参考于：
https://www.luogu.org/problemnew/solution/P4723 shadowice1984 （太难）
https://www.cnblogs.com/zhgyki/p/9671855.html （玄学）
居然还有解释：
https://www.cnblogs.com/zhouzhendong/p/Berlekamp-Massey.html
线性递推BM算法，把前面的8～10项push进去？
下面是板子，求第n项调用参数是n-1。
#include<bits/stdc++.h>
using namespace std;
#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 long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head
ll n;
namespace linear_seq {
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
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]+...
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*powmod(b,mod-2)%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*powmod(b,mod-2)%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)));
}
};
int main() {
/*push_back 进去前 8~10 项左右、最后调用 gao 得第 n 项*/
vector<int>v;
v.push_back(3);
v.push_back(9);
v.push_back(20);
v.push_back(46);
v.push_back(106);
v.push_back(244);
v.push_back(560);
v.push_back(1286);
v.push_back(2956);
v.push_back(6794);
int nCase;
scanf("%d", &nCase);
while(nCase--){
scanf("%lld", &n);
printf("%lld\n",1LL * linear_seq::gao(v,n-1) % mod);
}
}仿照上面的思路瞎搞通过。
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
typedef vector<int> vi;
typedef long long ll;
const ll mod=1000000007;
ll qpow(ll x,ll n) {
ll res=1;
for(; n; x=x*x%mod,n>>=1)
if(n&1)
res=res*x%mod;
return res;
}
namespace Linear_Recurrence {
//空间无所谓，开大一点
const int N=10010;
ll res[N],base[N],c[N],md[N];
int umd[N],sizumd;
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,sizumd) c[i-k+umd[j]]=(c[i-k+umd[j]]-c[i]*md[umd[j]])%mod;
rep(i,0,k) a[i]=c[i];
}
int solve(ll n,vi a,vi b) {
//a是系数向量，b是初值向量
//递推公式形如b[n+1]=a[0]*b[n]+a[1]*b[n-1]+...
ll ans=0,pnt=0;
int k=a.size();
rep(i,0,k) md[k-1-i]=-a[i];
md[k]=1;
sizumd=0;
rep(i,0,k) {
if (md[i]!=0)
umd[sizumd++]=i;
}
rep(i,0,k) res[i]=0,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,sizumd) res[umd[j]]=(res[umd[j]]-res[k]*md[umd[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;
int sizs=s.size();
rep(n,0,sizs) {
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;
int sizB=B.size();
while (C.size()<sizB+m)
C.push_back(0);
ll t=mod-d*qpow(b,mod-2)%mod;
rep(i,0,sizB) C[i+m]=(C[i+m]+t*B[i])%mod;
L=n+1-L,B=T,b=d,m=1;
} else {
int sizB=B.size();
while (C.size()<sizB+m)
C.push_back(0);
ll t=mod-d*qpow(b,mod-2)%mod;
rep(i,0,sizB) C[i+m]=(C[i+m]+t*B[i])%mod;
++m;
}
}
return C;
}
int gao(vi A,ll n) {
vi C=BM(A);
C.erase(C.begin());
int sizC=C.size();
rep(i,0,sizC) C[i]=(mod-C[i])%mod;
return solve(n,C,vi(A.begin(),A.begin()+sizC));
}
};
ll dp[3005];
ll solve(int k,ll n) {
ll invk=qpow(k,mod-2);
dp[0]=1;
//至少取2倍以上
int c=k*2;
for(int i=1; i<=c+1; i++) {
dp[i]=0;
for(int j=max(0,i-k); j<=i-1; j++) {
dp[i]+=dp[j];
}
dp[i]=(dp[i]%mod)*invk%mod;
}
vi v;
for(int i=0; i<=c+1; i++)
v.push_back((int)dp[i]);
//从0开始，到n，这里其实是第n+1项
return Linear_Recurrence::gao(v,n);
}
int main() {
#ifdef Yinku
freopen("Yinku.in","r",stdin);
#endif // Yinku
int t,k;
ll n;
scanf("%d", &t);
while(t--) {
scanf("%d%lld", &k,&n);
if(n==-1)
printf("%lld\n",2ll*qpow(k+1,mod-2)%mod);
else
printf("%lld\n",solve(k,n));
}
}好像前2K项比较保险，前K项就WA了，前1.5K项也WA了，临界值是多少呢？反正大概取2～3倍递推可能就够了吧。