bzoj4161 Shlw loves matrixI（常系数线性递推）

$Description$
给定数列 {hn}前k项，其后每一项满足
$hn=a1\ast h(n-1)+a2\ast h(n-2)+...+ak\ast h(n-k)$
其中 a1,a2…ak 为给定数列。请计算 h(n)，并将结果对 1000000007 取模输出。
$Input$
第 1 行包含两个整数 $n,k$
第 2 行包含 k 个整数 $a1,a2...ak$
第 3 行包含 k 个整数$h[0],h[1],…,h[k-1] $
$n<=10^9;k<=2000;abs(hi)<=10^9;abs(ai)<=10^9$
$Output$
一行一个整数 hn mod 1000000007
$SampleInput$
6 4
3 -1 0 4
-2 3 1 5
$SampleOutput$
73
$HINT$
$Source$
By submittersubmitter

---

常系数线性递推模板

#include
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair
#define Pil pair
#define Pli pair
#define Pll pair
#define pb push_back 
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k))
#define ll long long
int rd()
{
    int sum = 0;char c = getchar();bool flag = true;
    while(c < '0' || c > '9') {if(c == '-') flag = false;c = getchar();}
    while(c >= '0' && c <= '9') sum = sum * 10 + c - 48,c = getchar();
    if(flag) return sum;
    else return -sum;
}
const int p = 1e9+7;
int n,k;
int a[4010],B[4010],g[4010];
int f[4010],c[4010];
inline int calc(int a,int b){return (a+b)%p;}
inline int mul(int a,int b){return 1ll*a*b%p;}
inline void mo(int &x){x = (x%p+p)%p;}
void change(int *a,int *b)
{
	rep(i,0,2*k-2) c[i] = 0;
	rep(i,0,k-1) rep(j,0,k-1) c[i+j] = calc(c[i+j],mul(a[i],b[j]));
	repp(i,2*k-2,k)
	if(c[i])
	{
		rep(j,i-k,i-1) c[j] = calc(c[j],mul(c[i],B[j-i+k+1]));
		c[i] = 0;
	}
	rep(i,0,k-1) a[i] = c[i];
}
void work(int n)
{
	for(int i = 1;i <= n;i <<= 1)
	{
		if(n&i) change(f,g);
		change(g,g);
	}
}
int main() 
{
	n = rd();k = rd();
	repp(i,k,1) B[i] = rd();rep(i,0,k-1) a[i] = rd();
	rep(i,1,k) mo(a[i]),mo(B[i]);
    f[0] = 1;g[1] = 1;
	work(n);
	int ans = 0;
	rep(i,0,k-1) ans = calc(ans,mul(f[i],a[i]));
	printf("%d\n",ans);
	return 0;
}