Reversing Encryption (模拟水题)_by a string s of length n-优快云博客

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/81568621

本文介绍了一种特殊的字符串解密算法，该算法通过反转操作来还原加密字符串。输入为加密后的字符串，任务是找到原始字符串。文章包含了一个示例性的C++代码实现。

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Reversing Encryption

A string s of length n

can be encrypted by the following algorithm:

iterate over all divisors of n

in decreasing order (i.e. from n to 1
),
for each divisor d
, reverse the substring s[1…d] (i.e. the substring which starts at position 1 and ends at position d

).

For example, the above algorithm applied to the string s
=”codeforces” leads to the following changes: “codeforces” → “secrofedoc” → “orcesfedoc” → “rocesfedoc” → “rocesfedoc” (obviously, the last reverse operation doesn’t change the string because d=1

You are given the encrypted string t
. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s

always exists and is unique.

Input

The first line of input consists of a single integer n

(1≤n≤100) — the length of the string t. The second line of input consists of the string t. The length of t is n

, and it consists only of lowercase Latin letters.

Output

Print a string s

such that the above algorithm results in t

Examples
Input

10
rocesfedoc

Output

codeforces

Input

16
plmaetwoxesisiht

Output

thisisexampletwo

Input

1
z

Output

z

Note

The first example is described in the problem statement.

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
vector<int>fac;
int main(){
    int n;
    char s[150];
    cin >> n >> s;
    for(int i = 1; i <= n; i++){
        if(n % i == 0) fac.push_back(i);
    }
    for(int i = 0; i < fac.size(); i++){
        reverse(s,s+fac[i]);
    }
    cout << s << endl;
    return 0;
}