【洛谷 P4921】—情侣?给我烧了!(容斥+组合数学)

传送门


gig_i表示至少ii对情侣坐一起的方案数

显然gj=(nj)2j!2j(2n2j)!g_j={nchoose j}^2j!2^j(2n-2j)!

ansians_i表示恰好ii对情侣坐一起的方案
显然有gi=j=in(ji)ansjg_i=sum_{j=i}^n {jchoose i}ans_j

二项式反演得
ansi=j=in(1)ji(ji)gj=j=in(1)ji(ji)(nj)2j!2j(2n2j)!=j=in(1)jij!i!(ji)!n!2j!2(nj)!2j!2j(2n2j)!=j=0ni(1)j1i!j!2i+jn!2(nji)!2(2n2j2i)!=n!22ii!j=0ni(1)j1j!2j1(nji)!2(2n2j2i)! ans_i=sum_{j=i}^n(-1)^{j-i}{jchoose i}g_j\ =sum_{j=i}^n(-1)^{j-i}{jchoose i}{nchoose j}^2j!2^j(2n-2j)!\ =sum_{j=i}^{n}(-1)^{j-i}frac{j!}{i!(j-i)!}frac{n!^2}{j!^2(n-j)!^2}j!2^j(2n-2j)!\ =sum_{j=0}^{n-i}(-1)^jfrac{1}{i!j!}2^{i+j}frac{n!^2}{(n-j-i)!^2}(2n-2j-2i)!\ =frac{n!^22^i}{i!}sum_{j=0}^{n-i}(-1)^jfrac 1 {j!}2^jfrac{1}{(n-j-i)!^2}(2n-2j-2i)!

里面只和nin-i有关系

定义f(n)=j=0n(1)j1j!2j1(nj)!2(2n2j)!f(n)=sum_{j=0}^n(-1)^jfrac{1}{j!}2^jfrac{1}{(n-j)!^2}(2n-2j)!

ansi=n!22ii!f(ni)ans_i=frac{n!^22^i}{i!}f(n-i)

O(n2)O(n^2)预处理ff可以O(n)O(n)回答每个询问

#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define poly vector<int>
#define bg begin
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){
	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=2005;
int fac[N],ifac[N],bin[N];
int n,f[N];
#define P(x) mul((x),(x))
inline int calc(int n){
	int res=0;
	for(int j=0;j<=n;j++)
		if(j&1)Dec(res,mul(bin[j],mul(ifac[j],mul(P(ifac[n-j]),fac[2*n-2*j]))));
		else Add(res,mul(bin[j],mul(ifac[j],mul(P(ifac[n-j]),fac[2*n-2*j]))));
	return res;
}
inline int C(int n,int m){
	if(n<m)return 0;
	return mul(fac[n],mul(ifac[m],ifac[n-m]));
}
inline void init(int len=N-5){
	bin[0]=fac[0]=ifac[0]=1;
	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
	ifac[len]=ksm(fac[len],mod-2);
	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
	for(int i=1;i<=len;i++)bin[i]=mul(bin[i-1],2);
	for(int i=0;i<=1000;i++)f[i]=calc(i);
}
int main(){
	init();
	int T=read();
	while(T--){
		int n=read();
		for(int i=0;i<=n;i++)
		cout<<mul(f[n-i],mul(ifac[i],mul(P(fac[n]),bin[i])))<<'
';
	}
}
原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328661.html