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leetcode-cn/problem (Chinese)/最小可整除数位乘积 I [smallest-divisible-digit-product-i].html Normal file
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<p>给你两个整数&nbsp;<code>n</code> 和&nbsp;<code>t</code>&nbsp;。请你返回大于等于&nbsp;<code>n</code>&nbsp;的&nbsp;<strong>最小</strong>&nbsp;整数，且该整数的&nbsp;<strong>各数位之积</strong>&nbsp;能被&nbsp;<code>t</code>&nbsp;整除。</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 = 10, t = 2</span></p>
<p><span class="example-io"><b>输出：</b>10</span></p>
<p><strong>解释：</strong></p>
<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>
<p><strong class="example">示例 2：</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入：</b>n = 15, t = 3</span></p>
<p><span class="example-io"><b>输出：</b>16</span></p>
<p><strong>解释：</strong></p>
<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>
<p>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 &lt;= n &lt;= 100</code></li>
<li><code>1 &lt;= t &lt;= 10</code></li>
</ul>