poj 2785 4 Values whose Sum is 0

二分枚举；

也可以将两个和数组都排序，这样可以在查找时保持沿一个方向，最坏情况下复杂度为O(n)，不如二分查找；

计算cpd[]时，c[i] + d[j]错写成c[i]+d[i]查了半天才发现。。。

 1 # include <stdio.h>
 2 # include <stdlib.h>
 3 
 4 # define MAXN 4001
 5 
 6 int apb[MAXN*MAXN], cpd[MAXN*MAXN];
 7 int a[MAXN], b[MAXN], c[MAXN], d[MAXN];
 8 
 9 int cmp(const void *a, const void *b)
10 {
11     return (*(int *)a - *(int *)b);
12 }
13 
14 int bin_search(int *a, int key, int n);
15 
16 int main()
17 {
18     int n, i, j, ans, t;
19         
20     while (~scanf("%d", &n))
21     {
22         for (i = 0; i < n; ++i)
23             scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
24         for (i = 0; i < n; ++i)
25         for (j = 0; j < n; ++j)
26         {
27             apb[i*n + j] = a[i] + b[j];
28             cpd[i*n + j] = c[i] + d[j];
29            }
30            qsort(apb, n*n, sizeof(int), cmp);
31            t = n*n;
32            for (ans = i = 0; i < t; ++i)
33                ans += bin_search(apb, -cpd[i], t);
34            printf("%d\n", ans);
35     }
36     
37     return 0;
38 }
39 
40 int bin_search(int *a, int key, int n)
41 {
42     int cnt, up, low, mid, t;
43     
44     cnt = 0;
45     up = n-1;
46     low = 0;
47     while (low <= up)
48     {
49         mid = (up + low) >> 1;
50         if (key == a[mid]) break;
51         if (key < a[mid]) up = mid - 1;
52         else low = mid + 1;
53     }
54     if (key == a[mid]) 
55     {
56         ++cnt;
57         t = mid;
58         while (t > 0 && key == a[--t]) ++cnt;
59         while (mid < n-1 && key == a[++mid]) ++cnt;
60     }
61     return cnt;
62 }