LOJ #138. 类欧几里得算法

题目
推导方式比较牛逼，一劳永逸
推公式最详细的博客
消去公式：
$$\begin{array}{rl}F(k_1,k_2,n,a,b,c)& =\sum _{i=0}^n{i}^{k_1}{\lfloor \frac{ai+b}{c}\rfloor }^{k_2}\\ & =\sum _{i=0}^n{i}^{k_1}(\lfloor \frac{a}{c}\rfloor i+\lfloor \frac{b}{c}\rfloor +\lfloor \frac{(a\mathrm{m}\mathrm{o}\mathrm{d}c)i+(b\mathrm{m}\mathrm{o}\mathrm{d}c)}{c}\rfloor {)}^{k_2}\\ & =\sum _{i=0}^n{i}^{k_1}\sum _{{\sum }_{p=1}^3{j}_p=k_2}\frac{k!}{{\prod }_{p=1}^3{j}_p!}\times \\ & {\left(\lfloor \frac{a}{c}\rfloor i\right)}^{j_1}{\lfloor \frac{b}{c}\rfloor }^{j_2}{\lfloor \frac{(a\mathrm{m}\mathrm{o}\mathrm{d}c)i+(b\mathrm{m}\mathrm{o}\mathrm{d}c)}{c}\rfloor }^{j_3}\\ & =\sum _{{\sum }_{p=1}^3{j}_p=k_2}\frac{k!{\lfloor \frac{a}{c}\rfloor }^{j_1}{\lfloor \frac{b}{c}\rfloor }^{j_2}}{{\prod }_{p=1}^3{j}_p!}F(k_1+j_1,j_3,n,a\%c,b\%c,c)\end{array}$$

反转公式：
令$m=\lfloor \frac{an+b}{c}\rfloor -1$
$$\begin{array}{rl}F(k_1,k_2,n,a,b,c)& =\sum _{i=0}^n{i}^{k_1}{\lfloor \frac{ai+b}{c}\rfloor }^{k_2}\\ & =\sum _{i=0}^n{i}^{k_1}\sum _{\forall p\in [1,k_2],j_p\in [0,m]}\prod _{p=1}^{k_2}[j_p<\lfloor \frac{ai+b}{c}\rfloor ]\\ & (枚举j_p的最大值)\\ & =\sum _{i=0}^n{i}^{k_1}\sum _{q=0}^m\left[q<\lfloor \frac{ai+b}{c}\rfloor \right]\left[(q+1{)}^{k_2}-q^{k_2}\right]\\ & (q<\lfloor \frac{ai+b}{c}\rfloor 等价于i>\lfloor \frac{cq+c-b-1}{a}\rfloor )\\ & =\sum _{q=0}^m\left[(q+1{)}^{k_2}-q^{k_2}\right](\sum _{i=0}^n{i}^k-\sum _{i=0}^{\lfloor \frac{cq+c-b-1}{a}\rfloor }{i}^k)\\ & =\left[(m+1{)}^{k_2}-0^{k_2}\right]\sum _{i=0}^n{i}^k-\\ & \sum _{q=0}^m\sum _{r=0}^{k_2-1}\left(\genfrac{}{}{0ex}{}{k_2}{r}\right)q^r\sum _{i=0}^{\lfloor \frac{cq+c-b-1}{a}\rfloor }{i}^k\\ & (将函数S_k(x)=\sum _{i=0}^x{i}^k插值后得到其k+1次表达式写作\\ & S_k(x)=\sum _{i=0}^{k+1}{a}_{k,i}{x}^i)\\ & =\left[(m+1{)}^{k_2}-0^{k_2}\right]\sum _{i=0}^n{i}^k-\\ & \sum _{r=0}^{k_2-1}\sum _{i=0}^{k_1+1}\left(\genfrac{}{}{0ex}{}{k_2}{i}\right)a_{k_1,i}\sum _{q=0}^m{q}^r{\lfloor \frac{cq+c-b-1}{a}\rfloor }^i\\ & =\left[(m+1{)}^{k_2}-0^{k_2}\right]\sum _{i=0}^n{i}^k-\\ & \sum _{r=0}^{k_2-1}\sum _{i=0}^{k_1+1}\left(\genfrac{}{}{0ex}{}{k_2}{i}\right)a_{k_1,i}F(r,i,m,c,c-b-1,a)\end{array}$$
这个公式真的一点都不吓人。
但是我们需要注意我们在第二行做了的这个操作。
它在前后并不是完全相等（完全恒等）的。
当$k_2=0$时会出现我们${\sum }_{\forall p\in [1,k_2],j_p\in [0,m]}{\prod }_{p=1}^{k_2}[j_p<\lfloor \frac{ai+b}{c}\rfloor ]$为$0$的情况，但实际上我们公式中这个位置应该是$1$才是正确的。
但是我们经过简单的计算和验证。
我们只需要让
$$\begin{gathered} F(k_1,k_2,n,a,b,c)=(m+1{)}^{k_2}\sum _{i=0}^n{i}^k- \\ \sum _{r=0}^{k_2-1}\sum _{i=0}^{k_1+1}(\genfrac{}{}{0ex}{}{k_2}{i})a_{k_1,i}F(r,i,m,c,c-b-1,a) \end{gathered}$$
即可解决问题。
注：实际上边界好像只用判$a=0$，别的判了也没多快，记忆化是真的快。

$\mathrm{A}\mathrm{C}\mathrm{C}\mathrm{o}\mathrm{d}\mathrm{e}$

#include
#define maxn 15
#define mod 1000000007
#define LL long long
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
#define per(i,j,k) for(int i=(j),LIM=(k);i>=LIM;i--)
using namespace std;

int inv[maxn]={1,1},fac[maxn]={1,1},invf[maxn]={1,1};
int Pow(int b,int k){
	int r = 1;
	for(;k;k>>=1,b=1ll*b*b%mod)
		if(k&1)
			r=1ll*r*b%mod;
	return r;
}

int C(int n,int m){
	static int c[maxn][maxn]={},f=0;
	if(!f){
		c[0][0]=f=1;
		rep(i,1,maxn-1) rep(j,c[i][0]=1,i) c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod; 
	}
	return c[n][m];
}

int a[maxn][maxn];
int S(int n,int m){
	n %= mod;
	LL r = 0 , pw = 1;
	rep(j,0,m+1) r = (r + 1ll * a[m][j] * pw) , pw = 1ll * pw * n % mod;	
	return (r % mod + mod) % mod; 
}

int f[maxn][maxn][50];
int F(int k1,int k2,int n,int a,int b,int c,int dep=0){
	if(~f[k1][k2][dep]) return f[k1][k2][dep];
	if(a == 0) return f[k1][k2][dep] = (1ll * Pow(b/c,k2) * S(n,k1) % mod + mod) % mod;
	if(a>=c||b>=c){
		int r = 0 , A = a % c , B = b % c;a /= c , b /= c;
		int pwa[maxn]={},pwb[maxn]={};
		rep(j,pwa[0]=pwb[0]=1,k2) 
			pwa[j]=1ll*pwa[j-1]*a%mod,
			pwb[j]=1ll*pwb[j-1]*b%mod;
		rep(j1,0,k2) rep(j2,0,k2-j1){ int j3 = k2 - j1 - j2;
			r = (r + 1ll * pwa[j1] * pwb[j2] % mod * invf[j1] % mod * invf[j2] % mod * invf[j3] % mod * F(k1+j1,j3,n,A,B,c,dep+1)) % mod;
		}
		return f[k1][k2][dep] = ((1ll * r * fac[k2] % mod)+mod)%mod;
	}
	int m = (1ll * a * n + b) / c - 1 , r = 1ll * Pow(m+1,k2) * S(n,k1) % mod;
	rep(i,0,k2-1) rep(j,0,k1+1) r = (r - 1ll * C(k2 , i) * ::a[k1][j] % mod * F(i,j,m,c,c-b-1,a,dep+1)) % mod;
	return f[k1][k2][dep]=(r+mod)%mod;
}

int main(){
	rep(i,2,maxn-1) 
		inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod,
		fac[i]=1ll*fac[i-1]*i%mod,
		invf[i]=1ll*invf[i-1]*inv[i]%mod;
	a[0][1] = a[0][0] = 1;
	rep(i,1,maxn-2){
		rep(j,0,i+1) a[i][j] = C(i+1,j);
		rep(j,0,i-1) rep(k,0,j+1) 
			a[i][k] = (a[i][k] - 1ll * C(i+1,j) * a[j][k]) % mod;
		rep(j,0,i+1) a[i][j] = 1ll * a[i][j] * inv[i+1] % mod;
	}
	int T,n,a,b,c,k1,k2;
	for(scanf("%d",&T);T--;){
		scanf("%d%d%d%d%d%d",&n,&a,&b,&c,&k1,&k2);
		memset(f,-1,sizeof f);
		printf("%d\n",(F(k1,k2,n,a,b,c)+mod)%mod);
	}
}