PAT 1031：Hello World for U

Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as:

h  d
e  l
l  r
lowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

Sample Output:

h   !
e   d
l   l
lowor

------------------------------------------------ 我是分割线--------------------------------------------

此题关键是理解n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N这句数学表达式，简单描述就是满足n1 + n2 + n3 - 2 = N的情况下，n1（= n3)取 <= n2的最大值，用python描述就是n1 = n3 = (N + 2) // 3，n2 = N + 2 - 2 * n1。用代码实现就很简单了：

def ustring(seq):
    n = len(seq)
    n1 = (n + 2) // 3
    snum = n - n1 * 2        # U形每一行之间的空格数，等价于 n2 -2

    for i in range(n1 - 1):
        print(seq[i], seq[n - 1 - i], sep=' ' * snum)
    print(seq[n1 - 1:-(n1 - 1)])

测试一下：

ustring('123456789000987654321')

结果：

1       1
2       2
3       3
4       4
5       5
6       6
789000987

中间加个0再试：

ustring('1234567890000987654321')

看看有什么变化：

1      1
2      2
3      3
4      4
5      5
6      6
7      7
89000098