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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.NoSuchElementException;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().solve();
	}

	private void solve() throws IOException {
		try {
			//			solveA();
			//			solveB();
			solveC();
			//			 solveD();
			//			 solveE();
			//			 solveF();
		} finally {
			if (in != null) {
				in.close();
			}
			if (out != null) {
				out.flush();
				out.close();
			}
		}

	}

	private void solveA() {
		int numN = nextInt();
		int numA = nextInt();

		out.println(numN * numN - numA);
	}

	private void solveB() {
		int numN = nextInt();
		int numK = nextInt();

		int[] wk = IntStream.range(0, numN).map(i -> nextInt()).toArray();

		int res = Arrays.stream(wk).reduce(0, (sum, num) -> {
			return sum += Math.min(num * 2, (numK - num) * 2);
		});

		out.println(res);

	}

	private void chkC(int numA, int numB, int numC, int numD, int numE, int numF, int currentW,
			int currentS, int[][] memoC) {

		int index = getNoudo(currentW, currentS);

		if (currentW + currentS > numF) {
			return;
		}

		if (currentS > numE * (currentW / 100)) {
			return;
		}

		if (currentW != 0 && currentS != 0) {
			memoC[index][0] = currentW;
			memoC[index][1] = currentS;
		}

		chkC(numA, numB, numC, numD, numE, numF, currentW + numA * 100, currentS, memoC);
		chkC(numA, numB, numC, numD, numE, numF, currentW + numB * 100, currentS, memoC);
		chkC(numA, numB, numC, numD, numE, numF, currentW, currentS + numC, memoC);
		chkC(numA, numB, numC, numD, numE, numF, currentW, currentS + numD, memoC);

	}

	private int getNoudo(int w, int s) {
		if (s + w == 0) {
			return 0;
		}
		return (100 * s) / (s + w);
	}

	private void solveC() {
		int numA = nextInt();
		int numB = nextInt();
		int numC = nextInt();
		int numD = nextInt();
		int numE = nextInt();
		int numF = nextInt();

		int[][] memoC = new int[101][2];

		double noudoMax = -1;
		int water = 0;
		int sugar = 0;
		for (int i = 0; (numA * 100) * i < numF; i++) {
			int wA = (numA * 100) * i;
			if (wA > numF) {
				break;
			}
			for (int j = 0; wA + ((numB * 100) * j) < numF; j++) {
				int wB = (numB * 100) * j;
				if (wA + wB > numF) {
					break;
				}
				if (wA + wB == 0) {
					continue;
				}
				for (int k = 0; wA + wB + (numC * k) < numF; k++) {
					int sC = numC * k;
					if (numE < 100 * k / (wA + wB)) {
						break;
					}
					for (int l = 0; wA + wB + sC + (numD * l) < numF; l++) {
						int sD = numD * l;
						if (wA + wB + sC + sD == 0) {
							continue;
						}
						//						if (sC + sD == 660 || sC + sD == 448) {
						if (wA + wB + sC + sD > numF) {
							break;
						}
						if (numE * (wA + wB) < 100 * (sC + sD)) {
							break;
						}
						double noudo = 1000000 * (sC + sD) / (wA + wB + sC + sD);

						if (noudoMax < noudo) {
							noudoMax = noudo;
							water = wA + wB;
							sugar = sC + sD;
						}
					}
				}
			}
		}

		out.println(water + sugar + " " + sugar);

	}

	private void solveC2() {
		int numA = nextInt();
		int numB = nextInt();
		int numC = nextInt();
		int numD = nextInt();
		int numE = nextInt();
		int numF = nextInt();

		int[][] memoC = new int[101][2];

		chkC(numA, numB, numC, numD, numE, numF, 0, 0, memoC);

		Arrays.sort(memoC, (x, y) -> Integer.compare(x[0], y[0]));
		int p1w = memoC[100][0];
		int p1s = memoC[100][1];
		int p1n = getNoudo(p1w, p1s);
		out.println(p1w + p1s + " " + p1s);
		//		memoC = new int[101][2];
		//		chkC(numA, numB, numC, numD, numE, numF, numB * 100, 0);
		//		Arrays.sort(memoC, (x, y) -> Integer.compare(x[0], y[0]));
		//		int p2w = memoC[100][0];
		//		int p2s = memoC[100][1];
		//		int p2n = getNoudo(p2w, p2s);
		//		memoC = new int[101][2];
		//
		//		if (p1n > p2n) {
		//			out.println(p1w + p1s + " " + p1s);
		//		} else {
		//			out.println(p2w + p2s + " " + p2s);
		//		}
	}

	private void solveD() {
		int numN = nextInt();

		out.println("");
	}

	private void solveE() {
		int numN = nextInt();

		out.println("");
	}

	private void solveF() {
		int numN = nextInt();

		out.println("");
	}

	private final PrintWriter out = new PrintWriter(System.out);
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}