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feat: add biweekly contest 143 and weekly contest 423 (#3736)
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solution/3300-3399/3345.Smallest Divisible Digit Product I/README.md
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| --- | ||
| comments: true | ||
| difficulty: 简单 | ||
| edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3345.Smallest%20Divisible%20Digit%20Product%20I/README.md | ||
| --- | ||
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| <!-- problem:start --> | ||
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| # [3345. 最小可整除数位乘积 I](https://leetcode.cn/problems/smallest-divisible-digit-product-i) | ||
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| [English Version](/solution/3300-3399/3345.Smallest%20Divisible%20Digit%20Product%20I/README_EN.md) | ||
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| ## 题目描述 | ||
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| <!-- description:start --> | ||
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| <p>给你两个整数 <code>n</code> 和 <code>t</code> 。请你返回大于等于 <code>n</code> 的 <strong>最小</strong> 整数，且该整数的 <strong>各数位之积</strong> 能被 <code>t</code> 整除。</p> | ||
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| <p> </p> | ||
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| <p><strong class="example">示例 1：</strong></p> | ||
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| <div class="example-block"> | ||
| <p><span class="example-io"><b>输入：</b>n = 10, t = 2</span></p> | ||
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| <p><span class="example-io"><b>输出：</b>10</span></p> | ||
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| <p><strong>解释：</strong></p> | ||
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| <p>10 的数位乘积为 0 ，可以被 2 整除，所以它是大于等于 10 且满足题目要求的最小整数。</p> | ||
| </div> | ||
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| <p><strong class="example">示例 2：</strong></p> | ||
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| <div class="example-block"> | ||
| <p><span class="example-io"><b>输入：</b>n = 15, t = 3</span></p> | ||
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| <p><span class="example-io"><b>输出：</b>16</span></p> | ||
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| <p><strong>解释：</strong></p> | ||
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| <p>16 的数位乘积为 6 ，可以被 3 整除，所以它是大于等于 15 且满足题目要求的最小整数。</p> | ||
| </div> | ||
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| <p> </p> | ||
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| <p><strong>提示：</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= n <= 100</code></li> | ||
| <li><code>1 <= t <= 10</code></li> | ||
| </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：枚举 | ||
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| 我们注意到，每 $10$ 个数里一定会出现数位乘积为 $0$ 的整数，因此我们可以直接枚举大于等于 $n$ 的整数，直到找到一个数位乘积能被 $t$ 整除的整数。 | ||
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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 | ||
| class Solution: | ||
| def smallestNumber(self, n: int, t: int) -> int: | ||
| for i in count(n): | ||
| p = 1 | ||
| x = i | ||
| while x: | ||
| p *= x % 10 | ||
| x //= 10 | ||
| if p % t == 0: | ||
| return i | ||
| ``` | ||
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| #### Java | ||
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| ```java | ||
| class Solution { | ||
| public int smallestNumber(int n, int t) { | ||
| for (int i = n;; ++i) { | ||
| int p = 1; | ||
| for (int x = i; x > 0; x /= 10) { | ||
| p *= (x % 10); | ||
| } | ||
| if (p % t == 0) { | ||
| return i; | ||
| } | ||
| } | ||
| } | ||
| } | ||
| ``` | ||
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| #### C++ | ||
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| int smallestNumber(int n, int t) { | ||
| for (int i = n;; ++i) { | ||
| int p = 1; | ||
| for (int x = i; x > 0; x /= 10) { | ||
| p *= (x % 10); | ||
| } | ||
| if (p % t == 0) { | ||
| return i; | ||
| } | ||
| } | ||
| } | ||
| }; | ||
| ``` | ||
| #### Go | ||
| ```go | ||
| func smallestNumber(n int, t int) int { | ||
| for i := n; ; i++ { | ||
| p := 1 | ||
| for x := i; x > 0; x /= 10 { | ||
| p *= x % 10 | ||
| } | ||
| if p%t == 0 { | ||
| return i | ||
| } | ||
| } | ||
| } | ||
| ``` | ||
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| #### TypeScript | ||
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| ```ts | ||
| function smallestNumber(n: number, t: number): number { | ||
| for (let i = n; ; ++i) { | ||
| let p = 1; | ||
| for (let x = i; x; x = Math.floor(x / 10)) { | ||
| p *= x % 10; | ||
| } | ||
| if (p % t === 0) { | ||
| return i; | ||
| } | ||
| } | ||
| } | ||
| ``` | ||
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| <!-- tabs:end --> | ||
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| <!-- solution:end --> | ||
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| <!-- problem:end --> |
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