POJ 2785 4 Values whose Sum is 0

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 13069		Accepted: 3669
Case Time Limit: 5000MS

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).

题目大意： 输入n  表示a b c d 四个集合都有n个元素之后每行输入4个集合中的一个元素，求这四个集合每个集合中拿出一个数 相加等于0的组数。

解题方法：哈希表开放定址法。

#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <iostream>
using namespace std;

#define MAX_VAL 20000000

typedef struct
{
    int nCount;
    int x;
}Hash;

Hash HashTable[MAX_VAL];
int a[4005], b[4005], c[4005], d[4005];

void InsertHT(int n)
{
    int addr = n % MAX_VAL;
    if (addr < 0)
    {
        addr += MAX_VAL;
    }
    while(HashTable[addr].nCount != 0 && HashTable[addr].x != n)
    {
        addr = (addr + 1) % MAX_VAL;
    }
    HashTable[addr].nCount++;
    HashTable[addr].x = n;
}

int SearchHT(int n)
{
    int addr = n % MAX_VAL;
    if (addr < 0)
    {
        addr += MAX_VAL;
    }
    while (HashTable[addr].nCount != 0 && HashTable[addr].x != n)
    {
        addr = (addr + 1) % MAX_VAL;
    }
    return HashTable[addr].nCount;
}

int main()
{
    int n, ans = 0;
    scanf("%d", &n);
    memset(HashTable, 0, sizeof(HashTable));
    for (int i = 0; i < n; i++)
    {
        scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            InsertHT(a[i] + b[j]);
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            ans += SearchHT(-c[i] - d[j]);
        }
    }
    printf("%d\n", ans);
    return 0;
}