题目：

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 x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

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 2 ²⁸ ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -38 -10 62-32 -54 -6 45

Sample Output

5

Hint

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).

 

 

题意：

给一个n*4的矩阵，输入n*4个数，在每一列找一个数，使得四个数的和为0；

 

分析：

先分别求出a和b，c和d两列任意两个数的和存放到相应的数组，将cd的和进行排序后，再用二分法进行查找；二分查找的时候注意，倘若中间的数据符合条件的话要再往两边进行查找，因为不能排除有多个数字相等的情况

 

注意：

求第二组数据的时候，根据提交的结果是不需要初始化total的；

 

 

AC代码：

#include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int N=4005;int a[N],b[N],c[N],d[N];int ab[N*N],cd[N*N];int main(){   int n,total=0,i,j;   while (cin>>n)   {         for (i=0;i
          
           >a[i]>>b[i]>>c[i]>>d[i];           int num1=0,num2=0;         for (i=0;i
           
            =low;j--) { if (ab[i]==cd[j]) total++; else break; } break; } else { if (ab[i]>cd[mid]) low=mid+1; else up=mid-1; } } } cout << total << endl; } return 0;}