# hdu 4965 Fast Matrix Calculation (矩阵快速幂，2014多校#9)

Posted by 111qqz on Wednesday, October 19, 2016

## TOC

Step 2: Calculate M = C^(N*N).
Step 3: For each element x in M, calculate x % 6. All the remainders form a new matrix M’.
Step 4: Calculate the sum of all the elements in M’.

M = (AB)^(NN) = A * (BA)^(NN-1) * B

``````/* ***********************************************
Author :111qqz
Created Time :Wed 19 Oct 2016 08:39:49 PM CST
File Name :code/hdu/4965.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E3+5;
int n,K;
struct Mat
{
int mat[10][10];
void clear()
{
ms(mat,0);
}
}C,M;
Mat operator * ( Mat a, Mat b)
{
Mat ans;
ans.clear();
for ( int i = 0 ; i < K ; i++)
{
for ( int j = 0 ; j < K ; j++)
{
for ( int k = 0 ; k < K ; k++)
{
ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j])%6;
}
}
}
return ans;
}
Mat operator ^ ( Mat a,int b)
{
Mat ans;
ans.clear();
for ( int i = 0 ; i < K ; i++) ans.mat[i][i]  = 1;
while (b>0)
{
if (b&1) ans = ans * a;
b = b >> 1;
a = a * a;
}
return ans;
}
int A[N][10],B[10][N],ans[2][N][N];
int main()
{
#ifndef  ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif
while (~scanf("%d%d",&n,&K))
{
if (n==0&&K==0) break;
ms(A,0);
ms(B,0);
C.clear();
for ( int i = 0 ; i < n ; i++)
for ( int j = 0 ; j < K ; j++)
scanf("%d",&A[i][j]);
for ( int i = 0 ; i < K ; i++)
for ( int j = 0 ; j < n ; j++)
scanf("%d",&B[i][j]);
for ( int i = 0 ; i < K ; i++)
for ( int j = 0 ; j < K ; j++)
for ( int k = 0 ; k < n ; k++)
C.mat[i][j] = (C.mat[i][j] + B[i][k] * A[k][j])%6;
M = C^(n*n-1);
ms(ans,0);
for ( int i = 0  ; i < n ; i++)
for ( int j = 0 ; j < K ;j++)
for ( int k = 0 ; k < K ; k++)
ans[0][i][j] =( ans[0][i][j] + A[i][k] * M.mat[k][j])%6;
for ( int i = 0 ; i < n ; i++)
for ( int j = 0 ; j < n ; j++)
for ( int k = 0 ; k < K ; k++)
ans[1][i][j] =( ans[1][i][j] + ans[0][i][k] * B[k][j])%6;
LL sum = 0 ;
for ( int i = 0 ; i < n ; i++)
for ( int j = 0 ; j < n ; j++)
sum = sum + ans[1][i][j];
printf("%lld\n",sum);
}
#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}
``````

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