The Sum of 0 for four numbers（拆解加二分思想）

个人心得：单纯用二分法一直超时，后面发现我的那种方法并没有节省多少时间，后面看了大神的代码，真的是巧妙，

俩个数组分别装a+b，c+d。双指针一个指向最后，从第一个开始想加，加到刚好大于0停止，再看是否存在和为0的情况。

很巧妙，因为此时i，j所指想加刚好大于0，因为是排完序的，所以i往后面走的时候，大于j的数相加一定大于0，所以卡的非常好；

就没有再指针跳转回去了，佩服！

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).

#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
#define maxn 4004
using namespace std;
int map1[maxn*maxn];
int map2[maxn*maxn];
int a[maxn],b[maxn],c[maxn],d[maxn];
int main()
{
      int n,i,j,k,sum,p;
      scanf("%d",&n);
      for(i=0;i<n;i++)
      {
         scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
      }
      for(i=0;i<n;i++)
        for(j=0;j<n;j++)
         map1[i*n+j]=a[i]+b[j];
      for(i=0;i<n;i++)
        for(j=0;j<n;j++)
          map2[i*n+j]=c[i]+d[j];
      sort(map1,map1+n*n);
      sort(map2,map2+n*n);
      sum=0;
      p=n*n-1;
      for(i=0;i<n*n;i++)
      {
        while(p>=0&&map1[i]+map2[p]>0) p--;
        if(p<0) break;
        int temp=p;
        while(temp>=0&&map1[i]+map2[temp]==0)
        {
            sum++; temp--;
        }
      }
      printf("%d\n",sum);
      //system("pause");
       return 0;
}