P3612 [USACO17JAN]Secret Cow Code S【分治】

最新推荐文章于 2024-11-01 10:19:04 发布

ln2037

最新推荐文章于 2024-11-01 10:19:04 发布

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题目描述

The cows are experimenting with secret codes, and have devised a method for creating an infinite-length string to be used as part of one of their codes.

Given a string sss, let F(s)F(s)F(s) be sss followed by sss “rotated” one character to the right (in a right rotation, the last character of sss rotates around and becomes the new first character). Given an initial string sss, the cows build their infinite-length code string by repeatedly applying FFF; each step therefore doubles the length of the current string.

Given the initial string and an index NNN, please help the cows compute the character at the NNNth position within the infinite code string.

奶牛正在试验秘密代码，并设计了一种方法来创建一个无限长的字符串作为其代码的一部分使用。

给定一个字符串，让后面的字符旋转一次（每一次正确的旋转，最后一个字符都会成为新的第一个字符）。也就是说，给定一个初始字符串，之后的每一步都会增加当前字符串的长度。

给定初始字符串和索引，请帮助奶牛计算无限字符串中位置N的字符。
输入格式

The input consists of a single line containing a st