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 $2$)
$(1)$
$$\{\begin{array}{l}{x}_0+x_3=y_0\\ x_3+x_6=y_3\\ x_6+x_9=y_6\\ x_9+x_0=y_9\end{array}$$
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 $x_1,x_4,x_7,x_{10}$
$(2)$
$$\{\begin{array}{l}{x}_1+x_4=y_1\\ x_4+x_7=y_4\\ x_7+x_{10}=y_7\\ x_{10}+x_1=y_{10}\end{array}$$
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.
$(3)$
$$\{\begin{array}{l}{y}_0+y_3+y_6+y_9=0\\ y_1+y_4+y_7+y_{10}=0\\ y_2+y_5+y_8+y_{11}=0\end{array}$$
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 $O(n\cdot d(n))=O(n^{\frac{3}{2}})$ 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
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