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HDU 4965 Fast Matrix Calculation (矩阵快速幂

Posted on Edited on In ,

题目链接：
https://vjudge.net/contest/71746#problem/E

题目大意：
现有一个nk的矩阵A，以及一个kn的矩阵B，在模6域下，求得矩阵C = A*B，并对Cn∗nC^{n*n}的每一个元素求和，输出结果

分析：
直接暴力对C矩阵做快速幂，用类会爆空间，会数组写比较繁琐，且可能超时，所以可以得到以下式子

Cn∗n=(A∗B)n∗n=A∗(B∗A)∗(B∗A)⋯∗(B∗A)∗BC^{n*n} = (A*B)^{n*n}=A*(B*A)*(B*A)\cdots*(B*A)*B

即可写作

Cn∗n=A∗(B∗A)n∗n−1∗BC^{n*n} = A*(B*A)^{n*n-1}*B

而B∗AB*A是一个6*6的矩阵，对它求快速幂便很容易了

代码：
(因为调了好多遍，写的很丑，调试输出也很多，只要#define ONLINE即可不调试输出

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdlib>
//#define ONLINE
using namespace std;
const int N = 1200;
struct mart
{
   int n,m;
   int a[12][12];
};

int n,m,x,y,q;
mart now;



int a[N][10],b[10][N],c[10][N],d[N][N];
typedef long long ll;
const int mod = 1000000007;



mart unit;


mart multiply(mart A,mart B)
{
    mart C;
    C.n = A.n;
    C.m = B.m;
#ifndef ONLINE
    cout<<"A.n = "<<A.n<<"   B.m = "<<B.m<<endl;
#endif
    for (int i = 1 ; i <= m ; i ++)
    {
        for (int j = 1  ; j <= m ; j ++ )
        {
            C.a[i][j] = 0;
            for (int k = 1 ; k <= m ; k ++)
            {
                C.a[i][j] += A.a[i][k]*B.a[k][j];
                C.a[i][j] %= 6;
            }
        }
    }
    return C;
}

mart quickpow(mart p,int t)
{
    mart ans = unit;

    while (t)
    {
        if (t&1)
            ans = multiply(ans,p);
        p = multiply(p,p);
        t >>= 1;
    }
#ifndef ONLINE
    for (int i = 1 ; i <= m ; i ++)
    {
        for (int j = 1  ; j<= m ; j ++)
            cout<< ans.a[i][j]<<'\t';
        cout<<endl;
    }
    cout<<">>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>"<<endl;
#endif // ONLINE
    return ans;

}


void init()
{
    memset(now.a,0,sizeof(now.a));
    memset(c,0,sizeof(c));
    memset(d,0,sizeof(d));
    memset(unit.a,0,sizeof(unit.a));
    unit.m=m;
    unit.n=m;
    now.n = m;
    now.m = m;
    for (int i = 0 ; i < 12 ; i ++)
        unit.a[i][i] = 1;          //单位矩阵
    return;
}


void read() //读入
{
    for (int i = 1 ; i <= n ; i ++)
        for (int j = 1; j <= m ; j ++)
            scanf("%d",&a[i][j]);
    for (int i = 1 ; i <= m ; i ++)
        for (int j = 1; j <= n ; j ++)
            scanf("%d",&b[i][j]);
    return;
}

int main()
{
    init();
    while (~scanf("%d%d",&n,&m)&&(n||m)){
        if (n==0&&m==0)
            return 0;
        init();
        read();
                
                
        for (int i = 1 ; i <= m ; i ++)
        {
            for (int j = 1 ; j <= m ; j ++)
            {
                now.a[i][j] = 0;
                for (int k = 1 ; k <= n ; k ++)
                {
                    now.a[i][j] += b[i][k] * a[k][j];
                    now.a[i][j] %= 6;
                }
            }
        }
        
#ifndef ONLINE
        cout<<"---------------------------------------------------"<<endl;
        for (int i = 1 ; i <= m ; i ++)
        {
            for (int j = 1  ; j<= m ; j ++)
                cout<< now.a[i][j]<<'\t';
            cout<<endl;
        }
        cout<<"---------------------------------------------------"<<endl;
#endif // ONLINE

        mart Ans = quickpow(now,n*n-1);
        for (int i = 1 ; i <= m ; i ++)
        {
            for (int j = 1 ; j <= n ; j ++)
            {
                for (int k = 1 ; k <= m ; k ++)
                {
                    c[i][j] += Ans.a[i][k]*b[k][j];
                    c[i][j] %= 6;
                }
            }
        }
        
        for (int i = 1 ; i <= n ; i ++)
        {
            for (int j = 1 ; j <= n ; j ++)
            {
                for (int k = 1 ; k <= m ; k ++)
                {
                    d[i][j] += a[i][k] * c[k][j];
                    d[i][j] %= 6;
                }
            }
        }
        
        int ans = 0;
        for (int i = 1 ; i <= n ; i ++)
        {
            for (int j = 1 ; j <= n ; j ++)
            {
#ifndef ONLINE
                cout<< d[i][j]<<'\t';
#endif // ONLINE
                ans += d[i][j];
            }
#ifndef ONLINE
            cout<<endl;
#endif // ONLINE
        }
        printf("%d\n",ans);
    }

}