HDU 4965 Fast Matrix Calculation (Matrix Fast power)

HDU 4965 Fast Matrix Calculation (Matrix Fast power)
HDU 4965 Fast Matrix Calculation

Question Link

Multiply the matrix by AxBxAxB... by nn, it can be changed to Ax (BxAxBxA ...) xB: returns nn-1 in the middle. In this way, a matrix in the middle is only 6x6, so that we can use the matrix's quick power.

Code:

#include 
 
  #include 
  
   const int N = 1005;const int M = 10;int n, m;int A[N][M], B[M][N], C[M][M], CC[N][N];int ans[M][M];void tra() {    memset(CC, 0, sizeof(CC));    for (int i = 0; i < m; i++) {    for (int j = 0; j < m; j++) {        CC[i][j] = 0;        for (int k = 0; k < m; k++) {        CC[i][j] = (CC[i][j] + C[i][k] * C[k][j]) % 6;        }    }    }    for (int i = 0; i < m; i++)    for (int j = 0; j < m; j++)        C[i][j] = CC[i][j];}void mul() {    for (int i = 0; i < m; i++) {    for (int j = 0; j < m; j++) {        CC[i][j] = 0;        for (int k = 0; k < m; k++) {        CC[i][j] = (CC[i][j] + ans[i][k] * C[k][j]) % 6;        }    }    }    for (int i = 0; i < m; i++)    for (int j = 0; j < m; j++)        ans[i][j] = CC[i][j];}void pow_mod(int k) {    memset(ans, 0, sizeof(ans));    for (int i = 0; i < m; i++)    ans[i][i] = 1;    while (k) {    if (k&1) mul();    tra();    k >>= 1;    }}void init() {    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]);}int solve() {    for (int i = 0; i < m; i++) {    for (int j = 0; j < m; j++) {        C[i][j] = 0;        for (int k = 0; k < n; k++) {        C[i][j] = (C[i][j] + B[i][k] * A[k][j]) % 6;        }    }    }    pow_mod(n * n - 1);    for (int i = 0; i < m; i++) {    for (int j = 0; j < m; j++) {        C[i][j] = ans[i][j];    }    }    for (int i = 0; i < n; i++) {    for (int j = 0; j < m; j++) {        CC[i][j] = 0;        for (int k = 0; k < m; k++) {        CC[i][j] = (CC[i][j] + A[i][k] * C[k][j]) % 6;        }    }    }    for (int i = 0; i < n; i++)    for (int j = 0; j < m; j++)        A[i][j] = CC[i][j];    int ans = 0;    for (int i = 0; i < n; i++) {    for (int j = 0; j < n; j++) {        int sum = 0;        for (int k = 0; k < m; k++) {        sum = (sum + A[i][k] * B[k][j]) % 6;        }        ans += sum;    }    }    return ans;}int main() {    while (~scanf("%d%d", &n, &m) && n || m) {    init();    printf("%d\n", solve());    }    return 0;}