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"categoryTitle": "Algorithms",
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"title": "Ways to Express an Integer as Sum of Powers",
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"content": "<p>Given two <strong>positive</strong> integers <code>n</code> and <code>x</code>.</p>\n\n<p>Return <em>the number of ways </em><code>n</code><em> can be expressed as the sum of the </em><code>x<sup>th</sup></code><em> power of <strong>unique</strong> positive integers, in other words, the number of sets of unique integers </em><code>[n<sub>1</sub>, n<sub>2</sub>, ..., n<sub>k</sub>]</code><em> where </em><code>n = n<sub>1</sub><sup>x</sup> + n<sub>2</sub><sup>x</sup> + ... + n<sub>k</sub><sup>x</sup></code><em>.</em></p>\n\n<p>Since the result can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>For example, if <code>n = 160</code> and <code>x = 3</code>, one way to express <code>n</code> is <code>n = 2<sup>3</sup> + 3<sup>3</sup> + 5<sup>3</sup></code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 10, x = 2\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> We can express n as the following: n = 3<sup>2</sup> + 1<sup>2</sup> = 10.\nIt can be shown that it is the only way to express 10 as the sum of the 2<sup>nd</sup> power of unique integers.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 4, x = 1\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can express n in the following ways:\n- n = 4<sup>1</sup> = 4.\n- n = 3<sup>1</sup> + 1<sup>1</sup> = 4.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 300</code></li>\n\t<li><code>1 &lt;= x &lt;= 5</code></li>\n</ul>\n",
"translatedTitle": "将一个数字表示成幂的和的方案数",
"translatedContent": "<p>给你两个 <strong>正</strong>&nbsp;整数&nbsp;<code>n</code> 和&nbsp;<code>x</code>&nbsp;。</p>\n\n<p>请你返回将<em>&nbsp;</em><code>n</code>&nbsp;表示成一些&nbsp;<strong>互不相同</strong>&nbsp;正整数的<em>&nbsp;</em><code>x</code>&nbsp;次幂之和的方案数。换句话说,你需要返回互不相同整数&nbsp;<code>[n<sub>1</sub>, n<sub>2</sub>, ..., n<sub>k</sub>]</code>&nbsp;的集合数目,满足&nbsp;<code>n = n<sub>1</sub><sup>x</sup> + n<sub>2</sub><sup>x</sup> + ... + n<sub>k</sub><sup>x</sup></code>&nbsp;。</p>\n\n<p>由于答案可能非常大,请你将它对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;取余后返回。</p>\n\n<p>比方说,<code>n = 160</code> 且&nbsp;<code>x = 3</code>&nbsp;,一个表示&nbsp;<code>n</code>&nbsp;的方法是&nbsp;<code>n = 2<sup>3</sup> + 3<sup>3</sup> + 5<sup>3</sup></code><sup>&nbsp;</sup>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre><b>输入:</b>n = 10, x = 2\n<b>输出:</b>1\n<b>解释:</b>我们可以将 n 表示为n = 3<sup>2</sup> + 1<sup>2</sup> = 10 。\n这是唯一将 10 表达成不同整数 2 次方之和的方案。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre><b>输入:</b>n = 4, x = 1\n<b>输出:</b>2\n<b>解释:</b>我们可以将 n 按以下方案表示:\n- n = 4<sup>1</sup> = 4 。\n- n = 3<sup>1</sup> + 1<sup>1</sup> = 4 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 300</code></li>\n\t<li><code>1 &lt;= x &lt;= 5</code></li>\n</ul>\n",
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