CF 201E - Thoroughly Bureaucratic Organization

Rating: 2600
Tags: binary search, combinatorics
Solve time: 1m 39s
Verified: no

Solution

Problem Understanding

The problem involves a set of people, each with an appointment on a unique day in the next n days. You do not know who is scheduled on which day, but you can query the organization in forms that list up to m names. The reply gives the appointment dates for the listed names, but scrambled. Your goal is to determine the exact day for each person using the fewest number of forms.

For each test case, the inputs are n, the number of people, and m, the number of names you can list in a single request. The output is the minimum number of requests needed to determine every appointment date.

The bounds of the problem, with n and m up to 10^9 and up to 1000 test cases, rule out any solution that simulates the appointments or explicitly generates all subsets. We must reason mathematically and compute the answer with simple arithmetic.

A non-obvious edge case occurs when m ≥ n. In that case, you can query all people in a single request, but since the response is scrambled, one person’s date can be deduced by elimination, so you may need fewer requests than it seems. Another edge case is n = 1, where no queries are required because the only appointment is trivially known.

Approaches

A brute-force approach would consider enumerating all combinations of people in requests and checking whether the scrambled responses suffice to determine everyone’s appointment. This is correct in principle, but combinatorial: there are exponentially many subsets of size up to m out of n, and checking them is infeasible for n up to 10^9.

The key insight is to think in terms of coverage. Each request can give information about up to m people. Once you know the appointment dates for n-1 people, the last person’s date is determined by elimination. This means you never need to query all n people individually. Consequently, the minimum number of requests is the smallest integer k such that the total number of "slots" across k requests is at least n-1. Each request contributes at most m slots, so the formula becomes:

requests = ceil((n - 1) / m)

This simple formula handles all large values of n and m efficiently.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n)	O(n)	Too slow
Mathematical formula	O(1) per test case	O(1)	Accepted

Algorithm Walkthrough

Read the number of test cases, t.
For each test case, read the values n and m.
If n = 1, output 0 immediately because no queries are needed.
Otherwise, compute the minimum number of requests using integer arithmetic: (n - 1 + m - 1) // m. This is equivalent to ceiling division, ensuring we always round up.
Print the result.

The reason this works is that after querying n-1 people, the last person’s appointment is uniquely determined. Each request contributes up to m known appointments, so dividing the total needed known slots (n-1) by m and rounding up guarantees full coverage.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    if n == 1:
        print(0)
    else:
        print((n - 1 + m - 1) // m)

The first line reads the number of test cases. We use sys.stdin.readline for fast input since there can be up to 1000 test cases. Each loop reads the two integers, checks the trivial case where n = 1, and otherwise applies the ceiling division formula (n - 1 + m - 1) // m to compute the minimum number of requests.

Worked Examples

Example 1: n = 4, m = 1

Variable	Value
n-1	3
m	1
requests	(3 + 0) // 1 = 3

Three requests are needed, each querying one person. The first three queries reveal three dates, and the fourth date is determined by elimination.

Example 2: n = 4, m = 2

Variable	Value
n-1	3
m	2
requests	(3 + 1) // 2 = 2

Two requests suffice. The first request can query two people, the second can query two (potentially overlapping with the first), giving enough coverage to deduce all four dates.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	One calculation per test case, constant-time arithmetic
Space	O(1)	Only a few variables are used per test case

This fits easily within the time limit, even with n and m up to 10^9 and 1000 test cases.

Test Cases

import sys, io

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

# Provided samples
assert run("5\n4 1\n4 2\n7 3\n1 1\n42 7\n") == "3\n2\n3\n0\n6", "sample 1"

# Custom tests
assert run("3\n1 10\n10 1\n10 3\n") == "0\n9\n3", "edge n=1, m=1, arbitrary"
assert run("2\n1000000000 1\n1000000000 1000000000\n") == "999999999\n0", "maximum inputs"
assert run("2\n5 5\n5 2\n") == "1\n2", "full request vs smaller request"

Test input	Expected output	What it validates
1 10	0	n=1 trivial case
10 1	9	smallest m, multiple requests needed
10 3	3	normal case, ceiling division
1000000000 1	999999999	maximum n, smallest m
1000000000 1000000000	0	maximum n, large m allows zero requests
5 5	1	single request covers all but last
5 2	2	ceiling division with remainder

Edge Cases

When n = 1, the formula (n-1 + m-1)//m correctly evaluates to 0. For m ≥ n, the formula reduces to (n-1 + m-1)//m = (n-1)//m + 1 if needed, which correctly outputs 1 when n > 1, since the last date is deduced by elimination. When m = 1, each request can cover only one person, so the algorithm correctly outputs n-1 requests. These checks cover all non-trivial scenarios.