HDU 4965 Fast Matrix Calculation(矩阵高速幂)

题目大意：给你两个数字n和k，然后给你两个矩阵a是n*k的和b是k*n的，矩阵c = a*b，让你求c^(n*n)。

直接求的话c是n*n的矩阵所以是1000*1000。会超时的啊。

能够转化一下：(a*b)^(n*n)=a*(b*a)^(n*n-1)*b。b*a能够得到一个k*k的矩阵，k非常小所以不会超时。高速幂一下就能够了啊。

Fast Matrix Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 717    Accepted Submission(s): 355

Problem Description

One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her.

Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is not less than 0 and not greater than 5. Then he does similar thing again with a bit difference: he keeps throwing his dice for N times and each time repeat it for K times to write down a K*N matrix B, in which each element is not less than 0 and not greater than 5. With the two matrix A and B formed, Alice’s task is to perform the following 4-step calculation.

Step 1: Calculate a new N*N matrix C = A*B.
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’. 

Bob just made this problem for kidding but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math.

 

Input

The input contains several test cases. Each test case starts with two integer N and K, indicating the numbers N and K described above. Then N lines follow, and each line has K integers between 0 and 5, representing matrix A. Then K lines follow, and each line has N integers between 0 and 5, representing matrix B.

The end of input is indicated by N = K = 0.

 

Output

For each case, output the sum of all the elements in M’ in a line.

 

Sample Input

4 2
5 5
4 4
5 4
0 0
4 2 5 5
1 3 1 5
6 3
1 2 3
0 3 0
2 3 4
4 3 2
2 5 5
0 5 0
3 4 5 1 1 0
5 3 2 3 3 2
3 1 5 4 5 2
0 0

 

Sample Output

14
56

 

Author

SYSU

 

Source

2014 Multi-University Training Contest 9

 

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 1010;

#define mod 6

struct matrix
{
    int f[6][6];
};

int A[maxn][maxn];
int B[maxn][maxn];
int C[maxn][maxn];
int D[maxn][maxn];

matrix mul(matrix a, matrix b, int n)
{
    matrix c;
    memset(c.f, 0, sizeof(c.f));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            for(int k = 0; k < n; k++) c.f[i][j] += a.f[i][k]*b.f[k][j];
            c.f[i][j] %= 6;
        }
    }
    return c;
}

matrix pow_mod(matrix a, int b, int n)
{
    matrix s;
    memset(s.f, 0 , sizeof(s.f));
    for(int i = 0; i < n; i++) s.f[i][i] = 1;
    while(b)
    {
        if(b&1) s = mul(s, a, n);
        a = mul(a, a, n);
        b >>= 1;
    }
    return s;
}

int main()
{
    int n, k;
    while(cin >>n>>k)
    {
        if(!n && !k) break;
        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]);
        matrix c;
        memset(c.f, 0, sizeof(c.f));
        for(int i = 0; i < k; i++)
        {
            for(int j = 0; j < k; j++)
            {
                for(int jj = 0; jj < n; jj++) c.f[i][j] += B[i][jj]*A[jj][j];
                c.f[i][j] %= mod;
            }
        }
        c = pow_mod(c, n*n-1, k);
        memset(C, 0, sizeof(C));
        memset(D, 0, sizeof(D));
        for(int  i = 0; i < n; i++)
        {
            for(int j = 0; j < k; j++)
            {
                for(int jj = 0; jj < k; jj++) C[i][j] += A[i][jj]*c.f[jj][j];
                C[i][j] %= mod;
            }
        }
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                for(int jj = 0; jj < k; jj++) D[i][j] += C[i][jj]*B[jj][j];
                D[i][j] %= mod;
                ans += D[i][j];
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}