hdu - 4965
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.
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.OutputFor 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 题意 ：按照题目所给的要求，求最终答案
思路 ： 在一个结构体中去存一个二维矩阵时，最多可以开到 f[800][800],在往大就会直接停止运行了。
解决此问题有一个关键的地方就是 A*B^(n*n)， 这样求的话 A*B 就是1000*1000的矩阵，指定是 超时，如何展开 A*B*A*B*A*B……A*B,等于 A*(B*A)^(n*n-1),转变成为了一个 6 * 6 的矩阵。 代码 ：
int a[1005][10];
int b[10][1005];
struct mat
{
int a[6][6];
};
int k;
int c[1005][10]; mat mul(mat a, mat b){
mat r;
memset(r.a, 0, sizeof(r.a)); for(int i = 0; i < k; i++){
for(int f = 0; f < k; f++){
if(a.a[i][f]){
for(int j = 0; j < k; j++){
if(b.a[f][j]){
r.a[i][j] += a.a[i][f]*b.a[f][j];
r.a[i][j] %= 6;
}
}
}
}
}
return r;
} mat pow(mat a, int n){
mat b;
memset(b.a, 0, sizeof(b.a));
for(int i = 0; i < k; i++) b.a[i][i] = 1; while(n){
if(1 & n) b = mul(b, a);
a = mul(a, a);
n >>= 1;
}
return b;
} int main() {
int n; while(~scanf("%d%d", &n, &k)&& n+k ){
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]);
}
}
mat A;
memset(A.a, 0, sizeof(A.a));
for(int i = 0; i < k; i++){
for(int j = 0; j < k; j++){
for(int f = 0; f < n; f++){
A.a[i][j] += b[i][f]*a[f][j];
A.a[i][j] %= 6;
}
}
}
A = pow(A, n*n-1);
memset(c, 0, sizeof(c));
for(int i = 0; i < n; i++){
for(int j = 0; j < k; j++){
for(int f = 0; f < k; f++){
c[i][j] += a[i][f]*A.a[f][j];
c[i][j] %= 6;
}
}
}
int sum = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
int t = 0;
for(int f = 0; f < k; f++){
t += c[i][f]*b[f][j];
t %= 6;
}
sum += t;
}
}
printf("%d\n", sum);
} return 0;
}
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