4 Values whose Sum is 0 UVA

4 Values whose Sum is 0

 UVA - 1152

题意：给ｎ组４个数，问有多少种组合（每组选一个）使得四个数的和为零．

补个简单题．．．之前写的哈希表有问题一直没过就忘了，今天刚好看到～

#include <bits/stdc++.h>
using namespace std;
#define ll long long 
const int maxn = 4010;
const int H_SZ = 16131313;

int a[maxn][4];

struct HS{
    ll val[H_SZ], res[H_SZ];
    int head[H_SZ], nex[H_SZ];
    int cnt;

    void init(){
        memset(head, -1, sizeof(head));
        memset(res, 0, sizeof(res));
        cnt = 0;
    }

    void add(ll x) {
        x += 1e11;
        int u = x%H_SZ;
        val[cnt] = x;
        res[cnt] ++;
        nex[cnt] = head[u];
        head[u] = cnt++;
    }

    int query(ll x) {
        x += 1e11;
        int u = x%H_SZ;
        for(int i = head[u]; ~i; i = nex[i]) {
            if(val[i] == x) return i;
        }
        return H_SZ;
    }
}hs;

int main(){
    int t, kase = 0;
    //freopen("in.txt", "r", stdin);
    scanf("%d", &t);
    while(t--) {
        if(kase) puts("");
        kase++;
        hs.init();
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < 4; j++) {
                scanf("%d", &a[i][j]);
            }
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                int x = a[i][0] + a[j][1];
                int id = hs.query(x);
                if(id == H_SZ) hs.add(x);
                else hs.res[id]++;
            }
        }
        int ans = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                int x = -(a[i][2] + a[j][3]);
                int id = hs.query(x);
                if(id != H_SZ) ans += hs.res[id]; 
            }
        }
        printf("%d
", ans);
    }
}