poj 2785 4 Values whose Sum is 0

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 14474		Accepted: 4137
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 .

“和”的问题可总结如下：给定四个列表A，B，C，D里面是整数，计算出有多少四胞胎（A，B，C，D）∈ A x B x C x D，A + B+ C + D=0。在下文中，我们假设所有的列表具有相同的大小为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 .

输入文件的第一行包含列表n的大小（此值可以是大到4000）。然后，我们有n个行分别是A，B，C和D，每个列表4个整数值（以绝对值最大 2 ²⁸）。

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

输出有多少个四胞胎T_T

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

hash，一开始也知道就是用前两个的和做表，后两个和找表求和，因为跟前一题同一个思路。。。不过趁机学了一个新的hash方式，跟之前的线性探测不一样，就是用静态的链表，恩。。。这种直接模拟的链表，来解决冲突，然后直接用模大质数来做hash函数，这个方式应该效率蛮高的。主要最坏效率取决于链表长度呗。。。所以肯定用的质数越大，越不容易起冲突效率越高，然后就挑了跟学的那位的质数3999971，其实我印象里过4000000的数组基本就爆了。。。不过我的记性不咋地，如此看来也就能开这么大了。。。哦，由于加和的可能性是4000^4，所以要给输出的ans一个long long型，不然就爆了。。。就数据本身来说有的加和可能性也有4000*4000，所以hash表的长度咋地都不够，肯定要冲突。。。T_T

#include <iostream>
#include<algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#define MAX 4444444

using namespace std;
void insert_hash(int s);
int find_hash(int s);
int hash_i(int s);

struct LINK{
    int val;
    int next;
    int c;
}link[3999971];
int head[3999971];
int tot;

int main()
{
    int i, j, n;
    int a[4111], b[4111], c[4111], d[4111];
    long long ans;
    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout)；

    while(scanf("%d", &n)!= EOF){
        tot = 0;
        ans = 0;
        for (i = 0;i < n;i ++)
            scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);

        memset(head, -1, sizeof(head));
        for (i = 0;i < n;++ i){
            for (j = 0;j < n;++ j){
                insert_hash(a[i] + b[j]);
            }
        }
        for (i = 0;i < n;++ i){
            for (j = 0;j < n;++ j){
                ans += find_hash(-c[i]-d[j]);
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}

void insert_hash(int s){
    int pos = hash_i(s);
    int i;
    for (i = head[pos];i != -1; i = link[i].next){
        if (link[i].val == s){
            link[i].c ++;
            return;
        }
    }
    link[tot].val = s;
    link[tot].c = 1;
    link[tot].next = head[pos];
    head[pos] = tot ++;
}

int find_hash(int s){
    int pos = hash_i(s);
    int i;
    for (i = head[pos];i != -1; i = link[i].next){
        if (link[i].val == s){
            return link[i].c;
        }
    }
    return 0;
}
int hash_i(int s){
    return (s + 3899971) % 3999971;
}