POJ-2785 Values whose Sum is 0 Hash表

题目链接：https://cn.vjudge.net/problem/POJ-2785

题意

给出四组数，每组有n个数
现从每组数中取一个数作为a,b,c,d
问有几组这样的a+b+c+d=0

思路

首先把第一组和第二组的和添加在hash表里
再枚举三组四组的和，查找即可

代码

#include <cstdio>
#include <cstring>
const int hashSize=int(4e5), idxSize=int(1.6e7);
struct Data{
int value, next, cnt;
Data(int value=0, int next=-1, int cnt=0):
value(value),next(next),cnt(cnt) {}
}data[idxSize];
int head[hashSize];
struct Hash{
int size;
Hash(void):size(0) {
memset(head, -1, sizeof(head));
}

int hash(int num){
return (num+hashSize)%hashSize;
}
int find(int num){
int key=hash(num);
for (int i=head[key]; i!=-1; i=data[i].next)
if (data[i].value==num) return data[i].cnt;
return 0;
}
int insert(int num){
int key=hash(num);
for (int i=head[key]; i!=-1; i=data[i].next)
if (data[i].value==num) {data[i].cnt++; return i;}
data[size]=Data(num, head[key], 1);
return head[key]=size++;
}
};
int n;

int main(void){
while (scanf("%d", &n)==1){
int tmp[4000+5][4];
for (int i=0; i<n; i++)
scanf("%d%d%d%d", &tmp[i][0], &tmp[i][1], &tmp[i][2], &tmp[i][3]);

Hash hash;
for (int i=0; i<n; i++)
for (int j=0; j<n; j++)
hash.insert(tmp[i][2]+tmp[j][3]);

int ans=0;
for (int i=0; i<n; i++)
for (int j=0; j<n; j++) ans+=hash.find(-tmp[i][0]-tmp[j][1]);
printf("%d\n", ans);
}

return 0;
}