2019牛客暑期多校训练营（第二场） - B - Eddy Walker 2 - BM算法

参考于：

https://www.luogu.org/problemnew/solution/P4723 shadowice1984 （太难）

https://www.cnblogs.com/zhgyki/p/9671855.html （玄学）

居然还有解释：

https://www.cnblogs.com/zhouzhendong/p/Berlekamp-Massey.html

线性递推BM算法，把前面的8～10项push进去？

下面是板子，求第n项调用参数是n-1。

#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

ll n;
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)));
    }
};

int main() {
    /*push_back 进去前 8~10 项左右、最后调用 gao 得第 n 项*/
    vector<int>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);
    int nCase;
    scanf("%d", &nCase);
    while(nCase--){
        scanf("%lld", &n);
        printf("%lld\n",1LL * linear_seq::gao(v,n-1) % mod);
    }
}

仿照上面的思路瞎搞通过。

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
typedef vector<int> vi;
typedef long long ll;
const ll mod=1000000007;

ll qpow(ll x,ll n) {
    ll res=1;
    for(; n; x=x*x%mod,n>>=1)
        if(n&1)
            res=res*x%mod;
    return res;
}

namespace Linear_Recurrence {

//空间无所谓，开大一点
const int N=10010;
ll res[N],base[N],c[N],md[N];

int umd[N],sizumd;
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,sizumd) c[i-k+umd[j]]=(c[i-k+umd[j]]-c[i]*md[umd[j]])%mod;

    rep(i,0,k) a[i]=c[i];
}

int solve(ll n,vi a,vi b) {
    //a是系数向量，b是初值向量
    //递推公式形如b[n+1]=a[0]*b[n]+a[1]*b[n-1]+...
    ll ans=0,pnt=0;
    int k=a.size();
    rep(i,0,k) md[k-1-i]=-a[i];
    md[k]=1;
    sizumd=0;
    rep(i,0,k) {
        if (md[i]!=0)
            umd[sizumd++]=i;
    }
    rep(i,0,k) res[i]=0,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,sizumd) res[umd[j]]=(res[umd[j]]-res[k]*md[umd[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;
    int sizs=s.size();
    rep(n,0,sizs) {
        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;
            int sizB=B.size();
            while (C.size()<sizB+m)
                C.push_back(0);
            ll t=mod-d*qpow(b,mod-2)%mod;
            rep(i,0,sizB) C[i+m]=(C[i+m]+t*B[i])%mod;
            L=n+1-L,B=T,b=d,m=1;
        } else {
            int sizB=B.size();
            while (C.size()<sizB+m)
                C.push_back(0);
            ll t=mod-d*qpow(b,mod-2)%mod;
            rep(i,0,sizB) C[i+m]=(C[i+m]+t*B[i])%mod;
            ++m;
        }
    }
    return C;
}
int gao(vi A,ll n) {
    vi C=BM(A);
    C.erase(C.begin());
    int sizC=C.size();
    rep(i,0,sizC) C[i]=(mod-C[i])%mod;
    return solve(n,C,vi(A.begin(),A.begin()+sizC));
}
};

ll dp[3005];

ll solve(int k,ll n) {
    ll invk=qpow(k,mod-2);
    dp[0]=1;
    //至少取2倍以上
    int c=k*2;
    for(int i=1; i<=c+1; i++) {
        dp[i]=0;
        for(int j=max(0,i-k); j<=i-1; j++) {
            dp[i]+=dp[j];
        }
        dp[i]=(dp[i]%mod)*invk%mod;
    }
    vi v;
    for(int i=0; i<=c+1; i++)
        v.push_back((int)dp[i]);
    //从0开始，到n，这里其实是第n+1项
    return Linear_Recurrence::gao(v,n);
}

int main() {
#ifdef Yinku
    freopen("Yinku.in","r",stdin);
#endif // Yinku
    int t,k;
    ll n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%lld", &k,&n);
        if(n==-1)
            printf("%lld\n",2ll*qpow(k+1,mod-2)%mod);
        else
            printf("%lld\n",solve(k,n));
    }
}

好像前2K项比较保险，前K项就WA了，前1.5K项也WA了，临界值是多少呢？反正大概取2～3倍递推可能就够了吧。