BZOJ 4128 Matrix ——BSGS

矩阵的BSGS。

只需要哈希一下存起来就可以了。

也并不需要求逆。

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

#define F(i,j,k) for (ll i=j;i<=k;++i)
#define D(i,j,k) for (ll i=j;i>=k;--i)
#define ll long long
#define base 231
#define basemd 1000000007

ll n,p;

struct matrix{
    ll x[80][80];
    void init(){memset(x,0,sizeof x);}
    matrix operator * (matrix a){
        matrix ret;
        ret.init();
        F(i,1,n) F(j,1,n)
        {
            F(k,1,n) ret.x[i][j]=ret.x[i][j]+x[i][k]*a.x[k][j];
            ret.x[i][j]%=p;
        }
        return ret;
    }
    ll encode()
    {
        ll ret=0;
        F(i,1,n) F(j,1,n)
            ret=(ret*base+x[i][j])%basemd;
        return ret;
    }
    void read()
    {
        F(i,1,n) F(j,1,n)
            scanf("%lld",&x[i][j]),x[i][j]%=p;
    }
    void build()
    {init();F(i,1,n)x[i][i]=1;}
}A,B,E;

map <ll,ll> mp;

void BSGS()
{
    mp.clear();
    ll m=ceil(sqrt(p));
    matrix ans;
    F(i,0,m)
    {
        if (i==0){ans=B;mp[ans.encode()]=i;continue;}
        ans=ans*A;
        mp[ans.encode()]=i;
    }
    matrix tmp=E;
    F(i,1,m) tmp=tmp*A;  ans=E;
    F(i,1,m)
    {
        ans=ans*tmp;
        if (mp[ans.encode()])
        {
            ll ret=i*m-mp[ans.encode()];
            printf("%lld\n",(ret%p+p)%p);
            return ;
        }
    }
}

int main()
{
    scanf("%lld%lld",&n,&p);
    A.read();B.read();E.build();
    BSGS();
}