Poj2785 （4 Values whose Sum is 0）

最新推荐文章于 2019-12-28 09:32:36 发布

原创最新推荐文章于 2019-12-28 09:32:36 发布 · 550 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Poj2785 #ACM

ACM 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种解决四数之和为零问题的高效算法，通过使用哈希表来加速查找过程，降低了时间复杂度。具体步骤包括将前两个数求和并存储在哈希表中，然后查找后两个数的和是否等于前两个数和的相反数。通过实例解释了算法的应用，并提供了源代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

-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

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).

简要思路：

将四个数分成两个部分。一、将前两个数求和，创建哈希数组；二、求出后两个数的和的相反数，在哈希数组中查找。

源代码：（哈希）

#include <iostream>
using namespace std;
const int size=21835456;  //哈希表数组大小
const int M=1000000000;   //让取到的值皆有正数
const int key=1357;       //发生冲突时
int hash[size],mark[size];
int a[4001],b[4001],c[4001],d[4001];
void insert(int n)
{
      int t=n;
      n=(n+M)%size;
      while(hash[n] !=M && hash[n] != t)
      {
           n=(n+key)%size;
      }
      hash[n]=t;
      mark[n]++;
}
int find(int n)
{
      int t=n;
      n=(n+M)%size;
      while(hash[n] !=M && hash[n] != t)
      {
           n=(n+key)%size;
      }
      if(hash[n] == M) return 0;
      else return mark[n];
}
int main()
{
      int count = 0;
      int n,i,j;
      
      scanf("%d",&n);
      
      for(i=1; i<=n; i++)
      {
           scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
      }
      for(i=1;i<=size;i++)
      {
           hash[i] = M;
           mark[i]=0;
      }
      for(i = 1; i<=n; i++)
           for(j = 1; j<=n; j++)
                 insert(a[i]+b[j]);
      for(i = 1; i<=n; i++)
           for(j = 1; j<=n; j++)
           {
                 count +=find(-(c[i]+d[j]));
           }
      printf("%d\n",count);
      return 0;
}

大一的时候写的，现在从网易博客移过来~