欢迎来到起遇信息学
起遇信息学正处于上线筹建阶段,以下功能已全部开放免费体验: ✅ 完整题库浏览与代码提交评测(C / C++ / Python / Java 等) ✅ 入门到进阶的系列课程试读、作业与考试 ✅ AI 提示、AI 作业分析等智能助教功能 ✅ 赛事模拟与个人能力报告 ✅ 邮箱注册开放 ⏳ 付费课程订阅与微信/支付宝支付通道 ⏳ 手机号登录,微信扫码登录、微信公众号绑定 使用中如遇任何问题,欢迎通过页面底部 **"联系我们"** 与我们沟通。
CF999B.Reversing Encryption
Reversing Encryption
A string of length can be encrypted by the following algorithm:
- iterate over all divisors of in decreasing order (i.e. from to ),
- for each divisor , reverse the substring (i.e. the substring which starts at position and ends at position ).
For example, the above algorithm applied to the string ="codeforces" leads to the following changes: "codeforces" "secrofedoc" "orcesfedoc" "rocesfedoc" "rocesfedoc" (obviously, the last reverse operation doesn't change the string because ).
You are given the encrypted string . Your task is to decrypt this string, i.e., to find a string such that the above algorithm results in string . It can be proven that this string always exists and is unique.
Input
The first line of input consists of a single integer () — the length of the string . The second line of input consists of the string . The length of is , and it consists only of lowercase Latin letters.
Output
Print a string such that the above algorithm results in .
Note
The first example is described in the problem statement.
Samples
10
rocesfedoc
codeforces
16
plmaetwoxesisiht
thisisexampletwo
1
z
z
在线编程 IDE
建议全屏模式获得最佳体验
| 进入全屏编程 | Alt+E |
| 递交评测 | Ctrl+Enter |
| 注释/取消注释 | Ctrl+/ |
| 缩放字体 | Ctrl+滚轮 |