POJ 2785 4 Values whose Sum is 0 hash

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 14198 Accepted: 4027Case 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 -4626 -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

和HDOJ 1496 Equations hash 方法完全一样。不过网上很多人的解法。开的hash表比我的小很多。。。不知道为什么。话说4000个数与4000个数 两两相加不是最多有4000*4000钟可能么。。。

还有一种二分查找的做法，效率不如hash

#include <iostream>using namespace std;const int MAX=4000*4000+1;int h[MAX]={0},t[MAX]={0};int n,i,j,a[4000],b[4000],c[4000],d[4000],p,sum=0;int hash(int x){int y=x%MAX;if (y<0) y+=MAX;while (h[y]!=0&&t[y]!=x)y=(y+1)%MAX;return y;}int main(){cin>>n;for (i=0;i<n;++i)cin>>a[i]>>b[i]>>c[i]>>d[i];for (i=0;i<n;++i)for (j=0;j<n;++j){p=hash(a[i]+b[j]);h[p]++;t[p]=a[i]+b[j];}for (i=0;i<n;++i)for (j=0;j<n;++j){p=hash(-c[i]-d[j]);sum+=h[p];}cout<<sum<<endl;return 0;}

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