【CF888E】Maximum Subsequence（meet in the middle）

题面

CF
洛谷

题解

把所有数分一下，然后\(meet\ in\ the\ middle\)做就好了。
一侧的数正序，另一侧倒序，这样子指针单调就做完了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define ll long long
#define MAX 262222
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,m,k,a[50],ans;
int s1[MAX],s2[MAX],t1,t2;
void dfs1(int x,int s)
{
	if(x==k+1){s1[++t1]=s;return;}
	dfs1(x+1,s);dfs1(x+1,(s+a[x])%m);
}
void dfs2(int x,int s)
{
	if(x==n+1){s2[++t2]=s;return;}
	dfs2(x+1,s);dfs2(x+1,(s+a[x])%m);
}
int main()
{
	n=read();m=read();k=(n+1)/2;
	for(int i=1;i<=n;++i)a[i]=read();
	dfs1(1,0);dfs2(k+1,0);
	sort(&s1[1],&s1[t1+1]);t1=unique(&s1[1],&s1[t1+1])-s1-1;
	sort(&s2[1],&s2[t2+1]);t2=unique(&s2[1],&s2[t2+1])-s2-1;
	for(int i=1;i<=t2;++i)s2[t2+i]=s2[i];
	reverse(&s1[1],&s1[t1+1]);
	for(int i=1,pos=1;i<=t1;++i)
	{
		while((s1[i]+s2[pos])%m<(s1[i]+s2[pos+1])%m)++pos;
		ans=max(ans,(s1[i]+s2[pos])%m);
		if(pos>t2)pos-=t2;
	}
	printf("%d\n",ans);
	return 0;
}