HDU4965 Fast Matrix Calculation —— 矩阵乘法、快速幂

时间：2021-06-14 03:48:56

题目链接：https://vjudge.net/problem/HDU-4965

Fast Matrix Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2057    Accepted Submission(s): 954

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

题意：

A为矩阵n*k，B为矩阵k*n，其中n<=1e3， k<=6。求 (A*B)^(n*n) 矩阵中所有项模6之和。

题解：

1.如果先计算 A*B的矩阵，然后再快速幂，那么矩阵最大可达：1e3*1e3，计算量是十分庞大的。

2. (A*B)^(n*n) = A*B*A*B*A*B*A*B……*A*B = A*(B*A)^(n*n-1)*B，其中B*A最大只为6*6，因而可先用矩阵快速幂算出(B*A)^(n*n-1)，然后再计算A*(B*A)^(n*n-1)*B。

原理：矩阵乘法虽然不满足交换律，但是乘法的执行顺序却可以任意。

代码如下：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <cmath>

 #include <queue>

 #include <stack>

 #include <map>

 #include <string>

 #include <set>

 using namespace std;

 typedef long long LL;

 const int INF = 2e9;

 const LL LNF = 9e18;

 //const int MOD = 1000000007;

 const int MAXN = 1e6+;

 const int MOD = ;

 const int Size = ;

 struct MA

 {

     int mat[Size][Size];

     void init()

     {

         for(int i = ; i<Size; i++)

         for(int j = ; j<Size; j++)

             mat[i][j] = (i==j);

     }

 };

 MA mul(MA x, MA y)

 {

     MA ret;

     memset(ret.mat, , sizeof(ret.mat));

     for(int i = ; i<Size; i++)

     for(int j = ; j<Size; j++)

     for(int k = ; k<Size; k++)

         ret.mat[i][j] += x.mat[i][k]*y.mat[k][j]%MOD, ret.mat[i][j] %= MOD;

     return ret;

 }

 MA qpow(MA x, LL y)

 {

     MA s;

     s.init();

     while(y)

     {

         if(y&) s = mul(s, x);

         x = mul(x, x);

         y >>= ;

     }

     return s;

 }

 int a[][], b[][], c[][], M1[][], M2[][];

 int main()

 {

     int n, k;

     while(scanf("%d%d", &n,&k) &&(n||k))

     {

         memset(a, , sizeof(a));

         memset(b, , sizeof(b));

         memset(c, , sizeof(c));

         for(int i = ; i<n; i++)

         for(int j = ; j<k; j++)

             scanf("%d", &a[i][j]);

         for(int i = ; i<k; i++)

         for(int j = ; j<n; j++)

             scanf("%d", &b[i][j]);

         for(int i = ; i<k; i++)

         for(int j = ; j<k; j++)

         for(int t = ; t<n; t++)

             c[i][j] += b[i][t]*a[t][j]%MOD, c[i][j] %= MOD;

         MA s;

         memcpy(s.mat, c, sizeof(s.mat));

         s = qpow(s, n*n-);

         memcpy(c, s.mat, sizeof(c));

         memset(M1, , sizeof(M1));

         for(int i = ; i<n; i++)

         for(int j = ; j<k; j++)

         for(int t = ; t<k; t++)

             M1[i][j] += a[i][t]*c[t][j], M1[i][j] %= MOD;

         memset(M2, , sizeof(M2));

         for(int i = ; i<n; i++)

         for(int j = ; j<n; j++)

         for(int t = ; t<k; t++)

             M2[i][j] += M1[i][t]*b[t][j], M2[i][j] %= MOD;

         int ans = ;

         for(int i = ; i<n; i++)

         for(int j = ; j<n; j++)

             ans += M2[i][j];

         printf("%d\n", ans);

     }

 }

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