Subset

题目描述：

Given a list of N integers with absolute values no larger than 10 15, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.

输入：

The input contains multiple data sets, the first line of each data set contains N <= 35, the number of elements, the next line contains N numbers no larger than 10 15 in absolute value and separated by a single space. The input is terminated with N = 0

输出：

For each data set in the input print two integers, the minimum absolute sum and the number of elements in the optimal subset.

样例输入：

1
10
3
20 100 -100
0

样例输出：

10 1
0 2

题目大意：

让你从n个数里面找任意个数（>0），使他们的和的绝对值最小，如果有多组和一样最小，输出最小和的绝对值且最小个数。

2^35，会超时，所以考虑折半枚举，减少时间复杂度，并用二分，查找绝对值最小的和

代码如下：

#include<stdio.h>
#include<algorithm>
#include<iostream>
const int maxn=1000005;
long long a[maxn];
int qq;
using namespace std;
struct node{
	long long value;
	int len;
}nod[maxn],cun[maxn];
bool cmp(node a,node b)
{
	if(a.value==b.value)return a.len<b.len;
	else return a.value<b.value;
}
int findd(long long what)
{
	int lb=-1,ub=qq,mid;
	while(ub-lb>1)
	{
		mid=(lb+ub)/2;
		if(nod[mid].value>=what)ub=mid;
		else lb=mid;
	}
	return ub;
}
int main()
{
	int n;
	while(1)
	{
		scanf("%d",&n);
		long long ans=1e15+5,res=99999;//一开始写的是ans=9999999，然后一直wrong 
		if(n==0)
		{
			break;
		}
		int i,j,k;
		for(i=0;i<n;i++)
		{
			scanf("%lld",&a[i]);
		}
		int m=n/2;
		int s=1<<(n-m);
		for(i=1;i<s;i++)
		{
			int tt=i;
			long long temp=0;
			int cnt=0;
			for(j=m;j<n;j++)
			{
				if(tt&1)
				{
					temp+=a[j];
					cnt++;
				}
				tt>>=1;
			}
			cun[i].value=temp;
			cun[i].len=cnt;
		}
		sort(cun+1,cun+s,cmp);//后半段可能有的值 
		qq=2;
		nod[0].value=0;
		nod[0].len=0;//一定要有 
		nod[1].value=cun[1].value;
		nod[1].len=cun[1].len;
		for(i=2;i<s;i++)
		{
			if(cun[i].value!=nod[qq-1].value)
			{
				nod[qq].value=cun[i].value;
				nod[qq++].len=cun[i].len;
			}
		}
		sort(nod,nod+qq,cmp);
		//减少重复 
		s=1<<m;
		for(i=0;i<s;i++)
		{
			long long temp=0;
			int cnt=0,tt=i;
			for(j=0;j<m;j++)
			{
				if(tt&1)
				{
					temp+=a[j];
					cnt++;
				}
				tt>>=1;
			}
			long long temp1=-temp;
			int pos=findd(temp1);
			int small=max(0,pos-1);
			int big=min(pos+1,qq-1);
			for(j=small;j<=big;j++)
			{
				long long en=nod[j].value+temp;
				int tm=nod[j].len+cnt;
				if(en<0)en=-en;
				if(tm==0&&en==0)
					continue;
				if(en<ans||(en==ans&&tm<res))
				{
					ans=en;
					res=tm;
				 }
			}
		}//前半段和后半段的叠加 
		printf("%lld %lld\n",ans,res);
	}
	return 0;
}