ACM-二分-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 2  ²⁸ ) 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).

主要用到了二分的思想，让我们计算四个的和并不好计算，但是我们可以二分成两个两个的和，分而治之。

然后对两个的和进行排序，让大的和小的(负数)相加，计算结果。代码如下:

/**POJ2785 ------ 2015/4/16*/

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int max_n = 5000;
int a[max_n];
int b[max_n];
int c[max_n];
int d[max_n];
int pre[max_n*max_n];//用来记录前一部分的和
int next1[max_n*max_n];//用来记录后一部分的和
int n;
/**返回A是否大于B 按降序排列*/
bool com_max(int a,int b)
{
    return a>b;
}
int main()
{
    //freopen("POJ2785.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i=0 && pre[i]+next1[r]>0)
        {
            //说明在i不变的情况下，next太小，得增加
            r--;
        }
        if(r<0)
        {
            //查询完成
            break;
        }
        /**
        if((pre[i]+next1[r]) == 0)
        {
            res++;
            r--;
        }
        */
        /**一个i可能对应多个解*/
        int temp = r;
        while(temp >=0 && (pre[i]+next1[temp])==0)
        {
            res++;
            temp--;
        }
    }
    printf("%d\n",res);
}