B. Reversing Encryption

最新推荐文章于 2018-08-10 19:14:34 发布

原创最新推荐文章于 2018-08-10 19:14:34 发布 · 184 阅读

0 ·

CC 4.0 BY-SA版权

codeforces 专栏收录该内容

26 篇文章

订阅专栏

本文介绍了一种用于解密特定加密字符串的算法实现。通过逆向操作，该算法能准确还原被加密过的字符串，如将“rocesfedoc”还原为原始的“codeforces”。文章详细解释了算法的工作原理，并提供了一个C++实现示例。

B. Reversing Encryption

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A string ss of length nn can be encrypted by the following algorithm:

iterate over all divisors of nn in decreasing order (i.e. from nn to 11),
for each divisor dd, reverse the substring s[1…d]s[1…d] (i.e. the substring which starts at position 11 and ends at position dd).

For example, the above algorithm applied to the string ss="codeforces" leads to the following changes: "codeforces" →→"secrofedoc" →→ "orcesfedoc" →→ "rocesfedoc" →→ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1d=1).

You are given the encrypted string tt. Your task is to decrypt this string, i.e., to find a string ss such that the above algorithm results in string tt. It can be proven that this string ss always exists and is unique.

Input

The first line of input consists of a single integer nn (1≤n≤1001≤n≤100) — the length of the string tt. The second line of input consists of the string tt. The length of tt is nn, and it consists only of lowercase Latin letters.

Output

Print a string ss such that the above algorithm results in tt.

Examples

input

Copy

10
rocesfedoc

output

Copy

codeforces

input

Copy

16
plmaetwoxesisiht

output

Copy

thisisexampletwo

input

Copy

1
z

output

Copy

Note

The first example is described in the problem statement.

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

bool isdivisor(int a,int n)
{
    if(n%a==0) return true;
    else return false;
}

int main()
{
    int n;
    char s[110];
    cin>>n;
    scanf("%s",&s);
    for(int i=2;i<n;++i)
    {
        if(isdivisor(i,n))
        {
            for(int j=0;j<i/2;++j)
            {
                    char a;
                    a=s[j];
                    s[j]=s[i-1-j];
                    s[i-1-j]=a;
            }
        }
    }
    for(int i=n-1;i>=0;--i)
        cout<<s[i];
    cout<<endl;
    return 0;
}