Fast Matrix Calculation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072
K (Java/Others)
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.
has N integers between 0 and 5, representing matrix B.
The end of input is indicated by N = K = 0.
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
14 56
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1005;
#define mod 6
struct Matrix
{
int mat[10][10];
} ;
Matrix unit_matrix, c;
int A[N][10], B[10][N], aa[N][10], bb[N][N];
int n, k;
Matrix mul(Matrix a, Matrix b) //矩阵相乘
{
Matrix res;
for(int i = 0; i < k; i++)
for(int j = 0; j < k; j++)
{
res.mat[i][j] = 0;
for(int t = 0; t < k; t++)
{
res.mat[i][j] += a.mat[i][t] * b.mat[t][j];
res.mat[i][j] %= mod;
}
}
return res;
}
Matrix pow_matrix(Matrix a, int m) //矩阵快速幂
{
Matrix res = unit_matrix;
while(m != 0)
{
if(m & 1)
res = mul(res, a);
a = mul(a, a);
m >>= 1;
}
return res;
}
int main()
{
int i, j, t;
while(~scanf("%d%d",&n,&k) && (n+k))
{
for(i = 0; i < n; i++)
for(j = 0; j < k; j++)
scanf("%d",&A[i][j]);
for(i = 0; i < k; i++)
for(j = 0; j < n; j++)
scanf("%d",&B[i][j]);
//初始化单位矩阵
for(i = 0; i < k; i++)
for(j = 0; j < k; j++)
unit_matrix.mat[i][j] = 0;
for(i = 0; i < n; i++)
unit_matrix.mat[i][i] = 1;
for(i = 0; i < k; i++) //求B*A
{
for(j = 0; j < k; j++)
{
c.mat[i][j] = 0;
for(t = 0; t < n; t++)
{
c.mat[i][j] += B[i][t] * A[t][j];
c.mat[i][j] %= mod;
}
}
}
Matrix ans = pow_matrix(c, n*n-1);
for(i = 0; i < n; i++)
{
for(j = 0; j < k; j++)
{
aa[i][j] = 0;
for(t = 0; t < k; t++)
{
aa[i][j] += A[i][t] * ans.mat[t][j];
aa[i][j] %= mod;
}
}
}
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
bb[i][j] = 0;
for(t = 0; t < k; t++)
{
bb[i][j] += aa[i][t] * B[t][j];
bb[i][j] %= mod;
}
}
}
int sum = 0;
for(i = 0; i < n; i++)
for(j = 0; j <n; j++)
sum += bb[i][j];
printf("%d\n", sum);
}
return 0;
}