\(\mathcal{Description}\)
Link.
\(T\) 组询问，每次给出 \(n,a,b,c,k_1,k_2\)，求
\[\sum_{x=0}^nx^{k_1}\left\lfloor\frac{ax+b}{c}\right\rfloor^{k_2}\bmod(10^9+7) \]\(T=1000\)，\(n,a,b,c\le10^9\)，\(0\le k_1+k_2\le 10\)。
\(\mathcal{Solution}\)
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。
-
\(a=0\lor an+b<c\)
\[f(n,a,b,c,k_1,k_2)=\left\lfloor\frac{an+b}{c}\right\rfloor^{k_2}\sum_{x=0}^nx^{k_1} \]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 \(a\ge c\)
\[\begin{aligned} f(n,a,b,c,k_1,k_2)&=\sum_{x=0}^nx^{k_1} \left(\left\lfloor\frac{a}{c}\right\rfloor x+\left\lfloor\frac{(a\bmod c)x+b}{c}\right\rfloor \right)^{k_2}\\ &=\sum_{i=0}^{k_2} \binom{k_2}{i} \left\lfloor\frac{a}{c}\right\rfloor^i\sum_{x=0}^n x^{k_1+i}\left\lfloor\frac{(a\bmod c)x+b}{c}\right\rfloor^{k_2-i}\\ &=\sum_{i=0}^{k_2} \binom{k_2}{i} \left\lfloor\frac{a}{c}\right\rfloor^i f(n,a\bmod c,b,c,k_1+i,k_2-i) \end{aligned} \]递归处理。
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否则若 \(b\ge c\)
类似 2.，最终得到
\[f(n,a,b,c,k_1,k_2)=\sum_{i=0}^{k_2} \binom{k_2}{i} \left\lfloor\frac{b}{c}\right\rfloor^i f(n,a,b\bmod c,c,k_1,k_2-i) \]递归处理。
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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 \(\left\lfloor\frac{ax+b}{c}\right\rfloor^{k_2}\)，并交换求和顺序，得到
\[\begin{aligned} f(n,a,b,c,k_1,k_2)&=\sum_{j=0}^{\left\lfloor\frac{an+b}{c}\right\rfloor-1}[(j+1)^{k_2}-j^{k_2}]\sum_{x=0}^nx^{k_1}\left[x>\left\lfloor\frac{cj+c-b-1}{a}\right\rfloor \right]\\ &= \sum_{j=0}^{\left\lfloor\frac{an+b}{c}\right\rfloor-1} [(j+1)^{k_2}-j^{k_2}]\sum_{x=0}^n x^{k_1}-\sum_{j=0}^{\left\lfloor\frac{an+b}{c}\right\rfloor-1} [(j+1)^{k_2}-j^{k_2}]\sum_{x=0}^{\left\lfloor\frac{cj+c-b-1}{a}\right\rfloor} x^{k_1} \end{aligned} \]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 \(g_{k_1}\left(\left\lfloor\frac{cj+c-b-1}{a}\right\rfloor\right)\)，枚举次数将其展开：
\[\begin{aligned} &=\sum_{i=0}^{k_2-1}\binom{k_2}{i} \sum_{j=0}^{k_1+1}([x^j]g_{k_1})\sum_{x=0}^{\left\lfloor\frac{an+b}{c}\right\rfloor-1}x^i\left\lfloor\frac{cx+c-b-1}{a}\right\rfloor^j\\ &=\sum_{i=0}^{k_2-1}\binom{k_2}{i} \sum_{j=0}^{k_1+1}([x^j]g_{k_1})f\left(\left\lfloor\frac{an+b}{c}\right\rfloor-1,c,c-b-1,a,i,j\right) \end{aligned} \]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 \(\mathcal O(TK^4\log V)\)。
\(\mathcal{Code}\)
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} \rfloor^{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\n", solve( n, a, b, c )[k1][k2] );
}
return 0;
}