leetcode-problemset/leetcode-cn/originData/abbreviating-the-product-of-a-range.json

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        "question": {
            "questionId": "2222",
            "questionFrontendId": "2117",
            "categoryTitle": "Algorithms",
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            "title": "Abbreviating the Product of a Range",
            "titleSlug": "abbreviating-the-product-of-a-range",
            "content": "<p>You are given two positive integers <code>left</code> and <code>right</code> with <code>left &lt;= right</code>. Calculate the <strong>product</strong> of all integers in the <strong>inclusive</strong> range <code>[left, right]</code>.</p>\n\n<p>Since the product may be very large, you will <strong>abbreviate</strong> it following these steps:</p>\n\n<ol>\n\t<li>Count all <strong>trailing</strong> zeros in the product and <strong>remove</strong> them. Let us denote this count as <code>C</code>.\n\n\t<ul>\n\t\t<li>For example, there are <code>3</code> trailing zeros in <code>1000</code>, and there are <code>0</code> trailing zeros in <code>546</code>.</li>\n\t</ul>\n\t</li>\n\t<li>Denote the remaining number of digits in the product as <code>d</code>. If <code>d &gt; 10</code>, then express the product as <code>&lt;pre&gt;...&lt;suf&gt;</code> where <code>&lt;pre&gt;</code> denotes the <strong>first</strong> <code>5</code> digits of the product, and <code>&lt;suf&gt;</code> denotes the <strong>last</strong> <code>5</code> digits of the product <strong>after</strong> removing all trailing zeros. If <code>d &lt;= 10</code>, we keep it unchanged.\n\t<ul>\n\t\t<li>For example, we express <code>1234567654321</code> as <code>12345...54321</code>, but <code>1234567</code> is represented as <code>1234567</code>.</li>\n\t</ul>\n\t</li>\n\t<li>Finally, represent the product as a <strong>string</strong> <code>&quot;&lt;pre&gt;...&lt;suf&gt;eC&quot;</code>.\n\t<ul>\n\t\t<li>For example, <code>12345678987600000</code> will be represented as <code>&quot;12345...89876e5&quot;</code>.</li>\n\t</ul>\n\t</li>\n</ol>\n\n<p>Return <em>a string denoting the <strong>abbreviated product</strong> of all integers in the <strong>inclusive</strong> range</em> <code>[left, right]</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> left = 1, right = 4\n<strong>Output:</strong> &quot;24e0&quot;\n<strong>Explanation:</strong> The product is 1 &times; 2 &times; 3 &times; 4 = 24.\nThere are no trailing zeros, so 24 remains the same. The abbreviation will end with &quot;e0&quot;.\nSince the number of digits is 2, which is less than 10, we do not have to abbreviate it further.\nThus, the final representation is &quot;24e0&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> left = 2, right = 11\n<strong>Output:</strong> &quot;399168e2&quot;\n<strong>Explanation:</strong> The product is 39916800.\nThere are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with &quot;e2&quot;.\nThe number of digits after removing the trailing zeros is 6, so we do not abbreviate it further.\nHence, the abbreviated product is &quot;399168e2&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> left = 371, right = 375\n<strong>Output:</strong> &quot;7219856259e3&quot;\n<strong>Explanation:</strong> The product is 7219856259000.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= left &lt;= right &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "一个区间内所有数乘积的缩写",
            "translatedContent": "<p>给你两个正整数&nbsp;<code>left</code>&nbsp;和&nbsp;<code>right</code>&nbsp;，满足&nbsp;<code>left &lt;= right</code>&nbsp;。请你计算&nbsp;<strong>闭区间</strong>&nbsp;<code>[left, right]</code>&nbsp;中所有整数的&nbsp;<strong>乘积</strong>&nbsp;。</p>\n\n<p>由于乘积可能非常大，你需要将它按照以下步骤 <strong>缩写</strong>&nbsp;：</p>\n\n<ol>\n\t<li>统计乘积中&nbsp;<strong>后缀</strong> 0 的数目，并 <strong>移除</strong> 这些 0 ，将这个数目记为&nbsp;<code>C</code>&nbsp;。\n\n\t<ul>\n\t\t<li>比方说，<code>1000</code>&nbsp;中有 <code>3</code> 个后缀 0&nbsp;，<code>546</code>&nbsp;中没有后缀 0 。</li>\n\t</ul>\n\t</li>\n\t<li>将乘积中剩余数字的位数记为&nbsp;<code>d</code>&nbsp;。如果&nbsp;<code>d &gt; 10</code>&nbsp;，那么将乘积表示为&nbsp;<code>&lt;pre&gt;...&lt;suf&gt;</code>&nbsp;的形式，其中&nbsp;<code>&lt;pre&gt;</code>&nbsp;表示乘积最 <strong>开始</strong>&nbsp;的 <code>5</code>&nbsp;个数位，<code>&lt;suf&gt;</code>&nbsp;表示删除后缀 0 <strong>之后</strong>&nbsp;结尾的 <code>5</code>&nbsp;个数位。如果&nbsp;<code>d &lt;= 10</code>&nbsp;，我们不对它做修改。\n\t<ul>\n\t\t<li>比方说，我们将&nbsp;<code>1234567654321</code>&nbsp;表示为&nbsp;<code>12345...54321</code>&nbsp;，但是&nbsp;<code>1234567</code>&nbsp;仍然表示为&nbsp;<code>1234567</code>&nbsp;。</li>\n\t</ul>\n\t</li>\n\t<li>最后，将乘积表示为 <strong>字符串</strong>&nbsp;<code>\"&lt;pre&gt;...&lt;suf&gt;eC\"</code>&nbsp;。\n\t<ul>\n\t\t<li>比方说，<code>12345678987600000</code>&nbsp;被表示为&nbsp;<code>\"12345...89876e5\"</code>&nbsp;。</li>\n\t</ul>\n\t</li>\n</ol>\n\n<p>请你返回一个字符串，表示 <strong>闭区间</strong>&nbsp;<code>[left, right]</code>&nbsp;中所有整数&nbsp;<strong>乘积</strong>&nbsp;的&nbsp;<strong>缩写</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>left = 1, right = 4\n<b>输出：</b>\"24e0\"\n<strong>解释：</strong>\n乘积为 1 × 2 × 3 × 4 = 24 。\n由于没有后缀 0 ，所以 24 保持不变，缩写的结尾为 \"e0\" 。\n因为乘积的结果是 2 位数，小于 10 ，所欲我们不进一步将它缩写。\n所以，最终将乘积表示为 \"24e0\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>left = 2, right = 11\n<strong>输出：</strong>\"399168e2\"\n<strong>解释：</strong>乘积为 39916800 。\n有 2 个后缀 0 ，删除后得到 399168 。缩写的结尾为 \"e2\" 。 \n删除后缀 0 后是 6 位数，不需要进一步缩写。 \n所以，最终将乘积表示为 \"399168e2\" 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>left = 371, right = 375\n<strong>输出：</strong>\"7219856259e3\"\n<strong>解释：</strong>乘积为 7219856259000 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= left &lt;= right &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    string abbreviateProduct(int left, int right) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def abbreviateProduct(self, left, right):\n        \"\"\"\n        :type left: int\n        :type right: int\n        :rtype: str\n        \"\"\"",
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                    "code": "/**\n * @param {number} left\n * @param {number} right\n * @return {string}\n */\nvar abbreviateProduct = function(left, right) {\n\n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $left\n     * @param Integer $right\n     * @return String\n     */\n    function abbreviateProduct($left, $right) {\n\n    }\n}",
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                    "code": "class Solution {\n    func abbreviateProduct(_ left: Int, _ right: Int) -> String {\n\n    }\n}",
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                    "code": "class Solution {\n    fun abbreviateProduct(left: Int, right: Int): String {\n\n    }\n}",
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                    "code": "class Solution {\n  String abbreviateProduct(int left, int right) {\n    \n  }\n}",
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                    "lang": "Go",
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                    "code": "func abbreviateProduct(left int, right int) string {\n\n}",
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                    "code": "# @param {Integer} left\n# @param {Integer} right\n# @return {String}\ndef abbreviate_product(left, right)\n\nend",
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                    "code": "impl Solution {\n    pub fn abbreviate_product(left: i32, right: i32) -> String {\n\n    }\n}",
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                    "lang": "Erlang",
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                    "code": "defmodule Solution do\n  @spec abbreviate_product(left :: integer, right :: integer) :: String.t\n  def abbreviate_product(left, right) do\n    \n  end\nend",
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            "hints": [
                "Calculating the number of trailing zeros, the last five digits, and the first five digits can all be done separately.",
                "Use a prime factorization property to find the number of trailing zeros. Use modulo to find the last 5 digits. Use a logarithm property to find the first 5 digits.",
                "The number of trailing zeros C is nothing but the number of times the product is completely divisible by 10. Since 2 and 5 are the only prime factors of 10,  C will be equal to the minimum number of times 2 or 5 appear in the prime factorization of the product.",
                "Iterate through the integers from left to right. For every integer, keep dividing it by 2 as long as it is divisible by 2 and C occurrences of 2 haven't been removed in total. Repeat this process for 5. Finally, multiply the integer under modulo of 10^5 with the product obtained till now to obtain the last five digits.",
                "The product P can be represented as P=10^(x+y) where x is the integral part and y is the fractional part of x+y. Using the property \"if S = A * B, then log(S) = log(A) + log(B)\", we can write x+y = log_10(P) = sum(log_10(i)) for each integer i in [left, right]. Once we obtain the sum, the first five digits can be represented as floor(10^(y+4))."
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