ACM-ICPC 2018 焦作赛区网络预赛L(BM递推杜教版)

...讲道理，这板子是真的牛逼
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <unordered_map>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef vector<long long> VI;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define SZ(x) ((long long)(x).size())
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll powmod(ll a,ll b)
{
    ll res=1ll;
    while(b)
    {
        if(b&1) res=res*a%mod;
        a=a*a%mod,b>>=1;
    }
    return res;
}
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) {
        FOR(i,0,k+k-1) _c[i]=0;
        FOR(i,0,k-1) if (a[i]) FOR(j,0,k-1) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--) if (_c[i])
            FOR(j,0,SZ(Md)-1) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        FOR(i,0,k-1) 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]+...
//        printf("%d\n",SZ(b));
        ll ans=0,pnt=0;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        FOR(i,0,k-1) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        FOR(i,0,k-1) if (_md[i]!=0) Md.push_back(i);
        FOR(i,0,k-1) 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;
                FOR(j,0,SZ(Md)-1) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        FOR(i,0,k-1) 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;
        FOR(n,0,SZ(s)-1) {
            ll d=0;
            FOR(i,0,L) 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);
                FOR(i,0,SZ(B)-1) 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);
                FOR(i,0,SZ(B)-1) 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());
        FOR(i,0,SZ(c)-1) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};
int main()
{
    cin.tie(0);
    cout.tie(0);
    ll n,t;
    cin>>t;
    while(t--)
    {
        scanf("%lld",&n);
        printf("%lld\n",linear_seq::gao(VI{3,9,20,46,106,244,560,1286,2956,6794,15610},n-1)%mod);
    }
    return 0;
}
ACM-ICPC 2018 焦作赛区网络预赛L(BM递推杜教版)

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