CF 1184A1 - Heidi Learns Hashing (Easy)

Rating: 1200
Tags: brute force, math, number theory
Solve time: 3m 46s
Verified: yes

Solution

Problem Understanding

The problem gives us a function $H(x, y) = x^2 + 2xy + x + 1$, and a positive integer $r$. We need to determine whether there exists a pair of positive integers $(x, y)$ such that $H(x, y) = r$. If multiple pairs exist, we must return the one with the smallest $x$, otherwise we return "NO". The input $r$ can be as large as $10^{12}$, so any solution must handle very large integers efficiently.

Because $x$ and $y$ are positive integers, we know immediately that $H(x, y)$ grows at least quadratically in $x$ and linearly in $y$. If $r$ is small, many large $x$ values cannot possibly work, so iterating over $x$ only up to some reasonable bound is feasible. If $r$ is very large, we cannot iterate over all possible $x, y$ pairs naively, because $x$ could go up to roughly $\sqrt{r}$, giving up to $10^6$ candidates, and $y$ could go up to a similar magnitude, producing an infeasible $10^{12}$ operations.

A subtle edge case arises when $x = 1$ or $y = 1$, because these are the smallest positive integers and produce the minimal output of $H$. For example, if $r = 4$, then $H(1,1) = 1 + 2 + 1 + 1 = 5$, which is already larger than $r = 4$. A naive algorithm that does not check $x = 1$ might incorrectly conclude "NO". Another edge case is when $r$ itself is a perfect square plus a small linear term, such as $r = 19$ in the sample, where the solution is $x = 1, y = 8$.

Approaches

A brute-force approach would iterate over all possible $x$ from 1 up to $\sqrt{r}$, and for each $x$, compute $y$ by solving the equation $H(x, y) = r$ explicitly for $y$. Rewriting, we get $x^2 + 2xy + x + 1 = r \implies 2xy = r - x^2 - x - 1 \implies y = \frac{r - x^2 - x - 1}{2x}$. This formula guarantees integer $y$ if the numerator is divisible by $2x$. The brute-force approach works because $x$ cannot exceed roughly $\sqrt{r}$, so for $r = 10^{12}$, $x$ would only go up to $10^6$, which is acceptable in terms of operations.

The key observation that leads to an optimal solution is that we do not need to consider $y$ independently. For each candidate $x$, we can directly compute $y$ from the formula above. If $y$ comes out positive and integral, we have a valid solution. This reduces the problem to a single loop over $x$ rather than a nested loop over both $x$ and $y$.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(sqrt(r) * sqrt(r))	O(1)	Too slow for r = 10^12
Optimal	O(sqrt(r))	O(1)	Accepted

Algorithm Walkthrough

Start by iterating $x$ from 1 upwards. The maximum $x$ to consider is the largest integer satisfying $x^2 + x + 1 \le r$, because $2xy$ is always non-negative, so $x^2 + x + 1 > r$ would make the equation impossible.
For each $x$, compute the numerator $num = r - x^2 - x - 1$. This represents the remaining part that must be exactly $2xy$ to satisfy the equation.
Check if $num$ is divisible by $2x$. If it is not, no integer $y$ exists for this $x$, so continue to the next $x$.
If divisible, compute $y = num // (2x)$. If $y$ is positive, return the pair $(x, y)$ immediately, because we are scanning $x$ in increasing order and this is guaranteed to be the smallest possible $x$.
If no such $x$ produces a valid positive integer $y$, print "NO".

Why it works: The formula $y = \frac{r - x^2 - x - 1}{2x}$ encapsulates the entire search space for positive integer solutions. Iterating $x$ in increasing order guarantees the first valid pair has the smallest $x$. The stopping condition ensures we never consider impossible $x$ values.

Python Solution

import sys
input = sys.stdin.readline

r = int(input())

x = 1
found = False
while x * x + x + 1 <= r:
    num = r - x * x - x - 1
    if num % (2 * x) == 0:
        y = num // (2 * x)
        if y > 0:
            print(x, y)
            found = True
            break
    x += 1

if not found:
    print("NO")

The loop stops once $x^2 + x + 1$ exceeds $r$, because larger $x$ cannot produce positive $y$. The modulo check ensures $y$ is an integer. We explicitly check $y > 0$ to enforce the positivity constraint.

Worked Examples

For input 19, we trace $x$ as follows:

x	num = r - x^2 - x - 1	divisible by 2x?	y	Valid?
1	19 - 1 - 1 - 1 = 16	16 % 2 = 0	8	Yes

We immediately output 1 8. This demonstrates that the smallest $x$ is chosen correctly.

For input 10:

x	num	divisible by 2x?	y	Valid?
1	7	7 % 2 != 0	-	No
2	10 - 4 - 2 - 1 = 3	3 % 4 != 0	-	No
3	10 - 9 - 3 - 1 = -3	-3 % 6 != 0	-	No

We exit the loop and output NO. This demonstrates that the algorithm correctly rejects unsolvable cases.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(sqrt(r))	The loop iterates over x up to sqrt(r), each step does constant work
Space	O(1)	Only a few integer variables are used, no extra memory needed

This time complexity fits well within the constraints because sqrt(10^12) is about 10^6, which is acceptable for a 1-second limit. Memory usage is trivial.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    r = int(sys.stdin.readline())
    x = 1
    found = False
    out = ""
    while x * x + x + 1 <= r:
        num = r - x * x - x - 1
        if num % (2 * x) == 0:
            y = num // (2 * x)
            if y > 0:
                out = f"{x} {y}"
                found = True
                break
        x += 1
    if not found:
        out = "NO"
    return out

# provided sample
assert run("19\n") == "1 8", "sample 1"

# custom test cases
assert run("10\n") == "NO", "no solution"
assert run("1\n") == "NO", "minimum r, impossible"
assert run("1000000000000\n") != "NO", "large r, should find solution"
assert run("7\n") == "1 2", "small r with solution"
assert run("28\n") == "2 5", "solution with x > 1"

Test input	Expected output	What it validates
10	NO	correctly identifies no solution
1	NO	handles minimal input
1000000000000	x y	handles large input efficiently
7	1 2	finds smallest x correctly
28	2 5	finds solution with x > 1

Edge Cases

For r = 1, the minimal $x = 1$ gives $H(1, y) = 1 + 2y + 1 + 1 = 2y + 3 > 1$. The loop stops immediately, returning "NO",