You are given a string s consisting of digits and an integer k.
A round can be completed if the length of s is greater than k. In one round, do the following:
s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.k, repeat from step 1.Return s after all rounds have been completed.
\n
Example 1:
\n\n\nInput: s = "11111222223", k = 3\nOutput: "135"\nExplanation: \n- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".\n Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. \n So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.\n- For the second round, we divide s into "346" and "5".\n Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. \n So, s becomes "13" + "5" = "135" after second round. \nNow, s.length <= k, so we return "135" as the answer.\n\n\n
Example 2:
\n\n\nInput: s = "00000000", k = 3\nOutput: "000"\nExplanation: \nWe divide s into "000", "000", and "00".\nThen we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. \ns becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 1002 <= k <= 100s consists of digits only.给你一个由若干数字（0 - 9）组成的字符串 s ，和一个整数。
如果 s 的长度大于 k ，则可以执行一轮操作。在一轮操作中，需要完成以下工作：
s 拆分 成长度为 k 的若干 连续数字组 ，使得前 k 个字符都分在第一组，接下来的 k 个字符都分在第二组，依此类推。注意，最后一个数字组的长度可以小于 k 。\"346\" 会替换为 \"13\" ，因为 3 + 4 + 6 = 13 。k 则重复第一步。返回在完成所有轮操作后的 s 。
\n\n
示例 1：
\n\n输入：s = \"11111222223\", k = 3\n输出：\"135\"\n解释：\n- 第一轮，将 s 分成：\"111\"、\"112\"、\"222\" 和 \"23\" 。\n 接着，计算每一组的数字和：1 + 1 + 1 = 3、1 + 1 + 2 = 4、2 + 2 + 2 = 6 和 2 + 3 = 5 。 \n 这样，s 在第一轮之后变成 \"3\" + \"4\" + \"6\" + \"5\" = \"3465\" 。\n- 第二轮，将 s 分成：\"346\" 和 \"5\" 。\n 接着，计算每一组的数字和：3 + 4 + 6 = 13 、5 = 5 。\n 这样，s 在第二轮之后变成 \"13\" + \"5\" = \"135\" 。 \n现在，s.length <= k ，所以返回 \"135\" 作为答案。\n\n\n
示例 2：
\n\n输入：s = \"00000000\", k = 3\n输出：\"000\"\n解释：\n将 \"000\", \"000\", and \"00\".\n接着，计算每一组的数字和：0 + 0 + 0 = 0 、0 + 0 + 0 = 0 和 0 + 0 = 0 。 \ns 变为 \"0\" + \"0\" + \"0\" = \"000\" ，其长度等于 k ，所以返回 \"000\" 。\n\n\n
\n\n
提示：
\n\n1 <= s.length <= 1002 <= k <= 100s 仅由数字（0 - 9）组成。\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU99\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"\\/debug:pdbonly<\\/code>)\\u3002<\\/p>\"],\"javascript\":[\"JavaScript\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 5.5.2<\\/code><\\/p>\\r\\n\\r\\nGo 1.17<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n