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
z
Note
The first example is described in the problem statement.
题意看了好半天才理解，都是因为漏了规则的第一句话，认为他就是把字符串反转过来。
假如n为10，他的因子，你第一次反转区间1~10，第二次1~5.第三次1~2，第四次1，
你到过来反转就行，reserve大法好
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN=40010;
const int MAXM=100010;
const int M=110;
int n;
int main()
{
int n,m;
while(scanf("%d",&n)!=EOF)
{
char s[105];
cin>>s;
for(int i=1;i<=n;i++)
{
if(n%i==0)
reverse(s,s+i);
}
cout<<s<<endl;
}
return 0;
}
