Atcoder Beginner Contest 184F Programming Contest 题解

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Personally think it is a very good thinking question, using the idea of ​​two-way wide search

First of all, you should think of $O(2^{}$The violence of${}^N$$)$
Next, we divide the original sequence into lengths of$n,$The two paragraphs of$m$ , and use the above violence method to enumerate the two paragraphs separately, and put each combination of$A_i$The sum is stored in an array. Here set $a,The\; b$ array maintains two segments respectively.
Obviously, any plan in the first paragraph can be combined with any plan in the second paragraph.
So, now we only need to ask for the largest that does not exceed$T$ 的 $a_i+b_j$The maximum value of,
then we will $b$ array sorting. Then for each$a_i$, Use dichotomy to find the largest one that satisfies $a_i+b_j\le T$ 's$b_j$And update the answer. The correctness is obvious.

设$n\le m$ , the total complexity is$O(n\cdot 2^m)$
Obviously, when$n=m=\frac{N}{2}$When the complexity is optimal, it can reach $O(n\cdot \sqrt{2^N})$

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const long long Maxn=45;
vector <long long> a,b;
long long c[Maxn];
long long n,m,ans;
inline long long read()
{
    
    
	long long s=0,w=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
    
    if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
int main()
{
    
    
	n=read(),m=read();
	for(long long i=1;i<=n;++i)
	c[i]=read();
	long long cnt=(n>>1);
	for(long long x=0;x<(1<<cnt);++x)
	{
    
    
		long long ret=0;
		for(long long i=1;i<=cnt;++i)
		if((x>>(i-1)) & 1)ret+=c[i];
		a.push_back(ret);
	}
	cnt=n-cnt;
	long long tmp=(n>>1);
	for(long long x=0;x<(1<<cnt);++x)
	{
    
    
		long long ret=0;
		for(long long i=1;i<=cnt;++i)
		if((x>>(i-1)) & 1)ret+=c[i+tmp];
		b.push_back(ret);
	}
	sort(a.begin(),a.end()); // 比赛时没想那么多，直接两个都排序了
	sort(b.begin(),b.end());
	tmp=b.size();
	for(long long i=0;i<a.size();++i)
	{
    
    
		if(a[i]>m)break;
		if(a[i]+b[tmp-1]<=m)
		{
    
    ans=max(ans,a[i]+b[tmp-1]);continue;}
		long long l=0,r=b.size()-1;
		while(l<r)
		{
    
    
			long long mid=(l+r)>>1;++mid;
			if(a[i]+b[mid]<=m)l=mid;
			else r=mid-1;
		}
		ans=max(ans,a[i]+b[l]);
	}
	printf("%lld\n",ans);
	return 0;
}