BM线性递推

水深无声 2023-06-04 11:59 41阅读 0赞

#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));
            }
        }
    }

转载于:https://www.cnblogs.com/ACMerszl/p/11275755.html