feat: add solutions to lc problems: No.1925,1929,1930

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)
+
+[English Version](/solution/1900-1999/1925.Count%20Square%20Sum%20Triples/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<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>
+
+<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>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>n = 5
+<b>输出：</b>2
+<b>解释：</b>平方和三元组为 (3,4,5) 和 (4,3,5) 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>n = 10
+<b>输出：</b>4
+<b>解释：</b>平方和三元组为 (3,4,5)，(4,3,5)，(6,8,10) 和 (8,6,10) 。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 250</code></li>
+</ul>
+
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def countTriples(self, n: int) -> int:
+        res = 0
+        for a in range(1, n + 1):
+            for b in range(1, n + 1):
+                t = a ** 2 + b ** 2
+                c = int(sqrt(t))
+                if c <= n and c ** 2 == t:
+                    res += 1
+        return res
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int countTriples(int n) {
+        int res = 0;
+        for (int a = 1; a <= n; ++a) {
+            for (int b = 1; b <= n; ++b) {
+                int t = a * a + b * b;
+                int c = (int) Math.sqrt(t);
+                if (c <= n && c * c == t) {
+                    ++res;
+                }
+            }
+        }
+        return res;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int countTriples(int n) {
+        int res = 0;
+        for (int a = 1; a <= n; ++a) {
+            for (int b = 1; b <= n; ++b) {
+                int t = a * a + b * b;
+                int c = (int) sqrt(t);
+                if (c <= n && c * c == t) {
+                    ++res;
+                }
+            }
+        }
+        return res;
+    }
+};
+```
+
+### **Go**
+
+```go
+func countTriples(n int) int {
+	res := 0
+	for a := 1; a <= n; a++ {
+		for b := 1; b <= n; b++ {
+			t := a*a + b*b
+			c := int(math.Sqrt(float64(t)))
+			if c <= n && c*c == t {
+				res++
+			}
+		}
+	}
+	return res
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- 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)
+
+[中文文档](/solution/1900-1999/1925.Count%20Square%20Sum%20Triples/README.md)
+
+## Description
+
+<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>
+
+<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>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 5
+<strong>Output:</strong> 2
+<strong>Explanation</strong>: The square triples are (3,4,5) and (4,3,5).
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 10
+<strong>Output:</strong> 4
+<strong>Explanation</strong>: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 250</code></li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def countTriples(self, n: int) -> int:
+        res = 0
+        for a in range(1, n + 1):
+            for b in range(1, n + 1):
+                t = a ** 2 + b ** 2
+                c = int(sqrt(t))
+                if c <= n and c ** 2 == t:
+                    res += 1
+        return res
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int countTriples(int n) {
+        int res = 0;
+        for (int a = 1; a <= n; ++a) {
+            for (int b = 1; b <= n; ++b) {
+                int t = a * a + b * b;
+                int c = (int) Math.sqrt(t);
+                if (c <= n && c * c == t) {
+                    ++res;
+                }
+            }
+        }
+        return res;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int countTriples(int n) {
+        int res = 0;
+        for (int a = 1; a <= n; ++a) {
+            for (int b = 1; b <= n; ++b) {
+                int t = a * a + b * b;
+                int c = (int) sqrt(t);
+                if (c <= n && c * c == t) {
+                    ++res;
+                }
+            }
+        }
+        return res;
+    }
+};
+```
+
+### **Go**
+
+```go
+func countTriples(n int) int {
+	res := 0
+	for a := 1; a <= n; a++ {
+		for b := 1; b <= n; b++ {
+			t := a*a + b*b
+			c := int(math.Sqrt(float64(t)))
+			if c <= n && c*c == t {
+				res++
+			}
+		}
+	}
+	return res
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

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

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

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

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

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

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

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

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