poj 2785 4 Values whose Sum is 0

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).

 

思路：

这题要运用二分法。一共有4列数，前两列分一组，后两列分一组。把第一列每个数和第二列每个数相加形成一个新的数组apb[n*n],把第三列每个数和第四列每个数相加形成数组cpd[n*n],把apb[ ]进行从小到大排序。从cpd[ ]的第一个数开始二分查找apb[ ]中有没有符合相加之和为0的。

注意：

1.注意2^28用int就足够了。

2. 排序之后可能有值相同的情况，所以当apb[ ]+cpd[ ]=0的时候要考虑apb数组中当前位置下的前面数和后面的数是否同样满足满足apb[ ]+cpd[ ]=0。

3.mid+1/-1。eg 当前mid=4;high=5.下一步要让mid=5，但是因为mid low high都是整型,所以下一步mid=(4+5)/2还是4，这样就死循环了，所以mid要为4+1。

 

源代码：

#include <stdio.h>
#include <stdlib.h>
int a[4010],b[4010],c[4010],d[4010];
int apb[4000*4000+10],cpd[4000*4000+10];
int cmp(void const *a,void const *b)
{
    return ((*(int *)a>*(int *)b)?1:-1);
}
int main()
{
    int n,i,j,low,high,mid,ans,k;
    scanf("%d",&n);
    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++)
        {
            apb[i*n+j]=a[i]+b[j];
            cpd[i*n+j]=c[i]+d[j];
        }
    }
    qsort(apb,n*n,sizeof(int),cmp);
    for(i=0; i<n*n; i++)
               ans=0;
    for(i=0; i<n*n; i++)
    {
        high=n*n-1;
        low=0;
        mid=(high+low)/2;
        while(low<high)
        {
             if(cpd[i]+apb[mid]>0)
                high=mid-1;
            else if(cpd[i]+apb[mid]<0)
                low=mid+1;
            mid=(high+low)/2;
            if(cpd[i]+apb[mid]==0)
            {
                ans++;
                for(k=mid-1; k>=0; k--)
                {
                    if(apb[k]+cpd[i]==0)
                        ans++;
                    else break;
                }
                for(k=mid+1; k<n*n; k++)
                {
                    if(apb[k]+cpd[i]==0)
                        ans++;
                    else break;
                } break;
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}