Problem:
Score : 300 points

Problem Statement

Snuke is making sugar water in a beaker.
Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations.

Operation 1: Pour 100A grams of water into the beaker.

Operation 2: Pour 100B grams of water into the beaker.

Operation 3: Put C grams of sugar into the beaker.

Operation 4: Put D grams of sugar into the beaker.

In our experimental environment, E grams of sugar can dissolve into 100 grams of water.

Snuke will make sugar water with the highest possible density.

The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker.
Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it.
If there is more than one candidate, any of them will be accepted.

We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent.
Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.

Constraints

1 \leq A < B \leq 30

1 \leq C < D \leq 30

1 \leq E \leq 100

100A \leq F \leq 3 000

A, B, C, D, E and F are all integers.

Inputs

Input is given from Standard Input in the following format:

A B C D E F

Outputs

Print two integers separated by a space.
The first integer should be the mass of the desired sugar water, and the second should be the mass of the sugar dissolved in it.

Sample Input 1

1 2 10 20 15 200

Sample Output 1

110 10

In this environment, 15 grams of sugar can dissolve into 100 grams of water, and the beaker can contain at most 200 grams of substances.

We can make 110 grams of sugar water by performing Operation 1 once and Operation 3 once.
It is not possible to make sugar water with higher density.
For example, the following sequences of operations are infeasible:

If we perform Operation 1 once and Operation 4 once, there will be undissolved sugar in the beaker.

If we perform Operation 2 once and Operation 3 three times, the mass of substances in the beaker will exceed 200 grams.

Sample Input 2

1 2 1 2 100 1000

Sample Output 2

200 100

There are other acceptable outputs, such as:

400 200

However, the output below is not acceptable:

300 150

This is because, in order to make 300 grams of sugar water containing 150 grams of sugar, we need to pour exactly 150 grams of water into the beaker, which is impossible.

Sample Input 3

17 19 22 26 55 2802

Sample Output 3

2634 934

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <iomanip>
#include <vector>
#include <string.h>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <math.h>
#include <algorithm>
#include <numeric>


using namespace std;

// マクロ&定数&関数 ================================================
typedef unsigned int uint;
typedef long long ll;

typedef pair<ll, ll> pll;

typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<bool> vbool;
typedef vector<string> vstring;

typedef vector<pair<int, int>> vpint;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<double, double>> vpdouble;

typedef vector<vector<int>> vvint;
typedef vector<vector<ll>> vvll;
typedef vector<vpint> vvpint;
typedef vector<vpll> vvpll;
typedef vector<vector<double>> vvdouble;
typedef vector<vector<string>> vvstring;
typedef vector<vector<bool>> vvbool;

typedef vector<vector<vector<ll>>> vvvll;

const int INF = 1e9 + 1;
const ll LLINF = 1e17 + 1;
const int DX[9] = { 0,  0, 1, -1, 1,  1, -1, -1, 0 }; // 4;４近傍
const int DY[9] = { 1, -1, 0,  0, 1, -1,  1, -1, 0 }; // 8:８近傍 9:(0,0)を含む
const ll MOD = 1000000007; //10^9 + 7
const ll MAX = 1e9;
const double PI = 3.14159265358979323846264338327950288;



//---------------------------------------------------------------
//  オーバーフローチェック
//---------------------------------------------------------------
bool is_overflow(ll a, ll b)
{
	return ((a * b) / b != a);
}

//---------------------------------------------------------------
//  約数列挙
//---------------------------------------------------------------
vll divisor(ll n) {
	vll ret;
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(begin(ret), end(ret));
	return (ret);
}

//---------------------------------------------------------------
// N以下のすべての素数を列挙する(エラトステネスの篩)
//---------------------------------------------------------------
vbool searchSosuu(ll N) {
	vbool sosuu;
	for (ll i = 0; i < N; i++) {
		sosuu.emplace_back(true);
	}
	sosuu[0] = false;
	sosuu[1] = false;
	for (ll i = 2; i < N; i++) {
		if (sosuu[i]) {
			for (ll j = 2; i * j < N; j++) {
				sosuu[i * j] = false;
			}
		}
	}

	return sosuu;
}

//---------------------------------------------------------------
// 素因数分解　O(√N)
//---------------------------------------------------------------
vpll div_prime(ll n)
{
	vpll prime_factor;
	for (ll i = 2; i * i <= n; i++)
	{
		ll count = 0;
		while (n % i == 0)
		{
			count++;
			n /= i;
		}
		if (count)
		{
			pair<ll, ll> temp = { i, count };
			prime_factor.emplace_back(temp);
		}
	}
	if (n != 1)
	{
		pair<ll, ll> temp = { n, 1 };
		prime_factor.emplace_back(temp);
	}

	return prime_factor;
}

//---------------------------------------------------------------
// 素数判定
//---------------------------------------------------------------
bool is_sosuu(ll N) {

	if (N < 2)			 return false;
	else if (N == 2)	 return true;
	else if (N % 2 == 0) return false;

	for (ll i = 3; i <= sqrt(N); i += 2) {
		if (N % i == 0)
			return false;
	}

	return true;
}

//---------------------------------------------------------------
// 最大公約数（ユークリッドの互除法）
//---------------------------------------------------------------
ll gcd(ll a, ll b) {
	if (a < b) {
		ll tmp = a;
		a = b;
		b = tmp;
	}

	ll r = a % b;
	while (r != 0) {
		a = b;
		b = r;
		r = a % b;
	}

	return b;
}

//---------------------------------------------------------------
// 最小公倍数
//---------------------------------------------------------------
ll lcm(ll a, ll b) {
	ll temp = gcd(a, b);
	return temp * (a / temp) * (b / temp);
}

//---------------------------------------------------------------
// 階乗
//---------------------------------------------------------------
ll factorial(ll n)
{
	if (n <= 1)
		return 1;
	return (n * (factorial(n - 1))) % MOD;
}

//---------------------------------------------------------------
// 高速コンビネーション計算(前処理:O(N) 計算:O(1))
//---------------------------------------------------------------
// テーブルを作る前処理
ll comb_const = 200005;
vll fac(comb_const), finv(comb_const), inv(comb_const);
bool COMineted = false;
void COMinit() {

	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (ll i = 2; i < comb_const; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}

	COMineted = true;
}

// 二項係数計算
ll COM(ll n, ll k) {
	if (COMineted == false) COMinit();

	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//---------------------------------------------------------------
// 繰り返し２乗法 base^sisuu
//---------------------------------------------------------------
ll RepeatSquaring(ll base, ll sisuu) {
	if (sisuu < 0) {
		cout << "RepeatSquaring: 指数が負です!" << endl;
		return 0;
	}
	if (sisuu == 0) return 1;
	if (sisuu % 2 == 0) {
		ll t = RepeatSquaring(base, sisuu / 2) % MOD;
		return (t * t) % MOD;
	}
	return base * RepeatSquaring(base, sisuu - 1) % MOD;
}

//---------------------------------------------------------------
// 高速単発コンビネーション計算(O(logN))
//---------------------------------------------------------------
ll fast_com(ll a, ll b) {

	ll bunshi = 1;
	ll bunbo = 1;
	for (ll i = 1; i <= b; i++)
	{
		bunbo *= i;
		bunbo %= MOD;

		bunshi *= (a - i + 1);
		bunshi %= MOD;
	}

	ll ret = bunshi * RepeatSquaring(bunbo, MOD - 2);

	ret %= MOD;
	while (ret < 0)
	{
		ret += MOD;
	}

	return ret;
}

//---------------------------------------------------------------
// 2直線の交差判定(直線(x1, y1)->(x2, y2) と 直線(X1, Y1)->(X2, Y2))
//---------------------------------------------------------------
bool is_cross(ll x1, ll y1, ll x2, ll y2, ll X1, ll Y1, ll X2, ll Y2)
{
	ll dx_ai = X1 - x1;
	ll dy_ai = Y1 - y1;
	ll dx_bi = X1 - x2;
	ll dy_bi = Y1 - y2;

	ll dx_ai2 = X2 - x1;
	ll dy_ai2 = Y2 - y1;
	ll dx_bi2 = X2 - x2;
	ll dy_bi2 = Y2 - y2;

	ll si = dx_ai * dy_bi - dy_ai * dx_bi;
	ll si2 = dx_ai2 * dy_bi2 - dy_ai2 * dx_bi2;

	ll si3 = dx_ai * dy_ai2 - dy_ai * dx_ai2;
	ll si4 = dx_bi * dy_bi2 - dy_bi * dx_bi2;

	return (si * si2 < 0 && si3 * si4 < 0);

}

//---------------------------------------------------------------
// 最長増加部分列の長さ(O(NlogN))
//---------------------------------------------------------------
ll LSI(vll vec)
{
	ll size = vec.size();
	vll lsi(size + 1); // 長さjを作った時の右端の最小値
	for (ll i = 0; i <= size; i++)
	{
		lsi[i] = LLINF;
	}

	lsi[0] = 0;
	lsi[1] = vec[0];

	for (ll i = 1; i < size; i++)
	{
		// 初めてvec[i]の方が小さくなるところを探す
		auto Iter = lower_bound(lsi.begin(), lsi.end(), vec[i]);
		ll idx = Iter - lsi.begin();

		if (idx > 0 && lsi[idx - 1] < vec[i])
		{
			lsi[idx] = vec[i];
		}
	}

	for (ll i = size; i >= 0; i--)
	{
		if (lsi[i] < LLINF)
		{
			return i;
		}
	}
}

//---------------------------------------------------------------
// 木の根からの深さ
//---------------------------------------------------------------
vll tree_depth(vvll edge, ll start_node, ll n_node)
{
	vll dist(n_node, LLINF);
	dist[start_node] = 0;

	stack<pll> S;
	S.push({ start_node, 0 });

	while (!S.empty())
	{
		ll node = S.top().first;
		ll d = S.top().second;
		dist[node] = d;

		S.pop();

		for (int i = 0; i < edge[node].size(); i++)
		{
			if (dist[edge[node][i]] == LLINF)
			{
				S.push({ edge[node][i], d + 1 });
			}
		}
	}

	return dist;
}

//---------------------------------------------------------------
// ワーシャルフロイド法(O(N^3)) 任意の2点間の最短距離
//---------------------------------------------------------------
vvll warshall_floyd(vvll d) {
	ll n = d.size();
	for (int k = 0; k < n; k++) {       // 経由する頂点
		for (int i = 0; i < n; i++) {    // 始点
			for (int j = 0; j < n; j++) {  // 終点
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
			}
		}
	}
	return d;
}

//---------------------------------------------------------------
// Union Find
//---------------------------------------------------------------
class UnionFind {
public:
	/*
	UnionFind uf(要素の個数);
	for(int i = 0;i < 関係の個数; i++)
	{
		uf.merge(A[i], B[i]);
	}

	nを含む集合の大きさ = uf.size(n)
	nを含む集合の代表者 = uf.root(n)
	集合の個数 = uf.n_group
	*/
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)
	ll n_group; //集合の数

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
		n_group = sz_;
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		n_group--;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};


//---------------------------------------------------------------
// ダイクストラ(O(ElogV))   edge: 隣接リスト from->pair{to, cost}
//---------------------------------------------------------------
template< typename T >
vector< T > dijkstra(vector<vector<pair<ll, T>>> edge, int start) {


	vector< T > dist(edge.size(), LLINF);
	ll n = dist.size();

	using Pi = pair< T, int >;
	priority_queue< Pi, vector< Pi >, greater< Pi > > que;
	dist[start] = 0;
	que.emplace(dist[start], start);
	while (!que.empty()) {
		T cost;
		int idx;
		tie(cost, idx) = que.top();
		que.pop();
		if (dist[idx] < cost) continue;
		for (int i = 0; i < edge[idx].size(); i++)
		{
			auto next_cost = cost + edge[idx][i].second;
			if (dist[edge[idx][i].first] <= next_cost) continue;
			dist[edge[idx][i].first] = next_cost;
			que.emplace(dist[edge[idx][i].first], edge[idx][i].first);
		}
	}
	return dist;
}


//========================================================================






int main() {

	////==================================
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(30);
	////==================================


	/*
	～思いついたことはとりあえず絶対メモする！！～
	*/



	ll A, B, C, D, E, F;
	cin >> A >> B >> C >> D >> E >> F;

	ll best_all = LLINF;
	ll best_salt = 0;

	for (int a = 0; a * A * 100 <= F; a++)
	{
		for (int b = 0; a * A * 100 + b * B * 100 <= F; b++)
		{

			ll water = a * A * 100 + b * B * 100;
			ll can_salt = (a*A+b*B)*E;


			for (int c = 0; c * C + water <= F && c * C <= can_salt; c++)
			{
				for (int d = 0; d * D + c * C + water <= F && d * D + c * C <= can_salt; d++)
				{

					ll salt = c * C + d * D;

					double best_per = (double)best_salt / (double)best_all;
					double per = (double)salt / (double)(water + salt);

					if (per > best_per)
					{
						best_all = water + salt;
						best_salt = salt;
					}

				}
			}


		}
	}




	cout << best_all << " " << best_salt << endl;





}