Codeforces999B——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.

题目好像说的不是很好 不过看样例也能看出来
我的思路是首先暴力将所有因数存起来
然后就暴力反转即可

代码：

#include <cstdio>
#include <algorithm>
#include <string>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
string s;
int main(void){
    int n;
    scanf("%d",&n);
    cin >> s;
    priority_queue<int,vector<int>,greater<int>> divi;
    for(int i=1;i*i<=n;i++){
        if(n%i==0){
            divi.push(i);
            divi.push(n/i);
        }
    }
    int pre=0;
    while(divi.size()!=0){
        int d=divi.top();
        divi.pop();
        if(pre==d){
            continue;
        }
        pre=d;
        int i=0;
        int j=d-1;
        while(i<j){
            char t=s[i];
            s[i]=s[j];
            s[j]=t;
            i++;
            j--;
        }
    }
    cout << s << endl;
    return 0;
}