HDU 4965 Fast Matrix Calculation（矩阵快速幂)

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n
n - 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
[M], B[M]
, C[M][M], CC

;
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;
}