Overview

This problem asks you to find the minimum number of operations needed to reach exactly \(T\) cells from an initial count of \(S\) cells, where each “mitosis promotion” operation doubles the number of cells. However, the number of operations is limited to at most \(K\).

Analysis

The key to this problem is focusing on the property that cells “always double with each operation.”

1. Checking the ratio After performing \(n\) operations, the number of cells becomes \(S \times 2^n\). Therefore, to reach the target of \(T\) cells, it is a necessary condition that \(T\) is divisible by \(S\). If T % S != 0, then no sequence of operations can produce exactly \(T\) cells, so we can immediately output -1.

2. Checking for a power of 2 If \(T\) is divisible by \(S\), then the ratio \(ratio = T / S\) must be a power of \(2\) (\(2^0, 2^1, 2^2, \dots\)). For example:

   • When \(S=3, T=12\), the ratio is \(4\). Since \(4 = 2^2\), it can be reached in \(2\) operations.
   • When \(S=3, T=9\), the ratio is \(3\). Since \(3\) is not a power of \(2\), it is unreachable.

3. Constraint on the number of operations When the ratio can be expressed as \(2^n\), the required number of operations is \(n\). If this \(n\) is within the given limit \(K\) (i.e., \(n \leq K\)), output \(n\); otherwise, output -1 as it is unreachable.

Algorithm

1. Check whether \(T\) is divisible by \(S\).
2. If divisible, compute the ratio \(ratio = T // S\).
3. Check whether \(ratio\) is a power of \(2\).
   • Using bit manipulation properties, if ratio > 0 and (ratio & (ratio - 1)) == 0, then it is a power of \(2\).
   • Alternatively, you can verify by repeatedly dividing ratio by \(2\).
4. Find \(n\) such that \(ratio = 2^n\).
   • In Python, you can efficiently determine what power of \(2\) a number is by using ratio.bit_length() - 1.
5. Check whether \(n \leq K\), and output the result.

Complexity

• Time complexity: \(O(1)\) or \(O(\log(T/S))\)
  • Computations based on bit length and bit operations are very fast. Although \(T, S \leq 10^{18}\), the process completes with at most around \(60\) loop iterations or bit operations.
• Space complexity: \(O(1)\)
  • Only a constant number of variables are stored beyond the input values.

Implementation Notes

• Handling large numbers: Since \(S, T\) can be as large as \(10^{18}\), depending on the language, you may need to use 64-bit integer types (such as long long in C++). In Python, arbitrarily large integers are natively supported, so you can compute directly.

• Utilizing bit operations: ratio & (ratio - 1) is a commonly used technique in computer science for determining “whether a number is a power of \(2\).”

• Special case: When \(S = T\), the ratio is \(1\) (\(2^0\)), and the required number of operations is \(0\). This case is also correctly handled by the algorithm described above.

  Source Code

import sys

def solve():
    # 入力を読み込む
    line = sys.stdin.readline()
    if not line:
        return
    
    try:
        s, t, k = map(int, line.split())
    except ValueError:
        return

    # TがSで割り切れない場合は、2倍し続けてTにすることは不可能
    if t % s != 0:
        print("-1")
        return

    # 倍率を計算
    ratio = t // s
    
    # 倍率が2の累乗であるかを確認する
    # ratio & (ratio - 1) == 0 かつ ratio > 0 であれば2の累乗
    if ratio > 0 and (ratio & (ratio - 1)) == 0:
        # 2の何乗であるかを求める
        # bit_length() は 2^n のとき n + 1 を返す
        n = ratio.bit_length() - 1
        
        # 操作回数がK回以内であればその回数を出力
        if n <= k:
            print(n)
        else:
            print("-1")
    else:
        # 2の累乗でない場合は不可能
        print("-1")

if __name__ == '__main__':
    solve()

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