BM求线性递推式(板子整理)

心得

杜老师的奇技淫巧，绝大多数不懂原理，但只要套板子就行了

板子整理

以2019牛客多校B题为例，求出dp[0]到dp[2*k]塞入名为ans的vector

要求第n项时，直接linear_seq::(gao,n)即可，

如果不放第0项的话，就是linear_seq::(gao,n-1)

对于an=an-1+an-2+an-3这种依赖项为3项的线性递推式，似乎至少要塞入6项

常用于矩阵快速幂的dp线性递推式的问题

#include<bits/stdc++.h>
 
using namespace std;

typedef long long ll;

const ll mod=1e9+7;
const int N=1024;

ll modpow(ll a,ll b,ll mod) {ll res=1;a%=mod; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll inv(ll x){return modpow(x,mod-2,mod);}

namespace linear_seq {
	#define rep(i,a,n) for (int i=a;i<n;i++)
	#define per(i,a,n) for (int i=n-1;i>=a;i--)	
	#define pb push_back
	#define mp make_pair
	#define all(x) (x).begin(),(x).end()
	#define fi first
	#define se second
	#define SZ(x) ((int)(x).size())
	typedef vector<int> VI;
	typedef pair<int,int> PII;
	typedef long long ll;
	const ll mod=1e9+7;
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];
 	ll modpow(ll a,ll b,ll mod) {ll res=1;a%=mod; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    vector<int> Md;
    void mul(ll *a,ll *b,int k) {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(b));
        ll ans=0,pnt=0;
        int k=SZ(a);
       // assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (int p=pnt;p>=0;p--) {
            mul(res,res,k);
            if ((n>>p)&1) {
                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s) {
        VI C(1,1),B(1,1);
        int L=0,m=1,b=1;
        rep(n,0,SZ(s)) {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                VI T=C;
                ll c=mod-d*modpow(b,mod-2,mod)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*modpow(b,mod-2,mod)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    int gao(VI a,ll n) {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};
vector<int>ans;
ll dp[2*N],p,n;
int t,k;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%lld",&k,&n);
		if(n==-1)
		{
			printf("%lld\n",2*inv(k+1)%mod);//2/(k-1)
			continue;
		}
		p=inv(k);
		dp[0]=1;//dp[n]=1/k dp[n-1] +1/k dp[n-2]+...+1/k dp[n-k]
		ans.clear();
		ans.push_back(dp[0]); 
		for(int i=1;i<=2*k;++i)
		{
			dp[i]=0;
			for(int j=max(i-k,0);j<i;++j)
			{
				dp[i]=(dp[i]+dp[j])%mod;
			}
			dp[i]=dp[i]*p%mod;//1/k
			ans.push_back(dp[i]); 
		}
		printf("%d\n",linear_seq::gao(ans,n));
	}
	return 0;
}