HDU4965-Fast matrix calculation)

Question Link

N * k matrix A and K * n matrix B, C = a * B. M = (c) ^ (N * n), the sum of each element in matrix m (mod6 for each element)

Idea: Because N is up to 1000, we cannot directly use the Matrix to quickly calculate the N * n power of AB, but we can slightly convert the formula, M = AB * AB... * AB = A * (BA )*... * (BA) * B, so that the N * n-1 power of Ba can be solved using the matrix's rapid power, and then multiplied by A and B respectively.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1005;const int MOD = 6;struct mat{    int s[6][6];    mat() {        memset(s, 0, sizeof(s));     }    mat operator * (const mat& c) {        mat ans;         memset(ans.s, 0, sizeof(ans.s));        for (int i = 0; i < 6; i++)            for (int j = 0; j < 6; j++)                for (int k = 0; k < 6; k++)                    ans.s[i][j] = (ans.s[i][j] + s[i][k] * c.s[k][j]) % MOD;        return ans;    }};int a[N][6], b[6][N], temp[N][N], sum[N][N];int n, m;void init() {    memset(a, 0, sizeof(a));    memset(b, 0, sizeof(b));    for (int i = 0; i < n; i++)        for (int j = 0; j < m; j++)            scanf("%d", &a[i][j]);    for (int i = 0; i < m; i++)        for (int j = 0; j < n; j++)            scanf("%d", &b[i][j]);}mat pow_mod(mat c, int k) {    if (k == 1)        return c;    mat a = pow_mod(c, k / 2);    mat ans = a * a;    if (k % 2)        ans = ans * c;    return ans;}int main() {    while (scanf("%d%d", &n, &m)) {        if (n == 0 && m == 0)             break;        init();        mat c;        for (int i = 0; i < m; i++)            for (int j = 0; j < m; j++)                for (int k = 0; k < n; k++)                    c.s[i][j] = (c.s[i][j] + b[i][k] * a[k][j]) % MOD;        int cnt = n * n - 1;        mat ans = pow_mod(c, cnt);        memset(temp, 0, sizeof(temp));        for (int i = 0; i < n; i++)             for (int j = 0; j < m; j++)                for (int k = 0; k < m; k++)                     temp[i][j] = (temp[i][j] + a[i][k] * ans.s[k][j]) % MOD;        cnt = 0;        memset(sum, 0, sizeof(sum));        for (int i = 0; i < n; i++)             for (int j = 0; j < n; j++) {                for (int k = 0; k < m; k++)                     sum[i][j] = (sum[i][j] + temp[i][k] * b[k][j]) % MOD;                cnt += sum[i][j];            }        printf("%d\n", cnt);     }    return 0;}

HDU4965-Fast matrix calculation)