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  • 杜教筛BM(找规律)

    代码来自学长

    #include<bits/stdc++.h>
    using namespace std;
    #define rep(i,a,n) for (int i=a;i<n;i++)
    #define pb push_back
    typedef long long ll;
    #define SZ(x) ((ll)(x).size())
    typedef vector<ll> VI;
    typedef pair<ll,ll> 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
     
    ll _,n;
    namespace linear_seq {
        const ll N=10010;
        ll res[N],base[N],_c[N],_md[N];
     
        vector<ll> Md;
        void mul(ll *a,ll *b,ll 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 (ll 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];
        }
        ll solve(ll n,VI a,VI b) { 
    
            ll ans=0,pnt=0;
            ll 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 (ll p=pnt;p>=0;p--) {
                mul(res,res,k);
                if ((n>>p)&1) {
                    for (ll 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);
            ll 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;
        }
        ll 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() {
        int T;
        scanf("%d",&T);
        while (T--)
        {
            scanf("%lld",&n);
            vector<ll>v;
            v.push_back(3);
            v.push_back(9);
            v.push_back(20);
            v.push_back(46);
            v.push_back(106);
            v.push_back(244);
            v.push_back(560);
            v.push_back(1286);
            v.push_back(2956);
            v.push_back(6794);
             
            printf("%lld
    ",linear_seq::gao(v,n-1));
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/PushyTao/p/14507428.html
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