POJ-2785-4 Values whose Sum is 0

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
4
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
5
-56 45 34 78
-34 -67 45 -23
-12 -34 -56 46
45 34 -32 8
-23 -56 4 53

Sample Output

5
256
11

分析：

四列数字，然后给出每组的长度，从每组中抽出一个数字，满足这四个数字和是0的情况的总数。

1. 四列数字一个一个每组不存在，会超时。

2.先求出两列数字，（4000*4000种结果）然后对其中一种排序。

2.二分查找，找到一种算一种，还要考虑找到的周围有没有相等的，两个方向检测一下。

 

#include <bits/stdc++.h>
using namespace std;
int a[4001][4];
int sum1[16000002];
int sum2[16000002];
int main()
{
    int n,mid;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d%d%d%d",&a[i][0],&a[i][1],&a[i][2],&a[i][3]);
        }
        int k = 0;
        int m = 0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                sum1[k++] = a[i][0]+a[j][1];
                sum2[m++] = -a[i][2]-a[j][3];
            }
        }
        sort(sum2,sum2+m);
        int cnt = 0;
        for(int i=0;i<k;i++)
        {
            int left = 0;
            int right = k-1;
            while(left<=right)
            {
                mid = (left+right)/2;
                if(sum1[i]==sum2[mid])
                {
                    cnt++;
                    for(int j = mid-1;j>=0;j--)
                    {
                        if(sum1[i] == sum2[j])
                            cnt++;
                        else
                            break;
                    }
                    for(int j = mid+1;j<m;j++)
                    {
                        if(sum1[i] == sum2[j])
                            cnt++;
                        else
                            break;
                    }
                    break;
                }
                if(sum1[i]<sum2[mid])
                {
                    right = mid-1;
                }
                else
                    left = mid+1;
            }
        }
        printf("%d\n",cnt);
    }

}