update: 添加问题“2176.统计数组中相等且可以被整除的数对”的代码和题解(#885)

LetMeFly666 · LetMeFly666 · commit 49b548db26f2 · 2025-04-17T22:31:49.000+08:00
Signed-off-by: LetMeFly666 &lt;814114971@qq.com&gt;
diff --git a/Codes/2176-count-equal-and-divisible-pairs-in-an-array.cpp b/Codes/2176-count-equal-and-divisible-pairs-in-an-array.cpp
@@ -0,0 +1,24 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-17 20:13:54
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-17 20:28:40
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int countPairs(vector<int>& nums, int k) {
+        int ans = 0;
+        for (int i = 0; i < nums.size(); i++) {
+            for (int j = i + 1; j < nums.size(); j++) {
+                if (nums[i] == nums[j] && i * j % k == 0) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+};
diff --git a/Codes/2176-count-equal-and-divisible-pairs-in-an-array.go b/Codes/2176-count-equal-and-divisible-pairs-in-an-array.go
@@ -0,0 +1,18 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-17 20:19:13
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-17 20:21:29
+ */
+package main
+
+func countPairs(nums []int, k int) (ans int) {
+    for i := range nums {
+		for j := i + 1; j < len(nums); j++ {
+			if nums[i] == nums[j] && i * j % k == 0 {
+				ans++
+			}
+		}
+	}
+	return
+}
diff --git a/Codes/2176-count-equal-and-divisible-pairs-in-an-array.java b/Codes/2176-count-equal-and-divisible-pairs-in-an-array.java
@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-17 20:17:48
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-17 20:18:12
+ */
+class Solution {
+    public int countPairs(int[] nums, int k) {
+        int ans = 0;
+        for (int i = 0; i < nums.length; i++) {
+            for (int j = i + 1; j < nums.length; j++) {
+                if (nums[i] == nums[j] && i * j % k == 0) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+}
diff --git a/Codes/2176-count-equal-and-divisible-pairs-in-an-array.py b/Codes/2176-count-equal-and-divisible-pairs-in-an-array.py
@@ -0,0 +1,12 @@
+'''
+Author: LetMeFly
+Date: 2025-04-17 20:15:46
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-04-17 20:15:58
+'''
+from typing import List
+
+
+class Solution:
+    def countPairs(self, nums: List[int], k: int) -> int:
+        return sum(nums[i] == nums[j] and i * j % k == 0 for i in range(len(nums)) for j in range(i + 1, len(nums)))
diff --git a/README.md b/README.md
@@ -669,6 +669,7 @@
 |2132.用邮票贴满网格图|困难|<a href="https://leetcode.cn/problems/stamping-the-grid/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/14/LeetCode%202132.%E7%94%A8%E9%82%AE%E7%A5%A8%E8%B4%B4%E6%BB%A1%E7%BD%91%E6%A0%BC%E5%9B%BE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135002925" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/stamping-the-grid/solutions/2566644/letmefly-2132yong-you-piao-tie-man-wang-znigh/" target="_blank">LeetCode题解</a>|
 |2140.解决智力问题|中等|<a href="https://leetcode.cn/problems/solving-questions-with-brainpower/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/03/LeetCode%202140.%E8%A7%A3%E5%86%B3%E6%99%BA%E5%8A%9B%E9%97%AE%E9%A2%98/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146991317" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/solving-questions-with-brainpower/solutions/3639601/letmefly-2140jie-jue-zhi-li-wen-ti-ji-yi-pxc4/" target="_blank">LeetCode题解</a>|
 |2171.拿出最少数目的魔法豆|中等|<a href="https://leetcode.cn/problems/removing-minimum-number-of-magic-beans/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/01/18/LeetCode%202171.%E6%8B%BF%E5%87%BA%E6%9C%80%E5%B0%91%E6%95%B0%E7%9B%AE%E7%9A%84%E9%AD%94%E6%B3%95%E8%B1%86/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135682601" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/removing-minimum-number-of-magic-beans/solutions/2609831/letmefly-2171na-chu-zui-shao-shu-mu-de-m-jima/" target="_blank">LeetCode题解</a>|
+|2176.统计数组中相等且可以被整除的数对|简单|<a href="https://leetcode.cn/problems/count-equal-and-divisible-pairs-in-an-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/17/LeetCode%202176.%E7%BB%9F%E8%AE%A1%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9B%B8%E7%AD%89%E4%B8%94%E5%8F%AF%E4%BB%A5%E8%A2%AB%E6%95%B4%E9%99%A4%E7%9A%84%E6%95%B0%E5%AF%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147315213" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-equal-and-divisible-pairs-in-an-array/solutions/3654492/letmefly-2176tong-ji-shu-zu-zhong-xiang-643mr/" target="_blank">LeetCode题解</a>|
 |2178.拆分成最多数目的正偶数之和|中等|<a href="https://leetcode.cn/problems/maximum-split-of-positive-even-integers/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/06/LeetCode%202178.%E6%8B%86%E5%88%86%E6%88%90%E6%9C%80%E5%A4%9A%E6%95%B0%E7%9B%AE%E7%9A%84%E6%AD%A3%E5%81%B6%E6%95%B0%E4%B9%8B%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131567568" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-split-of-positive-even-integers/solutions/2332513/letmefly-2178chai-fen-cheng-zui-duo-shu-m44er/" target="_blank">LeetCode题解</a>|
 |2180.统计各位数字之和为偶数的整数个数|简单|<a href="https://leetcode.cn/problems/count-integers-with-even-digit-sum/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/01/06/LeetCode%202180.%E7%BB%9F%E8%AE%A1%E5%90%84%E4%BD%8D%E6%95%B0%E5%AD%97%E4%B9%8B%E5%92%8C%E4%B8%BA%E5%81%B6%E6%95%B0%E7%9A%84%E6%95%B4%E6%95%B0%E4%B8%AA%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128583772" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-integers-with-even-digit-sum/solutions/2048274/letmefly-2180tong-ji-ge-wei-shu-zi-zhi-h-g2ib/" target="_blank">LeetCode题解</a>|
 |2187.完成旅途的最少时间|中等|<a href="https://leetcode.cn/problems/minimum-time-to-complete-trips/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/05/LeetCode%202187.%E5%AE%8C%E6%88%90%E6%97%85%E9%80%94%E7%9A%84%E6%9C%80%E5%B0%91%E6%97%B6%E9%97%B4/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/142720764" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-time-to-complete-trips/solutions/2940132/letmefly-2187wan-cheng-lu-tu-de-zui-shao-kcpj/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 2176.统计数组中相等且可以被整除的数对.md b/Solutions/LeetCode 2176.统计数组中相等且可以被整除的数对.md
@@ -0,0 +1,148 @@
+---
+title: 2176.统计数组中相等且可以被整除的数对：两层遍历模拟
+date: 2025-04-17 20:23:19
+tags: [题解, LeetCode, 简单, 数组, 遍历, 模拟]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】2176.统计数组中相等且可以被整除的数对：两层遍历模拟
+
+力扣题目链接：[https://leetcode.cn/problems/count-equal-and-divisible-pairs-in-an-array/](https://leetcode.cn/problems/count-equal-and-divisible-pairs-in-an-array/)
+
+<p>给你一个下标从 <strong>0</strong>&nbsp;开始长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;，请你返回满足&nbsp;<code>0 &lt;= i &lt; j &lt; n</code>&nbsp;，<code>nums[i] == nums[j]</code> 且&nbsp;<code>(i * j)</code>&nbsp;能被&nbsp;<code>k</code>&nbsp;整除的数对&nbsp;<code>(i, j)</code>&nbsp;的&nbsp;<strong>数目</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>nums = [3,1,2,2,2,1,3], k = 2
+<b>输出：</b>4
+<strong>解释：</strong>
+总共有 4 对数符合所有要求：
+- nums[0] == nums[6] 且 0 * 6 == 0 ，能被 2 整除。
+- nums[2] == nums[3] 且 2 * 3 == 6 ，能被 2 整除。
+- nums[2] == nums[4] 且 2 * 4 == 8 ，能被 2 整除。
+- nums[3] == nums[4] 且 3 * 4 == 12 ，能被 2 整除。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>nums = [1,2,3,4], k = 1
+<b>输出：</b>0
+<b>解释：</b>由于数组中没有重复数值，所以没有数对 (i,j) 符合所有要求。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i], k &lt;= 100</code></li>
+</ul>
+
+
+    
+## 解题方法：模拟
+
+第一层使用变量$i$从$0$到$n-1$遍历，第二层使用变量$j$从$i+1$到$n-1$遍历。
+
+如果$nums[i]==nums[j]$相等且$i\times j \% k==0$，则答案数量加一。
+
++ 时间复杂度$O(len(nums)^2)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-17 20:13:54
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-17 20:28:40
+ */
+class Solution {
+public:
+    int countPairs(vector<int>& nums, int k) {
+        int ans = 0;
+        for (int i = 0; i < nums.size(); i++) {
+            for (int j = i + 1; j < nums.size(); j++) {
+                if (nums[i] == nums[j] && i * j % k == 0) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-04-17 20:15:46
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-04-17 20:15:58
+'''
+from typing import List
+
+
+class Solution:
+    def countPairs(self, nums: List[int], k: int) -> int:
+        return sum(nums[i] == nums[j] and i * j % k == 0 for i in range(len(nums)) for j in range(i + 1, len(nums)))
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-17 20:17:48
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-17 20:18:12
+ */
+class Solution {
+    public int countPairs(int[] nums, int k) {
+        int ans = 0;
+        for (int i = 0; i < nums.length; i++) {
+            for (int j = i + 1; j < nums.length; j++) {
+                if (nums[i] == nums[j] && i * j % k == 0) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-17 20:19:13
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-17 20:21:29
+ */
+package main
+
+func countPairs(nums []int, k int) (ans int) {
+    for i := range nums {
+		for j := i + 1; j < len(nums); j++ {
+			if nums[i] == nums[j] && i * j % k == 0 {
+				ans++
+			}
+		}
+	}
+	return
+}
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/147315213)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/2025/04/17/LeetCode%202176.%E7%BB%9F%E8%AE%A1%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9B%B8%E7%AD%89%E4%B8%94%E5%8F%AF%E4%BB%A5%E8%A2%AB%E6%95%B4%E9%99%A4%E7%9A%84%E6%95%B0%E5%AF%B9/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -1103,6 +1103,8 @@ categories: [自用]
 |mackintosh|n. 防水胶布，胶布雨衣（麦金托什）|
 |||
 |ferrous|adj. 含铁的，铁的|
+|||
+|stammer|v. 口吃，结结巴巴地说<br.>n. 口吃，结巴|
 
 <p class="wordCounts">单词收录总数</p>
 
diff --git a/newSolution.py b/newSolution.py
@@ -2,7 +2,7 @@
 Author: LetMeFly
 Date: 2022-07-03 11:21:14
 LastEditors: LetMeFly.xyz
-LastEditTime: 2025-04-14 00:27:31
+LastEditTime: 2025-04-17 20:24:00
 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
@@ -60,7 +60,7 @@ def getPlatform():
         return 'MacOS'
     else:
         return 'Linux(or others)'
-issueTitle = f'Who can add 1 more problem of LeetCode {num}'  # (#872)
+issueTitle = f'[newSolution]Who can add 1 more problem of LeetCode {num}'  # (#872)
 alreadyRelatedIssueLists = os.popen(f'gh issue list --search "{issueTitle}"').read()
 alreadyRelatedIssueListsSplited = alreadyRelatedIssueLists.split('\n')
 print(alreadyRelatedIssueLists)
