[BSGS 矩阵 随机化] BZOJ 4128 Matrix

其实完全不必求逆矩阵

判相同用了随机化

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=75;
int n,p,m;

struct Matrix{
int a

,n;
Matrix(int in=0,int f=0){
n=in; cl(a);
if (f==-1) for (int i=1;i<=n;i++) a[i][i]=1;
}
void read(){
for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) ::read(a[i][j]);
}
int *operator [](int x){ return a[x]; }
friend Matrix operator *(Matrix& A,Matrix &B) {
int n=A.n; Matrix ret(A.n);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
for (int k=1;k<=n;k++)
ret[i][j]+=A[i][k]*B[k][j]%p,ret[i][j]=ret[i][j]>=p?ret[i][j]-=p:ret[i][j];
return ret;
}
friend Matrix operator *(Matrix &A,int *B){
int n=A.n; Matrix ret(n);
for (int i=1;i<=n;i++)
for (int k=1;k<=n;k++)
ret[i][1]+=A[i][k]*B[k]%p,ret[i][1]=ret[i][1]>=p?ret[i][1]-=p:ret[i][1];
return ret;
}
friend Matrix Pow(Matrix& a,int b){
int n=a.n; Matrix ret(n,-1);
for (;b;b>>=1,a=a*a)
if (b&1)
ret=ret*a;
return ret;
}
friend bool Jud(Matrix &A,Matrix &B){
for (int i=1;i<=::n;i++)
if (A[i][1]!=B[i][1])
return 0;
return 1;
}
}A,B,H[205],X,Y;

int C
;

int main(){
srand(10086);
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); read(p); m=sqrt(p);
A.n=n; A.read();
B.n=n; B.read();
for (int i=1;i<=n;i++) C[i]=rand()%p;
for (int i=0;i<m;i++){
H[i]=B*C;
B=B*A;
}
A=Pow(A,m); Y=A;
for (int i=1;i<=m;i++){
X=Y*C;
for (int j=0;j<m;j++)
if (Jud(X,H[j]))
return printf("%d\n",i*m-j),0;
Y=Y*A;
}
return 0;
}