#NTT，DP#U138580 简单的打击

题目

给出两个等长的序列\(a,b\)，
重排序列\(b\)，使得\(a+b\)众数出现的次数最多

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分析

设\(f[i]\)表示众数为\(i\)的贡献，那么
\(f[i]=\sum_{j<i}min(A[j],B[i-j])\)
其中大写的\(a,b\)表示次数，但是这个东西很难做，
考虑把它变成正常的卷积的形式，那么就转换成判定，
判定\(f[i]\)是否能够达到阈值，那么显然就变成了只有\(0,1\)的卷积，
但是这样会超时，考虑阈值仅限于不超过一个常数就可以了

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代码

#include <cstdio>
#include <cctype>
#include <queue>
#define rr register
using namespace std;
const int mod=998244353,N=100011,inv3=332748118; priority_queue<pair<int,int> >q;
int a[N],b[N],A[N],B[N],ff[N<<2],n,TOT[2],lim,gg[N<<2],ans[N<<2],Ans,Gmi[31],Imi[31];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans; 
}
inline signed min(int a,int b){return a<b?a:b;}
inline signed mo1(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline signed mo2(int x,int y){return x<y?x-y+mod:x-y;}
inline signed ksm(int x,int y){
	rr int ans=1;
	for (;y;y>>=1,x=1ll*x*x%mod)
	    if (y&1) ans=1ll*ans*x%mod;
	return ans;
}
namespace Theoretic{
	int rev[N<<2],LAST,tt[N<<2];
	inline void Pro(int n){
		if (LAST==n) return; LAST=n;
		for (rr int i=0;i<n;++i)
		    rev[i]=(rev[i>>1]>>1)|((i&1)?n>>1:0);
	}
	inline void NTT(int *f,int n,int op){
		Pro(n);
		for (rr int i=0;i<n;++i)
	        if (i<rev[i]) swap(f[i],f[rev[i]]);
	    rr int p=2,len=1;
	    for (rr int o=1;p<=n;++o){
	    	rr int W=(op==1)?Gmi[o]:Imi[o];
	    	for (rr int i=0;i<n;i+=p){
			    rr int t=1;
	    		for (rr int j=i;j<i+len;++j){
	    			rr int z=1ll*f[len+j]*t%mod;
	    			f[len+j]=mo2(f[j],z),f[j]=mo1(f[j],z);
	    			t=1ll*t*W%mod;
				}
			}
			p<<=1,len<<=1;
		}
    }
	inline void Cb(int *f,int *g,int n){
		for (rr int i=0;i<n;++i) f[i]=1ll*f[i]*g[i]%mod;
	}
	inline void times(int *f,int *g,int len,int lim){
		rr int n=1,invn; for (;n<lim;n<<=1);
		for (rr int i=0;i<len;++i) tt[i]=g[i];
		for (rr int i=len;i<n;++i) tt[i]=0;
		NTT(f,n,1),NTT(tt,n,1),Cb(f,tt,n),NTT(f,n,-1);
		for (rr int i=lim;i<n;++i) f[i]=0;
		for (rr int i=0;i<n;++i) tt[i]=0;
		invn=ksm(n,mod-2);
		for (rr int i=0;i<lim;++i)
		    f[i]=1ll*f[i]*invn%mod;
	}
}
inline void GmiImi(){
	for (rr int i=0;i<31;++i) Gmi[i]=ksm(3,(mod-1)/(1<<i));
	for (rr int i=0;i<31;++i) Imi[i]=ksm(inv3,(mod-1)/(1<<i));		
}
inline void Prep(int lim){
	for (rr int i=1;i<=n;++i) ff[i]=a[i]>=lim;
	for (rr int i=1;i<=n;++i) gg[i]=b[i]>=lim;
	Theoretic::times(ff,gg,n,n*2);
	for (rr int i=1;i<=n;++i) ans[i]+=ff[i];
}
signed main(){
	n=iut(),GmiImi();
	for (rr int i=1;i<=n;++i) ++a[iut()];
	for (rr int i=1;i<=n;++i) ++b[iut()];
	for (rr int i=1;i<N;++i) if (a[i]) q.push(make_pair(a[i],0));
	for (rr int i=1;i<N;++i) if (b[i]) q.push(make_pair(b[i],1));
	while (!q.empty()){
		rr int t=q.top().second; q.pop();
		if ((++TOT[t])*TOT[t^1]>1e8) break;
	}
	if (!q.empty()) lim=q.top().first+1;
	for (rr int i=1;i<=lim;++i) Prep(i);
	for (rr int i=1;i<=n;++i) if (a[i]>lim) A[++A[0]]=i;
	for (rr int i=1;i<=n;++i) if (b[i]>lim) B[++B[0]]=i;
	for (rr int i=1;i<=A[0];++i)
	for (rr int j=1;j<=B[0];++j)
	    ans[A[i]+B[j]]+=min(a[A[i]],b[B[j]])-lim;
	for (rr int i=1;i<=2*n;++i)
	    if (Ans<ans[i]) Ans=ans[i];
	return !printf("%d",Ans);
}