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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
 
#define BIG 2000000007
 
#define MOD 1000000007
typedef unsigned long long ull;
typedef   signed long long sll;
 
typedef struct {
	int a;
	int b;
} hw;
 
#define N_MAX 1000
#define M_MAX 100000
 
ull n, m;
ull h, w;
int k;
ull va, vb, vc, vd, ve, vf;
// ull a[N_MAX];
// sll a[N_MAX];
// ull b[N_MAX];
// ull dp[N_MAX][M_MAX + 1];
char s[N_MAX + 1];
// char t[N_MAX + 1];
// hw arr[N_MAX];
ull digitdp[102][   2][    2];
//          pos  less  carry
 
void swap_adj(ull *a, ull *b){
	ull tmp = *b;
	*b = *a;
	*a = tmp;
	return;
}

ull divide(ull a, ull b){
	ull x = MOD - 2;
	ull ans = 1;
	while (x) {
		if (x & 1) ans = (ans * b) % MOD;
		b = (b * b) % MOD;
		x /= 2;
	}
	return (a * ans) % MOD;
}
 
int digits(ull x){
	int i = 1;
	while (x >= 10) {
		x /= 10;
		i++;
	}
	return i;
}
 
int min(ull x, ull y){
	return (x < y) ? x : y;
}
 
ull gcd(ull x, ull y){
	if (x < y) {
		return gcd(y, x);
	} else if (y == 0) {
		return x;
	} else {
		return gcd(y, x % y);
	}
}
 
ull bitpow(ull a, ull x){
	ull result = 1;
	while (x) {
		if (x & 1) {
			result *= a;
			result %= MOD;
		}
		x /= 2;
		a = (a * a) % MOD;
	}
	return result;
}

int comp(const void *left, const void *right){
	if ((*(int*)left) < (*(int*)right)) {
		return -1;
	} else if ((*(int*)left) > (*(int*)right)) {
		return +1;
	} else {
		return 0;
	}
}

// int nextroute(int arr[]){
// 	int i = n - 1;
// 	int j, x;
// 	while (arr[i - 1] > arr[i]) i--;

// 	x = n;
// 	for (j = i; j < n; j++) {
// 		if (arr[j] < arr[i - 1]) continue;
// 		if (x == n || arr[x] > arr[j]) x = j;
// 	}
// 	arr[i - 1] ^= arr[x];
// 	arr[x] ^= arr[i - 1];
// 	arr[i - 1] ^= arr[x];

// 	qsort(&arr[i], n - i, sizeof(int), comp);
// 	return 0;
// }

int nibutan_target(ull target){
	ull maxdist = (target * (target + 1) / 2); // 時刻targetまでに到着できる距離は[-maxdist, maxdist]
	return (n <= maxdist);
}

int targetdig(ull x, int index /* 1-indexed */){
	// static...?
	int posmax = digits(x);
	if (posmax < index) return -1;
	while (posmax > index) {
		posmax--;
		x /= 10;
	}
	return x % 10;
}

ull solve(){
	int i, j;
	// int pos, less, carry;
	// const int posmax = strlen(s);
	ull answer = 0;
	int bunshi[3001] = {1}, bunbo[3001] = {1};
	int underindex;
	int resultover = 0, resultunder = 0;

	for (i = 0; i * 100 < vf; i++) {
		if ((i >= va && bunbo[i - va]) || (i >= vb && bunbo[i - vb])) {
			bunbo[i] = 1;
			// printf("%d ", i);
		}
	}
	// puts("");

	for (i = 0; i < vf; i++) {
		if ((i >= vc && bunshi[i - vc]) || (i >= vd && bunshi[i - vd])) {
			bunshi[i] = 1;
			// printf("%d  ", i);
		}
	}
	// puts("");

	underindex = vf;
	for (i = 0; i < vf; i++) {
		if (!bunshi[i]) continue;
		while (i + underindex * 100 > vf || !bunbo[underindex]) underindex--;
		// printf("  %d... %d\n", i, underindex * 100);
		if (underindex == 0) break;
		if (i > ve * underindex) continue;

		// printf("  ? %d / %d <= %d / 100\n", i, underindex * 100, ve);

		if (underindex * resultover <= i * resultunder) {
			resultover = i;
			resultunder = underindex;
			// printf("    ! %d / %d <= %d / 100\n", resultover, resultunder * 100, ve);
		}

	}

	printf("%d %d\n", resultover + resultunder * 100, resultover);

	return answer;
}
 
int main(void){
	int i, j;
	int x;

	scanf("%llu%llu%llu%llu%llu%llu", &va, &vb, &vc, &vd, &ve, &vf);

	solve();


	return 0;
}