空白处最初形成一个固定的<代码>n-gon。 视之为复杂数目。 如果不丧失一般性,rad为1,轴线就位于斜面上,斜线为n。

空白之处为<代码>n-th origin of Unity, the tip of bladecode>k。

z_k = e^(2pi i * k/n)

z_k1 + z_k2 + ... + z_kr = 0

该战略将试图通过一套互不关联的定期<代码>d_i-gons,其中d_i为n的分母。 因此,在清单中发现,其指数因<代码><>n/code>的分数而异。

我的初步想法是,由于需要找回大众中心,需要四舍五入到数目相同的非中间点。

页: 1 推进器必须轮流测量。 如果一刀切的推进剂被打碎,就没有任何东西可以 recent。

要点是:

1. The experts assert that when all the blades were intact their endpoints formed a regular n-gon.

2. The integer n has at most two distinct prime divisors.

开始的是符合这两种特性的简单东西,例如 de。 Ask yourself: How do polygons and symmetry associated? 那么,10位分母如何与多角和高金属有关? 为了简化工作,你能否用粗体字而不是两维点来代表这些多角点? Hint: 模块算术对解决方案的影响。

lade形构造(未固定和固定)应有所帮助。 例如:

        2                     2                     2                   2           
    3       1                     1                                                 
  4           0         4           0         4           0       4           0     

  5           9         5                                 9       5           9     
    6       8             6       8             6       8                   8       
        7                     7                     7                   7           
   whole 10-gon        2 broken blades       3 broken blades     3 broken blades

对2个破碎的空白点和头3个破碎的空白点有两种可能的解决办法,但只有一种是最佳的。 第二名3名经纪人有一个解决办法。 看一角。

这个问题在数学上确实非常复杂,但可以通过对如何平衡工作的良好理解加以简化。

• If the number of blades "n" is a prime number, there is no solution other than removing every blade. One can t rebalance even one missing blade when n is a prime. If there is no solution for one blade, there won t be a solution for multiple missing.
• It the number of blades "n" is not a prime, by definition (and according to the rule of this problem) it is the produce of two primes "p1" and "p2", Example: for 21 blades, p1=3, p2=7.

If one blade "k1" is missing, balance will be obtained when a total of "N/p2 (example=3) blades equally spaced have been removed (visualize the Mercedes Benz sign). If two blades "k1" and "k2" are missing, balance will be obtained by doing exactly for "k2" as the preceding case, EXCEPT if k1 and k2 are spaced following the "p1" or "p2" symmetry

Examples with 12 and 20 blade make things difficult to understand, because they are multiple of 4, which means two symmetries of an order 2. I prefer examples with 21 blades, which don t have this problem.

1. 21 Blades, one missing blade k1 = 1 -> Balance is obtained by removing blade 8 and 15 (Symmetry p1 is order 3)
2. 21 Blades, two missing blades k1=1, k2=2 -> Balance is obtained by removing blade 8 and 15 for k1, and 8 and 16 for k2
3. 21 Blades, two missing blades k1=1, k2=8 -> since k2-k1=7, you just need to remove blade 15 to complete the balance
4. 21 Blades, two missing blades k1=1, k2=7->since k2-k1=6 = 2*3, you can remark that the blades are on the symmetry of order 7, therefore you could complete the pattern by removing the blades 4,10,13,16,19 to get perfect balance. But interestingly, you can also treat it as the example #2, and use two times the symmetry of order 3; so there is also a solution by removing blade 8 and 15 to balance k1, and 14 and 21 to balance k2.

Therefore to solve your problem, your algorithm should do these things: - If n is a prime, simply output "-1" - no solution - Starting wit the first missing blade, scan for groups of missing blade spaced by "n/p1 (= 7)" and "n/p2(=3)" . Count how many groups of each symmetry group have, and how many blades you need to replace.

• In the example, your first order solution is simply:
  The number of blades to remove will be (number of groups of symmetry p1) *7 + (Number of groups of symmetry p2 )*3 - number of blades missing. You ll have to treat the exceptions like the one shown in example 4 where it is better to treat blades individually, but you get the picture.