“数字总和”是指一个数字中所有数字的总和。

例如，的数字总和1324为10，因为1+3+2+4 = 10。

面临的挑战是编写一个程序/函数来计算比数字总和为输入的输入大的最小数字。

演练示例

例如，以数字9作为输入：

9 = 1+8 -> 18
9 = 2+7 -> 27
9 = 3+6 -> 36
...
9 = 8+1 -> 81
9 = 9+0 -> 90

有效输出将是上面的最小数字，即18。

眼镜

请注意，这9不是此示例的有效输出，因为反转的数字必须大于原始数字。

请注意，输入将为正。

测试用例：

 2 => 11      (2 = 1 + 1)
 8 => 17      (8 = 1 + 7)
12 => 39     (12 = 3 + 9)
16 => 79     (16 = 7 + 9)
18 => 99     (18 = 9 + 9)
24 => 699    (24 = 6 + 9 + 9)
32 => 5999   (32 = 5 + 9 + 9 + 9)

参考文献：

这是OEIS A161561。

编辑：添加了一个额外的测试用例（18）

感谢Martin Ender提供的排行榜摘要

var QUESTION_ID=81047,OVERRIDE_USER=31373;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;

body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

05AB1E，19 17 8字节

码：

[>DSO¹Q#

解释：

[            # start infinite loop
 >           # increase loop variable, will initially be input
  DSO        # make a copy and sum the digits
     ¹Q#     # if it equals the input, break
             # else return to start of loop
             # implicitly print

在线尝试

编辑：由于@Adnan，节省了9个字节

Python 2，33个字节

lambda n:[n+9,`n%9`+n/9*'9'][n>9]

直接表达。生成一个数字字符串，该数字字符串的末尾有9，其余部分在开始。除了，对于一位数n，给出n+9。

一些输出有前导零（099对18）。

视网膜，39 31字节

r`1{1,9}
$.&
T`d`_d`^.$
^.$
1$&

接受一元输入。

在线尝试！（前两行允许一次运行多个测试用例，并为方便起见从十进制转换为一进制。）

这实际上不是线性搜索结果，而是显式计算结果：

• 如果输入n大于9，我们将其替换为（n % 9后跟n / 9）（加水的）9。
• 否则，我们将其替换为n + 9。

使用!（或其他任何非1）作为一进制数字，我可以使用以下方法再保存一个字节：

^!(?!!{9})
1
r`!{0,9}
$.&
0\B

但是，我认为这种输入格式有点麻烦。

Pyth，8个字节

fqQsjT;h

测试套件。

fqQsjT;h

f      h first number T from (input+1) onward where:
 qQ          the input is equal to
   s         the sum of
    jT;      the base-10 representation of T

Java 7，68 61字节

int f(int n){return n>9?-~(n%9)*(int)Math.pow(10,n/9)-1:n+9;}

与这里的许多其他答案大致相同。想要展示Java方法而不使用字符串操作和循环。

感谢FryAmTheEggman提醒我我很傻;）

MATL，10 9字节

`QtV!UsG-

在线尝试！

说明

`        % Do...while
  Q      %   Add 1. Takes input implicitly the first time
  t      %   Duplicate
  V!Us   %   To string, transpose, to number, sum. Gives sum of digits
  G-     %   Subtract input. If 0, the loop ends and the stack is implicitly displayed

JavaScript（ES7），32个字节

n=>(n%9+1)*10**(n/9|0)-(n>9||-8)

38字节作为ES6：

n=>parseFloat(n%9+1+'e'+n/9)-(n>9||-8)

Python 3，128 94 84 74字节

没有输出，直接处理，初学者;）

def r(n):
 m=n
 while 1:
  m+=1
  if sum(map(int,str(m)))==n:return(m)

其实是17个位元组

╗1`;$♂≈Σ╜;)=)>*`╓

在线尝试！

说明：

╗1`;$♂≈Σ╜;)=)>*`╓
╗                  save input to reg0
 1`;$♂≈Σ╜;)=)>*`╓  first integer n (>= 0) where:
   ;$♂≈Σ╜;)=         the base-10 digital sum equals the input and
            )>*      is greater than the input

C 73 65字节

具有辅助功能的宏。

e(y){return y?10*e(y-1):1;}
#define F(n) n<9?n+9:(1+n%9)*e(n/9)-1

该e函数仅计算10的幂，并且F宏使用与此ruby和此python答案相同的求解方法。可悲的是，它的长度大于这两个答案的总和。但这是第一个C答案。

（Lynn的删除技巧节省了8个字节int。）

Brachylog，8个字节（不竞争）

<.=:ef+?

= 在发布此挑战后进行了修改，因此它现在可以在可能无限的域上工作，在这种情况下就是这样。

说明

<.       Output > Input
  =      Label the Output (i.e. unify it with an integer)
   :ef   Get the list of digits of the Output
      +? Input is the sum of all those digits

这将=一直追溯到Output的值使整个谓词为true为止。

TSQL（sqlserver 2012），107 99字节

DECLARE @ INT = 32

,@2 char(99)WHILE @>0SELECT
@2=concat(x,@2),@-=x FROM(SELECT IIF(@>9,9,IIF(@2>0,@,@-1))x)y PRINT @2

在线尝试！

Python 2，39个字节

lambda n:[n+9,(1+n%9)*10**(n/9)-1][n>9]

纯整数运算。

带输出的完整程序

f=lambda n:[n+9,(1+n%9)*10**(n/9)-1][n>9]

print(f(2))
print(f(8))
print(f(12))
print(f(16))
print(f(17))
print(f(18))
print(f(24))
print(f(32))

输出：

11
17
39
79
89
99
699
5999

PowerShell v2+, 62 bytes

param($n)for($a=$n+1;([char[]]"$a"-join'+'|iex)-ne$n;$a++){}$a

Takes input $n then executes a for loop. We initialize the loop by setting our target number, $a, to be one larger than $n (since it has to be bigger, plus this ensures 1..9 work correctly). Each loop we increment $a++. Nothing happens in the loop proper, but the conditional is where the program logic happens. We're literally taking the target number as a string, casting it as a char-array, -joining the array with + and then piping it to iex (similar to eval). We test whether that's equal to our input number or not, and continue looping accordingly. Once we've exited the loop, we've reached where our target number is digit-sum equal to our input number, so $a is placed on the pipeline and output is implicit.

For reference, here's the "construct a string with the appropriate number of 9's" method that other folks have done, at 67 bytes

param($n)(($n+9),(+(""+($n%9)+'9'*(($n/9)-replace'\..*'))))[$n-gt9]

or the "pure integer arithmetic" method that other folks have done, at 70 bytes

param($n)(($n+9),("(1+$n%9)*1e$(($n/9)-replace'\..*')-1"|iex))[$n-gt9]

Neither of which are shorter, but both of which are more interesting.

Pyke, 9 8 7 bytes, non-competing - uses more recent version

.fhsq)h

Try it here!

.f      - first where true:
  h     - n+1
   sq    - digital_root(^) == input()
      h - that number +1

JavaScript (ES2015), 45 39 33 bytes

Saved another 6 bytes thanks to @Conor O'Brien and @Shaun H.
I think I'll leave it as is, because this version differs from @Neil's answer by using String.repeat().

v=>+(v>9?v%9+'9'.repeat(v/9):v+9)

Previous version (saved 6 bytes thanks to @Qwertiy):

f=v=>+(v/9>1?v%9+'9'.repeat(v/9|0):v+9)

First version:

f=v=>+(v/9>1?v%9+'9'.repeat(~~(v/9)):'1'+v-1)

Lua, 52 bytes

n=...+0
print(n>9 and(n%9)..string.rep(9,n/9)or n+9)

Meant to be saved in a file and run with the Lua interpreter, e.g. lua <file> <input number>

You can also try it here: https://repl.it/CXom/1

(On repl.it the input number is hard-coded for ease of testing)

Racket 70 characters, 71 bytes

Same algorithm as most of the others pretty much. Pretty sad about not having % for modulo, or ** for expt, or integer division by default, otherwise this could be a lot shorter and I could've outgolfed C and Java. Still love the language though

(λ(x)(if(> x 9)(-(*(+(modulo x 9)1)(expt 10(floor(/ x 9))))1)(+ x 9)))

Hexagony, 40 31 30 bytes

<_:->.(.+><.'!.\@"9!%>!/{.}|.?

Or, if you prefer your code to be a little less linear and a little more polygonal:

    < _ : -
   > . ( . +
  > < . ' ! .
 \ @ " 9 ! % >
  ! / { . } |
   . ? . . .
    . . . .

Try it online!

Thanks to @FryAmTheEggman for some ideas and inspiration :o)

Previous version: <.:->+_.!(..'!.\><9!%>@.{.}|.?"

Previouser version: <><.-_|@"'!{|(.9+!8=@>{/".'/:!?$.%\1$..\

Perl 6,  38  29 bytes

{$_>9??(1+$_%9)*10**Int($_/9)-1!!$_+9}
{first *.comb.sum==$_,$_^..*}

_{^{( apparently the direct approach is shorter )}}

Explanation:

{
  first
    *.comb.sum == $_, # lambda that does the check
    $_ ^.. *          # Range.new: $_, Inf, :excludes-min
}

Test:

#! /usr/bin/env perl6

use v6.c;
use Test;

my @tests = (
   2 => 11,
   8 => 17,
   9 => 18,
  12 => 39,
  16 => 79,
  18 => 99,
  24 => 699,
  32 => 5999,
);

plan +@tests;

my &next-digital-sum = {first *.comb.sum==$_,$_^..*}

for @tests -> $_ ( :key($input), :value($expected) ) {
  is next-digital-sum($input), $expected, .gist;
}

1..8
ok 1 - 2 => 11
ok 2 - 8 => 17
ok 3 - 9 => 18
ok 4 - 12 => 39
ok 5 - 16 => 79
ok 6 - 18 => 99
ok 7 - 24 => 699
ok 8 - 32 => 5999

Java 10, 114 62 bytes

n->{var r="";for(int i=0;i++<n/9;r+=9);return(n>9?n%9:n+9)+r;}

Try it online.

EDIT: 130 73 bytes without leading zeros (Thanks to @levanth`):

n->{var r="";for(int i=0;i++<n/9;r+=9);return new Long((n>9?n%9:n+9)+r);}

Try it online.

Explanation:

n->                           // Method with integer parameter and long return-type
  var r="";                   //  Result-String, starting empty
  for(int i=0;i++<n/9;r+=9);  //  Append `n` integer-divided by 9 amount of 9's to `r`
  return new Long(            //  Cast String to number to remove any leading zeroes:
    (n>9?                     //   If the input `n` is 10 or larger
      n%9                     //    Use `n` modulo-9
     :                        //   Else (`n` is smaller than 10):
      n+9)                    //    Use `n+9`
    +r);}                     //   And append `r`

Ruby, 33 bytes

This is a int arithmetic version that just happens to be the same as xnor's python answer. It is an anonymous function that takes and returns an int.

->n{n<10?n+9:(1+n%9)*10**(n/9)-1}

MathGolf, 8 7 bytes

Æ)_Σk=▼

Try it online!

Implicit input yay.

Explanation:

          Implicit input
 Æ     ▼  Do while false loop that pops the condition
  )       Increment top of stack
   _      Duplicate
    Σ     Get the digit sum
     k    Get input
      =   Is equal?
          Implicit output

Ruby, 38 bytes

f=->n{n<11?n+9:"#{n<19?n-9:f.(n-9)}9"}

This answer returns a string or int depending on input size. It is a recursive solution that asks for a solution for 9 less then adds a "9" to the end.

Ruby, 39 bytes

f=->n{n<11?n+9:(n<19?n-9:f.(n-9))*10+9}

For one more byte, this answer always returns an int. same algorithm as above but with numbers.

C, 80 bytes

i;s;g(j){s=0;for(;j;j/=10)s+=j%10;return s;}f(n){i=n;while(n-g(i))i++;return i;}

Ungolfed try online

int g(int j)
{
    int s=0;
    for(;j;j/=10) s += j%10;
    return s;
}

int f(int n)
{
    int i=n;
    for(;n-g(i);i++);
    return i;
}

PHP, 77 characters

$n=$argv[1];$m=$n+1;while(1){if(array_sum(str_split($m))==$n)die("$m");$m++;}

Oracle SQL 11.2, 165 bytes

SELECT l FROM(SELECT LEVEL l,TO_NUMBER(XMLQUERY(REGEXP_REPLACE(LEVEL,'(\d)','+\1')RETURNING CONTENT)) s FROM DUAL CONNECT BY 1=1)WHERE s=:1 AND l!=s AND rownum=1;

Un-golfed

SELECT l   
FROM   (
         SELECT LEVEL l, -- Number to evaluate
                XMLQUERY( 
                          REGEXP_REPLACE(LEVEL,'(\d)','+\1')  -- Add a + in front of each digit 
                          RETURNING CONTENT
                        ).GETNUMBERVAL()s                     -- Evaluate le expression generated by the added + 
                FROM DUAL 
                CONNECT BY 1=1 -- Run forever            
       )
WHERE s=:1      -- The sum must be equal to the input
  AND l!=s      -- The sum must not be the input 
  AND rownum=1  -- Keep only the first result

Python 3 55 Bytes

Takes two arguments that are the same

f=lambda a,b:a if sum(map(int,str(a)))==b else f(-~a,b)

i.e. to call it you would use f(x,x)

Japt, 9 bytes

@¶Xìx}aUÄ

Try it

Powershell, 54 bytes

$args|%{(($_+9),+(''+$_%9+'9'*(($_-$_%9)/9)))[$_-gt9]}

Test script:

$f = {

$args|%{(($_+9),+(''+$_%9+'9'*(($_-$_%9)/9)))[$_-gt9]}

}

@(
    ,(  1,  10   )
    ,(  2 , 11   )
    ,(  8 , 17   )
    ,(  9 , 18   )
    ,( 10,  19   )
    ,( 11,  29   )
    ,( 12 , 39   )
    ,( 16 , 79   )
    ,( 18 , 99   )
    ,( 19 , 199  )
    ,( 24 , 699  )
    ,( 27 , 999  )
    ,( 32 , 5999 )
    ,( 52 , 799999 )
    ,( 128, 299999999999999 )
) | % {
    $a,$e = $_
    $r = &$f $a
    "$($r-eq$e): $r"
}

Output:

True: 1 : 10
True: 2 : 11
True: 8 : 17
True: 9 : 18
True: 10 : 19
True: 11 : 29
True: 12 : 39
True: 16 : 79
True: 18 : 99
True: 19 : 199
True: 24 : 699
True: 27 : 999
True: 32 : 5999
True: 52 : 799999
True: 128 : 299999999999999

Expalantion

• if [$_-gt9] returns a first digit and a tail
  • a first digit is the difference between $_ and the sum of 9 ($_%9)
  • a tail is a few nines - '9'*(($_-$_%9)/9))
  • finally, converts a result string into a number to remove the leading 0
• else returns ($_+9)