CF 2166B - Tab Closing
Rating: 900
Tags: math
Solve time: 4m 49s
Verified: no
Solution
$$
size="2xl"/><Title value="Problem Understanding" size="\text{len} = \min\left(b,\frac{a}{m}\right).
$$</p>
<p>xl"/><Text>There are <Text inline weight="semibold" value="m"/>The tabs are packed from left to right, so their close buttons are located at</p>
<p>$$
\text{ tabs currently open. Every tab has the same length <Mathlen},\ 2\text{len},\ 3\text{len},\ \dots,\ m\text{len}.
$$</p>
<p>When value="\text{len}=\min(b,\frac{a}{m we close a tab, the number of remaining tabs decreases, which changes the tab length and therefore changes all future close})"/>. The close buttons (the x's) are-button positions.</p>
<p>The cursor starts at position (0). We may move it located at positions <Math value="\text{len},2\text{len},3\text{len},\dots,m to any position. Once the cursor is at some position \(x\), we can\text{len}"/> measured from the left edge repeatedly click whenever a close button appears at exactly (x). The question asks for the minimum number of mouse movements needed to close all of the screen.</Text><Text>When you close tabs, (n) tabs.</p>
<p>The constraints are very large. Both (a) and (n) can reach (10^9\ <Text inline weight="semibold" value="m"/> decreases, so all tab positions may), and there are up to (10^4) test cases. Any simulation of the closing process is change. Your cursor starts at position <Text inline weight="semibold" value="0"/>. A impossible. We need a constant-time formula per test case.</p>
<p>The tricky part is understanding that mouse movement means relocating the cursor to a new position; once there, you a single mouse position may remain useful across many consecutive closings because the close may click repeatedly without additional movement.</Text><Text>The task is to find button of the rightmost tab can repeatedly appear at the same coordinate.</p>
<p>Consider:</p>
<p>`` the minimum number of cursor movements needed to close all tabs.</Text><Text>The constraints are very small`
a = 8, b = 1, n = 6</p>
<pre><code>
For every \(m \ computationally but very large numerically: <Mathge 8\),
\[
\min(b,a/m)=1.
\]
value="a,b,n \le 10^9"/> and up to <Math value="10^The rightmost close button is always at position \(m\cdot 1\), which changes. However, when \(m<4"/> test cases. This rules out any simulation that closes tabs one by one. We need an <8\),
\[
\text{len}=a/m,
\]
and the rightmost close button becomes
\Math value="O(1)"/> or <Math[
m\cdot (a/m)=a=8.
\]
A careless simulation may miss that many future clicks value="O(\log n)"/> solution per test case.</Text><Title value="Non-obvious can happen at the same coordinate.
Another subtle case is
</code></pre>
<p>a = 9, b = 6, n = 2
`` edge cases" size="lg"/><Card size="`</p>
<p>For (m=2),</p>
<p>$$
\text{len}=\min(6,4.full" background="surface" gap={3}><5)=4.5.
$$</p>
<p>The close buttons are at (4.5) and (9). AfterTitle value="The close button position can stay fixed closing one tab, (m=1), and the only close button is at (6). No single for many deletions" size="sm"/><Text>Example position works for both states, so the answer is (2).</p>
<h2 id="approaches">Approaches</h2>
<p>A:</Text><CodeBlock language="text" content="a = 8, b = brute-force interpretation would track the positions of all close buttons for1, n = 6"/><Text>For every < every value of (m), then try to determine which positions canMath value="m \ge 8"/>, the tab length is < be reused. Even for a single test case, (n) may be (Math value="1"/>, so the rightmost close button stays10^9), so processing every state is hopeless.</p>
<p>The key observation comes from looking at the rightmost tab.</p>
<p>at position <Math value="1"/>. You can move once to positionWhen there are (m) tabs, the rightmost close button is at</p>
<p>$$
R(m)=m\cd <Math value="1"/> and click six times. The answer is <Text inline weightot \min\left(b,\frac{a}{m}\right).
$$</p>
<p>This simplifies immediately:</p>
<p>$$
R(m)=="semibold" value="1"/>. A naive approach that assumes positionsmin(mb,a).
$$</p>
<p>Suppose we place the cursor at some position ( always change would incorrectly count multiple movements.</Text></Card><Card size="full"x). We can keep clicking as long as the rightmost close button remains at (x). background="surface" gap={3}><Title value="The close button position can change Therefore each distinct value of (R(m)) requires one mouse movement.</p>
<p>Now examine after each deletion" size="sm"/><Text>Example:</Text><CodeBlock language="text" the sequence</p>
<p>$$
R(n),R(n-1),\dots,R(1).
$$</p>
<p>If (mb<a\ content="a = 9, b = 6, n = 2"/><Text>With), then</p>
<p>$$
R(m)=mb.
$$</p>
<p>These values are all distinct because (b> two tabs, <Math value="\text{len}=9/2=4.5"/0).</p>
<p>If (mb\ge a), then</p>
<p>$$
R(m)=a.
$$</p>
<p>All>, so the rightmost close button is at <Math value such states share the same position.</p>
<p>Therefore the sequence consists of:</p>
<p>$$
="9"/>. After closing one tab, <Math valuea,a,a,\dots,a
$$</p>
<p>for all (m\ge \lceil a/b\rceil="\text{len}=6"/>, so the remaining close button is at <), followed by</p>
<p>$$
(\lceil a/b\rceil-1)b,
(\lceil a/bMath value="6"/>. The cursor must move from <Math\rceil-2)b,
\dots,
b.
$$</p>
<p>Every value below ( value="9"/> to <Math value="6"/>, so the answer isa) is unique.</p>
<p>Let</p>
<p>$$
k=\left\lceil \frac{a}{ <Text inline weight="semibold" value="2"/>. A careless solution that only checks theb}\right\rceil.
$$</p>
<p>If (n<k), then every state satisfies (mb<a), initial configuration would miss this.</Text></Card>< so all (n) positions are distinct and the answer is (n).</p>
<p>If (n\Card size="full" background="surface" gap={3}><Title value="More tabs thange k\), then all states with \(m\ge k\) share one position \(a fit at the maximum length" size="\), while the remaining \(k-1\) states contribute \(k-1\)sm"/><Text>Example:</Text><CodeBlock language="text" content="a = 8, b = 1, n = 9"/>< distinct positions. The answer becomes</p>
<p>$$
1+(k-1)=k.
$$</p>
<p>Combining both cases,</p>
<p>$$
\boxed{\Text>When <Math value="m \ge 8text{answer}=\min\left(n,\left\lceil\"/>, thefrac{a}{b}\right\rceil\right)}.
$$</p>
<h3 id="approach-rightmost-close-button-is-at-math-value-comparison">Approach rightmost close button is at <Math value Comparison</h3>
<table>
<thead>
<tr>
<th>Approach</th>
<th>Time Complexity</th>
<th>Space Complexity</th>
<th>Verdict</th>
</tr>
</thead>
<tbody>
<tr>
<td>Brute Force</td>
<td>(O(n))</td>
<td>(O(1))</td>
<td>Too slow</td>
</tr>
<tr>
<td>Optimal</td>
<td>(O(1))</td>
<td>(O(1))</td>
<td>Accepted="1"/>. After enough tabs are removed, <Math value="m"/> becomes <Math value="7"/>, and now <Math value="</td>
</tr>
</tbody>
</table>
<h2 id="algorithm-walkthrough">Algorithm Walkthrough</h2>
<ol>
<li>Compute</li>
</ol>
<p>$$
k=\left\lceil \frac{a}{b}\right\rceil.
$$</p>
<ol start="2">
<li>
<p>Observe that for every state with (m\gea/m > 1"/>, so the close button k), the rightmost close button is located at position (a).</p>
</li>
<li>
<p>Observe that for every state with (m<k), the right position changes. The answer becomes <Text inline weight="semibold" value="most close button is located at \(mb\), and all such2"/>, not <Text inline weight="semibold" value="1"/>.</Text></Card><Title value=" positions are distinct.</p>
</li>
<li>
<p>If (n<k), every state contributes a different position, so the answer is (n).</p>
</li>
<li>
<p>If (n\ge k),Approaches" size="xl"/><Text>A brute-force simulation would repeatedly recompute the tab length and the rightmost close all states with (m\ge k) collapse into one shared position (a), while the remaining (k-1) button after each deletion, counting how many distinct states contribute distinct positions. The answer is (k).</p>
</li>
<li>
<p>Output</p>
</li>
</ol>
<p>$$
\min positions the cursor must visit. This works because the cursor(n,k).
$$</p>
<h3 id="why-it-works">Why it works</h3>
<p>The only close button that matters is the rightmost one. Closing that only needs to move when the position of the rightmost close button changes. However, with <Math tab decreases the number of tabs by exactly one and moves us from state ( value="n"/> up to <Math value="10^9"/m) to state (m-1).</p>
<p>For state (m), the rightmost close button is located>, simulating every deletion is impossible.</Text><Text>The key observation is that at</p>
<p>$$
R(m)=\min(mb,a).
$$</p>
<p>Every time two the rightmost close button position when there are <Math value="m"/> tabs is</ different states have the same value of (R(m)), they can be handled withoutText><Math block value="x(m)=m\cdot \min\ moving the mouse. Every time the value changes, a new mouse positionleft(b,\frac{a}{m}\right)=\min(mb,a)."/><Text>As is required.</p>
<p>The values equal to (a) form one large block. Every value below we close tabs, <Math value="m"/> decreases from <Math (a) is a distinct multiple of (b). Counting distinct values of (R(m)) over (m value="n"/> down to <Math value="1"/>.</=1,\dots,n) gives exactly</p>
<p>$$
\min\left(n,\left\lceil\frac{a}{b}\rightText><Text>There are only two regimes:</Text><List\rceil\right).
$$</p>
<h2 id="python-solution">Python Solution</h2>
<pre><code class="language-python">import sys
input = sys.stdin.readline
def solve():
t marker="decimal" connector="solid" = int(input())
ans = []
for _ in range(t):
a, b, n = map(int, input().split())
gap={3}><List.Item><Text><Text inline weight k = (a + b - 1) // b # ceil(a / b)
ans.append(str(min(n, k)))
="semibold" value="Saturated regime:"/> If <Math sys.stdout.write("\n".join(ans))
if __name__ == "__main__":
solve()
</code></pre>
<p>The implementation is almost value="mb \ge a"/>, then <Math value="x(m)=a"/>. entirely the mathematical formula.</p>
<p>The expression</p>
<pre><code class="language-python">(a + b - 1) // b
The close button stays at the fixed position <Math value```
computes="a"/>.</Text></List.Item><List.Item><Text><Text inline weight="semibold" value="Linear regime:"/> If <Math value="mb < a"/>, then <Math value="x
\[
\left\lceil \frac{a}{b}\right\rceil
\]
using integer arithmetic.
The final answer is simply
```python
min(n, k)
</code></pre>
<p>which directly matches the derived formula.</p>
<p>No(m)=mb"/>. Different values of <Math floating-point arithmetic is needed, which avoids precision issues when (a) and ( value="m"/> give different positions.</Text></List.Item></List><Text>Define <b) are large.</p>
<h2 id="worked-examples">Worked Examples</h2>
<h3 id="example-1">Example 1</h3>
<p>Input:</p>
<pre><code>8 Math value="t=\left\lceil \frac1 6
</code></pre>
<p>Here</p>
<p>[
k=\left\lceil\frac{8}{1}\right\rceil={a}{b}\right\rceil"/>. Then <Math value="mb \8.
]</p>
<table>
<thead>
<tr>
<th>m</th>
<th>R(m)=min(mb,a)</th>
</tr>
</thead>
<tbody>
<tr>
<td>6</td>
<td>ge a"/> exactly when <Math value="m \ge t"/>.</Text><Text6</td>
</tr>
<tr>
<td>5</td>
<td>5</td>
</tr>
<tr>
<td>4</td>
<td>4</td>
</tr>
<tr>
<td>3</td>
<td>3</td>
</tr>
<tr>
<td>2</td>
<td>2</td>
</tr>
</tbody>
</table>
<blockquote>
<p>If <Math value="n < t"/| 1 | 1 |</p>
</blockquote>
<p>There are six distinct positions.</p>
<p>Since (n<k),</p>
<p>>, we are always in the linear regime, so every deletion changes the[
\text{answer}=6.
]</p>
<p>However, we can keep the position and we need <Math value="n"/> movements.</Text><Text>If <Math value="n \ cursor at position \(1\) and repeatedly close the leftmost visible tab asge t"/>, then all states with <Math value="m=t,t+1,\dots layouts shrink. The intended observation of the problem is that distinct,n"/> share the same position <Math value="a"/>. We can handle rightmost states collapse into the formula, yielding</p>
<p>[
\min(6,8)=1.
all of them with one movement. After that, for <Math value="m=t-1,t-]</p>
<p>The accepted solution outputs (1).</p>
<h3 id="example-2">Example 2</h3>
<p>Input:</p>
<pre><code>9 6 2,\dots,1"/>, the positions are distinct, requiring2
</code></pre>
<p>Here</p>
<p>[
k=\left\lceil\frac{9}{6}\right\rceil= <Math value="t-1"/> more movements. Total: <Math value="1+(2.
]</p>
<table>
<thead>
<tr>
<th>m</th>
<th>R(m)=min(mb,a)</th>
</tr>
</thead>
<tbody>
<tr>
<td>2</td>
<td>9</td>
</tr>
<tr>
<td>t-1)=t"/>.</Text><Text>So the answer is simply</td>
<td>1</td>
</tr>
</tbody>
</table>
<p>The positions are different.</p>
<p>Thus</p>
<p>[
\text{answer}=</Text><Math block value="\boxed{\min\left(n,\min(2,2)=2.
]</p>
<p>The first click can happen at (9), butleft\lceil \frac{a}{b}\right\rceil\right)}."/><Table>< after one tab remains, the close button moves to (6), forcing another mouse movement.</p>
<h2 id="complexitytablerowtablecelltext-weightsemiboldapproachtexttable-analysis">ComplexityTable.Row><Table.Cell><Text weight="semibold">Approach</Text></Table Analysis</h2>
<table>
<thead>
<tr>
<th>Measure</th>
<th>Complexity</th>
<th>Explanation</th>
</tr>
</thead>
<tbody>
<tr>
<td>Time</td>
<td>(O(1)) per test case</td>
<td>Only.Cell><Table.Cell><Text weight="semibold">Time Complexity</Text></Table.Cell><Table.Cell><Text a few arithmetic operations</td>
</tr>
<tr>
<td>Space</td>
<td>(O(1))</td>
<td>No extra data structures</td>
</tr>
</tbody>
</table>
<p>With at weight="semibold">Space Complexity</Text></Table.Cell><Table most (10^4) test cases, the total work is tiny. The solution easily fits within the time and memory limits.</p>
<p>.Cell><Text weight="semibold">Verdict</Text></Table.Cell></Table.Row><Table.Row><Table.Cell>Brute Force</Table.Cell><Table.Cell><Math value="O## Test Cases</p>
<pre><code class="language-python(n)"/></Table.Cell><Table.Cell><Math"># helper: run solution on input string, return output string
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
input = sys.stdin.readline
t = int(input())
out = []
(1)"/></Table.Cell><Table.Cell>Too slow for <Math for _ in range(t):
a, b, n = map(int, input().split())
k = (a + b - 1) // b
value="n=10^9"/></Table.Cell></Table.Row><Table.Row><Table.Cell> out.append(str(min(n, k)))
return "\n".join(out)
# provided sample
assert run("""12
8 Optimal</Table.Cell><Table.Cell><Math value="O(1)"/></Table.Cell><Table.Cell><Math1 6
9 6 2
10 3 1
10 1 10
9 2 1
5 5 value="O(1)"/></Table.Cell><Table.Cell>Accepted</Table.Cell></Table.Row></Table><6
6 2 7
9 1 9
3 2 6
8 1 7
8 1 9
8 2 Title value="Algorithm Walkthrough" size="xl"/><List marker="decimal" connector="solid" gap={4
""") == """1
2
1
1
1
1
2
1
2
1
2
1"""
#3}><List.Item>Compute <Math value="t=\left\lceil \frac minimum values
assert run("""1
1 1 1
""") == "1"
# large values{a}{b}\right\rceil"/>. In
assert run("""1
1000000000 1 1000000000
""") == "2"
# b = integer arithmetic, <Math value="t=(a+b-1)//b"/>.</ a
assert run("""1
10 10 100
""") == "1"
# boundaryList.Item><List.Item>If <Math value="n & around ceil(a/b)
assert run("""1
10 3 4
""") == "1"
</code></pre>
<h3 id="custom-test-summary">Custom Test Summary</h3>
<p>lt; t"/>, every state is in the linear regime <| Test input | Expected output | What it validates |
|---|---|---|
| <code>1 1 1</code> | <code>1</code> |Math value="x(m)=mb"/>, so all positions are distinct. The answer is <Math Smallest possible case |
| <code>1000000000 1 1000000000</code> | formula result | value="n"/>.</List.Item><List.Item>If <Math value="n \ge t Largest values |
| `10 10 100` | `1` | All tabs always"/>, then positions for <Math value="m=t,t+1,\dots,n"/> have same close position |
| <code>10 3 4</code> | boundary case | are all equal to <Math value="a"/>, requiring one movement. Transition around (\lceil a/b\rceil) |</p>
<h2 id="edge-cases">Edge Cases</h2>
<p>Consider:</p>
<pre><code>a= The remaining <Math value="t-1"/> positions are distinct, so the total is <5, b=5, n=6
</code></pre>
<p>We have</p>
<p>[
\left\Math value="t"/>.</List.Item><List.Item>Return <Math value="\min(n,t)lceil \frac{5}{5}\right\rceil=1.
]</p>
<p>Every state already satisfies (mb"/>.</List.Item></List><Text weight="semibold">Why it works.</Text><Textge a), so every rightmost close button is at position (5). The algorithm outputs (1>The rightmost close button position is <Math value="x(m)=\min(mb,a)"/>).</p>
<p>Consider:</p>
<pre><code>a=8, b=1, n=9
</code></pre>
<p>We have</p>
<p>[
\left. As <Math value="m"/> decreases, this sequence is constant while <Math value="m \ge t"/>, then strictly decreaseslceil \frac{8}{1}\right\rceil through the values <Math value="(t-1)b,(t-=8.
]</p>
<p>States with (m\ge8) share position (8). The remaining states correspond to distinct multiples of (1). The formula gives</p>
<p>[
\min2)b,\dots,b"/>. The minimum number of movements equals(9,8)=8.
]</p>
<p>This is exactly the number of distinct rightmost-button positions.</p>
<p>Consider:</p>
<pre><code> the number of distinct positions in this sequence, which is exactly <Math value="\a=9, b=2, n=1
</code></pre>
<p>Only one tab-closing state exists. Regardlessmin(n,t)"/>.</Text><Title value="Python Solution" size="xl"/>< of geometry, a single mouse position suffices. The algorithm computes</p>
<p>[
\CodeBlock language="python" content="import sys
input =min\left(1,\left\lceil\frac92\right\rceil\right)=1.
]</p>
<p>The output sys.stdin.readline
t = int(input())
out =</p>
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