矩阵快速幂代码:
int n; // 所有矩阵都是 n * n 的矩阵
struct matrix {
int a[100][100];
};
matrix matrix_mul(matrix A, matrix B, int mod) {
// 2 个矩阵相乘
matrix C;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
C.a[i][j] = 0;
for (int k = 0; k < n; ++k) {
C.a[i][j] += A.a[i][k] * B.a[k][j] % mod;
C.a[i][j] %= mod;
}
}
}
return C;
}
matrix unit() {
// 返回一个单位矩阵
matrix res;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j) {
res.a[i][j] = 1;
} else {
res.a[i][j] = 0;
}
}
}
return res;
}
matrix matrix_pow(matrix A, int n, int mod) {
// 快速求矩阵 A 的 n 次方
matrix res = unit(), temp = A;
for (; n; n /= 2) {
if (n & 1) {
res = matrix_mul(res, temp, mod);
}
temp = matrix_mul(temp, temp, mod);
}
return res;
}