The fourth week of training (high efficiency algorithm design) J Question (Midway meeting method)

Description

The SUM problem can formulated as Follows:given four lists A, B, C, D<tex2h  Tml_verbatim_mark> of integer values, compute how many quadruplet (a, b, c, D ) A x B x C x d<tex2html_verbatim_mark> is such that a + b + c + D = 0<tex2html_verbatim_mark>. In the following, we assume this all lists has the same size n<tex2html_verbatim_mark>.

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<tex2html_verbatim_mark> (this value C An IS as large as 4000). We then had n<tex2html_verbatim_mark> lines containing four integer values (with absolute value as Lar GE as 228<tex2html_verbatim_mark>) that belong respectively to A, B, C< Tex2html_verbatim_mark> and D<tex2html_verbatim_mark>.

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

16-45 22 42-16-41-27 56 30-36 53-37 77-36 30-75-4626-38-10 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).

If according to the original method, the four-cycle, there is no doubt will be timed out, can only use the method to turn the four-cycle double, but each of the largest number of up to 2 of the 28-square, using the short open array is not feasible, do a large integer hash. No, you can only use the array to save the values, and then look for ...

#include"Iostream"#include"CString"#include"Algorithm"#include"Map"#include"Cmath"UsingNamespace Std;Constint MAXN=4000+10;int book[16000010];int a[MAXN],b[MAXN],c[MAXN],d[MAXN];IntMain(){int T; Cin>>t;While(T--){int n; Cin>>n;Memset(Book,0,sizeof(Book));For(int I=0; I<n; I+ +) CIN>>a[I]>>b[I]>>c[I]>>d[I];int F=0;For(int IA=0; IA<n; IA++)For(int IB=0; IB<n; IB++){Book[F++]=-(A[IA]+b[IB]);}int sum=0;Sort(Book, book+f);For(int IC=0=i%<n=i%++)For(int ID=0; ID<n; ID++){int Temp=c[IC]+d[ID];int x1=Lower_bound(Book, book+f, temp)-bookint x2=upper_bound  (Book,book+f Temp+=x2 -x1;  Cout<<sum<<endlif (T< <endl return 0;< Span class= "Sh-cbracket" >             

The fourth week of training (high efficiency algorithm design) J Question (Midway meeting method)