Problem 265

Binary Circles

2^N binary digits can be placed in a circle so that all the N-digit clockwise subsequences are distinct.

For N=3, two such circular arrangements are possible, ignoring rotations:

For the first arrangement, the 3-digit subsequences, in clockwise order, are: 000, 001, 010, 101, 011, 111, 110 and 100.

Each circular arrangement can be encoded as a number by concatenating the binary digits starting with the subsequence of all zeros as the most significant bits and proceeding clockwise. The two arrangements for N=3 are thus represented as 23 and 29:

00010111 ₂ = 23 00011101 ₂ = 29

Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52.

Find S(5).

二进制圈

2^N个二进制位可以排成一圈，使得所有按顺时针顺序得到的N位二进制子串都不一样。

对于N=3，不考虑旋转，有两种可行的排圈方式：

在第一种排圈方式中，按顺时针顺序得到的3位二进制子串分别是：000、001、010、101、011、111、110和100。

每一种排圈方式都可以按如下方式编码：以全为0的子串作为起点，将所有的二进制位连接起来。N=3的两种排圈方式于是就可以编码为23和29：

00010111 ₂ = 23 00011101 ₂ = 29

记S(N)是所有排圈方式的编码之和，我们可以看到S(3) = 23 + 29 = 52。

求S(5)。