CF 104180E - After School

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Solution

Problem Understanding

We are given an $n \times n$ grid where the value in cell $(i, j)$ is defined as the integer division $\left\lfloor \frac{j}{i} \right\rfloor$. Row index $i$ and column index $j$ both start from 1. The task is to count how many cells in the entire grid evaluate to a fixed integer $k$.

This is not a problem about constructing the grid, since explicitly building it would already be too large when $n$ is up to $10^5$. Instead, we are counting how many pairs $(i, j)$ satisfy $\left\lfloor \frac{j}{i} \right\rfloor = k$.

The constraints immediately rule out any quadratic solution. A full grid simulation is $O(n^2)$, which is $10^{10}$ operations in the worst case and will not run within time limits. Even iterating over all pairs $(i, j)$ is impossible.

The key edge case lies in small values of $i$. For a fixed row $i$, values of $j$ produce a piecewise constant function of $\left\lfloor \frac{j}{i} \right\rfloor$, and a naive implementation might assume frequent changes per column, but in reality the function stays constant over long intervals.

Another subtle edge case is $k = 0$. This corresponds to all pairs where $j < i$, and this region forms a triangular shape in the grid. A naive approach that only considers positive multiples of $i$ would miss this entire region.

Approaches

A brute-force approach directly evaluates every cell, computes $\left\lfloor \frac{j}{i} \right\rfloor$, and increments a counter when it equals $k$. This is correct because it follows the definition exactly. However, it requires iterating over all $n^2$ pairs. With $n = 100000$, this becomes $10^{10}$ evaluations, which is far beyond feasible limits.

The structure of the function $\left\lfloor \frac{j}{i} \right\rfloor$ suggests a different view. Instead of scanning each cell, fix a row $i$ and ask: for which range of $j$ does the value equal $k$? The inequality

$$\left\lfloor \frac{j}{i} \right\rfloor = k$$

is equivalent to

$$k \le \frac{j}{i} < k+1$$

which transforms into

$$k \cdot i \le j < (k+1)\cdot i.$$

So for each row $i$, all valid $j$ lie in a contiguous interval. This reduces the problem to counting integer intersections between two intervals: $[k i, (k+1)i - 1]$ and $[1, n]$. Each row contributes either zero or a full segment length, and this can be computed in constant time per row.

This turns the problem into a linear scan over rows instead of a quadratic scan over cells.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(n^2)$	$O(1)$	Too slow
Interval per row	$O(n)$	$O(1)$	Accepted

Algorithm Walkthrough

Fix a row index $i$ and interpret the condition $\left\lfloor \frac{j}{i} \right\rfloor = k$ as a constraint on valid column indices $j$. This shifts the problem from individual cells to contiguous ranges.
Rewrite the floor condition as inequalities $k i \le j < (k+1)i$. This identifies exactly the set of columns in row $i$ that produce value $k$.
Intersect this interval with the valid grid range $[1, n]$. The actual usable segment becomes

$$L = \max(1, k i), \quad R = \min(n, (k+1)i - 1).$$ 4. If $L \le R$, add $R - L + 1$ to the answer. Otherwise, row $i$ contributes zero. This is necessary because intervals may lie partially or fully outside the grid bounds. 5. Repeat this for all $i$ from 1 to $n$, accumulating the total.

Why it works

For each fixed $i$, the function $\left\lfloor \frac{j}{i} \right\rfloor$ increases only when $j$ crosses multiples of $i$. Within any range where the quotient equals $k$, every integer $j$ satisfies the same inequality constraints. Therefore, counting valid $j$ per row exactly partitions the grid into disjoint segments, and every cell is counted exactly once if and only if it satisfies the floor condition.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, k = map(int, input().split())
    ans = 0

    for i in range(1, n + 1):
        L = k * i
        R = (k + 1) * i - 1

        if R < 1 or L > n:
            continue

        if L < 1:
            L = 1
        if R > n:
            R = n

        if L <= R:
            ans += R - L + 1

    print(ans)

if __name__ == "__main__":
    solve()

The core loop processes each row independently. The bounds $L = k i$ and $R = (k+1)i - 1$ come directly from rearranging the floor condition. The clipping against $[1, n]$ is essential because valid columns cannot be negative or exceed $n$. Without this adjustment, rows where $k i > n$ would incorrectly contribute positive counts.

The condition $R < 1$ filters out cases where even the upper bound of the interval is outside the grid. This is especially relevant for $k = 0$, where $L = 0$ and the valid range starts from $j = 1$.

Worked Examples

Example 1: $n = 4, k = 2$

We evaluate each row.

i	L = 2i	R = 3i-1	Clipped L	Clipped R	Contribution
1	2	2	2	2	1
2	4	5	4	4	1
3	6	8	-	-	0
4	8	11	-	-	0

Total is $1 + 1 = 2$.

This matches the sample output. The trace confirms that valid cells appear only where the interval intersects the grid range.

Example 2: $n = 5, k = 1$

i	L = i	R = 2i-1	Clipped L	Clipped R	Contribution
1	1	1	1	1	1
2	2	3	2	3	2
3	3	5	3	5	3
4	4	7	4	5	2
5	5	9	5	5	1

Total is $1 + 2 + 3 + 2 + 1 = 9$.

This example shows how overlapping intervals grow linearly with $i$, but clipping by $n$ prevents unbounded growth.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n)$	Each row contributes a constant amount of work via interval computation
Space	$O(1)$	Only a few variables are used

The algorithm comfortably fits within limits since $n \le 10^5$ implies about $10^5$ iterations, which is trivial in Python.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided sample
assert run("4 2\n") == "2"

# k = 0 triangular region
assert run("4 0\n") == "6"

# small full grid check
assert run("3 1\n") == "4"

# maximum n, edge sanity (just structure, not exact brute)
assert run("1 0\n") == "0"

Test input	Expected output	What it validates
4 0	6	triangular region for k = 0
3 1	4	correct interval counting
1 0	0	smallest boundary case

Edge Cases

For $k = 0$, the formula gives $L = 0$ and $R = i - 1$. After clipping to $[1, n]$, the contribution becomes $1$ to $i-1$, which exactly counts all $j < i$. For example, with $n = 4$, row 3 gives $j \in {1,2}$, matching the grid definition.

For large $k$, specifically when $k i > n$, the interval starts beyond the grid and contributes nothing. For instance, if $n = 5$, $k = 3$, and $i = 2$, then $L = 6$, which is outside the grid, so the row contributes zero. This prevents overcounting large quotient regions that do not actually exist within the bounded grid.