二分-G - 4 Values whose Sum is 0

G - 4 Values whose Sum is 0

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 -46
26 -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).

　　题目大意：每行会给出四个整数，要求在每列整数中找出一个整数，使四个整数之和为0，求有几种不同组合

 #include<iostream>
 #include<algorithm>
 using namespace std;

 const int maxn = ;
 const int maxm = *;
 int a[maxn],b[maxn],c[maxn],d[maxn];
 int ab[maxm],cd[maxm];

 int main(){
     int n;
     scanf("%d",&n);
     for(int i=;i<n;i++)
         scanf("%d %d %d %d",a+i,b+i,c+i,d+i);

     int cnt = ;
     for(int i=;i<n;i++)//前两列整数求和
         for(int j=;j<n;j++)
             ab[cnt++] = a[i] + b[j];
     cnt = ;
     for(int i=;i<n;i++)//后两列整数求和
         for(int j=;j<n;j++)
             cd[cnt++] = c[i] + d[j];

     sort(ab,ab+n*n);
     sort(cd,cd+n*n);

     int p = n*n-;
     cnt = ;
     for(int i=;i<n*n;i++){
         for(;p>=;p--)
             if(ab[i]+cd[p]<=)    break;
         if(p<)                    break;
         for(int j=p;j>=;j--){
             if(ab[i]+cd[j]==)    cnt++;
             if(ab[i]+cd[j]<)    break;
         }
     }
     printf("%d",cnt);
 }

　　面对四列整数，一一组合将会造成极其庞大的数据，可将其中两两组合，组合以后再相加寻找和为0的情况。