BM线性递推

#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

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)));

    }

};

ll dp[3000];

int main() {

    int _;

    for(scanf("%d",&_);_;_--) {

        int k;

        ll n;

        scanf("%d%lld",&k,&n);

        if(n==-1) {

            printf("%lld\n",2*powmod(k+1,mod-2)%mod);

        } else {

            VI t;

            dp[0]=1;

            t.push_back(dp[0]);

            for(int i=1;i<=2*k;i++) {

                dp[i]=0;

                for(int j=max(0,i-k);j<i;j++) {

                    dp[i]=(dp[i]+dp[j])%mod;

                }

                dp[i]=dp[i]*powmod(k,mod-2)%mod;

                t.push_back(dp[i]);

            }

            printf("%lld\n",linear_seq::gao(t,n));

        }

    }

}
巴特西

BM线性递推

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