poj2549（折半枚举）

/*
translation:
	给出一列数列，求满足等式a+b+c=d的最大d是多少？其中abcd都是数列中不同的数字。
solution:
	这类问题多半用折半枚举法就可以解决了
note:
date:
	2016.11.11
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;
const int maxn = 1000 + 5;

struct Node
{
	int id1, id2, val;
	Node(int d1, int d2, int v):id1(d1),id2(d2),val(v){}
	Node(){}
	bool operator < (const Node& rhs) const{
		return val < rhs.val;
	}
};

int a[maxn], n;
vector<Node> v;

bool inside(int x)
{
	return x >= 0 && x < v.size();
}

bool check(int id, int x, int y)
{
	int x0 = v[id].id1, y0 = v[id].id2;
	return (x == x0 || y == y0 || x == y0 || y == x0);
}

int main()
{
	//freopen("in.txt", "r", stdin);
    while(~scanf("%d", &n) && n){
		for(int i = 0; i < n; i++)	scanf("%d", &a[i]);

		sort(a, a + n);

		v.clear();
		for(int i = 0; i < n; i++){	//预处理
			for(int j = i+1; j < n; j++){
				v.push_back(Node(i, j, a[i] + a[j]));
			}
		}
		sort(v.begin(), v.end());

		bool noSolution = true;
		for(int i = n-1; i >= 0; i--){	//d
			for(int j = i-1; j >= 0; j--){	//c
				int id = lower_bound(v.begin(), v.end(), Node(i, j, a[i] - a[j])) - v.begin();

				if(inside(id) && !check(id, i, j) && v[id].val == a[i] - a[j]){
					printf("%d\n", a[i]);
					noSolution = false;
					break;
				}
			}
			if(!noSolution)	break;
		}

		if(noSolution)	printf("no solution\n");
    }
    return 0;
}