//杜教BM
#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=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;
}
ll _,n;
namespace linear_seq
{
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<ll> 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)
{
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 f[205];
int main()
{
ll n,m;
scanf("%lld%lld",&n,&m);
for(int i=1;i<=m;i++) f[i]=1;
for(int i=m;i<=200;i++)
f[i]=(f[i-1]+f[i-m])%mod;
vector<int>v;
n++;
for(int i=1;i<=200;i++)
v.push_back(f[i]);
printf("%lld\n",linear_seq::gao(v,n-1)%mod);
}
原文地址：https://blog.51cto.com/14093713/2361531