\(\mathcal{Description}\)

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。
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。
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;

}

巴特西

Solution -「LOJ #138」「模板」类欧几里得算法

\(\mathcal{Description}\)

\(\mathcal{Solution}\)

\(\mathcal{Code}\)

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