[TJOI2010]分金币

嘟嘟嘟


看数据范围，就能想到折半搜索。
但怎么搜，必须得想清楚了。
假设金币总数为1000，有20个人，首先搜前10个人，把答案记下来。然后如果在后十个人中搜到了4个人，价值为120，那么我们应该在记录的答案中的6个人中找价值最接近380的。
luogu的第一篇题解写的特别好，没有用set，而是以人数为第一关键字，价值为第二关键字排序。这样保证了同一人数的金币是单调的，就可以二分查找了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<set>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const ll INF = 1e18;
const db eps = 1e-8;
const int maxn = 35;
inline ll read()
{
	ll ans = 0;
	char ch = getchar(), last = ' ';
	while(!isdigit(ch)) last = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(last == '-') ans = -ans;
	return ans;
}
inline void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, a[maxn];
struct Node
{
	int num; ll sum;
	In bool operator < (const Node& oth)const
	{
		return num < oth.num || (num == oth.num && sum < oth.sum);
	}
}t[1 << (maxn >> 1)];
ll tp[1 << (maxn >> 1)];
int cnt = 0, l[maxn >> 1];

ll Sum = 0, ans = INF;
In void calc(int num, ll sum)
{
	if(num < 0) return;
	int pos = lower_bound(tp + l[num], tp + l[num + 1], (Sum >> 1) - sum) - tp;
	if(pos < l[num + 1]) ans = min(ans, abs(Sum - ((sum + tp[pos]) << 1)));
	if(pos > l[num]) ans = min(ans, abs(Sum - ((sum + tp[pos - 1]) << 1)));
}

int main()
{
	int T = read();
	while(T--)
	{
		ans = INF; cnt = 0; Sum = 0;
		n = read(); m = n >> 1;
		for(int i = 1; i <= n; ++i) a[i] = read(), Sum += a[i];
		for(int i = 0; i < (1 << m); ++i)
		{
			int num = 0; ll sum = 0;
			for(int j = 0; j < m; ++j) 
				if((i >> j) & 1) ++num, sum += a[j + 1];
			t[++cnt] = (Node){num, sum};
		}
		sort(t + 1, t + cnt + 1);
		for(int i = 1; i <= cnt; ++i) tp[i] = t[i].sum;
		for(int i = 1; i <= cnt; ++i) if(t[i].num ^ t[i - 1].num) l[t[i].num] = i;
		l[m + 1] = cnt + 1;
		for(int i = 0; i < (1 << (n - m)); ++i)
		{
			int num = 0; ll sum = 0;
			for(int j = 0; j < n - m; ++j)
				if((i >> j) & 1) ++num, sum += a[j + m + 1];
			calc(m - num, sum);
		}
		write(ans), enter;
	}
	return 0;
}