poj 2549 折半枚举+二分

三重循环肯定TLE，所以采用“折半枚举”的方法+二分查找来提高速度，不同的是需要保存两个下标用来判定是否有重复元素。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 1000;
int a[N];
int n, cnt;

struct Node
{
    int ai, aj, v;
    bool operator < ( const Node & o ) const
    {
        return v < o.v;
    }
} node[N * N];

void solve()
{
    for ( int i = n - 1; i > 0; i-- )
    {
        for ( int j = i - 1; j >= 0; j-- )
        {
            Node tmp;
            tmp.v = a[i] - a[j];
            int pos = lower_bound( node, node + cnt, tmp ) - node;
            while ( node[pos].v == tmp.v )
            {
                if ( node[pos].ai != a[i] && node[pos].ai != a[j]
                  && node[pos].aj != a[i] && node[pos].aj != a[j] )
                {
                    printf("%d
", a[i]);
                    return ;
                }
                pos++;
            }
        }
    }
    printf("no solution
");
}

int main ()
{
    while ( scanf("%d", &n) != EOF )
    {
        if ( n == 0 ) break;
        cnt = 0;
        for ( int i = 0; i < n; i++ )
        {
            scanf("%d", a + i);
            for ( int j = i - 1; j >= 0; j-- )
            {
                node[cnt].ai = a[i];
                node[cnt].aj = a[j];
                node[cnt].v = a[i] + a[j];
                cnt++;
            }
        }
        sort( node, node + cnt );
        node[cnt].v = 999999999;
        sort( a, a + n );
        solve();
    }
    return 0;
}