4 Values whose Sum is 0

 

Time Limit:15000MS     Memory Limit:228000KB     64bit IO Format:%I64d & %I64u

Submit Status

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 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<cstdio>
 #include<string.h>
 #include<algorithm>
 #define MAXN 4400
 using namespace std;
 int A[MAXN],B[MAXN],C[MAXN],D[MAXN];
 int S[MAXN*MAXN];
 int lower_bound1(int low,int high,int num,int a[])
  {
      int mid;
      while(low<high)
      {
          mid=low+(high-low)/;
          if(a[mid]>=num) high=mid;
          else low=mid+;
      }
      return low;
  }
 int upper_bound1(int low,int high,int num,int a[])
 {
     int mid;
     while(low<high)
     {
         mid=low+(high-low)/;
         if(a[mid]<=num) low=mid+;
         else
             high=mid;
     }
     return low;
 }
 int main()
 {
     int n,i;
     int p;
     int cout=;
     int l,r,j;
     while(scanf("%d",&n)!=EOF)
     {
         cout=;
         for(i=;i<n;i++)
             scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);
        p=;
        for(i=;i<n;i++)
             for(j=;j<n;j++)
              S[p++]=A[i]+B[j];
         sort(S,S+p);
        for(i=;i<n;i++)
          for(j=;j<n;j++)
        {
            int t=C[i]+D[j];
            l=lower_bound1(,p,-t,S);
            r=upper_bound1(,p,-t,S);
             cout+=(r-l);
        }
        printf("%d\n",cout);
     }
     return ;
 }