poj 2785 求4个数相加和为0 的个数 hash 或者二分

4 Values whose Sum is 0
Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 12654 Accepted: 3559
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 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
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).
Source

Southwestern Europe 2005

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

#include
#include
#include
using namespace std;
int a[4][4444],n;
int num1[16100000],num2[16100000],c;
int erfen(int left,int right,int k)
{
    int i;
    while(left<=right)
    {
      int mid=(left+right)/2;
      int num=0;
      if(num2[mid]==k)
       {
        num=1;
         for(i=mid-1;i>=0&&num2[i]==k;i--)  num++;
         for(i=mid+1;ik)
        right=mid-1;
       else left=mid+1;
    }
    return 0;

}
int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        int num=0;
        c=0;
        for(i=0;i 
 
#include 
#include
using namespace std;
#define MAX 1000000000
#define size 20345677
#define key 745
using namespace std;

int n,a[4040],b[4040],c[4040],d[4040],ans;
int hash[size],sum[size];
void Insert(int num)
{
	int tmp=num;
	num=(num+MAX)%size;
	while(hash[num]!=MAX && hash[num]!=tmp)
		num=(num+key)%size;
	hash[num]=tmp;
	sum[num]++;
}
int Find(int num)
{
	int tmp=num;
	num=(num+MAX)%size;
	while(hash[num]!=MAX && hash[num]!=tmp)
		num=(num+key)%size;
	if(hash[num]==MAX)
		return 0;
	else
		return sum[num];
}
int main()
{
	int i,j;
	scanf("%d",&n);
	for(i=0;i