【模板】矩阵快速幂

题意简述

给定 ( n*n )的矩阵A，求 ( A^k )

代码

#include <cstdio>
typedef long long ll;
const int mod = 1000000007;
int n;
ll k;
struct Matrix
{
	int a[200][200];
	Matrix& operator =(const Matrix& x)
	{
		for (register int i = 1; i <= n; ++i)
			for (register int j = 1; j <= n; ++j)
				a[i][j] = x.a[i][j];
		return *this;
	}
};
Matrix a, b;
Matrix Mul(const Matrix& x, const Matrix& y)
{
	Matrix s;
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j)
			s.a[i][j] = 0;
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j)
			for (register int k = 1; k <= n; ++k)
				s.a[i][j] = (s.a[i][j] + (ll)x.a[i][k] * y.a[k][j] % mod) % mod;
	return s;
}
Matrix _pow(Matrix x, ll y)
{
	Matrix s;
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j)
			s.a[i][j] = (i == j);
	for (; y; y >>= 1, x = Mul(x, x)) if (y & 1) s = Mul(s, x);
	return s;
}
int main()
{
	scanf("%d%lld", &n, &k);
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j)
			scanf("%d", &a.a[i][j]);
	b = _pow(a, k);
	for (register int i = 1; i <= n; ++i)
	{
		for (register int j = 1; j < n; ++j)
			printf("%d ", b.a[i][j]);
		printf("%d
", b.a[i][n]);
	}
}