feat: add solutions for lc No.3792 (#4941)

Author: yanglbme

README.md

@@ -52,7 +52,7 @@
 - [乘积小于 K 的子数组](/solution/0700-0799/0713.Subarray%20Product%20Less%20Than%20K/README.md) - `双指针`
 - [位 1 的个数](/solution/0100-0199/0191.Number%20of%201%20Bits/README.md) - `位运算`、`lowbit`
 - [合并区间](/solution/0000-0099/0056.Merge%20Intervals/README.md) - `区间合并`
-    <!-- 排序算法、待补充 -->
+  <!-- 排序算法、待补充 -->
 
 ### 2. 数据结构
 

solution/3700-3799/3792.Sum of Increasing Product Blocks/README.md

@@ -0,0 +1,196 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3792.Sum%20of%20Increasing%20Product%20Blocks/README.md
+---
+
+<!-- problem:start -->
+
+# [3792. 递增乘积块之和 🔒](https://leetcode.cn/problems/sum-of-increasing-product-blocks)
+
+[English Version](/solution/3700-3799/3792.Sum%20of%20Increasing%20Product%20Blocks/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定一个整数&nbsp;<code>n</code>。</p>
+
+<p>一个序列的形成如下：</p>
+
+<ul>
+	<li>第&nbsp;<code>1</code>&nbsp;块包含&nbsp;<code>1</code>。</li>
+	<li>第&nbsp;<code>2</code>&nbsp;块包含&nbsp;<code>2 * 3</code>。</li>
+	<li>第&nbsp;<code>i</code>&nbsp;块是之后&nbsp;<code>i</code>&nbsp;个连续整数的乘积。</li>
+</ul>
+
+<p>令&nbsp;<code>F(n)</code>&nbsp;为前 <code>n</code>&nbsp;块之和。</p>
+
+<p>返回一个整数表示&nbsp;<code>F(n)</code> <strong>模上</strong>&nbsp;<code>10<sup>9</sup> + 7</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 3</span></p>
+
+<p><span class="example-io"><b>输出：</b>127</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>块 1：<code>1</code></li>
+	<li>块 2：<code>2 * 3 = 6</code></li>
+	<li>块 3：<code>4 * 5 * 6 = 120</code></li>
+</ul>
+
+<p><code>F(3) = 1 + 6 + 120 = 127</code></p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 7</span></p>
+
+<p><span class="example-io"><b>输出：</b>6997165</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>块 1：<code>1</code></li>
+	<li>块 2：<code>2 * 3 = 6</code></li>
+	<li>块 3：<code>4 * 5 * 6 = 120</code></li>
+	<li>块 4：<code>7 * 8 * 9 * 10 = 5040</code></li>
+	<li>块 5：<code>11 * 12 * 13 * 14 * 15 = 360360</code></li>
+	<li>块 6：<code>16 * 17 * 18 * 19 * 20 * 21 = 39070080</code></li>
+	<li>块 7：<code>22 * 23 * 24 * 25 * 26 * 27 * 28 = 5967561600</code></li>
+</ul>
+
+<p><code>F(7) = 6006997207 % (10<sup>9</sup> + 7) = 6997165</code></p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们可以直接模拟每一块的乘积并累加到答案中。需要注意的是，由于乘积可能会非常大，我们需要在每一步计算时对结果取模。
+
+时间复杂度 $O(n^2)$，空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def sumOfBlocks(self, n: int) -> int:
+        ans = 0
+        mod = 10**9 + 7
+        k = 1
+        for i in range(1, n + 1):
+            x = 1
+            for j in range(k, k + i):
+                x = (x * j) % mod
+            ans = (ans + x) % mod
+            k += i
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int sumOfBlocks(int n) {
+        final int mod = (int) 1e9 + 7;
+        long ans = 0;
+        int k = 1;
+        for (int i = 1; i <= n; ++i) {
+            long x = 1;
+            for (int j = k; j < k + i; ++j) {
+                x = x * j % mod;
+            }
+            ans = (ans + x) % mod;
+            k += i;
+        }
+        return (int) ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int sumOfBlocks(int n) {
+        const int mod = 1e9 + 7;
+        long long ans = 0;
+        int k = 1;
+        for (int i = 1; i <= n; ++i) {
+            long long x = 1;
+            for (int j = k; j < k + i; ++j) {
+                x = x * j % mod;
+            }
+            ans = (ans + x) % mod;
+            k += i;
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func sumOfBlocks(n int) (ans int) {
+	const mod int = 1e9 + 7
+	k := 1
+	for i := 1; i <= n; i++ {
+		x := 1
+		for j := k; j < k+i; j++ {
+			x = x * j % mod
+		}
+		ans = (ans + x) % mod
+		k += i
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function sumOfBlocks(n: number): number {
+    const mod = 1000000007;
+    let k = 1;
+    let ans = 0;
+    for (let i = 1; i <= n; i++) {
+        let x = 1;
+        for (let j = k; j < k + i; j++) {
+            x = (x * j) % mod;
+        }
+        ans = (ans + x) % mod;
+        k += i;
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->