4 Values whose Sum is 0(POJ-2785)

Problem 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.

Examples

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

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).

题意:给出 n 行,每行有 4 个数,每一列看做一个组,现在在每个组中选出一个数,问有多少种组合使得选出的 4 个数和为 0

思路:分治法

由于计算 4 个数的和并不好计算,因此可以使用分治的思想,将 4 组数分为两组,然后每组再分别计算和,最后对两组合进行排序,让正数与负数相加判断是否为 0 即可

Source Program

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define PI acos(-1.0)
#define E 1e-6
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
const int MOD=998244353;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,-1,1};
using namespace std;
int a[N],b[N],c[N],d[N];
int sumBe[N*N],sumLa[N*N];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i>a[i]>>b[i]>>c[i]>>d[i];

    //分组求和
    int cnt=0;
    for(int i=0;i=0&&(sumBe[left]+sumLa[right])>0)
            right--;

        if(right<0)
            break;

        //一个left可能对应多组解
        int newRight=right;
        while(newRight>=0&&(sumBe[left]+sumLa[newRight])==0){
            res++;
            newRight--;
        }
    }
    cout<

 

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