J Midway meeting Method, summation

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J-Midway Meeting MethodTime limit:9000MS Memory Limit:0KB 64bit IO Format:%lld &%llu SubmitStatus

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).

Topic Ideas: To find a number for each column, and to get a sequence of 0, there are several different scenarios analysis: If the four-cycle, there is no doubt that the time-out, you need to use the enumeration case, the number of 1, 2 columns for one and, 3, 4 columns of the number of a and, and then the binary search program code

#include <cstdio>#include<algorithm>using namespacestd;inta[16000010];intb[16000010];intp[4010][4];intT;intErfen (intx) {    intCnt=0; intL=0, r=t-1, Mid;  while(r>l) {mid= (l+r) >>1; if(b[mid]>=x) r=mid; ElseL=mid+1; }     while(b[l]==x&&l<t) {cnt++; L++; }    returnCNT;}intMain () {intn,i,j,t; Long LongRes; scanf ("%d",&T);  while(t--) {scanf ("%d",&N); Res=0;  for(i=0; i<n;i++)             for(j=0;j<4; j + +) scanf ("%d",&P[i][j]); T=0;  for(i=0; i<n;i++)             for(j=0; j<n;j++) A[t++]=p[i][0]+p[j][1]; Sort (A,a+t); T=0;  for(i=0; i<n;i++)             for(j=0; j<n;j++) B[t++]=p[i][2]+p[j][3]; Sort (b,b+t);  for(i=0; i<t;i++) Res+=erfen (-A[i]); printf ("%d\n", RES); if(T) printf ("\ n"); }    return 0;}

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J Midway meeting Method, summation