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

题目链接

题意:Step 1: Calculate a new NN matrix C = AB. 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’.

思路: 水题。。就一个trick...

朴素的矩阵乘法的复杂度是n^3的。。。按照题目的顺序求的话。。。。求M矩阵时会超时。。。(而且1000*1000的数组也不能放到结构体里...?

我们可以根据矩阵乘法的结合律。

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

然后就可以搞了。

  1/* ***********************************************
  2Author :111qqz
  3Created Time :Wed 19 Oct 2016 08:39:49 PM CST
  4File Name :code/hdu/4965.cpp
  5************************************************ */
  6#include <cstdio>
  7#include <cstring>
  8#include <iostream>
  9#include <algorithm>
 10#include <vector>
 11#include <queue>
 12#include <set>
 13#include <map>
 14#include <string>
 15#include <cmath>
 16#include <cstdlib>
 17#include <ctime>
 18#define fst first
 19#define sec second
 20#define lson l,m,rt<<1
 21#define rson m+1,r,rt<<1|1
 22#define ms(a,x) memset(a,x,sizeof(a))
 23typedef long long LL;
 24#define pi pair < int ,int >
 25#define MP make_pair
 26using namespace std;
 27const double eps = 1E-8;
 28const int dx4[4]={1,0,0,-1};
 29const int dy4[4]={0,-1,1,0};
 30const int inf = 0x3f3f3f3f;
 31const int N=1E3+5;
 32int n,K;
 33struct Mat
 34{
 35    int mat[10][10];
 36    void clear()
 37    {
 38	ms(mat,0);
 39    }
 40}C,M;
 41Mat operator * ( Mat a, Mat b)
 42{
 43    Mat ans;
 44    ans.clear();
 45    for ( int i = 0 ; i < K ; i++)
 46    {
 47	for ( int j = 0 ; j < K ; j++)
 48	{
 49	    for ( int k = 0 ; k < K ; k++)
 50	    {
 51		ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j])%6;
 52	    }
 53	}
 54    }
 55    return ans;
 56}
 57Mat operator ^ ( Mat a,int b)
 58{
 59    Mat ans;
 60    ans.clear();
 61    for ( int i = 0 ; i < K ; i++) ans.mat[i][i]  = 1;
 62    while (b>0)
 63    {
 64	if (b&1) ans = ans * a;
 65	b = b >> 1;
 66	a = a * a;
 67    }
 68    return ans;
 69}
 70int A[N][10],B[10][N],ans[2][N][N];
 71int main()
 72{
 73	#ifndef  ONLINE_JUDGE 
 74	freopen("code/in.txt","r",stdin);
 75  #endif
 76	while (~scanf("%d%d",&n,&K))
 77	{
 78	    if (n==0&&K==0) break;
 79	    ms(A,0);
 80	    ms(B,0);
 81	    C.clear();
 82	    for ( int i = 0 ; i < n ; i++)
 83		for ( int j = 0 ; j < K ; j++)
 84		    scanf("%d",&A[i][j]);
 85	    for ( int i = 0 ; i < K ; i++)
 86		for ( int j = 0 ; j < n ; j++)
 87		    scanf("%d",&B[i][j]);
 88	    for ( int i = 0 ; i < K ; i++)
 89		for ( int j = 0 ; j < K ; j++)
 90		    for ( int k = 0 ; k < n ; k++)
 91			C.mat[i][j] = (C.mat[i][j] + B[i][k] * A[k][j])%6;
 92	    M = C^(n*n-1);
 93	    ms(ans,0);
 94	    for ( int i = 0  ; i < n ; i++)
 95		for ( int j = 0 ; j < K ;j++)
 96		    for ( int k = 0 ; k < K ; k++)
 97			ans[0][i][j] =( ans[0][i][j] + A[i][k] * M.mat[k][j])%6;
 98	    for ( int i = 0 ; i < n ; i++)
 99		for ( int j = 0 ; j < n ; j++)
100		    for ( int k = 0 ; k < K ; k++)
101			ans[1][i][j] =( ans[1][i][j] + ans[0][i][k] * B[k][j])%6;
102	    LL sum = 0 ;
103	    for ( int i = 0 ; i < n ; i++)
104		for ( int j = 0 ; j < n ; j++)
105		    sum = sum + ans[1][i][j];
106	    printf("%lld\n",sum);
107	}
108  #ifndef ONLINE_JUDGE  
109  fclose(stdin);
110  #endif
111    return 0;
112}