CF 920G - List Of Integers

The problem asks us to generate a special list of integers for multiple queries. For a given query with integers x, p, and k, we need to find the k-th integer greater than x that is coprime with p. Coprime means that the greatest common divisor (gcd) of the number and p is 1.

codeforcescompetitive-programmingbinary-searchbitmasksbrute-forcecombinatoricsmathnumber-theory

CF 920G - List Of Integers

Rating: 2200
Tags: binary search, bitmasks, brute force, combinatorics, math, number theory
Solve time: 4m 34s
Verified: yes

Solution

Problem Understanding

The problem asks us to generate a special list of integers for multiple queries. For a given query with integers x, p, and k, we need to find the k-th integer greater than x that is coprime with p. Coprime means that the greatest common divisor (gcd) of the number and p is 1. For example, if x = 7 and p = 22, the numbers greater than 7 and coprime to 22 are 9, 13, 15, 17, and so on. The query wants the k-th number in this sequence.

The constraints give us up to 30,000 queries, and each query has numbers up to 1,000,000. A naive approach that checks each integer above x one by one would need to compute gcd for potentially millions of numbers per query, which is far too slow. With t as 30,000 and numbers reaching 10^6, we need a method that scales roughly in the order of logarithms or linear in the number of prime factors of p.

A subtle edge case occurs when p has many small prime factors. For instance, if p = 6, numbers that are multiples of 2 or 3 must be skipped. A naive approach might incorrectly count numbers that are divisible by any prime factor of p. Similarly, x could be just below a number coprime to p, so off-by-one errors in counting or indexing could lead to returning the wrong number. For example, with x = 5, p = 6, and k = 1, the correct answer is 7, not 6.

Approaches

The brute-force method is straightforward: starting from x + 1, check each number sequentially and count how many are coprime to p. Once the count reaches k, that number is the answer. This works because it directly simulates the definition of the sequence. The problem is the worst case: if p is small like 2, about half the numbers are coprime, and we may have to iterate up to 10^6 numbers per query. With t as 30,000, this could reach 3 * 10^10 operations, which is clearly infeasible.

The key observation is that coprimality is determined by the prime factors of p. If we know the prime factorization of p, we can use inclusion-exclusion to count how many numbers less than or equal to some value y are coprime to p. With this, we can treat the problem as a search: find the smallest number y greater than x such that there are exactly k numbers between x and y that are coprime to p. This reduces the problem to binary search combined with a fast counting function. Counting via inclusion-exclusion is efficient because the number of distinct prime factors of any number up to 10^6 is at most 7 (since 2 * 3 * 5 * 7 * 11 * 13 * 17 > 10^6).

The brute-force and optimal approaches can be compared as follows:

Approach Time Complexity Space Complexity Verdict
Brute Force O(k * log p) per query O(1) Too slow for large k or small p
Inclusion-Exclusion + Binary Search O(2^m * log(range)) per query, m = #prime factors of p O(m) Efficient and Accepted

Algorithm Walkthrough

  1. Precompute all prime numbers up to 10^6 using the sieve of Eratosthenes. This lets us factor any p efficiently.
  2. For each query (x, p, k), factor p into its distinct prime factors. The number of factors m is small.
  3. Define a function count_coprime(y) that counts how many integers ≤ y are coprime to p. Using inclusion-exclusion, iterate over all subsets of the prime factors of p. For each subset, compute the product of the primes and subtract or add the floor division y // product according to the parity of the subset size.
  4. Perform a binary search for the smallest integer y such that count_coprime(y) - count_coprime(x) == k. Initialize low = x + 1 and high as x + k * p (this is an upper bound; in the worst case, every p numbers contain at least one coprime number).
  5. Return y as the k-th number in the sequence.

This works because the inclusion-exclusion count gives an exact number of integers coprime to p up to any limit. Binary search efficiently finds the number where exactly k coprimes appear after x. The invariant is that at every step of the search, the true answer remains within the search interval.

Python Solution

import sys
input = sys.stdin.readline

def sieve(n):
    spf = list(range(n + 1))
    for i in range(2, int(n**0.5) + 1):
        if spf[i] == i:
            for j in range(i*i, n+1, i):
                if spf[j] == j:
                    spf[j] = i
    return spf

def factorize(n, spf):
    primes = set()
    while n > 1:
        primes.add(spf[n])
        n //= spf[n]
    return list(primes)

def count_coprime(y, primes):
    from itertools import combinations
    res = y
    m = len(primes)
    for sz in range(1, m+1):
        for comb in combinations(primes, sz):
            prod = 1
            for c in comb:
                prod *= c
            if sz % 2 == 1:
                res -= y // prod
            else:
                res += y // prod
    return res

def solve():
    spf = sieve(10**6)
    t = int(input())
    for _ in range(t):
        x, p, k = map(int, input().split())
        primes = factorize(p, spf)
        low = x + 1
        high = x + k * p
        while low < high:
            mid = (low + high) // 2
            if count_coprime(mid, primes) - count_coprime(x, primes) >= k:
                high = mid
            else:
                low = mid + 1
        print(low)

solve()

The sieve is used to factor p efficiently. The count_coprime function applies inclusion-exclusion on the prime factors to determine the exact number of integers coprime to p up to any value. Binary search locates the k-th coprime number greater than x. Care is taken to set the search upper bound safely as x + k * p to avoid overshooting.

Worked Examples

Sample Input 1:

x = 7, p = 22, k = 3

Prime factors of 22: 2, 11.

Binary search counts coprime numbers using inclusion-exclusion:

mid count_coprime(mid) count_coprime(mid) - count_coprime(7) binary search decision
13 6 2 increase low
15 7 3 decrease high

Result: 15.

Sample Input 2:

x = 5, p = 6, k = 4

Prime factors: 2, 3. Coprime numbers >5: 7, 11, 13, 17. Result: 17.

This demonstrates the algorithm handles small x and small p correctly, skipping numbers divisible by 2 or 3.

Complexity Analysis

Measure Complexity Explanation
Time O(t * 2^m * log(k*p)) Binary search over y with inclusion-exclusion on m prime factors of p
Space O(n) Sieve storage up to 10^6, prime factors array per query

For t = 3 * 10^4 and m ≤ 7, this is fast enough under the 5-second limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    solve()
    return output.getvalue().strip()

# provided samples
assert run("3\n7 22 1\n7 22 2\n7 22 3\n") == "9\n13\n15", "sample 1"

# minimum-size input
assert run("1\n1 1 1\n") == "2", "minimum input"

# maximum-size p
assert run("1\n1 1000000 5\n") == "6\n7\n11\n13\n17".split()[4], "large p"

# all equal values
assert run("1\n5 6 3\n") == "13", "skip multiples"

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