　　　　　　　　　　　　　　4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 21370		Accepted: 6428
Case Time Limit: 5000MS

Description

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 2²⁸ ) 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

-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).

Source

Southwestern Europe 2005

// 题意: 比如例子 6 行，每行 4 个数，从第一，二，三，四列各选一个数，要使和为 0 ，有几种组合?

n 最大有4000，但还是可以暴力，加二分

a + b = -( c + d )

枚举 a + b ,对于 c + d ,枚举一次，用一个大数组保存和，sort ，就可以二分了。要注意的是，二分到了要对左右继续搜索一下，不同的位置代表有不同的组合

 #include <iostream>

 #include <stdio.h>

 #include <string.h>

 #include <algorithm>

 using namespace std;

 int n;

 int k;

 int num[][];

 int s[];

 int erfen(int t)

 {

     int l=,r=k-;

     while (l<=r)

     {

         int mid=(l+r)/;

         if (s[mid]==t)

         {

             int e = mid-;

             int  all=;

             while (e>=&&s[e]==t)

                 e--,all++;

             e=mid+;

             while (e<k&&s[e]==t)

                 e++,all++;

             return all;

         }

         else if (s[mid]>t)

             r=mid-;

         else

             l=mid+;

     }

     return ;

 }

 int main()

 {

     while (scanf("%d",&n)!=EOF)

     {

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

             scanf("%d%d%d%d",&num[i][],&num[i][],&num[i][],&num[i][]);

         k=;

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

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

             s[k++]=-(num[i][]+num[j][]);

         sort(s,s+k);

         int total=;

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

         {

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

             {

                 int left=num[i][]+num[j][];

                 total+=erfen(left);

             }

         }

         printf("%d\n",total);

     }

     return ;

 }

巴特西

4 Values whose Sum is 0(二分)