |3567.子矩阵的最小绝对差|中等|<a href="https://leetcode.cn/problems/minimum-absolute-difference-in-sliding-submatrix/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/03/20/LeetCode%203567.%E5%AD%90%E7%9F%A9%E9%98%B5%E7%9A%84%E6%9C%80%E5%B0%8F%E7%BB%9D%E5%AF%B9%E5%B7%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/159291000" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-absolute-difference-in-sliding-submatrix/solutions/3932222/letmefly-3567zi-ju-zhen-de-zui-xiao-jue-4ux0g/" target="_blank">LeetCode题解</a>|
 |3573.买卖股票的最佳时机V|中等|<a href="https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-v/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/17/LeetCode%203573.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAV/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156029259" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-v/solutions/3859674/letmefly-3573mai-mai-gu-piao-de-zui-jia-592yz/" target="_blank">LeetCode题解</a>|
 |3577.统计计算机解锁顺序排列数|中等|<a href="https://leetcode.cn/problems/count-the-number-of-computer-unlocking-permutations/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/10/LeetCode%203577.%E7%BB%9F%E8%AE%A1%E8%AE%A1%E7%AE%97%E6%9C%BA%E8%A7%A3%E9%94%81%E9%A1%BA%E5%BA%8F%E6%8E%92%E5%88%97%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/155791805" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-the-number-of-computer-unlocking-permutations/solutions/3854206/letmefly-3577tong-ji-ji-suan-ji-jie-suo-5b6b8/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 2906.构造乘积矩阵.md b/Solutions/LeetCode 2906.构造乘积矩阵.md
@@ -55,7 +55,7 @@ p[0][2] = grid[0][0] * grid[0][1] = 12345 * 2 = 24690. 24690 % 12345 = 0 ，所
 
 ## 解题方法一：前后缀分解 O(mn)空间
 
-由于模数并非质数，所以不方便全部求积后通过模逆元做除法，因此可以使用前缀和和后缀和两个数组，分别记录每个位置前面元素的乘积以及后面元素的乘积。
+由于模数$12345$并非质数，所以不方便全部求积后通过模逆元做除法，因此可以使用前缀和和后缀和两个数组，分别记录每个位置前面元素的乘积以及后面元素的乘积。
 
 二维数组怎么定义“前面的元素”和“后面的元素”呢？很简单，把$n$行$m$列的二维数组看成$n\times m$大小的一维数组就好了，一维数组的下标$t$在原始二维数组中的下标是$(\lfloor\frac{t}{m}\rfloor, t\ \%\ m)$。
 
diff --git a/Solutions/LeetCode 3546.等和矩阵分割I.md b/Solutions/LeetCode 3546.等和矩阵分割I.md
@@ -0,0 +1,130 @@
+---
+title: 3546.等和矩阵分割 I：记总记当前（遍历）
+date: 2026-03-25 22:08:15
+tags: [题解, LeetCode, 中等, 数组, 枚举, 矩阵, 前缀和, 遍历]
+categories: [题解, LeetCode]
+index_img: https://pic.leetcode.cn/1746839596-kWigaF-lc.jpeg
+---
+
+# 【LetMeFly】3546.等和矩阵分割 I：记总记当前（遍历）
+
+力扣题目链接：[https://leetcode.cn/problems/equal-sum-grid-partition-i/](https://leetcode.cn/problems/equal-sum-grid-partition-i/)
+
+<p>给你一个由正整数组成的 <code>m x n</code> 矩阵 <code>grid</code>。你的任务是判断是否可以通过&nbsp;<strong>一条水平或一条垂直分割线&nbsp;</strong>将矩阵分割成两部分，使得：</p>
+
+<ul>
+	<li>分割后形成的每个部分都是&nbsp;<strong>非空&nbsp;</strong>的。</li>
+	<li>两个部分中所有元素的和&nbsp;<strong>相等&nbsp;</strong>。</li>
+</ul>
+
+<p>如果存在这样的分割，返回 <code>true</code>；否则，返回 <code>false</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> grid = [[1,4],[2,3]]</p>
+
+<p><strong>输出：</strong> true</p>
+
+<p><strong>解释：</strong></p>
+
+<p><img alt="" src="https://pic.leetcode.cn/1746839596-kWigaF-lc.jpeg" style="height: 200px; width: 200px;" /></p>
+
+<p>在第 0 行和第 1 行之间进行水平分割，得到两个非空部分，每部分的元素之和为 5。因此，答案是 <code>true</code>。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> grid = [[1,3],[2,4]]</p>
+
+<p><strong>输出：</strong> false</p>
+
+<p><strong>解释：</strong></p>
+
+<p>无论是水平分割还是垂直分割，都无法使两个非空部分的元素之和相等。因此，答案是 <code>false</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= m == grid.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= n == grid[i].length &lt;= 10<sup>5</sup></code></li>
+	<li><code>2 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<p>&nbsp;</p>
+
+
+    
+## 解题方法：三次遍历
+
+第一次遍历计算整个grid的元素之和，若非偶数直接返回`false`，若为偶数则直接除以$2$。
+
+接着从上往下遍历grid，每次累加一整行的元素，若恰好等于除以二后的总和则返回true，大于则break。
+
+最后从左往右遍历，方法和从上往下遍历同理。
+
++ 时间复杂度$O(nm)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2026-03-25 22:05:04
+ */
+#define CHECK_AND_BREAK(now, sum) \
+    if ((now) == (sum)) {         \
+        return true;              \
+    }                             \
+    if ((now) > (sum)) {          \
+        break;                    \
+    }
+typedef long long ll;
+class Solution {
+public:
+    bool canPartitionGrid(vector<vector<int>>& grid) {
+        ll sum = 0;
+        for (vector<int>& row : grid) {
+            for (int& v : row) {
+                sum += v;
+            }
+        }
+        if (sum % 2) {
+            return false;
+        }
+        sum /= 2;
+
+        ll now = 0;
+        for (vector<int>& row : grid) {
+            for (int& v : row) {
+                now += v;
+            }
+            CHECK_AND_BREAK(now, sum)
+        }
+
+        now = 0;
+        for (int j = 0; j < grid[0].size(); j++) {
+            for (int i = 0; i < grid.size(); i++) {
+                now += grid[i][j];
+            }
+            CHECK_AND_BREAK(now, sum)
+        }
+
+        return false;
+    }
+};
+
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/159476515)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/2026/03/25/LeetCode%203546.%E7%AD%89%E5%92%8C%E7%9F%A9%E9%98%B5%E5%88%86%E5%89%B2I/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/chat.sh b/chat.sh
@@ -0,0 +1,11 @@
+###
+ # @Author: LetMeFly
+ # @Date: 2026-03-24 23:16:39
+ # @LastEditors: LetMeFly.xyz
+ # @LastEditTime: 2026-03-24 23:18:55
+### 
+我linux想默认4534端口ssh登录，而22端口使用一个蜜罐程序监听
+这个蜜罐程序在用户尝试连接时不论输入什么密码都会连接成功，然后提示一句Welcome，you idot，之后是一个假shell。
+我想让这个shell尽可能真实，你帮我设计一下。
+
+使用go实现，尽量不要使用第三方依赖。
diff --git a/html-encryptor.py b/html-encryptor.py

Original file line number	Diff line number	Diff line change
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	`1`	`+2906: AC.cpp ($1463) - AC,92.86%,83.33%`
	`2`	`+`
	`3`	`+中断了一阵子，其实写不了这么久`