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)或(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)

令(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))的

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)或(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)

记(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;
}