您的老板希望您编写如下代码：

public static boolean isPowerOfTen(long input) {
  return
    input == 1L
  || input == 10L
  || input == 100L
  || input == 1000L
  || input == 10000L
  || input == 100000L
  || input == 1000000L
  || input == 10000000L
  || input == 100000000L
  || input == 1000000000L
  || input == 10000000000L
  || input == 100000000000L
  || input == 1000000000000L
  || input == 10000000000000L
  || input == 100000000000000L
  || input == 1000000000000000L
  || input == 10000000000000000L
  || input == 100000000000000000L
  || input == 1000000000000000000L;
}

（马丁·史密斯（Martin Smith），https：//codereview.stackexchange.com/a/117294/61929）

这很有效，但键入起来却没有那么有趣。由于要尽量减少按键操作的次数，因此编写了一个较短的程序或函数（或方法），可以为您输出此函数（或返回要输出的字符串）。而且，由于您拥有自己的自定义全范围unicode键盘，其中包含所有unicode 8.0所需的全部120,737个按键，因此我们将计算unicode字符，而不是按键。如果您的语言不使用unicode源代码，则为字节或字节。

程序或函数的任何输入都将计入分数，因为显然您也必须输入该分数。

澄清和修改：

• 最后一个之后删除了3个尾随空格 }
• 之后删除了一个尾随空格 return
• 从函数/方法返回输出字符串是可以的

PostgreSQL，158个字符

select'public static boolean isPowerOfTen(long input) {
  return
   '||string_agg(' input == 1'||repeat('0',x)||'L','
  ||')||';
}'from generate_series(0,18)x

Vim 97按键

ipublic static boolean isPowerOfTen(long input) {
  return
  || input == 1L<esc>qyYpfLi0<esc>q16@yo}<esc>3Gxx

好吧，今天我与vim制作Java进行了交流，所以为什么不继续这种趋势呢！

05AB1E，99 97 96 94 93 87字节

码：

“‚Æ£‹ÒŒ€ˆPowerOfTen(“?“¢„î®) {
 «‡
   “?19FN0›i"  ||"?}’ î® == ’?N°?'L?N18Qi';,"}"?}"",

在线尝试！

使用CP-1252编码。

CJam，52个字符

YA#_("𐀑򀺸󆚜񸎟񜏓񞍁򛟯󩥰󾐚򉴍􍼯𹾚򶗜򳙯󭧐񹷜񊽅𸏘򴂃򦗩󧥮𤠐𰑈򶂤𘏧󔆧򇃫󡀽򊠑񊩭򯐙񛌲񊚩𤱶𻺢"f&bY7#b:c~

在线尝试！

阶段1

使用Unicode字符U + 10000到U + 10FFFF，我们可以在单个字符中编码20位。CJam内部使用16位字符，因此每个字符将被编码为一对代理，一个在U + D800到U + DBFF的范围内，然后在一个从U + DC00到U + DFFF的范围内。

通过将每个代理的位与1023相加，我们获得其编码的10位信息。我们可以将结果数组从1024转换为128，以将BMP之外的任意Unicode字符字符串解码为ASCII字符串。

该代码执行以下操作：

YA#    e# Push 1024 as 2 ** 10.
_(     e# Copy and decrement to push 1023.

"𑅰󻢶񹱨񉽌񍍎񄆋򎿙򧃮񑩹󠷽􂼩􉪦񭲣񶿝򭁩󭰺􄔨񍢤𘎖񮧗򦹀𹀠񐢑񜅈𠟏򘍎󾇗򲁺􅀢򅌛񎠲򦙤򃅒𹣬񧵀򑀢"

f&     e# Apply bitwise AND with 1023 to each surrogate character.
b      e# Convert the string from base 1024 to integer.
Y7#    e# Push 128 as 2 ** 7.
b      e# Convert the integer to base 128.
:c     e# Cast each base-128 to an ASCII character.
~      e# Evaluate the resulting string.

第二阶段

从上面进行解码的过程将产生以下源代码（98字节）。

"public static boolean isPowerOfTen(long input) {
  return
   ""L
  || input == ":S6>AJ,f#S*"L;
}"

在线尝试！

该代码执行以下操作：

e# Push the following string.

"public static boolean isPowerOfTen(long input) {
  return
   "

e# Push the following string and save it in S.

"L
  || input == ":S

e# Discard the first 6 characters of S. The new string begins with " input".

6>

e# Elevate 10 (A) to each exponent below 19 (J).

AJ,f#

e# Join the resulting array, using the string L as separator.

S*

e# Push the following string.

"L;
}"

Java中，217个 215 220 219 192字节

打高尔夫球：

public static String b(){String s="public static boolean isPowerOfTen(long input) {\n  return\n    input == 1L",z="";for(int i=0;i++<18;){z+="0";s+="\n  || input == 1"+z+"L";}return s+";\n}";}

取消高尔夫：

  public static String a(){
    String s = "public static boolean isPowerOfTen(long input) {\n  return\n    input == 1L", z="";
    for (int i=0; i++ < 18;) {
        z += "0";
        s += "\n  || input == 1"+z+"L";
    }
    return s + ";\n}";
  }

（第一个答案，芜湖）

谢谢！
-2个字节：user902383
-1个字节：Denham Coote

变化：

• 用制表符代替空格
• 错过了输出的最后一行：18-> 19
• 删除内部循环
• 从打印更改为返回字符串

Pyth，118个 106 103字节

s[."
{Z-L¡JxÙÿ
LæÝ<­í?¢µb'¥ÜA«Ç}h¹äÚÏß"\nb*4dj"\n  || "ms[." uøs|ÀiÝ"*d\0\L)U19\;b\}

在线尝试！

所有这些字符串硬编码确实占用了很多字节，但是对此我无能为力。

更新：使用打包的字符串节省了3个字节。感谢@ user81655的提示！

C# (CSI) 181 180 179 byte

string i=" input == 1",e="public static bool";Console.Write(e+@"ean isPowerOfTen(long input) {
  return
   "+i+string.Join(@"L
  ||"+i,e.Select((_,x)=>new string('0',x)))+@"L;
}")

There is only one little trick involved. The straight forward way to write this would be:

string.Join("L\n  || input == 1",Enumerable.Range(0,18).Select(x=>new string('0',x)))

by using the string with the first 18 characters of the text which I need anyways I can get rid off the lengthy Enumerable.Range. This works because string implements IEnumerable and there is a version of Select that hands the item (not needed) and the index which we want to the lambda function.

PowerShell, 120 bytes

'public static boolean isPowerOfTen(long input) {'
'  return'
"   $((0..18|%{" input == 1"+"0"*$_})-join"L`n  ||")L;`n}"

The first two lines are simply string literals, which are output as-is.

The third line starts with three spaces, and ends with L;`n}" to finish off the last couple bytes. The middle bit inside the script block $(...) is constructed by for-looping % from 0 to 18 and each iteration constructing a string that starts with input == 1 concatenated with the corresponding number of zeros. This will spit out an array of strings. We then -join each element of the array with L`n || to achieve the newline-pipes. That big string is the output of the script block, which gets inserted automatically into the middle and output.

PS C:\Tools\Scripts\golfing> .\go-generate-some-java.ps1
public static boolean isPowerOfTen(long input) {
  return
    input == 1L
  || input == 10L
  || input == 100L
  || input == 1000L
  || input == 10000L
  || input == 100000L
  || input == 1000000L
  || input == 10000000L
  || input == 100000000L
  || input == 1000000000L
  || input == 10000000000L
  || input == 100000000000L
  || input == 1000000000000L
  || input == 10000000000000L
  || input == 100000000000000L
  || input == 1000000000000000L
  || input == 10000000000000000L
  || input == 100000000000000000L
  || input == 1000000000000000000L;
}

Javascript, 172 157 152 150 148 bytes

p=>`public static boolean isPowerOfTen(long input) {
  return${[...Array(19)].map((x,i)=>`
  ${i?'||':' '} input == 1${'0'.repeat(i)}L`).join``};
}`

f=p=>`public static boolean isPowerOfTen(long input) {
  return${[...Array(19)].map((x,i)=>`
  ${i?'||':' '} input == 1${'0'.repeat(i)}L`).join``};
}`

document.body.innerHTML = '<pre>' + f() + '</pre>'

C, 158 155 bytes

i;main(){for(puts("public static boolean isPowerOfTen(long input) {\n  return");i<19;)printf("  %s input == 1%0.*dL%s\n",i++?"||":" ",i,0,i<18?"":";\n}");}

Try it online here.

Jelly, 75 bytes

(These are bytes in Jelly's custom codepage.)

0r18⁵*;@€⁶j“¢œḤḅg^NrÞḢ⁷ẉ»“⁵®UẆƓḃÐL⁴ṖịṛFþẈ¹9}¶ ƁḋȮ¦sẒẆd€Ḟɼ¿ỌṀP^µ\f@»;;“L;¶}”

Try it here.

Explanation

0r18      Range [0..18]
⁵*        Take the 10^ of each number
;@€⁶      Prepend a space to each number
j“...»    Join by compressed string "L\n  || input =="
“...»;    Prepend compressed string "public static ... =="
;“L;¶}”   Append "L;\n}"

Vimscript, 120 bytes

Might as well use the right tool for the job.

This assumes that autoindent, etc have not been set. ^[ and ^M are escape characters for the ESC and CR characters respectively.

The a macro duplicates the current line and adds a 0 to the copy. The :norm line generates all the boilerplate and the indent == 1L line, then uses a to create the others.

:let @a='yyp$i0^['
:norm ipublic static boolean isPowerOfTen(long input) {^M  return^M  || input == 1L^[18@a$a;^M}
:3s/||/ /

In case the trailing spaces the sample output had on two lines weren't typos, here's a 126 byte version that includes them.

:let @a='yyp/L^Mi0^['
:norm ipublic static boolean isPowerOfTen(long input) {^M  return ^M  || input == 1L^[18@a$a;^M}   
:3s/||/ /

Oracle SQL 9.2, 311 bytes

SELECT REPLACE(REPLACE('public static boolean isPowerOfTen(long input) {'||CHR(10)||'  return'||c||';'||'}', 'n  ||', 'n'||CHR(10)||'   '),CHR(10)||';', ';'||CHR(10)) FROM(SELECT LEVEL l,SYS_CONNECT_BY_PATH('input == '||TO_CHAR(POWER(10,LEVEL-1))||'L'||CHR(10),'  || ')c FROM DUAL CONNECT BY LEVEL<20)WHERE l=19

Perl 5 - 130 141

@s=map{'input == 1'.0 x$_."L\n  ||"}0..18;$s[$#s]=~s/\n  \|\|/;\n}/g;print"public static boolean isPowerOfTen(long input){\n  return\n    @s"

EDIT: fixed to have exact indentation

ES6, 139 bytes

_=>"0".repeat(19).replace(/./g,`
 || input == 1$\`L`).replace(`
 ||`,`public static boolean isPowerOfTen(long input) {
  return\n  `)+`;
}`

I do so love these triangle generation questions.

Kotlin, 194 193 characters

fun main(u:Array<String>){var o="public static boolean isPowerOfTen(long input) {\n\treturn"
var p:Long=1
for(k in 0..18){
o+="\n\t"
if(k>0)o+="||"
o+=" input == ${p}L"
p*=10
}
print("$o;\n}")}

Test it at http://try.kotlinlang.org/

Ruby, 125 119 bytes

$><<'public static boolean isPowerOfTen(long input) {
  return
   '+(0..19).map{|i|" input == #{10**i}L"}*'
  ||'+';
}'

Thanks to manatwork for -6 bytes!

jq, 123 characters

(121 characters code + 2 characters command line option.)

"public static boolean isPowerOfTen(long input) {
  return
   \([range(19)|" input == 1\("0"*.//"")L"]|join("
  ||"));
}"

Sample run:

bash-4.3$ jq -nr '"public static boolean isPowerOfTen(long input) {
>   return
>    \([range(19)|" input == 1\("0"*.//"")L"]|join("
>   ||"));
> }"'
public static boolean isPowerOfTen(long input) {
  return
    input == 1L
  || input == 10L
  || input == 100L
  || input == 1000L
  || input == 10000L
  || input == 100000L
  || input == 1000000L
  || input == 10000000L
  || input == 100000000L
  || input == 1000000000L
  || input == 10000000000L
  || input == 100000000000L
  || input == 1000000000000L
  || input == 10000000000000L
  || input == 100000000000000L
  || input == 1000000000000000L
  || input == 10000000000000000L
  || input == 100000000000000000L
  || input == 1000000000000000000L;
}

On-line test (Passing -r through URL is not supported – check Raw Output yourself.)

Javascript 175 bytes

Let's do this regularly

var s = "public static boolean isPowerOfTen(long input) {\n\treturn\n\t\tinput == 1";
for (var i = 1; i < 20; i++) {
    s += "\n\t|| input == 1";
    for (var j = 0; j < i; j++) {
        s += "0";
    }
    s += "L" ;
}
s += ";\n}";
alert(s);

Pretty small. Now, some javascript magic, like no semicolons needed, or no var's, etc.:

k="input == 1"
s="public static boolean isPowerOfTen(long input) {\n\treturn\n\t\t"+k+"L"
for(i=1;i<20;i++){s+="\n\t|| "+k
for(j=0;j<i;j++)s+="0";s+="L"}s+=";\n}"
alert(s)

Python (3.5) 137 136 bytes

print("public static boolean isPowerOfTen(long input) {\n  return\n   ",'\n  || '.join("input == %rL"%10**i for i in range(19))+";\n}")

Previous version

print("public static boolean isPowerOfTen(long input) {\n  return\n   ",'\n  || '.join("input == 1"+"0"*i+"L"for i in range(19))+";\n}")

ANSI-SQL, 252 characters

WITH t as(SELECT '   'x,1 c,1 l UNION SELECT'  ||',c*10,l+1 FROM t WHERE l<19)SELECT 'public static boolean isPowerOfTen(long input) {'UNION ALL SELECT'  return 'UNION ALL SELECT x||' input == '||c||'L'||SUBSTR(';',1,l/19)FROM t UNION ALL SELECT'}   ';

Ungolfed:

WITH t as (SELECT '   ' x,1 c,1 l UNION
           SELECT '  ||',c*10,l+1 FROM t WHERE l<19)
SELECT 'public static boolean isPowerOfTen(long input) {' UNION ALL
SELECT '  return ' UNION ALL
SELECT x||' input == '||c||'L'||SUBSTR(';',1,l/19) FROM t UNION ALL
SELECT '}   ';

Not a serious attempt, just poking at the Oracle SQL/T-SQL entries.

JavaScript (Node.js), 156 bytes

s="public static boolean isPowerOfTen(long input) {\n  return "
for(i=1;i<1e19;i*=10)s+="\n  "+(i-1?"||":" ")+" input == "+i+"L"
console.log(s+";\n}   \n")

The i-1 will only be 0 (and thus falsey) on the very first round (it's just slightly shorter than i!=1.

Suggestions welcome!

Perl 5, 137 bytes

Not based on the previous Perl answer, but it is somehow shorter. I believe it can be shortened down again by taking care of the first "input" inside the loop, but I didn't try anything yet (at work atm)

$i="input";for(1..18){$b.="  || $i == 1"."0"x$_."L;\n"}print"public static boolean isPowerOfTen(long $i) {\n  return\n    $i == 1L;\n$b}"

CJam, 112 chars

"public static boolean isPowerOfTen(long input) {
  return
    input == 1"19,"0"a19*.*"L
  || input == 1"*"L;
}"

AWK+shell, 157 bytes

echo 18|awk '{s="input == 1";printf"public static boolean isPowerOfTen(long input) {\n return\n    "s"L";for(;I<$1;I++)printf"\n  ||%sL",s=s"0";print";\n}"}'

The question did say to count everything you would have to type. This does have the added bonus of being able to select how many lines would be placed in the isPowersOfTen method when the boss inevitably changes his mind.

T-SQL 289, 277, 250, 249 bytes

SELECT'public static boolean isPowerOfTen(long input){return '+STUFF((SELECT'||input=='+N+'L 'FROM(SELECT TOP 19 FORMAT(POWER(10.0,ROW_NUMBER()OVER(ORDER BY id)),'F0')N FROM syscolumns)A FOR XML PATH(''),TYPE).value('.','VARCHAR(MAX)'),1,2,'')+';}'

Update: Thanks @Bridge, I found a few more spaces too :)

Update2: Changed CTE to subquery -27 chars :) Update3: Another space bites the dust @bridge :)

R, 185 bytes

Golfed

options(scipen=999);p=paste;cat(p("public static boolean isPowerOfTen(long input) {"," return",p(sapply(0:19,function(x)p(" input == ",10^x,"L",sep="")),collapse="\n ||"),"}",sep="\n"))

Ungolfed

options(scipen=999)
p=paste
cat(
  p("public static boolean isPowerOfTen(long input) {",
        " return",
        p(sapply(0:19,function(x)p(" input == ",10^x,"L",sep="")),collapse="\n ||"),
        "}",
        sep="\n")
)

Perl 6 (115 bytes)

say "public static boolean isPowerOfTen(long input) \{
  return
   {join "L
  ||",(" input == "X~(10 X**^19))}L;
}"

X operator does list cartesian product operation, for example 10 X** ^19 gives powers of ten (from 10 to the power of 0 to 19, as ^ is a range operator that counts from 0). Strings can have code blocks with { (which is why I escape the first instance of it).

Java, 210 / 166

Score is depending on whether returning the input from a function meets the definition of 'output'.

Console output (210):

class A{public static void main(String[]z){String a=" input == 1",t="L\n  ||"+a,s="public static boolean isPowerOfTen(long input) {\n  return\n   "+a;for(int i=0;++i<19;)s+=t+="0";System.out.print(s+"L;\n}");}}

String return (166):

String a(){String a=" input == 1",t="L\n  ||"+a,s="public static boolean isPowerOfTen(long input) {\n  return\n   "+a;for(int i=0;++i<19;)s+=t+="0";return s+"L;\n}";}

Legible version:

String a() {
    String a=" input == 1", t = "L\n  ||"+a,
        s = "public static boolean isPowerOfTen(long input) {\n  return\n   "+a;
    for (int i = 0; ++i < 19;)
        s += t += "0";
    return s + "L;\n}";
}

Batch, 230 208 206 205 bytes

@echo off
echo public static boolean isPowerOfTen(long input) {
echo   return
set m=input == 1
echo    %m%L
for /l %%a in (1,1,17)do call:a
call:a ;
echo }
exit/b
:a
set m=%m%0
echo  ^|^| %m%L%1

Edit: Saved 22 bytes by avoiding repeating input == and reusing the subroutine for the line with the extra semicolon. Saved 2 3 bytes by removing unnecessary spaces.