BZOJ4128: Matrix 矩阵BSGS

A^x=B(mod C)

令x=im-j

A^(im)=BA^j(mod C)

这就不用求逆元了

#include<bits/stdc++.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<iostream>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<iomanip>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pb push_back
#define FOR(a) for(int i=1;i<=a;i++)
const int inf=0x3f3f3f3f;
const ll Linf=9e18;
const int maxn=1e6+7; 
const ll mod=1e9+7;
const double eps=1e-6;

#define base 231
ll n,p;
struct MATRIX{
	ll x[80][80];
	void init(){memset(x,0,sizeof x);}
	MATRIX operator * (MATRIX a){
		MATRIX ret;
		ret.init();
		FOR(n)for(int j=1;j<=n;j++){
			for(int k=1;k<=n;k++)ret.x[i][j]=ret.x[i][j]+x[i][k]*a.x[k][j];
			ret.x[i][j]%=p;
		}
		return ret;
	}

	ll Hash(){
		ll ret=0;
		FOR(n)for(int j=1;j<=n;j++){
			ret=(ret*base+x[i][j])%mod;
		}
		return ret;
	}
	void read(){
		FOR(n)for(int j=1;j<=n;j++)
			scanf("%lld",&x[i][j]),x[i][j]%=p;
	}
	void build(){
		init();FOR(n)x[i][i]=1;	
	}
}A,B,E;

map<ll,ll>mp;

void BSGS(){
	mp.clear();
	ll m=ceil(sqrt(p));
	MATRIX ans;
	for(int i=0;i<=m;i++){
		if(i==0){ans=B;mp[ans.Hash()]=i;continue;}
		ans=ans*A;
		mp[ans.Hash()]=i;
	}
	MATRIX tmp=E;
	FOR(m)tmp=tmp*A;	//A^(im)=BA^j(mod C)
	ans=E;
	FOR(m){
		ans=ans*tmp;
		if(mp[ans.Hash()]){
			ll ret=i*m-mp[ans.Hash()];
			printf("%lld\n",(ret%p+p)%p);
			return;
		}
	}
}

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