2. 无限递归字符串查询

Description
Consider a string A = “12345”. An infinite string s is built by performing infinite steps on A recursively. In i-th step, A is concatenated with ‘$’ i times followed by reverse of A. A=A|$…$|reverse(A), where | denotes concatenation.

Constraints:1<=Q<=10^5, 1<=POS<=10^12

Input
输入第一行为查询次数，后面为每次查询的具体字符位置。

Output
输出每一次查询位置上的字符。

Sample Input 1
2
3
10

Sample Output 1
3
2

def solution(number):
    # 除了这个迭代A和reverse(A)不一样外，后面的迭代都是一样的
    init = ['1', '2', '3', '4', '5', '$', '5', '4', '3', '2', '1']
    while number > len(init):
        val, iter = location(number)
        # 判断number是在中间的$符号，还是在后面的reverse中,temp为上一个A+$的长度
        temp = (val - iter) // 2 + iter
        if number <= temp:
            return "$"
        # 说明在reverse中，和上一个迭代A相同
        else:
            number -= temp
    return init[number - 1]


# 求第number个数所在字符串的位置，返回字符串的长度及中间的$个数
def location(number):
    val = 5
    iter = 0
    while number > val:
        iter += 1
        val = 2 * val + iter
    return val, iter


if __name__ == '__main__':
    n = int(input())
    for i in range(n):
        number = int(input())
        result = solution(number)
        print(result)

来源：CSDN

作者：cy求求你让我过吧

链接：https://blog.csdn.net/qq_26496077/article/details/103240755