• 类欧几里得


    参考资料:cz_xuyixuan

    要求这么一类问题

    [egin{aligned} f(a,b,c,n) &=sum_{i=0}^n leftlfloor{ai+bover c} ight floor end{aligned} ]

    开始颓柿子

    • (Case 1:a=0)

    [egin{aligned} f(a,b,c,n) &=(n+1)leftlfloor{bover c} ight floor\ end{aligned} ]

    • (Case 2:ageq c)(bgeq c)

    [egin{aligned} f(a,b,c,n) &=sum_{i=0}^n leftlfloor{ai+bover c} ight floor\ &=sum_{i=0}^n leftlfloor{(amod c)i+(bmod c)over c} ight floor+{n(n+1)over 2}leftlfloor{aover c} ight floor+(n+1)leftlfloor{bover c} ight floor\ &=f(amod c,bmod c,n)+{n(n+1)over 2}leftlfloor{aover c} ight floor+(n+1)leftlfloor{bover c} ight floor\ end{aligned} ]

    • (Case 3:Otherwise)

    (m=leftlfloor{an+bover c} ight floor),显然有(mleq n)

    [egin{aligned} f(a,b,c,n) &=sum_{i=0}^n leftlfloor{ai+bover c} ight floor\ &=sum_{i=0}^n sum_{j=1}^mleft[leftlfloor{ai+bover c} ight floorgeq j ight]\ &=sum_{i=0}^n sum_{j=0}^{m-1}left[leftlfloor{ai+bover c} ight floorgeq j+1 ight]\ &=sum_{i=0}^n sum_{j=0}^{m-1}left[ai>cj+c-b-1 ight]\ &=sum_{i=0}^n sum_{j=0}^{m-1}left[i>leftlfloor{cj+c-b-1over a} ight floor ight]\ &=sum_{j=0}^{m-1}n-leftlfloor{cj+c-b-1over a} ight floor\ &=nm-f(c,c-b-1,a,m-1)\ end{aligned} ]

    发现这个过程中(a,c)的迭代类似(gcd),所以这个算法的复杂度也是(O(log n))

    另外的两个扩展不学了……

    2019.11.4 upd:被jz姐姐欺负了之后只好去学一下扩展

    我们需要计算

    [egin{aligned} f(n,a,b,c) &=sum_{i=0}^n i^{k_1}leftlfloor{ai+bover c} ight floor^{k_2} end{aligned} ]

    • (Case 1:a=0)(leftlfloor{an+bover c} ight floor=0)

    此时(leftlfloor{ai+bover c} ight floor=0)的值始终相等,则

    [egin{aligned} f(n,a,b,c) &=leftlfloor{ai+bover c} ight floor^{k_2}sum_{i=0}^n i^{k_1} end{aligned} ]

    后面那个可以插值计算

    • (Case 2:ageq c)

    (q=leftlfloor{aover c} ight floor,r=a-qc),则

    [egin{aligned} f(n,a,b,c) &=sum_{i=0}^n i^{k_1}left(qi+leftlfloor{ri+bover c} ight floor ight)^{k_2}\ &=sum_{i=0}^n i^{k_1}sum_{j=0}^{k_2}{k_2choose j}(qi)^{k_2}leftlfloor{ri+bover c} ight floor^{k_2-j}\ &=sum_{j=0}^{k_2}{k_2choose j}q^{k_2}sum_{i=0}^n i^{k_1+j}leftlfloor{ri+bover c} ight floor^{k_2-j}\ end{aligned} ]

    递归算一下(f(n,r,b,c))即可

    • (Case 3:bgeq c)

    和上面类似,记(q=leftlfloor{bover c} ight floor,r=b-qc),则

    [egin{aligned} f(n,a,b,c) &=sum_{i=0}^n i^{k_1}left(q+leftlfloor{ri+bover c} ight floor ight)^{k_2}\ &=sum_{i=0}^n i^{k_1}sum_{j=0}^{k_2}{k_2choose j}q^{k_2}leftlfloor{ri+bover c} ight floor^{k_2-j}\ &=sum_{j=0}^{k_2}{k_2choose j}q^{k_2}sum_{i=0}^n i^{k_1}leftlfloor{ri+bover c} ight floor^{k_2-j}\ end{aligned} ]

    递归算一下(f(n,a,r,c))即可

    • (Case 4:Otherwise)

    我们令(m=leftlfloor{an+bover c} ight floor),则此时有(m<n)

    考虑将(leftlfloor{ai+bover c} ight floor^{k_2})变形,规定(0^0=0),则有

    [egin{aligned} leftlfloor{ai+bover c} ight floor^{k_2} &=sum_{j=0}^{leftlfloor{ai+bover c} ight floor-1}left((j+1)^{k_2}-j^{k_2} ight) end{aligned} ]

    则原式可变为

    [egin{aligned} f(n,a,b,c) &=sum_{i=0}^n i^{k_1}sum_{j=0}^{leftlfloor{ai+bover c} ight floor-1}left((j+1)^{k_2}-j^{k_2} ight)\ &=sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{i=0}^n i^{k_1}left[leftlfloor{ai+bover c} ight floorgeq j+1 ight]\ &=sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{i=0}^n i^{k_1}left[i>leftlfloor{cj+c-b-1over a} ight floor ight]\ &=sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{i=0}^n i^{k_1}-sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{i=0}^{leftlfloor{cj+c-b-1over a} ight floor} i^{k_1}\ end{aligned} ]

    前面的可以直接插值计算,现在考虑后面

    由于(sum_{i=0}^{leftlfloor{cj+c-b-1over a} ight floor} i^{k_1})是关于(leftlfloor{cj+c-b-1over a} ight floor)的一个(k_1+1)次多项式,我们假设这个多项式为(A(x)),则可以推得

    [egin{aligned} sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{i=0}^{leftlfloor{cj+c-b-1over a} ight floor} i^{k_1} &=sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{k=0}^{k_1+1} B_k{leftlfloor{cj+c-b-1over a} ight floor}^k end{aligned} ]

    再把前面(left((j+1)^{k_2}-j^{k_2} ight))的二项式展开,有

    [egin{aligned} sum_{j=0}^{m-1}left((j+1)^{k_2}-j^{k_2} ight)sum_{k=0}^{k_1+1} B_k{leftlfloor{cj+c-b-1over a} ight floor}^k &=sum_{j=0}^{k_2-1}{k_2choose j}sum_{k=0}^{k_1+1}B_ksum_{i=0}^{m-1}i^j{leftlfloor{cj+c-b-1over a} ight floor}^k end{aligned} ]

    递归计算(f(m-1,c,c-b-1,a))即可

    不难发现递归层数是(O(log n))级别的

    总复杂度为(O(k^4log n))

    传送门

    //quming
    #include<bits/stdc++.h>
    #define R register
    #define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
    #define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
    #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
    template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
    template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
    using namespace std;
    const int P=1e9+7;
    inline void upd(R int &x,R int y){(x+=y)>=P?x-=P:0;}
    inline int inc(R int x,R int y){return x+y>=P?x+y-P:x+y;}
    inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
    inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
    int ksm(R int x,R int y){
    	R int res=1;
    	for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
    	return res;
    }
    const int N=25;
    int bn[N][N],val[N],g[N],h[N],f[N];
    struct node{
    	int sz,v[N];
    	void init(int *f){
    		memset(g,0,sizeof(g));
    		g[0]=1;
    		int i,j;
    		for(i=0;i<21;++i){
    			for(j=i+1;j;--j) 
    			 g[j]=dec(g[j-1],1ll*g[j]*i%P);
    			g[0]=(P-1ll*g[0]*i%P)%P;
    		}
    		fp(i,0,20){
    			memcpy(h,g,sizeof(h));
    			fd(j,21,1)upd(h[j-1],mul(h[j],i));
    			R int res=1;
    			fp(j,0,i-1)res=mul(res,i-j);
    			fp(j,i+1,20)res=mul(res,P+i-j);
    			res=mul(f[i],ksm(res,P-2));
    			fp(j,0,20)upd(v[j],mul(res,h[j+1]));
    		}
    		for(sz=21;sz&&!v[sz];--sz);
    	}
    	inline int calc(R int x){
    		R int res=0,pw=1;
    		for(R int i=0;i<=sz;++i,pw=mul(pw,x))upd(res,mul(pw,v[i]));
    		return res;
    	}
    }L[N];
    struct Base{
    	int a[11][11];
    	inline Base(){memset(a,0,sizeof(a));}
    	inline int* operator [](const int &x){return a[x];}
    };
    Base func(int n,int a,int b,int c){
    	Base ans;
    	if(!a||1ll*a*n+b<c){
    		fp(k1,0,10){
    			R int res=L[k1].calc(n),pw=(1ll*a*n+b)/c;
    			fp(k2,0,10-k1)ans[k1][k2]=res,res=mul(res,pw);
    		}
    		return ans;
    	}
    	if(a>=c){
    		Base las=func(n,a%c,b,c);
    		fp(k1,0,10)fp(k2,0,10-k1){
    			R int res=1,pw=a/c;
    			fp(i,0,k2){
    				upd(ans[k1][k2],1ll*bn[k2][i]*res%P*las[k1+i][k2-i]%P);
    				res=mul(res,pw);
    			}
    		}
    		return ans;
    	}
    	if(b>=c){
    		Base las=func(n,a,b%c,c);
    		fp(k1,0,10)fp(k2,0,10-k1){
    			R int res=1,pw=b/c;
    			fp(i,0,k2){
    				upd(ans[k1][k2],1ll*bn[k2][i]*res%P*las[k1][k2-i]%P);
    				res=mul(res,pw);
    			}
    		}
    		return ans;
    	}
    	int m=(1ll*a*n+b)/c;
    	Base las=func(m-1,c,c-b-1,a);
    	fp(k1,0,10){
    		R int res=L[k1].calc(n);
    		fp(k2,0,10-k1){
    			ans[k1][k2]=mul(ksm(m,k2),res);
    			fp(i,0,k2-1)fp(j,0,k1+1){
    				R int coef=mul(bn[k2][i],L[k1].v[j]);
    				upd(ans[k1][k2],P-mul(coef,las[i][j]));
    			}
    		}
    	}
    	return ans;
    }
    void init(){
    	fp(i,0,20){
    		bn[i][0]=1;
    		fp(j,1,i)bn[i][j]=inc(bn[i-1][j],bn[i-1][j-1]);
    	}
    	fp(i,0,10){
    		R int res=0;
    		fp(j,0,20)upd(res,ksm(j,i)),f[j]=res;
    		L[i].init(f);
    	}
    }
    int main(){
    //	freopen("testdata.in","r",stdin);
    	init();
    	int T,n,a,b,c,k1,k2;
    	for(scanf("%d",&T);T;--T){
    		scanf("%d%d%d%d%d%d",&n,&a,&b,&c,&k1,&k2);
    		printf("%d
    ",func(n,a,b,c)[k1][k2]);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/yuanquming/p/11777184.html
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