题目大意是，一个集合中有N个元素，找出最大的S，满足条件A+B+C=S，并且这四个数都属于该集合，N不超过1000.

        如果直接枚举O(n^4)显然复杂度太高，将等式转化一下A+B=S-C，此时分别对左右两边的值进行枚举，这一步复杂度为O(n ^ 2)，接着就用二分法查找满足该等式的最大S值，

复杂度为O(n^2*log(n))。

#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>

using namespace std;

int values[1001];

struct MyStruct
{
	int i;
	int j;
	int res;
};

int compp(const void* a1, const void* a2)
{
	return *(int*)a1 - *(int*)a2;
}

int compp1(const void* a1, const void* a2)
{
	return ((MyStruct*)a1)->res - ((MyStruct*)a2)->res;
}

MyStruct S[2][1000000];

int main()
{
#ifdef _DEBUG
	freopen("e:\\in.txt", "r", stdin);
#endif
	int n;
	while (scanf("%d", &n) != EOF)
	{
		if (n == 0)
		{
			break;
		}
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &values[i]);
		}
		qsort(values, n, sizeof(int), compp);
		int count = 0;
		for (int i = n - 1; i >= 0; i--)
		{
			for (int j = i - 1; j >= 0; j--)
			{
				S[0][count].i = i;
				S[0][count].j = j;
				S[0][count].res = values[i] - values[j];
				count++;
			}
		}
		count = 0;
		for (int i = n - 1; i >= 0; i--)
		{
			for (int j = i - 1; j >= 0; j--)
			{
				S[1][count].i = i;
				S[1][count].j = j;
				S[1][count].res = values[i] + values[j];
				count++;
			}
		}
		qsort(S[1], count, sizeof(MyStruct), compp1);
		bool ans = false;
		for (int i = 0; i < count;i++)
		{
			if (ans)
			{
				break;
			}
			int l = 0;
			int r = count;
			while (r - l > 1)
			{
				int mid = (l + r) / 2;
				if (S[0][i].res <= S[1][mid].res)
				{
					r = mid;
				}
				else
					l = mid;
			}
		
			for (int j = l; j < count;j++)
			{
				if (S[0][i].res < S[1][j].res)
				{
					break;
				}
				if (S[0][i].i == S[1][j].i || S[0][i].i == S[1][j].j||
					S[0][i].j == S[1][j].i || S[0][i].j == S[1][j].j)
				{
					continue;
				}
				if (S[0][i].res == S[1][j].res)
				{
					ans = true;
					printf("%d\n", values[S[0][i].i]);
					break;
				}
			}
			
		}
		if (!ans)
		{
			printf("no solution\n");
		}
	}
	return 1;
}

POJ2549 Sumsets 折半枚举