poj 2785 求4個數相加和為0 的個數 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
 
輸入n  表示a b c d 四個集合都有n個元素    之後每行輸入4個集合中的一個元素  求這四個集合每個集合中拿出一個數 相加等於0的組數

#include<stdio.h>
#include<string.h>
#include<algorithm>
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;i<n*n&&num2[i]==k;i++)  num++;
         return num;
       }
      else if(num2[mid]>k)
        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<n;i++)
        {
            scanf("%d %d %d %d",&a[0][i],&a[1][i],&a[2][i],&a[3][i]);
        }
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            {
                 num1[c]=a[2][i]+a[3][j];
                 num2[c++]=-(a[0][i]+a[1][j]);
            }
            sort(num2,num2+c);
        for(i=0;i<c;i++)
            {
                num+=erfen(0,n*n-1,num1[i]);
            }
        printf("%d\n",num);
    }
    return 0;
}

#include <iostream>
#include<stdio.h>
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<n;i++)
		scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
	for(i=0;i<size;i++)
		hash[i]=MAX;
	for(i=0;i<n;i++)
		for(j=0;j<n;j++)
			Insert(a[i]+b[j]);
	for(i=0;i<n;i++)
		for(j=0;j<n;j++)
			ans+=Find(-(c[i]+d[j]));
	printf("%d\n",ans);
}