4 Values whose Sum is 0 （二分+排序）

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 -46

26 -38 -10 62

-32 -54 -6 45

Sample Output

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<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

using namespace std;

const int N=;

int a[N],b[N],c[N],d[N];

int ab[N*N],cd[N*N];

int main()

{

   int n,total=,i,j;

   while (cin>>n)

   {

         for (i=;i<n;i++)

           cin>>a[i]>>b[i]>>c[i]>>d[i];

           int num1=,num2=;

         for (i=;i<n;i++)

            for (j=;j<n;j++)

           {

                ab[num1++]=a[i]+b[j];

                cd[num2++]=-(c[i]+d[j]);

           }

          sort (cd,cd+num2);

         for (i=;i<num1;i++)

           {

              int mid,up=num2-,low=;

             while (low<=up)

             {

                mid=low+(up-low)/;

               if (ab[i]==cd[mid])

                {

                  total++;

                    for (j=mid+;j<=up;j++)

                  {

                       if (ab[i]==cd[j])

                          total++;

                       else

                           break;

                  }

                  for (j=mid-;j>=low;j--)

                  {

                       if (ab[i]==cd[j])

                         total++;

                       else

                         break;

                  }

                  break;

              }

              else

              {

                  if (ab[i]>cd[mid])

                    low=mid+;

                  else

                    up=mid-;

              }

        }

    }

      cout << total << endl;

   }

        return ;

}

巴特西

4 Values whose Sum is 0 （二分+排序）

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