[codeforces1184A2]Heidi Learns Hashing (Medium)

time limit per test : 2 seconds
memory limit per test : 256 megabytes
分数：2100 // 虽然我觉得思维难度还挺高的，像是2500的题，不像普通的2100那么简单

After learning about polynomial hashing, Heidi decided to learn about shift-xor hashing. In particular, she came across this interesting problem.

Given a bitstring $y\in \{0,1{\}}^n$ find out the number of different $k(0\le k<n)$ such that there exists $x\in \{0,1{\}}^n$ for which $y=x\oplus shif{t}^k(x)$.

In the above, ⊕ is the xor operation and shiftk is the operation of shifting a bitstring cyclically to the right k times. For example, 001⊕111=110 and shift3(00010010111000)=00000010010111.

Input

The first line contains an integer $n(1\le n\le 2\cdot 10^5)$, the length of the bitstring $y$.
The second line contains the bitstring $y$.

Output

Output a single integer: the number of suitable values of $k$.

Example
Input

4
1010

Output

3

Note

In the first example:

1100⊕shift1(1100)=1010
1000⊕shift2(1000)=1010
0110⊕shift3(0110)=1010

There is no $x$ such that x⊕x=1010, hence the answer is 3.

题意：
给一个长度为n的01串$y$，问有多少个$k(0<=k<n)$满足存在一个长度为n的01串x使得$x\oplus shif{t}^k(x)=y$
$shif{t}^k(x)$操作是将字符串x向右循环位移$k$位。

题解:
考虑位移k的情况
显然字符串满足以下等式

x[1]^x[1+k]=y[1]
x[2]^x[2+k]=y[2]
...
x[n]^x[k]=y[n]

将等式拆分：

//对于所有i<=k都满足如下情况
x[i]^x[i+k]=y[i]
x[i+k]^x[i+2*k]=y[i+k]
...

可以发现对于所有的$gcd(n,u)=k$的情况，都符合上面的方程组
将方程组化简可得：

y[i]^y[i+k]^y[i+2*k]^.... = x[i]^x[i+k]^x[i+k]^.....(相邻两两相消) = 0

那么我们可以预处理出所有$gcd(n,i)$是否能够成为解，最后直接枚举$k$统计答案即可。

#include
#define ll long long
using namespace std;
const int MAXN=200004;
int ok[MAXN],l;
char s[MAXN];
void work(int base){
    for(int i=1;i<=base;i++){
        ll sum=0;
        for(int j=i;j<=l;j+=base){
            sum^=(s[j]-'0');
        }
        if(sum)return ;
    }
    ok[base]=1;
}
int main(){
    scanf("%d",&l);
    scanf("%s",s+1);
    memset(ok,0,sizeof(ok));
    bool flag=0;
    for(int i=1;i<=l;i++){
        if(s[i]=='1')flag=1;
    }
    if(!flag)ok[0]=1;
    for(int k=1;k*k<=l;k++){
        if(l%k==0){
            work(k);
            work(l/k);
        }
    }
    int ans=0;
    if(!flag)ans++;
    for(int k=1;k<l;k++){
        ans+=ok[__gcd(l,k)];
    }
    printf("%d\n",ans);
    return 0;
}