Codeforces1184A2-Heidi Learns Hashing (Medium)

Here are the official tutorial：
A2. Medium
Let us first find a method to check whether a shift by a fixed number $k$ yields a solution or not. For the sake of simplicity let us work with the case of $n = 12$, let also $k = 3$. We can start by writing a set of equations
(all modulo $2$)
$(1)$
$\begin{cases} x_0 + x_3 = y_0\\ x_3 + x_6 = y_3\\ x_6 + x_9 = y_6\\ x_9 + x_0 = y_9\\ \end{cases}$
Note that for $x$ to be solution to the shift equation, these equations are necessary to be satisfied, but of course it is not yet sufficient. Also it is easy to see that the $3$ first equations can be always satisfied, and that by taking the sum of all these equations we obtain
$0 = y_0 + y_3 + y_6 + y_9$
which means that $y_0 + y_3 + y_6 + y_9$ should be even. This can be easily seen to be a necessary and sufficient condition for $1$ to have a solution $x$. Similarly we can write an analogous system for bits $x_1, x_4, x_7, x_{10}$
$(2)$
$\begin{cases} x_1 + x_4 = y_1\\ x_4 + x_7 = y_4\\ x_7 + x_{10} = y_7\\ x_{10} + x_1 = y_{10}\\ \end{cases}$
which is solvable if $y_1 + y_4 + y_7 + y_{10}$ is even. Finally we obtain that there is a solution x to the $k$-xor-shift equation.
$(3)$
$\begin{cases} y_0 + y_3 + y_6 + y_9 = 0\\ y_1 + y_4 + y_7 + y_{10} = 0\\ y_2 + y_5 + y_8 + y_{11} = 0\\ \end{cases}$
Note also that the set of equations corresponding to the value $k = 9$ would be exactly the same as above!This observation is quite easy to generalize for all values of $n$ and $k$. More precisely, whenever $gcd(n, k1) =gcd(n, k2) = d$ then one can write $d$ equations for $y$ that determine whether the xor-shift equations has a
solution for $k_1$ (and equivalently for $k_2$).
The complete solution is: first precompute the answer for every $k$ being a divisor of $n$, then for every other $k$, compute $gcd(n, k)$ to reduce to one of the precomputed cases. The complexity is $O(n · d(n)) = O(n^{\frac 3 2})$ where $d(n)$ is the number of divisors of $n$.
就是分层数1的个数，必须每一层都有偶数个1才合法。
$Code:$

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int f[N],n,a[N];
char ch[N];
int main()
{
	scanf("%d",&n);
	int t=1;a[1]=1;
	for(int i=2;i<=sqrt(n);i++)
		if(n%i==0)a[++t]=i,a[++t]=n/i;
	for(int i=1;i<n;i++)f[i]=1;
	scanf("%s",ch);
	for(int i=1;i<=t;i++)
		for(int j=0;j<a[i];j++)
		{
			int sum=0;
			for(int k=j;k<n;k+=a[i])sum+=ch[k]-'0';
			if(sum%2!=0)
			{
				f[a[i]]=0;
				break;
			}
		}
	int ans=0;
	bool flag=true;
	for(int i=0;i<n;i++)
		if(ch[i]!='0')flag=false;
	if(flag)ans++;
	for(int i=1;i<n;i++)
	{
		int k=__gcd(i,n);
		ans+=f[k];
	}
	printf("%d\n",ans);
	return 0;
}