2021牛客暑期多校训练营9 Cells(LGV引理,FFT)

2021牛客暑期多校训练营9

Cells(LGV引理,FFT)

题目链接

题意

在一个无限大的$xy$​坐标轴上，求从 A = { ( 0 , a 1 ) , ( 0 , a 2 ) , . . . , ( 0 , a n ) } A=\{(0,a_1),(0,a_2),...,(0,a_n)\} A={(0,a1​),(0,a2​),...,(0,an​)}​分别到 B = { ( 1 , 0 ) , ( 2 , 0 ) , . . . , ( n , 0 ) } B=\{(1,0),(2,0),...,(n,0)\} B={(1,0),(2,0),...,(n,0)}​​，不相交路径的方案数。

并且对于每个点$A_i(x,y)$​​​每一步只能移动到$(x+1,y)or(x,y-1)$。

思路

因为每一个格子只能向两个方向移动，并且路径不会存在环，则可以将表格视为一个有向无环图。

考虑使用LGV引理，则答案为：
$$M=\left[\begin{array}{cccc}e(A_1,B_1)& e(A_1,B_2)& \cdots & e(A_1,B_n)\\ e(A_2,B_1)& e(A_2,B_2)& \cdots & e(A_2,B_n)\\ ⋮& ⋮& \ddots & ⋮\\ e(A_n,B_1)& e(A_n,B_2)& \cdots & e(A_n,B_n)\end{array}\right]$$
$M_{i,j}=e(A_i,B_j)$​，当所有边权都设为$1$​时，$\forall P,w(P)=1$​​，则$e(A_i,B_j)$​等价于求$A_i$​到$B_j$​的路径数。（最好先去看看LGV引理的结论）

考虑如何求$A_i(0,a_i)$​到$B_j(j,0)$​的方案数：

$A_i$​​​​​到$B_j$​​​​​无论怎么走，都必须向右走$j$​​​​​步和向下走$a_i$​​​​​步，可以使用排列组合，排列哪一步是向右走的，$C_{j+a_i}^j$​​​​。
$$C_{j+a_i}^j=\frac{(j+a_i)!}{j!\times (j+a_i-j)!}=\frac{(j+a_i)!}{j!\times a_i!}$$

将上面排列组合得到的式子代入$M$​，然后进行化简：
$$\begin{array}{ccccc}M& =& \left[\begin{array}{cccc}\frac{(1+a_1)!}{1!\times a_1!}& \frac{(2+a_1)!}{2!\times a_1!}& \cdots & \frac{(n+a_1)!}{n!\times a_1!}\\ \frac{(1+a_2)!}{1!\times a_2!}& \frac{(2+a_2)!}{2!\times a_2!}& \cdots & \frac{(n+a_2)!}{n!\times a_2!}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{(1+a_n)!}{1!\times a_n!}& \frac{(2+a_n)!}{2!\times a_n!}& \cdots & \frac{(n+a_n)!}{n!\times a_n!}\end{array}\right]& ①& \\ & =& {\prod }_{j=1}^n\frac{1}{j!}\times \left[\begin{array}{cccc}\frac{(1+a_1)!}{a_1!}& \frac{(2+a_1)!}{a_1!}& \cdots & \frac{(n+a_1)!}{a_1!}\\ \frac{(1+a_2)!}{a_2!}& \frac{(2+a_2)!}{a_2!}& \cdots & \frac{(n+a_2)!}{a_2!}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{(1+a_n)!}{a_n!}& \frac{(2+a_n)!}{a_n!}& \cdots & \frac{(n+a_n)!}{a_n!}\end{array}\right]& ②& \\ & =& {\prod }_{j=1}^n\frac{1}{j!}\times \left[\begin{array}{cccc}(1+a_1)& (1+a_1)(2+a_1)& \cdots & (1+a_1)(2+a_1)...(n+a_1)\\ (1+a_2)& (1+a_2)(2+a_2)& \cdots & (1+a_2)(2+a_2)...(n+a_2)\\ ⋮& ⋮& \ddots & ⋮\\ (1+a_n)& (1+a_n)(2+a_n)& \cdots & (1+a_1)(2+a_n)...(n+a_n)\end{array}\right]& ③& \\ & =& {\prod }_{j=1}^n\frac{1}{j!}\times \left[\begin{array}{cccc}(1+a_1)& (1+a_1{)}^2& \cdots & (1+a_1{)}^n\\ (1+a_2)& (1+a_2{)}^2& \cdots & (1+a_2{)}^n\\ ⋮& ⋮& \ddots & ⋮\\ (1+a_n)& (1+a_n{)}^2& \cdots & (1+a_n{)}^n\end{array}\right]& ④& \\ & =& {\prod }_{j=1}^n\frac{1}{j!}\times {\prod }_{k=1}^n(1+a_k)\times \left[\begin{array}{cccc}1& (1+a_1)& \cdots & (1+a_1{)}^{(}n-1)\\ 1& (1+a_2)& \cdots & (1+a_2{)}^{(}n-1)\\ ⋮& ⋮& \ddots & ⋮\\ 1& (1+a_n)& \cdots & (1+a_n{)}^{(}n-1)\end{array}\right]& ⑥& \\ & =& {\prod }_{j=1}^n\frac{1}{j!}\times {\prod }_{k=1}^n(1+a_k)\times {\prod }_{1\le i^{\prime }<j^{\prime }\le n-1}((a_{i^{\prime }}+1)-(a_{j^{\prime }}+1))& ⑤& \\ & =& {\prod }_{j=1}^n\frac{1}{j!}\times {\prod }_{k=1}^n(1+a_k)\times {\prod }_{1\le i^{\prime }<j^{\prime }\le n}(a_{i^{\prime }}-a_{j^{\prime }})& ⑦& \end{array}$$
对矩阵化简的操作进行解释：

• 第②步操作是根据行列式的性质，将每一列的$j!$​都提到前面。

  行列式中的某一行(列)的所有元素的公因子都可以提到行列式记号的外边。

  $$\left[\begin{array}{cccc}{b}_1\times a_{1,1}& b_1\times a_{1,2}& \cdots & b_1\times a_{1,n}\\ b_2\times a_{2,1}& b_2\times a_{2,2}& \cdots & b_2\times a_{2,n}\\ ⋮& ⋮& \ddots & ⋮\\ b_n\times a_{n,1}& b_n\times a_{n,2}& \cdots & b_n\times a_{n,n}\end{array}\right]=\prod _{j=1}^n{b}_j\times \left[\begin{array}{cccc}{a}_{1,1}& a_{1,2}& \cdots & a_{1,n}\\ a_{2,1}& a_{2,2}& \cdots & a_{2,n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n,1}& a_{n,2}& \cdots & a_{n,n}\end{array}\right]$$

• 对第④步操作是根据行列式的性质。

  将行列式的某一行(列)的各元素乘以同一个数k后加到另一行(列)对应的元素上，行列式的值不变。

  举一个三阶的例子：
  $$\begin{array}{ccc}{M}^{\prime }& =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1)(2+a_1)& (1+a_1)(2+a_1)(3+a_1)\\ (1+a_2)& (1+a_2)(2+a_2)& (1+a_2)(2+a_2)(3+a_2)\\ (1+a_3)& (1+a_3)(2+a_3)& (1+a_3)(2+a_3)(3+a_3)\end{array}\right]\\ & =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1)(2+a_1)& (1+a_1)(2+a_1)(3+a_1)-(1+a_1)(2+a_1)\times 2\\ (1+a_2)& (1+a_2)(2+a_2)& (1+a_2)(2+a_2)(3+a_2)-(1+a_2)(2+a_2)\times 2\\ (1+a_3)& (1+a_3)(2+a_3)& (1+a_3)(2+a_3)(3+a_3)-(1+a_3)(2+a_3)\times 2\end{array}\right]\\ & =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1)(2+a_1)& (1+a_1{)}^2(2+a_1)\\ (1+a_2)& (1+a_2)(2+a_2)& (1+a_2{)}^2(2+a_2)\\ (1+a_3)& (1+a_3)(2+a_3)& (1+a_3{)}^2(2+a_3)\end{array}\right]\\ & =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1)(2+a_1)-(1+a_1)& (1+a_1{)}^2(2+a_1)\\ (1+a_2)& (1+a_2)(2+a_2)-(1+a_2)& (1+a_2{)}^2(2+a_2)\\ (1+a_3)& (1+a_3)(2+a_3)-(1+a_3)& (1+a_3{)}^2(2+a_3)\end{array}\right]\\ & =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1{)}^2& (1+a_1{)}^2(2+a_1)\\ (1+a_2)& (1+a_2{)}^2& (1+a_2{)}^2(2+a_2)\\ (1+a_3)& (1+a_3{)}^2& (1+a_3{)}^2(2+a_3)\end{array}\right]\\ & =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1{)}^2& (1+a_1{)}^2(2+a_1)-(1+a_1{)}^2\\ (1+a_2)& (1+a_2{)}^2& (1+a_2{)}^2(2+a_2)-(1+a_2{)}^2\\ (1+a_3)& (1+a_3{)}^2& (1+a_3{)}^2(2+a_3)-(1+a_3{)}^2\end{array}\right]\\ & =& \left[\begin{array}{ccc}(1+a_1)& (1+a_1{)}^2& (1+a_1{)}^3\\ (1+a_2)& (1+a_2{)}^2& (1+a_2{)}^3\\ (1+a_3)& (1+a_3{)}^2& (1+a_3{)}^3\end{array}\right]\end{array}$$

• 对第⑤步操作是根据范德蒙行列式的性质。

最后的结果就是求$M={\prod }_{j=1}^{n}j!\times {\prod }_{k=1}^n(1+a_k)\times {\prod }_{1\le i^{\prime }<j^{\prime }\le n}(a_{i^{\prime }}-a_{j^{\prime }})$​。

后边这个很明显是一个卷积，可以用FFT来做，和牛客第一场的Hash Function很像，可以先看下这个题解。

AC的代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long

const double PI = acos(-1);
const int P = 1e6+10;
const int N = 5e6+100, mod = 998244353;

int limit, bit;
ll R[N];

// 复数的板子
struct Complex {
	double x, y;
	Complex(double x = 0, double y = 0) : x(x), y(y) { }
} a[N], b[N];

Complex operator * (Complex J, Complex Q) {
	return Complex(J.x * Q.x - J.y * Q.y, J.x * Q.y + J.y * Q.x);
}
Complex operator - (Complex J, Complex Q) {
	return Complex(J.x - Q.x, J.y - Q.y);
}
Complex operator + (Complex J, Complex Q) {
	return Complex(J.x + Q.x, J.y + Q.y);
}


//FFT板子
void FFT(Complex * A, int type) {
	for (int i = 0; i < limit; ++i)
		if (i < R[i])
			swap(A[i], A[R[i]]);
	for (int mid = 1; mid < limit; mid <<= 1) {
		Complex wn(cos(PI / mid), type * sin(PI / mid));
		for (int len = mid << 1, pos = 0; pos < limit; pos += len) {
			Complex w(1, 0);
			for (int k = 0; k < mid; ++k, w = w * wn) {
				Complex x = A[pos + k];
				Complex y = w * A[pos + mid + k];
				A[pos + k] = x + y;
				A[pos + mid + k] = x - y;
			}
		}
	}
	if (type == 1) return;
	for (int i = 0; i <= limit; ++i)
		a[i].x /= limit, a[i].y /= limit;
}

ll power(ll a, ll b){
	ll res = 1;
	while(b){
		if(b & 1)	res = (res * a) % mod;
		b >>= 1;
		a = (a * a) % mod;
	}
	return res;
}

ll inv(ll a){
	return power(a, mod-2);
}

int main() {
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin.exceptions(ios::badbit | ios::failbit);
	//freopen("input.txt", "r", stdin);
	int n;
	cin>>n;
	ll res = 1;
	ll fac = 1;
	for(int i = 1; i <= n; i++){
		ll tmp;
		cin>>tmp;
		a[tmp] = 1;
		b[P - tmp] = 1;
		fac = fac * i % mod;
		res = (res * inv(fac)) % mod;
		res = (res * (1 + tmp)) % mod;
	}

	// =================================== 
	limit = 1; while (limit <= 2*P+1) limit <<= 1, bit++;

	// 预处理组合数
	for (int i = 0; i < limit; i++)
		R[i] = (R[i >> 1] >> 1) | ((i & 1) << (bit - 1));
		
	FFT(a, 1);
	FFT(b, 1);
	for (int i = 0; i <= limit; i++) {
		a[i] = a[i] * b[i];
	}
	FFT(a, -1);
	// ===================================
	for (int i = P+1; i < limit; i++) {
		int x = (int)(a[i].x + 0.5);
		if(x > 0){
			res = res * power(abs(i - P), x) % mod;
		}
	}
	
	cout<<res;
	return 0;
}