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

• iterate（叠代） over all divisors（除數） 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 d
  d).

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=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≤100

Output

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

Examples

10 
rocesfedoc

codeforces

16 
plmaetwoxesisiht

thisisexampletwo

1 z

z

Note

The first example is described in the problem statement.

題目意思：對所給的字符串進行加密得到一新的字符串，加密方式是從n的最小因子開始，反轉從開始到因子的字符串。

之前做這道題的時候理解錯意思了，一直以為是一半一半的反轉字符串，很是可惜，沒有做出來

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,i;
    char s[110];
    scanf("%d",&n);
    getchar();
    scanf("%s",&s);
    for(i=1;i<=n;i++)
    {
        if(n%i==0)
        {
            reverse(s,s+i);
        }
    }
    printf("%s\n",s);
    return 0;
}

這裏有我之前一篇關於字符串反轉函數的博客

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,i,j,count;
 8     char s[110];
 9     int a[110];
10     scanf("%d",&n);
11     getchar();
12     scanf("%s",&s);
13     count=0;
14     for(i=1;i<=n;i++)
15     {
16         if(n%i==0)
17         {
18             a[count++]=i;///記錄n因子
19         }
20     }
21     for(i=0;i<count;i++)
22     {
23         for(j=0;j<a[i]/2;j++)
24         {
25             swap(s[j],s[a[i]-j-1]);
26         }
27     }
28     printf("%s\n",s);
29     return 0;
30 }

Reversing Encryption