Solution

#include <cassert>
#include <cstdio>
#include <iostream>
#include <set>
#include <vector>
using namespace std;
using ll = int64_t;

// The possible amounts that could be left in the tank at the first barn
set<ll> possible;
void solve(ll trips_left, const vector<ll> &buckets_here,
           const vector<ll> &buckets_there, ll tank_here, ll tank_there) {
	if (trips_left == 0) {
		// We completed all the trips, so the amount currently in the tank at
		// the first barn is possible We started at the first barn and made four
		// trips, so we also end at the first barn, so @tank_here represents the
		// amount in the tank at the first barn
		possible.insert(tank_here);
	} else {
		for (ll taken_idx = 0; taken_idx < buckets_here.size(); taken_idx++) {
			ll bucket = buckets_here[taken_idx];
			// Leave the taken bucket at our destination
			vector<ll> new_buckets_dest(buckets_there);
			new_buckets_dest.push_back(bucket);
			// Remove the taken bucket from our current barn
			vector<ll> new_buckets_src;
			for (ll i = 0; i < buckets_here.size(); i++) {
				if (i != taken_idx) {
					new_buckets_src.push_back(buckets_here[i]);
				}
			}
			// Move @bucket gallons of milk from the current tank to the other
			// tank
			ll new_tank_dest = tank_there + bucket;
			ll new_tank_src = tank_here - bucket;
			// Since we went to the other barn, "here" now refers to the "dest"
			// barn and "there" now refers to the "src" barn
			solve(trips_left - 1, new_buckets_dest, new_buckets_src,
			      new_tank_dest, new_tank_src);
		}
	}
}

int main() {
	vector<ll> buckets_here;
	for (ll i = 0; i < 10; i++) {
		ll bucket;
		cin >> bucket;
		buckets_here.push_back(bucket);
	}
	vector<ll> buckets_there;
	for (ll i = 0; i < 10; i++) {
		ll bucket;
		cin >> bucket;
		buckets_there.push_back(bucket);
	}
	solve(4, buckets_here, buckets_there, 1000, 1000);
	cout << possible.size() << endl;
}