CF 83D - Numbers
We need to count integers inside the interval $[a,b]$ whose smallest divisor greater than $1$ is exactly $k$. For a number $x$, saying that its smallest divisor is $k$ means two things happen simultaneously: 1. $x$ is divisible by $k$. 2.
Rating: 2400
Tags: dp, math, number theory
Solve time: 2m 7s
Verified: yes
Solution
Problem Understanding
We need to count integers inside the interval $[a,b]$ whose smallest divisor greater than $1$ is exactly $k$.
For a number $x$, saying that its smallest divisor is $k$ means two things happen simultaneously:
- $x$ is divisible by $k$.
- $x$ is not divisible by any integer from $2$ to $k-1$.
The second condition immediately implies that $k$ itself must be prime. If $k$ were composite, one of its smaller divisors would also divide $x$, so $k$ could never be the smallest divisor.
The limits are large. Both $a$ and $b$ can reach $2 \cdot 10^9$, so iterating through the interval is impossible. Even checking every number individually with trial division would require billions of operations in the worst case.
The value of $k$ is also large, but an important observation saves us: if $k$ is prime, then every valid number has the form
$$x = k \cdot t$$
and $t$ must avoid all prime factors smaller than $k$.
Since the smallest divisor of any composite number is prime, it is enough to forbid divisibility by primes smaller than $k$. That transforms the problem into a counting problem with inclusion-exclusion and the Möbius function.
Several edge cases are easy to mishandle.
If $k$ is composite, the answer is always zero. For example:
Input:
1 100 6
Every multiple of $6$ is divisible by $2$ or $3$, so $6$ can never be the smallest divisor. The correct answer is:
0
A careless implementation that only checks divisibility by numbers smaller than $k$ after scaling might accidentally count multiples of $6$.
Another subtle case appears when $k^2 > b$. Then the only possible valid number is $k$ itself, because every other multiple $2k,3k,\dots$ already has a smaller divisor. Example:
Input:
10 20 13
The only multiple of $13$ in the interval is $13$, and its smallest divisor is indeed $13$. The answer is:
1
There is also an off-by-one trap when converting the interval $[a,b]$ into a range for the multiplier $t$. Suppose:
Input:
15 30 5
We rewrite numbers as $x=5t$. Then:
$$t \in \left[\left\lceil \frac{15}{5} \right\rceil,\left\lfloor \frac{30}{5} \right\rfloor\right] = [3,6]$$
Among $3,4,5,6$, only $5$ is divisible by no prime smaller than $5$. That gives $25$ as the only valid number. Forgetting the ceiling on the left boundary would incorrectly include $10$.
Approaches
The brute-force approach directly follows the definition. For every integer $x$ in $[a,b]$, we find its smallest divisor greater than $1$. If that divisor equals $k$, we count it.
This works logically because the smallest divisor uniquely determines whether the number satisfies the condition. The problem is the scale. The interval length can be almost $2 \cdot 10^9$. Even an $O(\sqrt{x})$ check per number becomes hopelessly large.
The key observation is that if the smallest divisor of $x$ equals $k$, then $k$ must be prime and $x$ must be divisible by $k$. Write:
$$x = k \cdot t$$
Now ask what conditions apply to $t$.
If some prime $p < k$ divides $t$, then $p$ also divides $x$, which contradicts $k$ being the smallest divisor. So $t$ must be coprime to the product of all primes smaller than $k$.
This converts the problem into counting integers in a range that are not divisible by a set of primes.
The number of primes smaller than $k$ cannot be large when their product stays below $2 \cdot 10^9$. The product of the first ten primes already exceeds this limit. That means we can safely use inclusion-exclusion over all subsets of these primes.
The brute-force checks every number individually. The optimal solution counts whole groups at once by subtracting multiples of forbidden primes.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | $O((b-a+1)\sqrt{b})$ | $O(1)$ | Too slow |
| Optimal | $O(\sqrt{k} + 2^{\pi(k)})$ | $O(\pi(k))$ | Accepted |
Algorithm Walkthrough
- Check whether $k$ is prime.
If $k$ is composite, immediately print $0$. A composite number can never be the smallest divisor of another number because one of its smaller prime divisors would appear first. 2. Convert the interval from numbers $x$ to multipliers $t$.
Every valid number has the form:
$$x = k \cdot t$$
So:
$$L = \left\lceil \frac{a}{k} \right\rceil$$
$$R = \left\lfloor \frac{b}{k} \right\rfloor$$
We now need to count integers $t \in [L,R]$ that are not divisible by any prime smaller than $k$. 3. Generate all primes smaller than $k$.
We only need primes up to $\sqrt{k}$ to test primality, but we also store every prime less than $k$ because these are the forbidden divisors for $t$. 4. Apply inclusion-exclusion.
Let the forbidden primes be:
$$p_1,p_2,\dots,p_m$$
For every subset of these primes, compute their product:
$$d = p_{i_1}p_{i_2}\cdots p_{i_s}$$
Count how many numbers in $[L,R]$ are divisible by $d$:
$$\left\lfloor \frac{R}{d} \right\rfloor - \left\lfloor \frac{L-1}{d} \right\rfloor$$
Add this count if the subset size is even, subtract it if the subset size is odd. 5. The final inclusion-exclusion result is the number of valid multipliers $t$, which equals the number of valid integers $x$.
Why it works
Every valid number can be uniquely written as $x = kt$. Since $k$ is prime, the only way for $x$ to have a smaller divisor is for $t$ to contain a prime factor smaller than $k$. The algorithm counts exactly those $t$ that avoid all forbidden primes.
Inclusion-exclusion is correct because it alternates between subtracting numbers divisible by at least one forbidden prime and adding back numbers counted multiple times. After processing all subsets, every invalid $t$ contributes zero net count, while every valid $t$ contributes exactly one.
Python Solution
import sys
input = sys.stdin.readline
def is_prime(x):
if x < 2:
return False
d = 2
while d * d <= x:
if x % d == 0:
return False
d += 1
return True
def primes_less_than(k):
primes = []
for x in range(2, k):
ok = True
d = 2
while d * d <= x:
if x % d == 0:
ok = False
break
d += 1
if ok:
primes.append(x)
return primes
def solve():
a, b, k = map(int, input().split())
if not is_prime(k):
print(0)
return
L = (a + k - 1) // k
R = b // k
if L > R:
print(0)
return
primes = primes_less_than(k)
ans = 0
m = len(primes)
for mask in range(1 << m):
bits = 0
prod = 1
for i in range(m):
if (mask >> i) & 1:
bits += 1
prod *= primes[i]
if prod > R:
break
else:
cnt = R // prod - (L - 1) // prod
if bits % 2 == 0:
ans += cnt
else:
ans -= cnt
print(ans)
solve()
The first part checks whether $k$ is prime. This is not an optimization, it is a correctness requirement. If $k$ is composite, no number can have smallest divisor exactly $k$.
Next we convert the original interval into multiplier space. Every candidate number is a multiple of $k$, so we only work with values of $t$. The ceiling division for the left boundary is easy to get wrong. Using:
(a + k - 1) // k
correctly computes the smallest integer $t$ such that $kt \ge a$.
The inclusion-exclusion loop iterates through every subset of forbidden primes. The variable prod stores the product of primes in the current subset. If that product already exceeds R, no number in the range can be divisible by it, so the subset contributes nothing.
The alternating signs come directly from inclusion-exclusion. Even-sized subsets add counts, odd-sized subsets subtract counts.
The implementation never risks overflow because Python integers are arbitrary precision, but the early break on prod > R keeps the subset processing small.
Worked Examples
Example 1
Input:
1 10 2
We have $k=2$, which is prime.
$$L = \left\lceil \frac{1}{2} \right\rceil = 1$$
$$R = \left\lfloor \frac{10}{2} \right\rfloor = 5$$
There are no primes smaller than $2$, so every $t$ in $[1,5]$ is valid.
| Step | Value |
|---|---|
| $L$ | 1 |
| $R$ | 5 |
| Forbidden primes | none |
| Valid $t$ | 1,2,3,4,5 |
| Valid $x$ | 2,4,6,8,10 |
| Answer | 5 |
This demonstrates the simplest case where inclusion-exclusion has only the empty subset.
Example 2
Input:
1 30 5
Since $5$ is prime:
$$L = 1,\quad R = 6$$
Forbidden primes are $2$ and $3$.
| Subset | Product | Count in $[1,6]$ | Contribution |
|---|---|---|---|
| {} | 1 | 6 | +6 |
| {2} | 2 | 3 | -3 |
| {3} | 3 | 2 | -2 |
| {2,3} | 6 | 1 | +1 |
Final answer:
$$6 - 3 - 2 + 1 = 2$$
The valid multipliers are $1$ and $5$, giving numbers $5$ and $25$.
This trace shows how inclusion-exclusion removes numbers divisible by forbidden primes and restores overlaps correctly.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(\sqrt{k} + 2^{\pi(k)})$ | primality testing plus inclusion-exclusion over primes smaller than $k$ |
| Space | $O(\pi(k))$ | storing forbidden primes |
The number of relevant primes is tiny because their product grows very quickly. Even though inclusion-exclusion is exponential in the number of primes, the actual subset count stays small enough for the time limit.
Test Cases
# helper: run solution on input string, return output string
import sys
import io
def solve():
input = sys.stdin.readline
def is_prime(x):
if x < 2:
return False
d = 2
while d * d <= x:
if x % d == 0:
return False
d += 1
return True
def primes_less_than(k):
primes = []
for x in range(2, k):
ok = True
d = 2
while d * d <= x:
if x % d == 0:
ok = False
break
d += 1
if ok:
primes.append(x)
return primes
a, b, k = map(int, input().split())
if not is_prime(k):
print(0)
return
L = (a + k - 1) // k
R = b // k
if L > R:
print(0)
return
primes = primes_less_than(k)
ans = 0
m = len(primes)
for mask in range(1 << m):
bits = 0
prod = 1
for i in range(m):
if (mask >> i) & 1:
bits += 1
prod *= primes[i]
if prod > R:
break
else:
cnt = R // prod - (L - 1) // prod
if bits % 2 == 0:
ans += cnt
else:
ans -= cnt
print(ans)
def run(inp: str) -> str:
backup_stdin = sys.stdin
backup_stdout = sys.stdout
sys.stdin = io.StringIO(inp)
sys.stdout = io.StringIO()
solve()
out = sys.stdout.getvalue()
sys.stdin = backup_stdin
sys.stdout = backup_stdout
return out.strip()
# provided sample
assert run("1 10 2\n") == "5", "sample 1"
# composite k
assert run("1 100 6\n") == "0", "composite k"
# single valid number
assert run("13 13 13\n") == "1", "single prime"
# no multiples of k in range
assert run("1 10 13\n") == "0", "empty multiplier interval"
# boundary case with ceil division
assert run("15 30 5\n") == "1", "left boundary correctness"
# larger interval
assert run("1 30 5\n") == "2", "inclusion exclusion correctness"
| Test input | Expected output | What it validates |
|---|---|---|
1 100 6 |
0 |
composite $k$ immediately fails |
13 13 13 |
1 |
interval containing exactly one valid number |
1 10 13 |
0 |
no multiples of $k$ exist |
15 30 5 |
1 |
correct ceiling division on left boundary |
1 30 5 |
2 |
inclusion-exclusion overlap handling |
Edge Cases
Consider the composite case:
Input:
1 100 6
The algorithm first checks primality. Since $6$ is divisible by $2$, it is not prime, so the algorithm immediately returns:
0
This is correct because every multiple of $6$ is already divisible by $2$, which is smaller than $6$.
Now consider the boundary case where only $k$ itself may qualify:
Input:
10 20 13
We compute:
$$L = \left\lceil \frac{10}{13} \right\rceil = 1$$
$$R = \left\lfloor \frac{20}{13} \right\rfloor = 1$$
The only multiplier is $t=1$, corresponding to $x=13$. Since $1$ is divisible by no forbidden primes, the answer becomes:
1
The algorithm handles this naturally without special branching.
Finally, consider the off-by-one trap:
Input:
15 30 5
The algorithm computes:
$$L = (15+5-1)//5 = 3$$
$$R = 30//5 = 6$$
The candidate multipliers are $3,4,5,6$. Inclusion-exclusion removes $3,4,6$ because they are divisible by $2$ or $3$. Only $5$ remains, giving the single valid number $25$.
If we had used ordinary floor division for the left boundary, we would incorrectly include multiplier $2$, corresponding to $10$, which lies outside the interval.