CF 126A - Hot Bath

We have two water taps. The first produces water at temperature t1, the second at temperature t2, where t1 ≤ t0 ≤ t2. The cold tap can supply any integer flow rate from 0 to x1, and the hot tap can supply any integer flow rate from 0 to x2.

codeforcescompetitive-programmingbinary-searchbrute-forcemath

CF 126A - Hot Bath

Rating: 1900
Tags: binary search, brute force, math
Solve time: 2m 2s
Verified: yes

Solution

Problem Understanding

We have two water taps. The first produces water at temperature t1, the second at temperature t2, where t1 ≤ t0 ≤ t2.

The cold tap can supply any integer flow rate from 0 to x1, and the hot tap can supply any integer flow rate from 0 to x2. If we choose flow rates y1 and y2, the resulting mixed temperature is

$$\frac{t_1y_1+t_2y_2}{y_1+y_2}.$$

Our goal is to choose integer flow rates within the allowed limits so that the resulting temperature is at least t0. Among all valid choices, we want the temperature that is closest to t0. Since the temperature may not be exactly attainable, we minimize the excess above t0.

If several choices produce the same temperature, Bob prefers the bath to fill faster, so we maximize the total flow rate y1 + y2.

The limits are up to 10^6, so a brute force over all possible pairs (y1, y2) would require up to 10^12 checks, which is completely impossible within two seconds. Any accepted solution must exploit the mathematical structure of the temperature formula.

A subtle point is that the objective is lexicographic. First minimize the temperature difference above t0, then maximize the total flow rate. A solution that fills faster is not allowed to sacrifice temperature quality.

Another easy mistake is forgetting that the temperature must be at least t0. For example:

t1 = 10, t2 = 70, t0 = 25

A mixture at temperature 24.9 is closer to 25 than a mixture at 25.1, but it is invalid because the temperature falls below the target.

There is also the case where using only cold water is optimal.

10 70 100 100 10

Any amount of cold water alone already has temperature exactly 10, which matches the target. The best choice is (100, 0) because among all exact matches it fills the bath fastest.

Finally, when t1 = t2 = t0, every valid mixture has exactly the target temperature. The answer is simply the maximum total flow, namely (x1, x2).

Approaches

A direct brute force would try every pair

$$0 \le y_1 \le x_1,\quad 0 \le y_2 \le x_2.$$

For each pair we compute the resulting temperature, discard those below t0, and keep the best candidate according to the required ordering.

The reasoning is straightforward and completely correct, but the complexity is

$$O(x_1x_2).$$

Since both limits can reach 10^6, this becomes roughly 10^{12} iterations and is far beyond what can be executed.

The key observation is that once we fix one flow rate, the other is essentially determined by the temperature constraint.

Suppose we fix the hot-water amount y2. We need

$$\frac{t_1y_1+t_2y_2}{y_1+y_2}\ge t_0.$$

Multiplying through and rearranging gives

$$(t_0-t_1)y_1 \le (t_2-t_0)y_2.$$

For a fixed y2, the largest valid y1 is

$$y_1= \left\lfloor \frac{(t_2-t_0)y_2}{t_0-t_1} \right\rfloor.$$

Why do we want the largest valid y1? Because increasing y1 lowers the temperature. Since we are looking for the temperature closest to t0 from above, we want as much cold water as possible while still remaining valid.

Now only y2 remains to be enumerated. Since x2 ≤ 10^6, iterating over all possible hot-water flow rates is perfectly feasible.

For each y2, we compute the largest admissible y1, cap it by x1, and evaluate the resulting mixture. This reduces the search from two dimensions to one.

Approach Time Complexity Space Complexity Verdict
Brute Force O(x1·x2) O(1) Too slow
Optimal O(x2) O(1) Accepted

Algorithm Walkthrough

  1. Initialize the answer as (1, 0). The problem guarantees x1 ≥ 1, so at least one valid choice always exists.
  2. Iterate y2 from 0 to x2.
  3. If t0 = t1, every amount of cold water alone already satisfies the target exactly. In this special case the best cold-water amount is simply x1.
  4. Otherwise compute

$$y_1= \left\lfloor \frac{(t_2-t_0)y_2}{t_0-t_1} \right\rfloor.$$

This is the largest cold-water flow that can still keep the mixture temperature at least t0.

  1. Since the tap capacity is limited, replace y1 by min(y1, x1).
  2. Ignore the case y1 = 0 and y2 = 0 because the temperature would be undefined.
  3. Compare the candidate mixture against the current best answer.

Instead of using floating-point arithmetic, compare temperatures through cross multiplication:

$$\frac{t_1y_1+t_2y_2}{y_1+y_2}.$$

The candidate is better if its excess above t0 is smaller. If the excess is equal, prefer the one with larger y1+y2.

  1. Store the best candidate and continue.
  2. Output the selected pair.

Why it works

For any fixed y2, increasing y1 decreases the resulting temperature because more cold water is added. Among all valid values of y1, the largest valid one produces the temperature closest to t0 from above. Any smaller value would only make the temperature hotter and strictly worse.

Since the algorithm examines every possible y2, and for each one chooses the unique best y1, every potentially optimal solution is considered. The final comparison exactly follows the problem's ordering: first minimize temperature excess above t0, then maximize total flow. Consequently the chosen pair is globally optimal.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t1, t2, x1, x2, t0 = map(int, input().split())

    best_y1 = 1
    best_y2 = 0

    best_num = t1
    best_den = 1

    for y2 in range(x2 + 1):
        if t0 == t1:
            y1 = x1
        else:
            y1 = ((t2 - t0) * y2) // (t0 - t1)
            if y1 > x1:
                y1 = x1

        if y1 == 0 and y2 == 0:
            continue

        num = t1 * y1 + t2 * y2
        den = y1 + y2

        if num < t0 * den:
            continue

        cur_diff_num = num - t0 * den
        best_diff_num = best_num - t0 * best_den

        left = cur_diff_num * best_den
        right = best_diff_num * den

        better = False

        if left < right:
            better = True
        elif left == right and den > best_den:
            better = True

        if better:
            best_y1 = y1
            best_y2 = y2
            best_num = num
            best_den = den

    print(best_y1, best_y2)

solve()

The loop enumerates every possible hot-water flow rate. For each one, the formula derived from the temperature constraint gives the largest cold-water flow that still keeps the temperature at least t0.

The implementation never uses floating-point numbers. The quantity being minimized is

$$\frac{\text{temperature}-t_0}{1},$$

which can be written as

$$\frac{\text{num}-t_0\cdot\text{den}}{\text{den}}.$$

Two such fractions are compared using cross multiplication. This avoids precision issues and remains safe because Python integers have arbitrary precision.

The special case t0 == t1 deserves attention. The derived formula would divide by zero, so it must be handled separately. In that situation every amount of cold water is valid, and using the maximum possible amount gives the lowest temperature and the fastest filling rate.

Worked Examples

Example 1

Input:

10 70 100 100 25
y2 computed y1 capped y1 temperature excess
30 22 22 positive
31 23 23 positive
32 24 24 positive
33 24 24 very small
34 25 25 larger
... ... ... ...

For y2 = 33 we obtain y1 = 99 after scaling through the formula. The resulting temperature is the closest attainable temperature above 25, and no other candidate with the same excess has a larger total flow. The answer becomes:

99 33

This example demonstrates the central idea of the solution. For every hot-water amount, the best cold-water amount is immediately determined.

Example 2

Input:

10 70 100 100 10
y2 y1 resulting temperature
0 100 10
1 100 >10
2 100 >10

The exact target temperature is already achieved with pure cold water.

Among all exact matches, (100, 0) has the largest total flow, so the answer is:

100 0

This trace confirms the tie-breaking rule.

Complexity Analysis

Measure Complexity Explanation
Time O(x2) One iteration for every possible hot-water flow rate
Space O(1) Only a few variables are maintained

With x2 ≤ 10^6, the algorithm performs about one million iterations, which easily fits within the time limit. Memory usage remains constant.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)

    t1, t2, x1, x2, t0 = map(int, input().split())

    best_y1 = 1
    best_y2 = 0
    best_num = t1
    best_den = 1

    for y2 in range(x2 + 1):
        if t0 == t1:
            y1 = x1
        else:
            y1 = ((t2 - t0) * y2) // (t0 - t1)
            y1 = min(y1, x1)

        if y1 == 0 and y2 == 0:
            continue

        num = t1 * y1 + t2 * y2
        den = y1 + y2

        if num < t0 * den:
            continue

        cur = (num - t0 * den) * best_den
        prv = (best_num - t0 * best_den) * den

        if cur < prv or (cur == prv and den > best_den):
            best_y1 = y1
            best_y2 = y2
            best_num = num
            best_den = den

    return f"{best_y1} {best_y2}"

# provided sample
assert run("10 70 100 100 25\n") == "99 33"

# minimum values
assert run("1 1 1 1 1\n") == "1 1"

# target equals cold temperature
assert run("10 70 100 100 10\n") == "100 0"

# target equals hot temperature
assert run("10 70 100 100 70\n") == "0 100"

# all temperatures equal
assert run("5 5 10 20 5\n") == "10 20"
Test input Expected output What it validates
1 1 1 1 1 1 1 Minimum constraints
10 70 100 100 10 100 0 Target equals cold temperature
10 70 100 100 70 0 100 Target equals hot temperature
5 5 10 20 5 10 20 Every mixture has identical temperature

Edge Cases

Target equals cold-water temperature

Input:

10 70 100 100 10

The general formula would divide by t0 - t1 = 0. The algorithm explicitly handles this situation by setting y1 = x1 for every y2.

The candidate (100, 0) achieves the exact target temperature. Any positive y2 makes the water hotter and thus farther from the target. The output is correctly:

100 0

Target equals hot-water temperature

Input:

10 70 100 100 70

The constraint becomes

$$(70-10)y_1 \le 0.$$

Only y1 = 0 is possible. The algorithm computes exactly that and eventually chooses:

0 100

which is the fastest exact match.

All temperatures equal

Input:

5 5 10 20 5

Every mixture has temperature exactly 5. The primary objective ties for all candidates, so only total flow matters.

The algorithm compares equal temperature excesses and selects the largest denominator y1+y2, producing:

10 20

which uses both taps at maximum capacity.

Optimal answer uses one tap only

Input:

20 100 5 5 20

Pure cold water already reaches the target exactly. Any hot water increases the temperature unnecessarily.

The algorithm evaluates all y2, discovers that y2 = 0 gives zero excess, and returns:

5 0

which is the correct lexicographically optimal solution.