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feat: add solutions to lc problems: No.1925,1929,1930 (doocs#508)
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‎solution/1900-1999/1925.Count Square Sum Triples/README.md

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# [1925. 统计平方和三元组的数目](https://leetcode-cn.com/problems/count-square-sum-triples)
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[English Version](/solution/1900-1999/1925.Count%20Square%20Sum%20Triples/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>一个 <strong>平方和三元组</strong> <code>(a,b,c)</code> 指的是满足 <code>a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup></code> 的 <strong>整数 </strong>三元组 <code>a</code>，<code>b</code> 和 <code>c</code> 。</p>
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<p>给你一个整数 <code>n</code> ，请你返回满足<em> </em><code>1 &lt;= a, b, c &lt;= n</code> 的 <strong>平方和三元组</strong> 的数目。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>n = 5
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<b>输出：</b>2
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<b>解释：</b>平方和三元组为 (3,4,5) 和 (4,3,5) 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>n = 10
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<b>输出：</b>4
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<b>解释：</b>平方和三元组为 (3,4,5)，(4,3,5)，(6,8,10) 和 (8,6,10) 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 250</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countTriples(self, n: int) -> int:
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res = 0
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for a in range(1, n + 1):
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for b in range(1, n + 1):
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t = a ** 2 + b ** 2
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c = int(sqrt(t))
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if c <= n and c ** 2 == t:
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res += 1
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countTriples(int n) {
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int res = 0;
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for (int a = 1; a <= n; ++a) {
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for (int b = 1; b <= n; ++b) {
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int t = a * a + b * b;
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int c = (int) Math.sqrt(t);
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if (c <= n && c * c == t) {
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++res;
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}
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}
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}
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return res;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countTriples(int n) {
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int res = 0;
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for (int a = 1; a <= n; ++a) {
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for (int b = 1; b <= n; ++b) {
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int t = a * a + b * b;
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int c = (int) sqrt(t);
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if (c <= n && c * c == t) {
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++res;
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}
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}
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}
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return res;
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}
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};
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```
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### **Go**
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```go
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func countTriples(n int) int {
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res := 0
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for a := 1; a <= n; a++ {
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for b := 1; b <= n; b++ {
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t := a*a + b*b
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c := int(math.Sqrt(float64(t)))
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if c <= n && c*c == t {
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res++
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}
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}
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}
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return res
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1900-1999/1925.Count Square Sum Triples/README_EN.md

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# [1925. Count Square Sum Triples](https://leetcode.com/problems/count-square-sum-triples)
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[中文文档](/solution/1900-1999/1925.Count%20Square%20Sum%20Triples/README.md)
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## Description
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<p>A <strong>square triple</strong> <code>(a,b,c)</code> is a triple where <code>a</code>, <code>b</code>, and <code>c</code> are <strong>integers</strong> and <code>a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup></code>.</p>
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<p>Given an integer <code>n</code>, return <em>the number of <strong>square triples</strong> such that </em><code>1 &lt;= a, b, c &lt;= n</code>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 5
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<strong>Output:</strong> 2
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<strong>Explanation</strong>: The square triples are (3,4,5) and (4,3,5).
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 10
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<strong>Output:</strong> 4
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<strong>Explanation</strong>: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 250</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def countTriples(self, n: int) -> int:
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res = 0
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for a in range(1, n + 1):
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for b in range(1, n + 1):
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t = a ** 2 + b ** 2
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c = int(sqrt(t))
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if c <= n and c ** 2 == t:
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res += 1
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return res
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```
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### **Java**
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```java
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class Solution {
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public int countTriples(int n) {
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int res = 0;
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for (int a = 1; a <= n; ++a) {
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for (int b = 1; b <= n; ++b) {
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int t = a * a + b * b;
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int c = (int) Math.sqrt(t);
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if (c <= n && c * c == t) {
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++res;
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}
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}
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}
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return res;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countTriples(int n) {
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int res = 0;
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for (int a = 1; a <= n; ++a) {
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for (int b = 1; b <= n; ++b) {
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int t = a * a + b * b;
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int c = (int) sqrt(t);
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if (c <= n && c * c == t) {
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++res;
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}
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}
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}
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return res;
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}
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};
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```
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### **Go**
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```go
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func countTriples(n int) int {
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res := 0
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for a := 1; a <= n; a++ {
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for b := 1; b <= n; b++ {
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t := a*a + b*b
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c := int(math.Sqrt(float64(t)))
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if c <= n && c*c == t {
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res++
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}
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}
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}
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return res
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1900-1999/1925.Count Square Sum Triples/Solution.cpp

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class Solution {
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public:
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int countTriples(int n) {
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int res = 0;
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for (int a = 1; a <= n; ++a) {
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for (int b = 1; b <= n; ++b) {
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int t = a * a + b * b;
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int c = (int) sqrt(t);
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if (c <= n && c * c == t) {
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++res;
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}
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}
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}
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return res;
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}
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};

‎solution/1900-1999/1925.Count Square Sum Triples/Solution.go

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func countTriples(n int) int {
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res := 0
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for a := 1; a <= n; a++ {
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for b := 1; b <= n; b++ {
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t := a*a + b*b
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c := int(math.Sqrt(float64(t)))
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if c <= n && c*c == t {
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res++
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}
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}
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}
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return res
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}

‎solution/1900-1999/1925.Count Square Sum Triples/Solution.java

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class Solution {
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public int countTriples(int n) {
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int res = 0;
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for (int a = 1; a <= n; ++a) {
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for (int b = 1; b <= n; ++b) {
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int t = a * a + b * b;
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int c = (int) Math.sqrt(t);
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if (c <= n && c * c == t) {
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++res;
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}
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}
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}
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return res;
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}
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}

‎solution/1900-1999/1925.Count Square Sum Triples/Solution.py

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class Solution:
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def countTriples(self, n: int) -> int:
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res = 0
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for a in range(1, n + 1):
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for b in range(1, n + 1):
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t = a ** 2 + b ** 2
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c = int(sqrt(t))
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if c <= n and c ** 2 == t:
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res += 1
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return res

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