(mathcal{Description})
Link.
(T) 组询问，每次给出 (n,a,b,c,k_1,k_2)，求
[sum_{x=0}^nx^{k_1}leftlfloorfrac{ax+b}{c}
ight
floor^{k_2}mod(10^9+7)
]
(T=1000)，(n,a,b,cle10^9)，(0le k_1+k_2le 10)。
(mathcal{Solution})
类欧模板题的集大成者。
令 (f(n,a,b,c,k_1,k_2)) 表示所求答案，尝试通过分类讨论递归到不同的情况求解。
-
(a=0lor an+b<c)
[f(n,a,b,c,k_1,k_2)=leftlfloorfrac{an+b}{c} ight floor^{k_2}sum_{x=0}^nx^{k_1} ]预处理 (k+1) 次多项式 (g_k(x)=sum_{i=0}^xi^k) 的各项系数，这个就能直接算出来。
-
否则若 (age c)
[egin{aligned} f(n,a,b,c,k_1,k_2)&=sum_{x=0}^nx^{k_1} left(leftlfloorfrac{a}{c} ight floor x+leftlfloorfrac{(amod c)x+b}{c} ight floor ight)^{k_2}\ &=sum_{i=0}^{k_2} inom{k_2}{i} leftlfloorfrac{a}{c} ight floor^isum_{x=0}^n x^{k_1+i}leftlfloorfrac{(amod c)x+b}{c} ight floor^{k_2-i}\ &=sum_{i=0}^{k_2} inom{k_2}{i} leftlfloorfrac{a}{c} ight floor^i f(n,amod c,b,c,k_1+i,k_2-i) end{aligned} ]递归处理。
-
否则若 (bge c)
类似 2.，最终得到
[f(n,a,b,c,k_1,k_2)=sum_{i=0}^{k_2} inom{k_2}{i} leftlfloorfrac{b}{c} ight floor^i f(n,a,bmod c,c,k_1,k_2-i) ]递归处理。
-
对于其余情况
考虑把高斯函数丢到求和指标上，注意到
[m^k=sum_{j=0}^{m-1}[(j+1)^k-j^k] ]以此替代 (leftlfloorfrac{ax+b}{c} ight floor^{k_2})，并交换求和顺序，得到
[egin{aligned} f(n,a,b,c,k_1,k_2)&=sum_{j=0}^{leftlfloorfrac{an+b}{c} ight floor-1}[(j+1)^{k_2}-j^{k_2}]sum_{x=0}^nx^{k_1}left[x>leftlfloorfrac{cj+c-b-1}{a} ight floor ight]\ &= sum_{j=0}^{leftlfloorfrac{an+b}{c} ight floor-1} [(j+1)^{k_2}-j^{k_2}]sum_{x=0}^n x^{k_1}-sum_{j=0}^{leftlfloorfrac{an+b}{c} ight floor-1} [(j+1)^{k_2}-j^{k_2}]sum_{x=0}^{leftlfloorfrac{cj+c-b-1}{a} ight floor} x^{k_1} end{aligned} ]前半部分直接计算，研究后半部分，发现最后一个求和符号是 (g_{k_1}left(leftlfloorfrac{cj+c-b-1}{a} ight floor ight))，枚举次数将其展开：
[egin{aligned} &=sum_{i=0}^{k_2-1}inom{k_2}{i} sum_{j=0}^{k_1+1}([x^j]g_{k_1})sum_{x=0}^{leftlfloorfrac{an+b}{c} ight floor-1}x^ileftlfloorfrac{cx+c-b-1}{a} ight floor^j\ &=sum_{i=0}^{k_2-1}inom{k_2}{i} sum_{j=0}^{k_1+1}([x^j]g_{k_1})fleft(leftlfloorfrac{an+b}{c} ight floor-1,c,c-b-1,a,i,j ight) end{aligned} ]其中 (i) 枚举 ((j+1)^{k_2}) 的次数；(j) 枚举 (g_{k_1}) 的次数；(x) 枚举原来的 (j)。这样也能递归求解了。
递归中指标变换形如欧几里得算法，设指标值域为 (V)，(K=k_1+k_2)，则复杂度为 (mathcal O(TK^4log V))。
(mathcal{Code})
实现时，可以对于 (n,a,b,c)，一次算出所有可能的 (f(n,a,b,c,k_1,k_2)) 的值。
/*~Rainybunny~*/
#include <cstdio>
#include <cassert>
#include <iostream>
#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )
typedef long long LL;
const int MOD = 1e9 + 7;
int comb[15][15];
inline void subeq( int& a, const int b ) { ( a -= b ) < 0 && ( a += MOD ); }
inline int sub( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int mul( const long long a, const int b ) { return int( a * b % MOD ); }
inline int add( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline void addeq( int& a, const int b ) { ( a += b ) >= MOD && ( a -= MOD ); }
inline int mpow( int a, int b ) {
int ret = 1;
for ( ; b; a = mul( a, a ), b >>= 1 ) ret = mul( ret, b & 1 ? a : 1 );
return ret;
}
struct PowerPoly {
int n, a[12];
PowerPoly() {}
PowerPoly( const int k ): n( k ) {
static int mat[15][15];
for ( int i = 0, p = 0; i <= n + 1; ++i ) {
addeq( p, mpow( i, n ) );
mat[i][n + 2] = p;
}
rep ( i, 0, n + 1 ) {
for ( int j = 0, p = 1; j <= n + 1; ++j, p = mul( p, i ) ) {
mat[i][j] = p;
}
}
rep ( i, 0, n + 1 ) {
int p = -1;
rep ( j, i, n + 1 ) if ( mat[j][i] ) { p = j; break; }
assert( ~p );
if ( p != i ) std::swap( mat[i], mat[p] );
int iv = mpow( mat[i][i], MOD - 2 );
rep ( j, i + 1, n + 1 ) {
int t = mul( iv, mat[j][i] );
rep ( l, i, n + 2 ) subeq( mat[j][l], mul( mat[i][l], t ) );
}
}
per ( i, n + 1, 0 ) {
a[i] = mul( mat[i][n + 2], mpow( mat[i][i], MOD - 2 ) );
rep ( j, 0, i - 1 ) subeq( mat[j][n + 2], mul( a[i], mat[j][i] ) );
}
}
inline int operator () ( LL x ) const {
int ret = 0; x %= MOD;
for ( int i = 0, p = 1; i <= n + 1; ++i, p = mul( x, p ) ) {
addeq( ret, mul( a[i], p ) );
}
return ret;
}
};
const PowerPoly pwr[11]
= { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
struct Atom {
int res[11][11];
Atom(): res{} {}
inline int* operator [] ( const int k ) { return res[k]; }
};
/* *
* calculate sum_{x=0}^n x^{k_1} lfloor frac{ax+b}{c}
floor^{k_2},
* where k_1+k_2 <= 10.
* */
inline Atom solve( const LL n, const LL a, const LL b, const LL c ) {
Atom ret; LL m = ( a * n + b ) / c;
if ( !a || 1ll * a * n + b < c ) {
rep ( k1, 0, 10 ) {
int t = int( m % MOD ), p = 1;
rep ( k2, 0, 10 - k1 ) {
ret[k1][k2] = mul( p, pwr[k1]( n ) );
p = mul( p, t );
}
}
return ret;
}
if ( a >= c ) {
Atom tmp( solve( n, a % c, b, c ) );
rep ( k1, 0, 10 ) rep ( k2, 0, 10 - k1 ) {
int t = int( ( a / c ) % MOD ), p = 1;
rep ( i, 0, k2 ) {
addeq( ret[k1][k2], mul( comb[k2][i],
mul( p, tmp[k1 + i][k2 - i] ) ) );
p = mul( p, t );
}
}
return ret;
}
if ( b >= c ) {
Atom tmp( solve( n, a, b % c, c ) );
rep ( k1, 0, 10 ) rep ( k2, 0, 10 ) {
int t = int( ( b / c ) % MOD ), p = 1;
rep ( i, 0, k2 ) {
addeq( ret[k1][k2], mul( comb[k2][i],
mul( p, tmp[k1][k2 - i] ) ) );
p = mul( p, t );
}
}
return ret;
}
Atom tmp( solve( m - 1, c, c - b - 1, a ) );
rep ( k1, 0, 10 ) {
int t = int( m % MOD ), p = 1;
rep ( k2, 0, 10 - k1 ) {
ret[k1][k2] = mul( p, pwr[k1]( n ) );
rep ( i, 0, k2 - 1 ) rep ( j, 0, k1 + 1 ) {
subeq( ret[k1][k2], mul( comb[k2][i],
mul( pwr[k1].a[j], tmp[i][j] ) ) );
}
p = mul( t, p );
}
}
return ret;
}
int main() {
comb[0][0] = 1;
rep ( i, 1, 10 ) {
comb[i][0] = 1;
rep ( j, 1, i ) comb[i][j] = add( comb[i - 1][j - 1], comb[i - 1][j] );
}
int n, a, b, c, k1, k2, T;
for ( scanf( "%d", &T ); T--; ) {
scanf( "%d %d %d %d %d %d", &n, &a, &b, &c, &k1, &k2 );
printf( "%d
", solve( n, a, b, c )[k1][k2] );
}
return 0;
}