K

K - 4 Values whose Sum is 0

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Submit Status Practice UVA 1152

Appoint description: System Crawler (2015-03-12)

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 ) AxBxCxD are such that a + b + c + d = 0 . In the following, we assume that all lists have 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 line is followed by a blank line, and there is also a blank line between two 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 have 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 two consecutive cases will be separated by a blank line.

For each input file, your program has 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), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cstdlib>
#include <sstream>
using namespace std;
typedef long long LL;
const int INF=0x5fffffff;
const double EXP=1e-8;
const int MS=4005;
int A[MS],B[MS],C[MS],D[MS],n,sum[MS*MS];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                sum[cnt++]=A[i]+B[j];
        sort(sum,sum+cnt);
        LL ans=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
        {
            ans+=upper_bound(sum,sum+cnt,-C[i]-D[j])-lower_bound(sum,sum+cnt,-C[i]-D[j]);
        }
        printf("%lld
",ans);
        if(T)
            printf("
");
    }
    return 0;
}