Fast Matrix Calculation HDU

(AB)(AB)(AB)=A(BA)(BA)B

用到线性代数的知识还是蛮有亲切感的。。毕竟上了大学就感觉这门课学到了东西。。

//#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#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;  
  
const double pi=acos(-1.0);  
#define ll long long  
#define pb push_back

#define sqr(a) ((a)*(a))
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))

const double eps=1e-10;
const int maxn=1e3+56;
const int inf=0x3f3f3f3f;
const ll mod=6;

ll arr[maxn];

ll n,m;

struct mat{  
    ll a[15][15];  
    ll n,m;  
    mat(){memset(a,0,sizeof a);n=0;m=0;}  
    mat(ll x,ll y){memset(a,0,sizeof a);n=x;m=y;}  
    mat operator* (const mat &rhs)const{  
        mat ans;  
        ans.n=n;ans.m=rhs.m;  
        for(int i=1;i<=n;i++){  
            for(int j=1;j<=rhs.m;j++){  
                for(int k=1;k<=m;k++){  
                    ans.a[i][j]=(ans.a[i][j]+a[i][k]*rhs.a[k][j])%mod;  
                }  
            }  
        }  
        return ans;  
    }  
    mat operator^ (ll rhs)const{  
        mat ans(n,n),b=*this;  
        for(int i=1;i<=n;i++)ans.a[i][i]=1;  
        for(;rhs;rhs>>=1,b=b*b)  
            if(rhs&1)ans=ans*b;  
        return ans;  
    }   
};

ll a[maxn][maxn],b[maxn][maxn],c1[maxn][maxn],c2[maxn][maxn];

int main(){
	while(~scanf("%lld%lld",&n,&m)&&n){
		for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
			scanf("%lld",&a[i][j]);
		for(int i=1;i<=m;i++)for(int j=1;j<=n;j++)
			scanf("%lld",&b[i][j]);
		
		mat tmp(m,m);
		for(int i=1;i<=m;i++)
			for(int j=1;j<=m;j++){
				tmp.a[i][j]=0;
				for(int k=1;k<=n;k++){
					tmp.a[i][j]+=b[i][k]*a[k][j];
					tmp.a[i][j]%=6;
				}
			}
		mat p=tmp^(n*n-1);
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++){
				c1[i][j]=0;
				for(int k=1;k<=m;k++){
					c1[i][j]=(c1[i][j]+a[i][k]*p.a[k][j])%6;
				}
			}
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++){
				c2[i][j]=0;
				for(int k=1;k<=m;k++){
					c2[i][j]=(c2[i][j]+c1[i][k]*b[k][j])%6;
				}
			}
		ll ans=0;
		for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)ans+=c2[i][j];
		printf("%lld
",ans);
	}
}