HDU 4965 Fast Matrix Calculation（矩阵快速幂)

来源：互联网发布：嬲知乎编辑：程序博客网时间：2024/05/21 09:28

HDU 4965 Fast Matrix Calculation

题目链接

矩阵相乘为AxBxAxB...乘nn次，可以变成Ax(BxAxBxA...)xB,中间乘nn - 1次，这样中间的矩阵一个只有6x6，就可以用矩阵快速幂搞了

代码：

#include <cstdio>#include <cstring>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;}

1 0