LOJ#6051. 「雅礼集训 2017 Day11」PATH 题解

注意:本题解数组下标从 \(1\) 开始

题目链接

不难发现概率为可行方案数/总方案数。

总方案数显然为 \(\large \frac {(\sum\limits_{i=1}^{n} a_i)!}{\prod\limits_{i=1}^{n} a_i!}\)

可行方案数如何计算？

考虑记 \(t_{i,j}\) 表示第 \(i\) 维 \(x_i\) 的值刚好加到 \(j\) 的时间。

不难发现只有 \(i \leq n\) 且 \(j \leq a_i\) 的时候, \(t_{i,j}\) 才有意义, 并且 \(t_{i,j} < t_{i,j+1}\)(如果存在) , \(t_{i,j} < t_{i+1,j}\)(如果存在)

那么这就是一个给定形状的杨表计数问题。

杨表计数问题的答案为 \(\large \frac {(\sum\limits_{i=1}^{n} a_i)!}{ \sum\limits_{i\leq n,j \leq a_i} Cnt(i,j)}\) , 其中 \(Cnt(i,j)\) 为 \((i,j)\) 这个格子的钩长+1.

记 \(b_x = \sum\limits_{i=1}^{x} [a_i \leq x]\) , 不难发现如果 \((i,j)\) 在杨表内部,其 \(Cnt(i,j) = (a_i - j) + (b_j - i) + 1\) ,否则 \((a_i - j) + (b_j - i) + 1\) 会小于 \(0\) .

那么直接把式子拆成 \((a_i - i) + (b_j - j) + 1\) 即可,可以直接 \(NTT\) 或者 \(FFT\) 解决，最后乘上 \(\prod\limits_{i=1}^{n} a_i!\) , 而 \((\sum\limits_{i=1}^{n} a_i)!\) 这个巨大的阶乘在两个方案数相除的时候已经被消掉了，不需要处理。

\(\Theta(N\log N)\) , 其中 \(N\) 为 \((n+\max\limits_{i=1}^{n} a_i)\)

code :

#include <bits/stdc++.h>
#define LL unsigned long long
#define uint unsigned
using namespace std;
template <typename T> void read(T &x){
	static char ch; x = 0;
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0',ch = getchar();
}
inline void write(uint x){if (x > 9) write(x/10); putchar(x%10+'0'); }

const uint P = 1004535809,N = 500005,L = 2097152;
inline uint power(uint x,int y){
	static uint r; r = 1; while (y){ if (y&1) r = (LL)r * x % P; x = (LL)x * x % P,y >>= 1; } return r;
}

uint R[L],F[L],G[L],wn[L<<1],iwn[L<<1];
inline int getR(int n){
	int Lim = 2,l = 1; while (Lim <= n) Lim <<= 1,++l; 
	for (register int i = 1; i < Lim; ++i) R[i] = (R[i>>1]>>1) | ((i&1)<<l-1);
	return Lim;
}
inline void NTT(uint *A,int n){
	register int i,j,k,v;
	for (i = 1; i < n; ++i) if (i < R[i]) swap(A[i],A[R[i]]); 
	for (i = 1; i < n; i <<= 1) for (j = 0; j < n; j += i << 1) for (k = j; k < i+j; ++k)
		v = (LL)wn[(i<<1)+k-j] * A[k+i] % P,A[k+i] = (A[k]<v)?(A[k]+P-v):(A[k]-v),A[k] = (A[k]+v>=P)?(A[k]+v-P):(A[k]+v);
}
inline void iNTT(uint *A,int n){
	register int i,j,k,v;
	for (i = 1; i < n; ++i) if (i < R[i]) swap(A[i],A[R[i]]); 
	for (i = 1; i < n; i <<= 1) for (j = 0; j < n; j += i << 1) for (k = j; k < i+j; ++k)
		v = (LL)iwn[(i<<1)+k-j] * A[k+i] % P,A[k+i] = (A[k]<v)?(A[k]+P-v):(A[k]-v),A[k] = (A[k]+v>=P)?(A[k]+v-P):(A[k]+v);
	for (i = 0,v = power(n,P-2); i < n; ++i) A[i] = (LL)A[i] * v % P; 
}
int n,m,a[N],b[N],fac[N];
int main(){
	int i,j,L; uint v;
	read(n);
	for (i = 1; i <= n; ++i) read(a[i]),m = max(a[i],m),++b[a[i]],++F[a[i]+n-i];
	for (i = m; i ; --i) b[i] += b[i+1],++G[b[i]+m+1-i];
	L = getR((n+m<<1)+1);
	for (i = 1,j = 2; i <= 21 && (1<<i) <= L; ++i,j <<= 1){
		int l = j + (j>>1);
		v = power(3,(P-1)/j),wn[j] = 1;
		for (register int k = j+1; k < l; ++k) wn[k] = (LL)wn[k-1] * v % P;
		v = power(v,P-2),iwn[j] = 1;
		for (register int k = j+1; k < l; ++k) iwn[k] = (LL)iwn[k-1] * v % P;
	}
	NTT(F,L),NTT(G,L); for (i = 0; i < L; ++i) F[i] = (LL)F[i] * G[i] % P; iNTT(F,L);
	for (v = 1,i = n+m+1; i < L; ++i) if (F[i]) v = (LL)v * power(i-n-m,F[i]) % P;
	v = power(v,P-2);
	for (fac[0] = i = 1; i <= m; ++i) fac[i] = (LL)fac[i-1] * i % P;
	for (i = 1; i <= n; ++i) v = (LL)v * fac[a[i]] % P; 
	write(v),putchar('\n');
	return 0;
}