（POJ 2549）Sumsets（折半枚举）

Description

Given S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S.

Input

Several S, each consisting of a line containing an integer 1 <= n <= 1000 indicating the number of elements in S, followed by the elements of S, one per line. Each element of S is a distinct integer between -536870912 and +536870911 inclusive. The last line of input contains 0.

Output

For each S, a single line containing d, or a single line containing "no solution".

Sample Input

5
2 
3 
5 
7 
12
5
2 
16 
64 
256 
1024
0

Sample Output

12
no solution


看数据范围想想就知道暴力不可行，这几天心血了另外一种枚举方法，折半枚举。
因为a+b+c=d;
所以式子可以化为a+b=d-c;
所以把暴力枚举的o(n^4)变成o（n^2）；
先把所有的a+b枚举出来，用map记录其是否出现过，再枚举d-c，看看其对应的数字有没有出现。因为还要判断abcd互不相同，所以枚举时要注意条件。
PS：由于a+b可以小于0，用map方便一点。

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<map>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#define inf 0x3f3f3f3f

using namespace std;
int a[1010];
int n;

class po
{
public:
    int a,b;
    po(int a1,int b1){a=a1,b=b1;}
};
 map<int,vector<po> > run;



int solve()
{
     for(int i=n;i>=1;i--)
            {
                for(int j=1;j<=n;j++)
                {
                    if(j==i)
                        continue;
                    int tem=a[j]-a[i];
                    for(int k=0;k<run[tem].size();k++)
                    {
                       // cout << a[i] << " " << a[j] << endl;
                        int a1=run[tem][k].a;
                        int b1=run[tem][k].b;
                        if(a[i]!=a1&&a[i]!=b1&&a[j]!=a1&&a[j]!=b1)
                            return a[j];
                    }
                }
            }
    return 536870912;
}

int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        run.clear();
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                po tem(a[i],a[j]);
                run[a[j]+a[i]].push_back(tem);
            }
        }
       int ans=solve();
       if(ans==536870912)
        printf("no solution
");
       else
        printf("%d
",ans);
    }
    return 0;
}