# BM Linear Recursive Formula (Plate Finishing)

### Experience

Teacher Du's magic is obscene. Most of them don't understand the principle, but they just need to set a board.

### Plate finishing

Taking the bullock multi-school B question in 2019 as an example, the vector named ans is inserted from dp[0] to dp[2*k].

When item n is required, direct linear_seq::(gao,n) is enough.

If you don't put item 0, it's linear_seq::(gao,n-1)

Generally, 7-8 guarantees of violence are absolutely correct. If you don't feel confident, you can make several more guarantees.

dp linear recurrence formulas commonly used for fast power of matrices

```#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) { // Initial value 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;
}```

Added by silentmajority3 on Tue, 01 Oct 2019 02:58:59 +0300