您的目标是确定数字是否可被3整除而不使用条件句。输入将是一个从0到255的无符号8位数字。鼓励创新！

您只能使用

• 平等/不等式（==，!=，>，<，>=，<=）

• 算术（+，-，x）

• 逻辑运算符（!非，&&和，|| 或）

• 位运算符（~未，&和，|或，^异或，<<，>>，>>>算术和逻辑左和右移位）

• 常量（如果保持较小的值会更好）

• 变量分配

0如果为假，1则输出，如果为真。

适用标准原子代码高尔夫球规则。如有任何疑问，请在评论中保留。示例方法在这里。令牌是以上任何一种，但不包括常量和变量。

C-2个代币

int div3(int x) {
    return x * 0xAAAAAAAB <= x;
}

似乎可以达到2 ³¹ -1。

归功于zalgo("nhahtdh")乘法逆思想。

Python，3 2个令牌

蛮力解决方案，但它可行。

0x9249249249249249249249249249249249249249249249249249249249249249>>x&1

感谢Howard减少了1个代币。

C- 5 4（？）令牌

int div3_m2(uint32_t n) {
    return n == 3 * (n * 0xAAAAAAABull >> 33);
}

适用于任何无符号的32位数字。

该代码利用除数的乘法逆模2 ³²将除法运算转换为乘法运算。

编辑

我的解决方案（发布2分钟后）与aittsu的解决方案具有相同的精神。归功于他的使用==，我的解决方案提高了1个令牌。

参考

• 部门劳动-ridiculousfish.com
• 黑客的喜悦-魔术数
• 维基百科-模乘逆

C-15（？）令牌

int div3_m1(unsigned int n) {
    n = (n & 0xf) + (n >> 4);
    n = (n & 0x3) + (n >> 2);
    n = (n & 0x3) + (n >> 2);
    return n == 0 || n == 3;
}

由于4≡1（MOD 3），我们有4 ^Ñ ≡1（模3）。数字求和规则不仅限于对数字求和，还允许我们在保持全等的同时将数字任意分解为数字序列并求和所有数字。

以10为底的示例，除数= 9：

1234≡12 + 34≡1 + 2 + 3 + 4≡123 + 4≡1（mod 9）

程序中的所有语句都使用此属性。实际上，可以将其简化为运行语句n = (n & 0x3) + (n >> 2);直到的循环n < 4，因为该语句只是将base-4中的数字在最低有效位处中断，然后将2个部分加起来。

Python (2 tokens?)

1&66166908135609254527754848576393090201868562666080322308261476575950359794249L>>x

Or

1&0x9249249249249249249249249249249249249249249249249249249249249249L>>x

Or

1&0b1001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001001>>x

JavaScript - 3 tokens

function div3(n) {
    var a = n * 0.3333333333333333;
    return (a | 0) == a;
}

This abuses the fact that using bitwise operators on a number truncates it to an integer in JavaScript.

C - 4 tokens

int div3(int x) {
    return ((x * 43) >> 7) * 3 == x;
}

Works up to 383.

Previous version (bigger constants):

int div3(int x) {
    return ((x * 171) >> 9) * 3 == x;
}

Works up to 1535

bash – ???

Not sure how to score this.

seq 0 85 | awk '{print $1 * 3}' | grep -w [number] | wc -l

e.g.

$ seq 0 85 | awk '{print $1 * 3}' | grep -w 11 | wc -l
0

$ seq 0 85 | awk '{print $1 * 3}' | grep -w 12 | wc -l
1

$seq 0 85 | awk '{print $1 * 3}' | grep -w 254 | wc -l
0

$seq 0 85 | awk '{print $1 * 3}' | grep -w 255 | wc -l
1

Befunge 93 - 5 tokens

Fixed - division removed.

v      @._1.@
         \   
         0   
         +   
         3   
>&>3-:0\`|   
  ^      <   

Gets input, keeps subtracting 3 until it's smaller than 0, direct the pointer up ('|'), then adds 3. If the value is 0 then the pointer moves right ("1.@" outputs '1') else moves left ("@." outputs '0'). '@' terminates the program.

Batch - 7 Tokens

I think

@echo off
for /L %%a in (0,3,%1) do set a=%%a
if %a%==%1 echo 1

Returns 1 if the given number (as stdin) is divisible by three.

Ruby, 6(?) tokens

I'm really not sure how to count tokens. OP, can you score me?

I think it's 6... 1, 0, 0, *, 255, x

Note that the * is not integer multiplication.

def div3(x)
  ([1,0,0]*255)[x]
end

Python 0

I posted eariler but I used conditionals. Here's to using no conditionals and no tokens, just keywords

def g(x): return ([[lambda : g(sum(int(y) for y in list(str(x)))),lambda: 0][[False,True].index(x in[0,1,2,4,5,7,8])], lambda: 1][[False,True].index((lambda y: y in[3,6,9])(x))])()

uses the trick that multiple of 3s have digits which add to 3

Edit: Removed unnecessary lambda

def g(x):return([[lambda: g(sum(int(y) for y in list(str(x)))),lambda:0][[False,True].index(x in[0,1,2,4,5,7,8])], lambda:1][[False,True].index(x in[3,6,9])])()

Edit: Golfed further (117 chars) still no tokens

exec"g=`x:(((`:g(sum(int(y)for y in str(x)),`:0)[x in[0,1,2,4,5,7,8]],`:1)[x in[3,6,9]])()".replace('`','lambda ')

Killed direct access for python's nifty getitem Longer at 132 char

exec"g={0}x:((({0}:g(sum(int(y)for y in str(x))),{0}:0{1}0,1,2,4,5,7,8]),{0}:1{1}3,6,9]))()".format('lambda ',').__getitem__(x in[')

http://www.codeskulptor.org/#user34_uUl7SwOBJb_0.py

Python - 25 tokens

To get things started, I have a lengthy solution that is a implementation of one of the answers in the link in my first post. n is input.

a = (n>>7)-((n&64)>>6)+((n&32)>>5)-((n&16)>>4)+((n&8)>>3)-((n&4)>>2)+((n&2)>>1)-(n&1)
print(a==0 or a==3)

or is equivalent to ||.

JavaScript - 3 Tokens

Test it on your browser's console:

a = prompt().split('');
sum = 0;

do {
  sum = a.reduce(function(p, c) {
     return parseInt(p) + parseInt(c); 
  });

  a = sum.toString().split('');

} while(a.length > 1)

alert([3, 6, 9].indexOf(+sum) > -1)

JavaScript
_{not sure about the token#}

function mod3 (i) { return {'undefined':'100','0':'0'}[[0][i]][i.toString (3).split('').pop ()]}

or if the output for 0 is allowed to be 1;

function mod3 (i) { return '100'[i.toString (3).split('').pop ()]}

Tcl, 83 bytes

proc T n {while \$n>9 {set n [expr [join [split $n ""] +]]};expr {$n in {0 3 6 9}}}

Try it online!