例题8-3　和为0的4个值（4 Values Whose Sum is Zero, ACM/ICPC SWERC 2005, UVa 1152）

#include <set>
#include <map>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <cstdio>
#include <string>
#include <vector>
#include <cctype>
#include <sstream>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define SF(a) scanf("%d", &a)
#define PF(a) printf("%d\n", a)  
#define SFF(a, b) scanf("%d%d", &a, &b)  
#define SFFF(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define CLEAR(a, b) memset(a, b, sizeof(a))
#define IN() freopen("in.txt", "r", stdin)
#define OUT() freopen("out.txt", "w", stdout)
#define FOR(i, a, b) for(int i = a; i < b; ++i)
#define LL long long
#define maxn 4005
#define maxm 205
#define mod 1000000007
#define INF 10000007
#define eps 1e-4
using namespace std;
//-------------------------CHC------------------------------//
int a[maxn], b[maxn], c[maxn], d[maxn];
int sum[maxn*maxn];

int main() {
	int t;
	SF(t);
	while (t--) {
		int n;
		SF(n);
		FOR(i, 0, n) SFFF(a[i], b[i], c[i]), SF(d[i]);
		int ns = 0;
		FOR(i, 0, n) FOR(j, 0, n) sum[ns++] = a[i] + b[j];
		LL ans = 0;
		sort(sum, sum + ns);
		FOR(i, 0, n) FOR(j, 0, n)
			ans += upper_bound(sum, sum + ns, -c[i] - d[j]) - lower_bound(sum, sum + ns, -c[i] - d[j]);
		printf("%lld\n", ans);
		if (t) puts("");
	}
	return 0;
}
#include <set>
#include <map