杜教bm筛
<pre><code class="language-cpp">#include <bits/stdc++.h>
#define rep(i,a,n) for (long long i=a;i<n;i++)
#define per(i,a,n) for (long long 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) ((long long)(x).size())
using namespace std;
typedef vector<long long> VI;
typedef long long ll;
typedef pair<long long,long long> PII;
const ll mod=1e9+7;
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
long long _,n;
namespace linear_seq
{
const long long N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<long long> Md;
void mul(ll *a,ll *b,long long 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 (long long 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];
}
long long 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;
long long 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 (long long p=pnt;p>=0;p--)
{
mul(res,res,k);
if ((n>>p)&1)
{
for (long long 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);
long long 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;
}
long long 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()
{
while(~scanf("%d", &n))
{ //printf("%I64d\n",linear_seq::gao(VI{1,5,11,36,95,281,781,2245,6336,18061, 51205},n-1));
printf("%d\n",linear_seq::gao(VI{1,2,2,3,3,4,4,4,5,5,5,6},n-1));
}
}</code></pre>