[BZOJ4128]Matrix（BSGS+矩乘）

题目描述

题解

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define N 75

struct data
{
int a

;
bool operator == (const data &b) const
{
for (int i=1;i<=70;++i)
for (int j=1;j<=70;++j)
if (a[i][j]!=b.a[i][j]) return 0;
return 1;
}
bool operator < (const data &b) const
{
for (int i=1;i<=70;++i)
for (int j=1;j<=70;++j)
{
if (a[i][j]<b.a[i][j]) return 1;
if (a[i][j]>b.a[i][j]) return 0;
}
return 0;
}
}zero,unit,A,B,A_m;
int n,ans,m,p;
map <data,int> hash;

data cheng(data a,data b)
{
data ans=zero;
for (int k=1;k<=n;++k)
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
ans.a[i][j]=(ans.a[i][j]+a.a[i][k]*b.a[k][j]%p)%p;
return ans;
}
data fast_pow(data a,int p)
{
data ans=unit;
for (;p;p>>=1,a=cheng(a,a))
if (p&1)
ans=cheng(ans,a);
return ans;
}
int main()
{
scanf("%d%d",&n,&p);
for (int i=1;i<=n;++i) unit.a[i][i]=1;
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
scanf("%d",&A.a[i][j]);
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
scanf("%d",&B.a[i][j]);
m=ceil(sqrt(p));
A_m=fast_pow(A,m);
data mi=B;
hash[mi]=1;
for (int i=1;i<=m;++i)
{
mi=cheng(mi,A);
hash[mi]=i+1;
}
mi=unit;
for (int i=1;i<=m;++i)
{
mi=cheng(mi,A_m);
if (hash[mi])
{
ans=i*m-hash[mi]+1;
break;
}
}
printf("%d\n",ans);
}