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。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6 
 7 const int N = 1000;
 8 int a[N];
 9 int n, cnt;
10 
11 struct Node
12 {
13     int ai, aj, v;
14     bool operator < ( const Node & o ) const
15     {
16         return v < o.v;
17     }
18 } node[N * N];
19 
20 void solve()
21 {
22     for ( int i = n - 1; i > 0; i-- )
23     {
24         for ( int j = i - 1; j >= 0; j-- )
25         {
26             Node tmp;
27             tmp.v = a[i] - a[j];
28             int pos = lower_bound( node, node + cnt, tmp ) - node;
29             while ( node[pos].v == tmp.v )
30             {
31                 if ( node[pos].ai != a[i] && node[pos].ai != a[j]
32                   && node[pos].aj != a[i] && node[pos].aj != a[j] )
33                 {
34                     printf("%d
", a[i]);
35                     return ;
36                 }
37                 pos++;
38             }
39         }
40     }
41     printf("no solution
");
42 }
43 
44 int main ()
45 {
46     while ( scanf("%d", &n) != EOF )
47     {
48         if ( n == 0 ) break;
49         cnt = 0;
50         for ( int i = 0; i < n; i++ )
51         {
52             scanf("%d", a + i);
53             for ( int j = i - 1; j >= 0; j-- )
54             {
55                 node[cnt].ai = a[i];
56                 node[cnt].aj = a[j];
57                 node[cnt].v = a[i] + a[j];
58                 cnt++;
59             }
60         }
61         sort( node, node + cnt );
62         node[cnt].v = 999999999;
63         sort( a, a + n );
64         solve();
65     }
66     return 0;
67 }