BZOJ 4161 Shlw loves matrixI ——特征多项式

矩阵乘法递推的新姿势。

叉姐论文里有讲到

利用特征多项式进行递推，然后可以做到k^2logn

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define p 1000000007
ll c[4010];
int n,k,u[2010],ans,x,cnt;
struct  P{int s[2010];}f,t;
void mult(P & a,const P & b)
{
    F(i,0,k+k-2) c[i]=0;
    F(i,0,k-1) F(j,0,k-1)
    {
        c[i+j]+=1LL*a.s[i]*b.s[j];
        if (c[i+j]>=1LL<<62) c[i+j]%=p; 
    }
    D(i,k+k-1,0) if (c[i]%=p,i>=k)
    {
        F(j,0,k-1)
        {
            c[i-1-j]+=c[i]*u[j];
            if (c[i-1-j]>=1LL<<62) c[i-1-j]%=p;
        }
        c[i]=0; 
    }
    F(i,0,k-1) a.s[i]=c[i];
}
int main()
{
    scanf("%d%d",&n,&k);
    F(i,0,k-1)
    {
        scanf("%d",&u[i]);
        u[i]=(u[i]%p+p)%p;
    }
    for (t.s[1]=f.s[0]=1;n;n>>=1,mult(t,t)) if (n&1) mult(f,t);
    F(i,0,k-1)
    {
        scanf("%d",&x);
        x=(x%p+p)%p;
        ans=(ans+1LL*x*f.s[i])%p;
    }
    printf("%d\n",ans);
}