feat: add weekly contest 477

solution/3700-3799/3754.Concatenate Non-Zero Digits and Multiply by Sum I/README.md

@@ -0,0 +1,184 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3754. 连接非零数字并乘以其数字和 I](https://leetcode.cn/problems/concatenate-non-zero-digits-and-multiply-by-sum-i)
+
+[English Version](/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数 <code>n</code>。</p>
+
+<p>将 <code>n</code> 中所有的&nbsp;<strong>非零数字&nbsp;</strong>按照它们的原始顺序连接起来，形成一个新的整数 <code>x</code>。如果不存在&nbsp;<strong>非零数字&nbsp;</strong>，则 <code>x = 0</code>。</p>
+
+<p><code>sum</code> 为 <code>x</code> 中所有数字的&nbsp;<strong>数字和&nbsp;</strong>。</p>
+
+<p>返回一个整数，表示 <code>x * sum</code> 的值。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">n = 10203004</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">12340</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>非零数字是 1、2、3 和 4。因此，<code>x = 1234</code>。</li>
+	<li>数字和为 <code>sum = 1 + 2 + 3 + 4 = 10</code>。</li>
+	<li>因此，答案是 <code>x * sum = 1234 * 10 = 12340</code>。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">n = 1000</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>非零数字是 1，因此 <code>x = 1</code> 且 <code>sum = 1</code>。</li>
+	<li>因此，答案是 <code>x * sum = 1 * 1 = 1</code>。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们可以通过对数字逐位处理来模拟题目要求的操作。在处理每一位数字时，我们将非零数字连接起来形成新的整数 $x$，同时计算数字和 $s$，最后返回 $x \times s$ 即可。
+
+时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def sumAndMultiply(self, n: int) -> int:
+        p = 1
+        x = s = 0
+        while n:
+            v = n % 10
+            s += v
+            if v:
+                x += p * v
+                p *= 10
+            n //= 10
+        return x * s
+```
+
+#### Java
+
+```java
+class Solution {
+    public long sumAndMultiply(int n) {
+        int p = 1;
+        int x = 0, s = 0;
+        for (; n > 0; n /= 10) {
+            int v = n % 10;
+            s += v;
+            if (v != 0) {
+                x += p * v;
+                p *= 10;
+            }
+        }
+        return 1L * x * s;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    long long sumAndMultiply(int n) {
+        int p = 1;
+        int x = 0, s = 0;
+        for (; n > 0; n /= 10) {
+            int v = n % 10;
+            s += v;
+            if (v != 0) {
+                x += p * v;
+                p *= 10;
+            }
+        }
+        return 1LL * x * s;
+    }
+};
+```
+
+#### Go
+
+```go
+func sumAndMultiply(n int) int64 {
+	p := 1
+	x := 0
+	s := 0
+	for n > 0 {
+		v := n % 10
+		s += v
+		if v != 0 {
+			x += p * v
+			p *= 10
+		}
+		n /= 10
+	}
+	return int64(x) * int64(s)
+}
+```
+
+#### TypeScript
+
+```ts
+function sumAndMultiply(n: number): number {
+    let p = 1;
+    let x = 0;
+    let s = 0;
+
+    while (n > 0) {
+        const v = n % 10;
+        s += v;
+        if (v !== 0) {
+            x += p * v;
+            p *= 10;
+        }
+        n = Math.floor(n / 10);
+    }
+
+    return x * s;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3700-3799/3754.Concatenate Non-Zero Digits and Multiply by Sum I/README_EN.md

@@ -0,0 +1,182 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3754. Concatenate Non-Zero Digits and Multiply by Sum I](https://leetcode.com/problems/concatenate-non-zero-digits-and-multiply-by-sum-i)
+
+[中文文档](/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer <code>n</code>.</p>
+
+<p>Form a new integer <code>x</code> by concatenating all the <strong>non-zero digits</strong> of <code>n</code> in their original order. If there are no <strong>non-zero</strong> digits, <code>x = 0</code>.</p>
+
+<p>Let <code>sum</code> be the <strong>sum of digits</strong> in <code>x</code>.</p>
+
+<p>Return an integer representing the value of <code>x * sum</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 10203004</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">12340</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The non-zero digits are 1, 2, 3, and 4. Thus, <code>x = 1234</code>.</li>
+	<li>The sum of digits is <code>sum = 1 + 2 + 3 + 4 = 10</code>.</li>
+	<li>Therefore, the answer is <code>x * sum = 1234 * 10 = 12340</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 1000</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The non-zero digit is 1, so <code>x = 1</code> and <code>sum = 1</code>.</li>
+	<li>Therefore, the answer is <code>x * sum = 1 * 1 = 1</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Simulation
+
+We can simulate the required operation by processing the number digit by digit. While processing each digit, we concatenate non-zero digits to form a new integer $x$ and calculate the digit sum $s$. Finally, we return $x \times s$.
+
+The time complexity is $O(\log n)$ and the space complexity is $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def sumAndMultiply(self, n: int) -> int:
+        p = 1
+        x = s = 0
+        while n:
+            v = n % 10
+            s += v
+            if v:
+                x += p * v
+                p *= 10
+            n //= 10
+        return x * s
+```
+
+#### Java
+
+```java
+class Solution {
+    public long sumAndMultiply(int n) {
+        int p = 1;
+        int x = 0, s = 0;
+        for (; n > 0; n /= 10) {
+            int v = n % 10;
+            s += v;
+            if (v != 0) {
+                x += p * v;
+                p *= 10;
+            }
+        }
+        return 1L * x * s;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    long long sumAndMultiply(int n) {
+        int p = 1;
+        int x = 0, s = 0;
+        for (; n > 0; n /= 10) {
+            int v = n % 10;
+            s += v;
+            if (v != 0) {
+                x += p * v;
+                p *= 10;
+            }
+        }
+        return 1LL * x * s;
+    }
+};
+```
+
+#### Go
+
+```go
+func sumAndMultiply(n int) int64 {
+	p := 1
+	x := 0
+	s := 0
+	for n > 0 {
+		v := n % 10
+		s += v
+		if v != 0 {
+			x += p * v
+			p *= 10
+		}
+		n /= 10
+	}
+	return int64(x) * int64(s)
+}
+```
+
+#### TypeScript
+
+```ts
+function sumAndMultiply(n: number): number {
+    let p = 1;
+    let x = 0;
+    let s = 0;
+
+    while (n > 0) {
+        const v = n % 10;
+        s += v;
+        if (v !== 0) {
+            x += p * v;
+            p *= 10;
+        }
+        n = Math.floor(n / 10);
+    }
+
+    return x * s;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3700-3799/3754.Concatenate Non-Zero Digits and Multiply by Sum I/Solution.cpp

@@ -0,0 +1,16 @@
+class Solution {
+public:
+    long long sumAndMultiply(int n) {
+        int p = 1;
+        int x = 0, s = 0;
+        for (; n > 0; n /= 10) {
+            int v = n % 10;
+            s += v;
+            if (v != 0) {
+                x += p * v;
+                p *= 10;
+            }
+        }
+        return 1LL * x * s;
+    }
+};