{"id":8385,"date":"2021-04-25T09:04:05","date_gmt":"2021-04-25T16:04:05","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8385"},"modified":"2021-04-25T09:23:19","modified_gmt":"2021-04-25T16:23:19","slug":"leetcode-1837-sum-of-digits-in-base-k","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/math\/leetcode-1837-sum-of-digits-in-base-k\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1837. Sum of Digits in Base K"},"content":{"rendered":"\n

Given an integer n<\/code> (in base 10<\/code>) and a base k<\/code>, return the sum<\/strong> of the digits of <\/em>n<\/code> after<\/strong> converting <\/em>n<\/code> from base <\/em>10<\/code> to base <\/em>k<\/code>.<\/p>\n\n\n\n

After converting, each digit should be interpreted as a base 10<\/code> number, and the sum should be returned in base 10<\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> n = 34, k = 6\nOutput:<\/strong> 9\nExplanation: <\/strong>34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> n = 10, k = 10\nOutput:<\/strong> 1\nExplanation: <\/strong>n is already in base 10. 1 + 0 = 1.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= n <= 100<\/code><\/li>
  • 2 <= k <= 10<\/code><\/li><\/ul>\n\n\n\n
    Solution: Base Conversion<\/strong><\/h2>\n\n\n\n
    Time complexity: O(logn)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \/\/ Author: Huahua\nclass Solution {\npublic:\n  int sumBase(int n, int k) {\n    int ans = 0;\n    while (n) {\n      ans += n % k;\n      n \/= k;\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given an integer n (in base 10) and a base k, return the sum of the digits of n after converting n from base 10 to base k. After converting, each digit should be interpreted as a base 10 number, and the…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[49],"tags":[379,222,31],"class_list":["post-8385","post","type-post","status-publish","format-standard","hentry","category-math","tag-base","tag-easy","tag-math","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8385","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=8385"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8385\/revisions"}],"predecessor-version":[{"id":8387,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8385\/revisions\/8387"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8385"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8385"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8385"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}