钟针Syzygy

16

（非常感谢El'endia Starman和Sp3000帮助我为此设计了测试用例！）

给定一个正整数n和一个多个时钟指针（以秒为单位）的正整数旋转周期的列表，输出最小的正整数x，其中x在启动时钟后的秒数中，所有n指针都对齐，并且指针的确切位置对齐。它们不必在起始位置对齐-任何位置都可以，只要x是整数且最小即可。此外，并非所有的手都必须在同一位置对齐-对于n=4，将2组2只手对齐的解决方案是有效的。组的大小必须为2或更大-两个不对齐的手不会构成2个1对齐的手的组，因此不是有效的解决方案。

您可能会假设仅n给出在整数秒后可能会完全对齐的2, [3,3,3]输入- 不是有效的输入，因为在任何秒数后，所有三只手都将对齐，因此这是不可能的精确地对齐2。

例子：

2, [3,4] -> 12
(the only option is a multiple of 12, so we pick 12 - 4 and 3 full rotations, respectively)

3, [3,5,6,9,29] -> 18
(picking 3, 6, and 9, the hands would align after 6, 3, and 2 rotations, respectively)

2, [1,1,4,5,10] -> 1
(picking 1 and 1 - note that 0 is not a valid answer because it is not a positive integer)

3, [2,2,6,7,11] -> 3
(picking 2, 2, and 6 - the 2s would be halfway through their second revolution, and the 6 would be halfway through its first revolution)

2, [2,7,5,3,3] -> 1
(picking 3 and 3, they are always aligned, so 1 is the minimum)

5, [4, 14, 36, 50, 63, 180, 210] -> 45
(after 45 seconds, the first, third, and sixth are aligned, as well as the second and seventh, for a total of 5)

测试数据：

7, [10, 22, 7, 6, 12, 21, 19] -> 87780
6, [25, 6, 2, 19, 11, 12] -> 62700
6, [23, 1, 8, 10, 9, 25] -> 41400
7, [6, 4, 1, 8, 10, 24, 23] -> 920
3, [18, 5, 23, 20, 21] -> 180
5, [10, 8, 14, 17, 5, 9] -> 2520
6, [1, 18, 12, 9, 8, 10, 23] -> 360
6, [12, 11, 6, 23, 25, 18, 13] -> 118404
4, [18, 11, 2, 9, 12, 8, 3] -> 8
7, [18, 25, 9, 13, 3, 5, 20] -> 11700
2, [17, 20, 15, 8, 23, 3] -> 15
3, [16, 3, 24, 13, 15, 2] -> 24
5, [7, 23, 24, 8, 21] -> 1932
6, [16, 10, 12, 24, 18, 2, 21] -> 720
6, [1, 17, 16, 13, 19, 4, 15] -> 53040
2, [3, 4, 20] -> 5
3, [9, 4, 16, 14, 1, 21] -> 16
5, [5, 17, 10, 20, 12, 11] -> 330
2, [21, 5, 22, 18] -> 90
4, [7, 25, 2, 8, 13, 24] -> 84
4, [13, 19, 2, 20, 7, 3] -> 420
5, [4, 14, 36, 50, 63, 180, 210] -> 45
5, [43, 69, 16, 7, 13, 57, 21] -> 27664
3, [22, 46, 92, 43, 89, 12] -> 276
4, [42, 3, 49, 88, 63, 81] -> 882
6, [2, 4, 7, 10, 20, 21, 52, 260] -> 65
6, [2, 3, 4, 7, 10, 20, 21, 52, 260] -> 35
2, [3, 4] -> 12
3, [3, 5, 6, 9, 29] -> 18
2, [1, 1, 4, 5, 10] -> 1
3, [2, 2, 6, 7, 11] -> 3
3, [41, 13, 31, 35, 11] -> 4433
3, [27, 15, 37, 44, 20, 38] -> 540
5, [36, 11, 14, 32, 44] -> 22176
3, [171, 1615, 3420] -> 3060
3, [46, 36, 12, 42, 28, 3, 26, 40] -> 36
5, [36, 25, 20, 49, 10, 27, 38, 42] -> 1350
4, [40, 28, 34, 36, 42, 25] -> 2142
5, [24, 26, 47, 22, 6, 17, 39, 5, 37, 32] -> 1248
4, [9, 27, 12, 6, 44, 10] -> 108

规则：

排行榜

这篇文章底部的堆栈摘录从答案a）生成了排行榜，a）是每种语言的最短解决方案列表，b）则是总体排行榜。

为了确保您的答案显示出来，请使用以下Markdown模板以标题开头。

## Language Name, N bytes

N您提交的文件大小在哪里。如果您提高了分数，可以将旧分数保留在标题中，方法是将它们打掉。例如：

## Ruby, <s>104</s> <s>101</s> 96 bytes

如果您想在标头中包含多个数字（例如，因为您的分数是两个文件的总和，或者您想单独列出解释器标志罚分），请确保实际分数是标头中的最后一个数字：

## Perl, 43 + 2 (-p flag) = 45 bytes

您还可以将语言名称设置为链接，然后该链接将显示在代码段中：

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

显示代码段

code-golf  math 
— 梅果
source
5
哇，我什至都不知道这是一个字。真棒，子手奖，拼字游戏奖！
— Digital Trauma 2015年
@DigitalTrauma祝你好运找到3个Y瓷砖。
— SuperJedi224 2015年

Answers:

7

Pyth，28 27 24字节

fqs-hMrS.RR7%R1cLTQ8 1vz

在Pyth编译器中在线尝试。

怎么运行的

fqs-hMrS.RR7%R1cLTQ8 1vz

                          (implicit) Save the input number in z (as string).
                          (implicit) Save the input list in Q.

f                         Find the first positive integer T such that:
               cLTQ         Compute T/α for each α in Q.
            %R1             Get the fractional part of each result.
        .RR7                Round each fractional part to 7 decimal digits.
       S                    Sort the resulting numbers.
      r            8        Perform run-length encoding.
    hM                      Get the lengths of the runs.
   -                 1      Discard runs of length 1.
  s                         Add the remaining runs.
 q                    vz    Check is the sum matches the input number.
                          If it does, break and return T.

测试用例

$ cat input
7\n[10, 22, 7, 6, 12, 21, 19]\n87780
6\n[25, 6, 2, 19, 11, 12]\n62700
6\n[23, 1, 8, 10, 9, 25]\n41400
7\n[6, 4, 1, 8, 10, 24, 23]\n920
3\n[18, 5, 23, 20, 21]\n180
5\n[10, 8, 14, 17, 5, 9]\n2520
6\n[1, 18, 12, 9, 8, 10, 23]\n360
6\n[12, 11, 6, 23, 25, 18, 13]\n118404
4\n[18, 11, 2, 9, 12, 8, 3]\n8
7\n[18, 25, 9, 13, 3, 5, 20]\n11700
2\n[17, 20, 15, 8, 23, 3]\n15
3\n[16, 3, 24, 13, 15, 2]\n24
5\n[7, 23, 24, 8, 21]\n1932
6\n[16, 10, 12, 24, 18, 2, 21]\n720
6\n[1, 17, 16, 13, 19, 4, 15]\n53040
2\n[3, 4, 20]\n5
3\n[9, 4, 16, 14, 1, 21]\n16
5\n[5, 17, 10, 20, 12, 11]\n330
2\n[21, 5, 22, 18]\n90
4\n[7, 25, 2, 8, 13, 24]\n84
4\n[13, 19, 2, 20, 7, 3]\n420
5\n[4, 14, 36, 50, 63, 180, 210]\n45
5\n[43, 69, 16, 7, 13, 57, 21]\n27664
3\n[22, 46, 92, 43, 89, 12]\n276
4\n[42, 3, 49, 88, 63, 81]\n882
6\n[2, 4, 7, 10, 20, 21, 52, 260]\n65
6\n[2, 3, 4, 7, 10, 20, 21, 52, 260]\n35
2\n[3, 4]\n12
3\n[3, 5, 6, 9, 29]\n18
2\n[1, 1, 4, 5, 10]\n1
3\n[2, 2, 6, 7, 11]\n3
3\n[41, 13, 31, 35, 11]\n4433
3\n[27, 15, 37, 44, 20, 38]\n540
5\n[36, 11, 14, 32, 44]\n22176
3\n[171, 1615, 3420]\n3060
3\n[46, 36, 12, 42, 28, 3, 26, 40]\n36
5\n[36, 25, 20, 49, 10, 27, 38, 42]\n1350
4\n[40, 28, 34, 36, 42, 25]\n2142
5\n[24, 26, 47, 22, 6, 17, 39, 5, 37, 32]\n1248
4\n[9, 27, 12, 6, 44, 10]\n108
$ while read -r; do echo -e "$REPLY" | pyth -c 'qvwfqs-hMrS.RR7%R1cLTQ8 1vz'; done < input
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
True
— 丹尼斯
source
3

果冻，19 16字节

P:×ⱮP%PĠẈḟ1Sʋ€iƓ

将数组作为参数，即STDIN中的整数。

对于大多数测试用例来说，速度太慢并且占用大量内存。

在线尝试！

备用版本，14字节

P÷€%1ĠẈḟ1Sʋ€iƓ

从理论上讲，这是可行的，但是由于浮点数的错误，它可能会失败。

在线尝试！

怎么运行的

P:×ⱮP%PĠẈḟ1Sʋ€iƓ  Main link. Argument: A (array)

P   P P           Yield the product of A.
 :                Divide the product by each n in A.
  ×Ɱ              Multiply the quotients by each k in [1, ..., prod(A)].
     %            Take the results modulo the product.
             €    Map the link to the left over the array of remainders.
            ʋ       Combine the links to the left into a dyadic chain.
       Ġ              Group indices of identical elements.
        Ẉ             Widths; yield the lengths of the groups.
         ḟ1           Filterfalse; remove all copies of 1.
           S          Take the sum.
               Ɠ  Read an integer j from STDIN.
              i   Find the first index of j in the array of sums.
— 丹尼斯
source
2

CJam，42 34 33字节

0{)_eas~@d\f/1f%7fmO$e`0f=1m1b-}g

在CJam解释器中尝试这种小提琴或此测试套件。

怎么运行的

0       e# Push 0 (accumulator).
{       e# Do:
  )_    e#   Increment the accumulator and push a copy.
  eas~  e#   Push the command-line args, flatten and evaluate.
        e#   This pushes a number and an array.
  @d    e#   Rotate the accumulator copy on top and cast to Double.
  \f/   e#   Divide it by each of the integers in the array.
  1f%   e#   Get the fractional part of each result.
  7fmO  e#   Round all fractional parts to seven decimal digits.
  $e`   e#   Sort and perform run-length encoding.
  0f=   e#   Select the lengths of the runs.
  X-    e#   Discard runs of length 1.
  Xb    e#   Compute the sum of the remaining runs.
  -     e#   Subtract the sum from the input number (target).
}g      e# If this pushes a non-zero value, we've missed the target;
        e# repeat the loop.
— 丹尼斯
source
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