思路：
折半枚舉，二分。

實現：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 int a[4][4005];
 6 int buf[16000005];
 7 int main()
 8 {
 9     int n;
10     cin >> n;
11     for (int i = 0; i < n; i++)
12     {
13         for (int j = 0; j < 4; j++)
 
14         {
15             cin >> a[j][i];
16         }
17     }
18     int cnt = 0;
19     for (int i = 0; i < n; i++)
20     {
21         for (int j = 0; j < n; j++)
22         {
23             buf[i * n + j] = a[0][i] + a[1][j];
24         }
25     }
26     sort(buf, buf + n * n);
27     for
 
 (int i = 0; i < n; i++)
28     {
29         for (int j = 0; j < n; j++)
30         {
31             int tmp = -a[2][i] - a[3][j];
32             if (binary_search(buf, buf + n * n, tmp))
33             {
34                 int l = lower_bound(buf, buf + n * n, tmp) - buf;
35                 int r = upper_bound(buf, buf + n * n, tmp) - buf;
 
36                 cnt += r - l;
37             }
38         }
39     }
40     cout << cnt << endl;
41     return 0;
42 }

poj2785 4 Values whose Sum is 0