UVA1152 4Values whose Sum is 0

Description

The SUM problem can formulated as Follows:given four lists A, B, C, D of an integer values, compute how many quadruplet ( A, B, C, D) $ \in$axbxcxd is such that A + B + c + d = 0. In the following, we assume this all lists has the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This was followed by a blank line, and there was also a blank line between the consecutive inputs.

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then had 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 test case, the output must follow the description below. The outputs of the consecutive cases would be separated to a blank line.

For each of the input file, your program have to write the number quadruplets whose sum is zero.
Sample Input
1

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

Sample Explanation:indeed, the sum of the five following quadruplets is zero: (-45,-27, 42, 30), (26, 30,-10,-46), (-3) 2, 22, 56,-46), (-32, 30,-75, 77), (-32,-54, 56, 30).

The best way to do this is not to timeout. Enumerates A, B, and then checks for the value of-(c+d), or the binary optimization.

AC Code:

#include <algorithm>#include<iostream>using namespacestd;Const intMax =4000+Ten;intA[max],b[max],c[max],d[max];intab[17000000];intTotal ;intMain () {intT; CIN>>T;  while(t--)    {        intN; CIN>>N;  for(intI=0; i<n; i++) {cin>>a[i]>>b[i]>>c[i]>>D[i]; }        intk=0;  for(intI=0;i<n; i++)        {             for(intj=0;j<n; J + +) {Ab[k]=a[i]+B[j]; K++; }} sort (Ab,ab+k); Total=0; intS,l,r,mid;  for(intI=0; i<n; i++)        {             for(intj=0; j<n; J + +)            {                intx=-c[i]-D[j]; L=0, r=k-1;  while(l<=r) {Mid= (l+r)/2; if(ab[mid]>x) R=mid-1; Else if(ab[mid]<x) L=mid+1; Else                    {                         for(s=mid;s>=0; s--)                        {                            if(ab[s]==x) Total++; Else                                Break; }                         for(s=mid+1; s<k; s++)                        {                            if(ab[s]==x) Total++; Else                                Break; }                         Break; } }}} cout<<total<<Endl; if(t>0) cout<<Endl; }    return 0;}

UVA1152 4Values whose Sum is 0