Poj2785: 4 values whose sum is 0 (Binary)

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) ε 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 -4626 -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 ).
Question: Find a number for each column to obtain a sequence with the sum of 0. There are several different solutions for finding an sum for the number of Column 1 and 2, and for the number of column 3 and 4, and then perform a binary search.

#include <stdio.h>#include <string.h>#include <algorithm>#include <map>using namespace std;#define up(i,x,y) for(i=x;i<=y;i++)#define down(i,x,y) for(i=x;i>=y;i--)#define mem(a,b) memset(a,b,sizeof(a))#define w(a) while(a)#define ll long longint n,a[4005],b[4005],c[4005],d[4005],sum1[16000005],sum2[16000005],len;int main(){    int i,j,ans,l,r,mid;    w(~scanf("%d",&n))    {        up(i,0,n-1)        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);        len=0;        up(i,0,n-1)        up(j,0,n-1)        sum1[len++]=a[i]+b[j];        len=0;        up(i,0,n-1)        up(j,0,n-1)        sum2[len++]=c[i]+d[j];        ans=0;        sort(sum2,sum2+len);        up(i,0,len-1)        {            l=0,r=len-1;            w(l<r)            {                mid=(l+r)>>1;                if(sum2[mid]<-sum1[i])                    l=mid+1;                else                    r=mid;            }            while(sum2[l]==-sum1[i]&&l<len)            {                ans++;                l++;            }        }        printf("%d\n",ans);    }    return 0;}