poj 2785 4 Values whose Sum is 0

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 10535		Accepted: 2904
Case Time Limit: 5000MS

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 ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

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 input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

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

Hint

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 <stdio.h>
#include <algorithm>
using namespace std;

const int N=4000;
int a[N],b[N],c[N],d[N],ab[N*N],cd[N*N];
 
int main()
{
    int n,pab,pcd,ans,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        pab=pcd=ans=0;
        for(i=0;i<n;i++)
            scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            {
                ab[pab++]=a[i]+b[j];
                cd[pcd++]=-c[i]-d[j];
            }
        sort(cd,cd+pcd);
        for(i=0;i<pab;i++)
        {
            int l=0,r=pcd-1,mid;
            while(l<=r)
            {
                mid=(l+r)/2;
                if(ab[i]==cd[mid])
                {
                    ans++;
                    for(k=mid+1;k<pcd;k++)
                        if(ab[i]==cd[k])
                            ans++;
                        else break;
                    for(k=mid-1;k>=0;k--)
                        if(ab[i]==cd[k])
                            ans++;
                        else break;
                    break;
                }
                else if(ab[i]<cd[mid])
                    r=mid-1;
                else l=mid+1;
            } 
        }
        printf("%d\n",ans); 
    } 
}