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  • 【HDU6172】—Array Challenge(BM+常系数齐次线性递推)

    传送门

    大胆猜测这个东西有递推式
    实际上也是有的

    BMBM打出来是
    f[i]=7f[i1]4f[i2]f[i]=7f[i-1]-4f[i-2]

    然后该咋搞咋搞

    #include<bits/stdc++.h>
    using namespace std;
    #define gc getchar
    inline int read(){
    	char ch=gc();
    	int res=0,f=1;
    	while(!isdigit(ch))f^=ch=='-',ch=gc();
    	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    	return f?res:-res;
    }
    #define re register
    #define pb push_back
    #define cs const
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define poly vector<int>
    #define bg begin
    cs int mod=1e9+7,G=3;
    inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
    inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
    inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
    inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
    inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
    inline void Mul(int &a,int b){a=mul(a,b);}
    inline int ksm(int a,int b,int res=1){
    	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
    }
    inline void chemx(ll &a,ll b){a<b?a=b:0;}
    inline void chemn(ll &a,ll b){a>b?a=b:0;}
    cs int N=(1<<19)|5;
    inline poly operator +(poly a,poly b){
    	int deg=max(a.size(),b.size());
    	a.resize(deg),b.resize(deg);
    	for(int i=0;i<deg;i++)Add(a[i],b[i]);
    	return a;
    }
    inline poly operator -(poly a,poly b){
    	int deg=max(a.size(),b.size());
    	a.resize(deg),b.resize(deg);
    	for(int i=0;i<deg;i++)Dec(a[i],b[i]);
    	return a;
    }
    inline poly operator *(cs poly &a,cs poly &b){
    	int deg=(int)a.size()+(int)b.size()-1;
    	poly c(deg,0);
    	for(int i=0;i<a.size();i++)
    	for(int j=0;j<b.size();j++)
    	Add(c[i+j],mul(a[i],b[j]));
    	return c;
    }
    inline poly operator %(poly a,poly b){
    	int n=(int)a.size()-1,m=(int)b.size()-1;
    	if(n<m)return a;
    	for(int i=n;i>=m;i--){
    		int x=a[i];
    		for(int j=i,k=m;~k;k--,j--)Dec(a[j],mul(b[k],x));
    	}
    	a.resize(m);return a;
    }
    inline poly ksm(poly a,ll b,poly res,poly Mod){
    	for(;b;b>>=1,a=a*a%Mod)if(b&1)res=res*a%Mod;
    	return res;
    }
    namespace Cas{
    	poly f;int n;
    	inline void init(poly coef){
    		n=coef.size()-1;
    		f.resize(n+1);
    		for(int i=1;i<=n;i++)f[n-i]=dec(0,coef[i]),cout<<coef[i]<<" ";
    		f[n]=1;
    	}
    	inline int calc(int *v,ll k){
    		if(k<n)return v[k+1];
    		poly g(2),res(1,1);g[1]=1;
    		res=ksm(g,k,res,f);
    		int anc=0;
    		for(int i=0;i<n;i++)Add(anc,mul(v[i+1],res[i]));
    		return anc;
    	}
    }
    namespace B_M{
    	poly r[N];
    	int fail[N],del[N],a[N],n,cnt;
    	inline void update(int i){
    		++cnt;
    		int MUL=mul(dec(a[i],del[i]),ksm(dec(a[fail[cnt-2]],del[fail[cnt-2]]),mod-2));
    		r[cnt].resize(i-fail[cnt-2],0);
    		r[cnt].pb(MUL);
    		for(int j=1;j<r[cnt-2].size();j++){
    			r[cnt].pb(mul(mod-r[cnt-2][j],MUL));
    		}
    		r[cnt]=r[cnt]+r[cnt-1];
    	}
    	inline void BM(){
    		for(int i=1;i<=9;i++){
    			for(int j=1;j<r[cnt].size();j++){
    				Add(del[i],mul(a[i-j],r[cnt][j]));
    			}
    			if(a[i]!=del[i]){
    				fail[cnt]=i;
    				if(!cnt)r[++cnt].resize(i+1);
    				else update(i);
    			}
    		}
    		Cas::init(r[cnt]);
    	}
    	ll h[N],b[N];
    	inline void init(){
    		h[0]=2,h[1]=3,h[2]=6;n=9;
    		for(int i=3;i<=12;++i)h[i]=h[i-1]*4+h[i-2]*17-h[i-3]*12-16;
    		for(int i=1;i<=11;++i)b[i]=h[i+1]*h[i]*3+h[i+1]*h[i-1]*9+h[i]*h[i]*9+h[i]*h[i-1]*27-h[i+1]*18-h[i]*126-h[i-1]*81+192;
    		for(int i=1;i<=n;i++)a[i]=((ll)sqrt(b[i]+(1ll<<(i*2))))%mod;
    		BM();
    	}
    }
    int main(){
    	B_M::init();
    	int T=read();
    	while(T--){
    		ll k;
    		scanf("%lld",&k);
    		cout<<Cas::calc(B_M::a,k-1)<<'
    ';
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328679.html
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