- // 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);
- }
来源: http://www.bubuko.com/infodetail-2984559.html