update

leetcode-cn/originData/smallest-integer-divisible-by-k.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给定正整数 <code>k</code>&nbsp;，你需要找出可以被 <code>k</code>&nbsp;整除的、仅包含数字 <code><strong>1</strong></code> 的最 <strong>小</strong> 正整数 <code>n</code>&nbsp;的长度。</p>\n\n<p>返回 <code>n</code>&nbsp;的长度。如果不存在这样的 <code>n</code>&nbsp;，就返回-1。</p>\n\n<p><strong>注意：</strong> <code>n</code>&nbsp;不符合 64 位带符号整数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>k = 1\n<strong>输出：</strong>1\n<strong>解释：</strong>最小的答案是 n = 1，其长度为 1。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>k = 2\n<strong>输出：</strong>-1\n<strong>解释：</strong>不存在可被 2 整除的正整数 n 。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>k = 3\n<strong>输出：</strong>3\n<strong>解释：</strong>最小的答案是 n = 111，其长度为 3。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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             "hints": [
                 "11111 = 1111 * 10 + 1\r\nWe only need to store remainders modulo K.",
                 "If we never get a remainder of 0, why would that happen, and how would we know that?"