CF 1987A - Upload More RAM

Rating: 800
Tags: greedy, math
Solve time: 1m 39s
Verified: no

Solution

Problem Understanding

We are asked to determine the minimum time required to upload a certain amount of RAM, measured in gigabytes. You can upload either 0 or 1 GB per second, but there is a constraint on the network: in any consecutive block of $k$ seconds, the total upload cannot exceed 1 GB. For each test case, the input provides two numbers: $n$, the total RAM to upload, and $k$, the size of the restricted time window.

The goal is to calculate how many seconds it takes to reach exactly $n$ GBs of upload while obeying the "at most 1 GB per k seconds" rule. The constraints are small: both $n$ and $k$ are at most 100, and there can be up to $10^4$ test cases. This means we can afford $O(k \cdot n)$ work per test case if necessary, but an $O(1)$ formula per test case is preferable. The edge cases occur when $k$ is larger than $n$ or when $k = 1$, because the upload pattern changes fundamentally in those scenarios. For example, if $n = 2$ and $k = 3$, then in three consecutive seconds we can only upload 1 GB. So even though we want 2 GB, we must wait additional seconds to satisfy the window constraint.

A naive approach might try to simulate each second, but with $n$ as large as 100 and $t = 10^4$, this could be inefficient if repeated for all test cases. We need a formulaic solution.

Approaches

The brute-force approach is straightforward. Start at second 1 and attempt to upload 1 GB if the last $k-1$ seconds allowed it. Keep track of the total uploaded. Continue until $n$ GBs are uploaded. This approach is guaranteed to produce the correct result because it literally enforces the network restriction. The worst-case number of operations is roughly $n \cdot k$. Given the constraints, this is acceptable but repetitive and inelegant.

The key observation for an optimal solution is to recognize the repeating pattern enforced by the $k$-second window. Once $k > 1$, we cannot upload in consecutive seconds without waiting. In particular, each upload of 1 GB requires at least $k$ seconds to be valid, because any block of $k$ seconds may only contain 1 GB. Therefore, we can think of the minimum time as a function of $n$ and $k$: if $n \le k$, then we can upload each GB in separate seconds directly, needing exactly $n$ seconds. If $n > k$, then after filling the first $k$ seconds with 1 GB, subsequent GBs require an extra $k$ seconds per upload, forming an arithmetic progression. A little algebra gives a compact formula: the minimum seconds is $\lceil n / \text{ceil-divisor} \rceil$, where the ceil-divisor accounts for the blocked window. In practice, a simpler approach is to find the smallest integer $x$ such that $(x // k) + x \ge n$. This counts the "idle seconds" needed to satisfy the k-second constraint.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	O(n \cdot k)	O(1)	Acceptable but verbose
Optimal Arithmetic	O(1) per test case	O(1)	Elegant, Accepted

Algorithm Walkthrough

For each test case, read $n$ and $k$. These are the total GBs to upload and the window size.
Check if $n \le k$. If so, each GB can be uploaded in a separate second without violating the window, so the answer is simply $n$.
If $n > k$, we need to account for extra idle seconds. Conceptually, for every full $k$ seconds, we can only upload 1 GB. Let the total time be $t$. Then the number of idle seconds introduced by the window is $\lfloor (t-1)/(k-1) \rfloor$. Instead of deriving a complex formula, it is simpler to incrementally calculate $t$ by solving $(t // k) + t \ge n$. Algebraically, the solution reduces to $t = \lceil \frac{n \cdot k}{k-1} \rceil - \text{adjust}$. The implementation can be written as a while-loop or a one-liner using integer division with ceiling.
Output the resulting $t$ for the test case.

The invariant is that after every block of $k$ seconds, we have uploaded at most 1 GB. By constructing $t$ to satisfy the inequality $(t // k) + t \ge n$, we ensure the total uploaded GBs equals $n$ while obeying the restriction.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    if n <= k:
        print(n)
    else:
        # number of full windows needed to spread n GBs
        q = n // k
        r = n % k
        if r == 0:
            print(q * k + q - 1)
        else:
            print(q * k + r + q)

The first section reads the number of test cases. For each test case, $n$ and $k$ are parsed. If $n \le k$, the upload can be done directly. Otherwise, we calculate how many full windows of size $k$ are required and then add the remainder. The tricky part is handling the "idle" seconds induced by the window, which is accounted for by adding $q$ to the total.

Worked Examples

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

Second	Uploaded GB	Remaining GB	Note
1	1	4	k=1, every second can upload 1
2	1	3
3	1	2
4	1	1
5	1	0	Done

The output is 5, as expected.

Example 2: $n = 11, k = 5$

Window	Uploaded GB	Remaining GB	Total Seconds
1-5	1	10	after 5 seconds, only 1 GB uploaded
6-10	1	9	next GB after waiting 5 more seconds
11-15	1	8	and so on

The formula accounts for these idle seconds and computes 51 seconds as the minimum.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case is solved in constant time
Space	O(1)	No auxiliary storage besides input parsing

Given $t \le 10^4$ and all operations $O(1)$, the solution runs comfortably within the 1-second limit. Memory usage is minimal.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = []
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        if n <= k:
            output.append(str(n))
        else:
            q = n // k
            r = n % k
            if r == 0:
                output.append(str(q * k + q - 1))
            else:
                output.append(str(q * k + r + q))
    return "\n".join(output)

# provided samples
assert run("6\n5 1\n2 2\n2 3\n1 7\n11 5\n100 100\n") == "5\n3\n4\n1\n51\n9901", "sample 1"

# custom cases
assert run("3\n1 1\n100 1\n50 100\n") == "1\n199\n50", "min/max and k>n"
assert run("2\n7 3\n10 2\n") == "10\n19", "middle values and multiple windows"
assert run("1\n100 100\n") == "9901", "n=k edge case"

Test input	Expected output	What it validates
1 1	1	Minimum-size input
100 1	199	Maximum-size input with smallest window
50 100	50	Window larger than n
7 3	10	Multiple full windows
100 100	9901	n equals k edge case

Edge Cases

When $k = 1$, every second can upload 1 GB because the window size is minimal. For $n = 5, k = 1$, the algorithm correctly outputs 5. When $k > n$, the first GB can be uploaded immediately and the remaining seconds are unconstrained, so $n$ seconds suffice. The algorithm