【BZOJ 4161】

1.题目链接。矩阵快速幂的多项式取模优化模板题。

#include
using namespace std;
const int maxn = 4000, mod = 1000000007;
int a[maxn + 5], p[maxn + 5], ans[maxn + 5], num[maxn + 5];
int h[maxn + 5], tmp[maxn + 5];
int n, k;

void mul(int* a, int* b, int* ans)
{
	for (int i = 0; i <= 2 * k; ++i) tmp[i] = 0;
	for (int i = 0; i < k; ++i)
		for (int j = 0; j < k; ++j)
			tmp[i + j] = (tmp[i + j] + 1LL * a[i] * b[j]) % mod;
	for (int i = 2 * k - 2; i >= k; --i)
	{
		for (int j = k - 1; j >= 0; --j)
			tmp[i - k + j] = (tmp[i - k + j] - 1LL * tmp[i] * p[j]) % mod, tmp[i - k + j] = (tmp[i - k + j] + mod) % mod;
		tmp[i] = 0;
	}
	for (int i = 0; i < k; ++i) ans[i] = tmp[i];
}
int main()
{
	/*
	   hn=a1*h[n-1]+a2*h[n-2]....ak*h[n-k]
	   这是一个线性地推，但是当k很大时，比如k>1000.这时已经不再使用矩阵快速幂解决。
	*/
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= k; ++i) scanf("%d", &a[i]);
	for (int i = 0; i < k; ++i) scanf("%d", &h[i]);
	p[k] = 1;
	for (int i = 1; i <= k; ++i) p[k - i] = mod - a[i];
	for (int i = k; i < 2 * k; ++i)
		for (int j = 1; j <= k; ++j)
		{
			h[i] = h[i] + 1LL * h[i - j] * a[j] % mod;
			h[i] %= mod;
		}
	if (n < 2 * k) return 0 * printf("%d\n", h[n]);
	int b = n - k + 1;
	num[1] = 1, ans[0] = 1;
	while (b)
	{
		if (b & 1) mul(ans, num, ans);
		mul(num, num, num);
		b >>= 1;
	}
	long long res = 0;
	for (int i = 0; i < k; ++i) res = (res + 1LL * ans[i] * h[i + k - 1]) % mod;
	printf("%lld\n", (res + mod) % mod);
	return 0;
}