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feat: add weekly contest 477 (#4857)
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‎solution/3700-3799/3754.Concatenate Non-Zero Digits and Multiply by Sum I/README.md‎

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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README.md
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---
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<!-- problem:start -->
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# [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)
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[English Version](/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你一个整数 <code>n</code>。</p>
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<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>
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<p><code>sum</code> 为 <code>x</code> 中所有数字的&nbsp;<strong>数字和&nbsp;</strong>。</p>
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<p>返回一个整数，表示 <code>x * sum</code> 的值。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">n = 10203004</span></p>
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<p><strong>输出：</strong> <span class="example-io">12340</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>非零数字是 1、2、3 和 4。因此，<code>x = 1234</code>。</li>
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<li>数字和为 <code>sum = 1 + 2 + 3 + 4 = 10</code>。</li>
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<li>因此，答案是 <code>x * sum = 1234 * 10 = 12340</code>。</li>
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</ul>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">n = 1000</span></p>
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<p><strong>输出：</strong> <span class="example-io">1</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>非零数字是 1，因此 <code>x = 1</code> 且 <code>sum = 1</code>。</li>
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<li>因此，答案是 <code>x * sum = 1 * 1 = 1</code>。</li>
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</ul>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一：模拟
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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。
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时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
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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 sumAndMultiply(self, n: int) -> int:
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p = 1
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x = s = 0
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while n:
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v = n % 10
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s += v
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if v:
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x += p * v
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p *= 10
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n //= 10
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return x * s
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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 long sumAndMultiply(int n) {
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int p = 1;
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int x = 0, s = 0;
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for (; n > 0; n /= 10) {
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int v = n % 10;
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s += v;
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if (v != 0) {
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x += p * v;
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p *= 10;
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}
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}
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return 1L * x * s;
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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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long long sumAndMultiply(int n) {
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int p = 1;
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int x = 0, s = 0;
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for (; n > 0; n /= 10) {
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int v = n % 10;
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s += v;
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if (v != 0) {
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x += p * v;
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p *= 10;
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}
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}
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return 1LL * x * s;
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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 sumAndMultiply(n int) int64 {
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p := 1
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x := 0
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s := 0
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for n > 0 {
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v := n % 10
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s += v
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if v != 0 {
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x += p * v
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p *= 10
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}
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n /= 10
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}
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return int64(x) * int64(s)
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}
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```
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#### TypeScript
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```ts
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function sumAndMultiply(n: number): number {
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let p = 1;
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let x = 0;
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let s = 0;
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while (n > 0) {
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const v = n % 10;
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s += v;
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if (v !== 0) {
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x += p * v;
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p *= 10;
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}
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n = Math.floor(n / 10);
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}
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return x * s;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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---
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comments: true
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difficulty: Easy
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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
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---
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<!-- problem:start -->
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# [3754. Concatenate Non-Zero Digits and Multiply by Sum I](https://leetcode.com/problems/concatenate-non-zero-digits-and-multiply-by-sum-i)
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[中文文档](/solution/3700-3799/3754.Concatenate%20Non-Zero%20Digits%20and%20Multiply%20by%20Sum%20I/README.md)
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## Description
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<!-- description:start -->
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<p>You are given an integer <code>n</code>.</p>
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<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>
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<p>Let <code>sum</code> be the <strong>sum of digits</strong> in <code>x</code>.</p>
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<p>Return an integer representing the value of <code>x * sum</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 10203004</span></p>
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<p><strong>Output:</strong> <span class="example-io">12340</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>The non-zero digits are 1, 2, 3, and 4. Thus, <code>x = 1234</code>.</li>
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<li>The sum of digits is <code>sum = 1 + 2 + 3 + 4 = 10</code>.</li>
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<li>Therefore, the answer is <code>x * sum = 1234 * 10 = 12340</code>.</li>
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</ul>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 1000</span></p>
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<p><strong>Output:</strong> <span class="example-io">1</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>The non-zero digit is 1, so <code>x = 1</code> and <code>sum = 1</code>.</li>
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<li>Therefore, the answer is <code>x * sum = 1 * 1 = 1</code>.</li>
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</ul>
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</div>
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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>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Simulation
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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 $x \times s$.
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The time complexity is $O(\log n)$ and the space complexity is $O(1)$.
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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 sumAndMultiply(self, n: int) -> int:
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p = 1
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x = s = 0
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while n:
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v = n % 10
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s += v
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if v:
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x += p * v
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p *= 10
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n //= 10
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return x * s
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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 long sumAndMultiply(int n) {
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int p = 1;
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int x = 0, s = 0;
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for (; n > 0; n /= 10) {
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int v = n % 10;
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s += v;
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if (v != 0) {
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x += p * v;
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p *= 10;
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}
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}
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return 1L * x * s;
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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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long long sumAndMultiply(int n) {
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int p = 1;
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int x = 0, s = 0;
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for (; n > 0; n /= 10) {
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int v = n % 10;
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s += v;
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if (v != 0) {
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x += p * v;
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p *= 10;
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}
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}
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return 1LL * x * s;
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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 sumAndMultiply(n int) int64 {
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p := 1
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x := 0
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s := 0
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for n > 0 {
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v := n % 10
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s += v
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if v != 0 {
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x += p * v
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p *= 10
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}
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n /= 10
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}
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return int64(x) * int64(s)
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}
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```
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#### TypeScript
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```ts
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function sumAndMultiply(n: number): number {
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let p = 1;
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let x = 0;
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let s = 0;
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while (n > 0) {
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const v = n % 10;
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s += v;
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if (v !== 0) {
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x += p * v;
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p *= 10;
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}
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n = Math.floor(n / 10);
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}
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return x * s;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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

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