bm求线性递推式(dls的板子)

发布于 2018-06-03  151 次阅读


dls的bm板子适合求一切递推式的题目,只要把递推式前几项求出来带入进去就行

附上板子:

#include <bits/stdc++.h>  

using namespace std;  
#define rep(i,a,n) for (long long i=a;i<n;i++)  
#define per(i,a,n) for (long long 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) ((long long)(x).size())  
typedef vector<long long> VI;  
typedef long long ll;  
typedef pair<long long,long long> 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;}  
// head  

long long _,n;  
namespace linear_seq  
{  
    const long long N=10010;  
    ll res[N],base[N],_c[N],_md[N];  

    vector<long long> Md;  
    void mul(ll *a,ll *b,long long 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 (long long 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];  
    }  
    long long 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;  
        long long 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 (long long p=pnt;p>=0;p--)  
        {  
            mul(res,res,k);  
            if ((n>>p)&1)  
            {  
                for (long long 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);  
        long long 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;  
    }  
    long long 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()  
{  
    while(~scanf("%I64d", &n))  
    {   printf("%I64d\n",linear_seq::gao(VI{1,5,11,36,95,281,781,2245,6336,18061, 51205},n-1));  //前几项带入进去就完事了, n-1就是求递推式的哪一项,**注意这里初始化的方式c++11不支持**
    } 
    return 0;
}  

附上徐州邀请赛H题题解:

#include <bits/stdc++.h>

using namespace std;
#define rep(i,a,n) for (long long i=a;i<n;i++)
#define per(i,a,n) for (long long 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) ((long long)(x).size())
typedef vector<long long> VI;
typedef long long ll;
typedef pair<long long,long long> PII;
const ll mod=1e9+7;
const int maxn=1e4+10;
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

long long _,n;
namespace linear_seq
{
    const long long N=10010;
    ll res[N],base[N],_c[N],_md[N];

    vector<long long> Md;
    void mul(ll *a,ll *b,long long 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 (long long 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];
    }
    long long 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;
        long long 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 (long long p=pnt;p>=0;p--)
        {
            mul(res,res,k);
            if ((n>>p)&1)
            {
                for (long long 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);
        long long 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;
    }
    long long 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 pp[maxn],qq[maxn];
int main()
{
    ll k;
    while(~scanf("%I64d%I64d", &n,&k))
    {
        VI aa;
        ll nn=n;
        for(int i=1;i<=n;i++)
            scanf("%lld",&pp[i]);
        for(int i=1;i<=nn;i++)
            scanf("%lld",&qq[i]);
        ll y=0,t=5;
        while(t--)
        {
            y=0;
            for(int i=1;i<=n;i++)
            y=(y+pp[i]*qq[nn-i+1])%mod;
        qq[++nn]=y;
        aa.push_back(y);
        //cout<<"y="<<y<<endl;
        }
        printf("%I64d\n",linear_seq::gao(aa,k-n-1));
    }
    return 0;
}

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