CF 1981A - Turtle and Piggy Are Playing a Game

Rating: 800
Tags: brute force, greedy, math
Solve time: 2m 13s
Verified: yes

Solution

Problem Understanding

We are given a range of integers $[l, r]$. Turtle chooses a number $x$ from this range. After that, Piggy repeatedly divides $x$ by one of its divisors greater than or equal to $2$, earning one point for each division, until the number becomes $1$.

Both players want the final score to be as large as possible. We need to determine that maximum possible score.

The key observation is that every operation replaces $x$ by $x / p$, where $p \ge 2$. Since $p$ is at least $2$, each operation reduces the current value by at least a factor of $2$. The score is simply the number of division steps before reaching $1$.

The constraints are extremely small from an algorithmic perspective. There are up to $10^4$ test cases and values are at most $10^9$. Any solution that iterates through the entire interval $[l,r]$ would be impossible because the interval length can approach $10^9$. We need a direct mathematical formula per test case.

A subtle mistake is to think that prime factorization itself determines the answer. For example, for $x=12$, the prime factorization is $2^2 \cdot 3$, giving three prime factors. The score is indeed $3$, but that is because Piggy can always choose prime divisors and remove only one prime factor at a time. The maximum score equals the total number of prime factors counted with multiplicity.

Another easy mistake is to search for the best number in the interval. Consider $l=114514$, $r=1919810$. Brute forcing millions of values is unnecessary. The optimal number is always the largest power of two that fits inside the interval.

Approaches

A brute-force solution would examine every number $x$ in $[l,r]$, factorize it, count the total number of prime factors, and keep the maximum. This is correct because the maximum score for a fixed number equals the number of prime factors with multiplicity. Unfortunately, the interval length can be nearly $10^9$, making such a search completely infeasible.

The crucial observation is that every operation divides the current value by at least $2$. Starting from $x$, after $s$ operations we must have

$$x \ge 2^s.$$

This immediately gives

$$s \le \lfloor \log_2 x \rfloor.$$

This bound is achievable whenever $x$ itself is a power of two. For example, $2^{10}$ can be divided by $2$ exactly ten times.

Since Turtle may choose any number in $[l,r]$, the largest possible score is achieved by choosing the largest power of two that does not exceed $r$. Let

$$2^k \le r < 2^{k+1}.$$

Then $k=\lfloor \log_2 r \rfloor$.

Because the statement guarantees $2l \le r$, the interval always contains at least one power of two. If $2^k$ is the largest power of two not exceeding $r$, then $2^k > r/2 \ge l$. Thus $2^k$ lies inside the interval and can be chosen.

The answer is simply $\lfloor \log_2 r \rfloor$.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O((r-l+1)\sqrt r)$	$O(1)$	Too slow
Optimal	$O(1)$ per test case	$O(1)$	Accepted

Algorithm Walkthrough

Read $l$ and $r$.
Compute the largest integer $k$ such that $2^k \le r$.
Output $k$.

The easiest way in Python is to use the bit length of $r$. For any positive integer,

$$\lfloor \log_2 r \rfloor = r.\text{bit_length()} - 1.$$

Why it works

Every division removes a factor at least equal to $2$, so after $s$ operations the original number must be at least $2^s$. No number not exceeding $r$ can produce more than $\lfloor \log_2 r \rfloor$ operations.

The interval always contains the largest power of two not exceeding $r$. Choosing that value achieves exactly $\lfloor \log_2 r \rfloor$ operations by repeatedly dividing by $2$. Since both an upper bound and a matching construction exist, the answer is exactly $\lfloor \log_2 r \rfloor$.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    ans = []

    for _ in range(t):
        l, r = map(int, input().split())
        ans.append(str(r.bit_length() - 1))

    sys.stdout.write("\n".join(ans))

solve()

The solution never uses $l$ after reading it. That may look surprising at first, but the guarantee $2l \le r$ is precisely what ensures the necessary power of two always lies inside the interval.

The expression r.bit_length() - 1 computes $\lfloor \log_2 r \rfloor$ without floating-point arithmetic, avoiding precision issues. Since $r \le 10^9$, integer overflow is never a concern.

Worked Examples

Example 1

Input:

2 15

The largest power of two not exceeding $15$ is $8$.

Value	Result
r	15
Largest power of two ≤ r	8
Score	3

Output:

Choosing $x=8$:

$$8 \to 4 \to 2 \to 1$$

which gives three operations.

Example 2

Input:

6 22

Value	Result
r	22
Largest power of two ≤ r	16
Score	4

Output:

Choosing $x=16$:

$$16 \to 8 \to 4 \to 2 \to 1$$

which gives four operations.

This example shows that the optimal choice is not necessarily $r$ itself. The best choice is the power of two inside the interval.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$ per test case	Only a bit-length computation
Space	$O(1)$	Constant extra memory

With at most $10^4$ test cases, the solution performs only a few integer operations per case and easily fits within the limits.

Test Cases

import sys
import io

def run(inp: str) -> str:
    lines = inp.strip().splitlines()
    t = int(lines[0])
    out = []

    idx = 1
    for _ in range(t):
        l, r = map(int, lines[idx].split())
        idx += 1
        out.append(str(r.bit_length() - 1))

    return "\n".join(out)

# provided sample
assert run(
"""5
2 4
3 6
2 15
6 22
114514 1919810
"""
) == "2\n2\n3\n4\n20"

# minimum valid range
assert run(
"""1
1 2
"""
) == "1"

# exact power of two
assert run(
"""1
8 16
"""
) == "4"

# interval containing many powers of two
assert run(
"""1
10 100
"""
) == "6"

# largest constraints style case
assert run(
"""1
500000000 1000000000
"""
) == "29"

Test input	Expected output	What it validates
`1 2`	`1`	Smallest valid interval
`8 16`	`4`	Upper endpoint is a power of two
`10 100`	`6`	Best choice is an interior power of two
`500000000 1000000000`	`29`	Large values near the limit

Edge Cases

Consider the smallest possible interval:

1
1 2

The only power of two available is $2$. We have

$$2 \to 1$$

so the score is $1$. The formula gives $\lfloor \log_2 2 \rfloor = 1$.

Consider an interval where the upper endpoint is itself a power of two:

1
8 16

The answer is $4$ because $16 = 2^4$. Repeatedly dividing by $2$ achieves four operations. The formula returns 16.bit_length() - 1 = 4.

Consider a range where the optimal number is not the largest number:

1
6 22

The largest value is $22$, whose prime factorization is $2 \cdot 11$, giving only two operations. The interval also contains $16$, which gives four operations. The algorithm correctly finds the answer from the largest power of two not exceeding $22$, producing $4$.