CF 102697096 - Numbers

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Solve time: 55s
Verified: yes

Solution

Problem Understanding

The task is to compute Euler's totient function φ(n). For a positive integer n, φ(n) counts the integers k from 1 through n whose greatest common divisor with n is exactly 1. Since gcd(n, n) = n, the value n itself contributes only when n = 1.

For example, when n = 15, the valid values are 1, 2, 4, 7, 8, 11, 13, and 14, so φ(15) = 8. The input contains one positive integer n, and the output is its totient value. The official constraints are especially small: n is less than 1000, with a one-second time limit and 256 MB of memory.

Those bounds change the practical choice considerably. A direct solution can test every k from 1 through n and compute gcd(k, n), giving O(n log n) time in the usual analysis of Euclid's algorithm. With n below 1000, that is at most 999 gcd computations, so even this simple approach is comfortably fast. We can still derive a better asymptotic solution by factoring n, which takes O(√n) trial divisions and uses O(1) extra space.

There are several small cases that can expose mistakes. The first is n = 1. The only positive integer not exceeding 1 is 1, and gcd(1, 1) = 1, so the correct output is 1. A program that blindly applies the formula φ(p) = p - 1 for a prime p without treating 1 separately can produce an incorrect result.

The second case is a prime number such as n = 7.

The correct output is 6, because every number from 1 through 6 is relatively prime to 7. A careless implementation that only counts proper divisors or only looks for composite numbers can get this wrong.

Repeated prime factors are another common source of mistakes. For

the prime factorization is 2² × 3 × 5, and the correct output is 16. Treating the two occurrences of 2 as two independent exclusions would incorrectly apply the same factor twice in the totient formula.

Approaches

The most direct approach follows the definition literally. For every integer k from 1 through n, compute gcd(k, n), and increment the answer whenever the gcd is 1. This is correct because the definition of φ(n) is exactly the number of such k. The worst case for the given input range is n = 999, so the program performs 999 gcd computations. In general the running time is O(n log n), while the extra space is O(1). For this problem that is already fast enough.

The more useful approach comes from looking at the prime factors of n instead of checking every candidate k. Suppose n has distinct prime factors p₁, p₂, ..., pᵣ. A number k is not relatively prime to n exactly when it is divisible by at least one of these primes. Among the n integers from 1 through n, exactly n / pᵢ are divisible by pᵢ. We could use inclusion-exclusion to count the union of these sets, but the same reasoning gives the compact formula

φ(n) = n × (1 - 1/p₁) × (1 - 1/p₂) × ... × (1 - 1/pᵣ).

The key point is that only distinct prime factors matter. Their exponents do not appear in the product.

For example, 60 = 2² × 3 × 5. Starting with 60, the factor 2 removes half of the candidates, giving 30. The factor 3 removes one third of the remaining candidates, giving 20. The factor 5 removes one fifth, giving 16. Thus φ(60) = 16.

We therefore only need to discover the distinct prime factors of n. Trial division up to √n is sufficient. Whenever a divisor p is found, we apply the totient update once and then divide p out completely. Removing it completely is what guarantees that the same prime is not processed again.

The brute-force method is simpler and accepted under these constraints, while the factorization method is the better general technique because its complexity depends on √n rather than n.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n log n)	O(1)	Accepted
Prime Factorization	O(√n)	O(1)	Accepted

Algorithm Walkthrough

Read n and initialize the answer to n. We start with n because the totient formula can be viewed as repeatedly removing the fraction of numbers divisible by each distinct prime factor.
Set a trial divisor p to 2 and test possible divisors while p² ≤ n. Once p² is greater than the remaining n, there cannot be another factor smaller than p, so any remaining value greater than 1 must itself be prime.
If p divides n, update the answer using answer -= answer // p. This applies the factor (1 - 1/p) using only integer arithmetic, avoiding floating-point calculations.
Divide n by p repeatedly until p no longer divides it. We do this because the formula needs each distinct prime factor exactly once. For example, if the factor is 2³, the update for 2 must happen once, not three times.
Move to the next possible divisor and continue until p² > n.
After the loop, if n is greater than 1, the remaining n is a prime factor that has not been processed. Apply one final totient update for this prime.
Print the resulting answer.

The invariant is that after every processed prime factor p, answer equals the number of integers from 1 through the original input that survive the divisibility restriction imposed by every prime factor processed so far. Removing all copies of p from n does not change which integers are divisible by p, so each prime contributes exactly one multiplicative reduction. When the factorization is complete, every prime that can make gcd(k, original_n) greater than 1 has been processed, so exactly the relatively prime integers remain.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())

    original = n
    ans = n
    p = 2

    while p * p <= n:
        if n % p == 0:
            ans -= ans // p

            while n % p == 0:
                n //= p

        p += 1

    if n > 1:
        ans -= ans // n

    print(ans)

if __name__ == "__main__":
    solve()

The variable ans starts at the original n and is updated once for every distinct prime factor. The expression ans -= ans // p is the integer form of multiplying by (p - 1) / p, so no floating-point arithmetic is involved.

The inner while loop removes every occurrence of the current prime factor. This is a subtle but necessary detail. For n = 60, after finding 2, the value of n becomes 15. The factor 2 has already been accounted for, so it must not be considered again.

The loop condition p * p <= n is also deliberate. Once the current divisor exceeds the square root of the remaining n, there can be at most one remaining factor, and that factor must be prime. Checking n > 1 after the loop handles precisely that situation.

Python integers do not have the fixed-width overflow issue that languages such as C++ can encounter, although no large intermediate values are needed here anyway.

Worked Examples

For Sample 1, the input is 15. The factorization is 15 = 3 × 5.

p	n before factoring	ans before update	ans after update	n after removing p
2	15	15	15	15
3	15	15	10	5
remaining 5	5	10	8	5

The divisor 2 does not occur, so it has no effect. Processing 3 changes 15 to 10, because one third of the candidates are divisible by 3. Processing 5 then changes 10 to 8. The final result is 8.

For Sample 2, the input is 60. Its factorization is 2² × 3 × 5.

p	n before factoring	ans before update	ans after update	n after removing p
2	60	60	30	15
3	15	30	20	5
remaining 5	5	20	16	5

The exponent of 2 is two, but the update for 2 occurs only once. After removing both copies of 2, the remaining value is 15, whose distinct prime factors are 3 and 5. The final result is 16.

For Sample 3, n = 7 is prime.

p	n before factoring	ans before update	ans after update	n after factoring
remaining 7	7	7	6	7

The trial loop never finds a divisor because 7 has no factor at most √7. The remaining value 7 is therefore prime, and one final update gives 6.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(√n)	Trial division checks divisors only while p² ≤ n, with each discovered factor removed completely.
Space	O(1)	Only a constant number of integer variables are stored.

With n < 1000, √n is below 32, so the optimal algorithm examines only a few dozen candidate divisors in the worst case. It is far below the one-second time limit and uses negligible memory.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    n = int(input())

    ans = n
    p = 2

    while p * p <= n:
        if n % p == 0:
            ans -= ans // p
            while n % p == 0:
                n //= p
        p += 1

    if n > 1:
        ans -= ans // n

    return str(ans)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)
    try:
        return solve() + "\n"
    finally:
        sys.stdin = old_stdin

# Provided samples
assert run("15\n") == "8\n", "sample 1"
assert run("60\n") == "16\n", "sample 2"
assert run("7\n") == "6\n", "sample 3"

# Minimum-size input
assert run("1\n") == "1\n", "n = 1"

# Prime input near the maximum
assert run("997\n") == "996\n", "997 is prime"

# Maximum allowed input
assert run("999\n") == "648\n", "999 = 3^3 * 37"

# Repeated prime factor
assert run("729\n") == "486\n", "729 = 3^6"

Test input	Expected output	What it validates
`1`	`1`	Minimum-size input and the special case n = 1
`997`	`996`	A prime near the upper boundary
`999`	`648`	Maximum-size input and a factor remaining after trial division
`729`	`486`	A prime power, verifying that repeated factors are processed only once

Edge Cases

The smallest possible input is n = 1.

The algorithm initializes ans = 1. Since there is no divisor to process and the remaining n is not greater than 1, the answer stays 1. This agrees with the definition because gcd(1, 1) = 1.

A prime input such as

never enters the factor-processing branch. After testing divisors up to the square root of 7, the remaining n is still 7. The final update computes 7 - 7 // 7 = 6, which is the correct totient of a prime.

A repeated prime factor can be seen with

The first discovered factor is 2. The answer changes from 60 to 30, and the inner loop removes both copies of 2, changing the working n from 60 to 15. The prime 2 is never applied again. The remaining factors 3 and 5 then reduce the answer to 20 and finally 16.

The maximum input is

which factors as 3³ × 37. The algorithm finds 3, applies its totient reduction once, and removes all three copies of 3. The remaining value 37 is greater than 1 after the trial loop, so it is recognized as a prime factor and processed once. The result is 648.

A prime power such as

is particularly useful for catching the mistake of applying the totient update once per exponent. Since 729 = 3⁶, the factor 3 must affect the formula only once. The correct calculation is 729 × (1 - 1/3) = 486, which the algorithm produces.