K-4 Values whose Sum is 0 (Midway encounter method)

K-4 Values whose Sum is 0 crawling in process ... crawling failed time limit:9000MS Memory Limit:0KB 64bit IO format:%lld &%llu Submit Status Practice UVA 1152 Appoint Description:System Crawler (2015-03-12)

Description

 

The SUM problem can formulated as Follows:given four lists A, B, C, D of Integ Er values, compute how many quadruplet (a, b, c, D ) ax b x C x D 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 re spectively 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

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).

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <string>7#include <vector>8#include <stack>9#include <queue>Ten#include <Set> One#include <map> A#include <list> -#include <iomanip> -#include <cstdlib> the#include <sstream> - using namespacestd; -typedefLong LongLL; - Const intinf=0x5fffffff; + Const Doubleexp=1e-8; - Const intms=4005; + inta[ms],b[ms],c[ms],d[ms],n,sum[ms*MS]; A  at intMain () - { -     intT; -scanf"%d",&T); -      while(t--) -     { inscanf"%d",&n); -          for(intI=0; i<n;i++) toscanf"%d%d%d%d",&a[i],&b[i],&c[i],&d[i]); +         intCnt=0; -          for(intI=0; i<n;i++) the              for(intj=0; j<n;j++) *sum[cnt++]=a[i]+B[j]; $Sort (sum,sum+CNT);Panax NotoginsengLL ans=0; -          for(intI=0; i<n;i++) the              for(intj=0; j<n;j++) +         { AAns+=upper_bound (Sum,sum+cnt,-c[i]-d[j])-lower_bound (sum,sum+cnt,-c[i]-d[j]); the         } +printf"%lld\n", ans); -         if(T) $printf"\ n"); $     } -     return 0; -}

K-4 Values whose Sum is 0 (Midway encounter method)