uva1152 4 Values whose Sum is 0（中途相遇法）

题解：

中途相遇法，巧妙利用二分查找

代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define maxn 10000
#define LL long long
int cas=1,T;
int a[maxn],b[maxn],c[maxn],d[maxn];
int e[4005*4005];
int main()
{
    scanf("%d",&T);
    while (T--)
    {
        int n;
        scanf("%d",&n);
        for (int i = 0;iscanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
        }
        int k = 0;
        for (int i = 0;ifor (int j = 0;jint ans = 0;
       for (int i = 0;ifor (int j = 0;jprintf("%d\n",ans);
       if (T)
           printf("\n");
    }

    return 0;
}

题目

4 Values whose Sum is 0

Time Limit:9000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute
how many quadruplet (a, b, c, d) ∈ A × B × C × D are such that a + b + c + d = 0. In the following, we
assume that all lists have the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.
The first line of the input file contains the size of the lists n (this value can be as large as 4000).
We then have n lines containing four integer values (with absolute value as large as 228) that belong
respectively to A, B, C and D.
Output
For each test case, your program has to write the number quadruplets whose sum is zero.
The outputs of two consecutive cases will be separated by a blank line.
Sample Input
1
6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45
Sample Output
5
Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30),
(26, 30, -10, -46), (-32, 22, 56, -46), (-32, 30, -75, 77), (-32, -54, 56, 30).