最大限度地减少将N根长度的棒切成N根单位长度的棒所需的切割次数

给定一个表示给定棒的长度的整数N ，任务是最小化将棒分割成单位长度的片段所需的时间，因为在任何时间实例都可以对棒的任何部分进行一次切割。

例子：

Input: N = 100 
Output: 7 
Explanation: 
(100 units) —> (2 portions of 50 units) —> (4 portions of 25 units) —> (4 portions of 12 units, 4 portions of 13 units) —> (12 portions of 6 units, 4 portions of 7 units) —> (28 portions of 3 units, 4 portions of 4 units) —> (28 portions of 1 unit, 36 portions of 2 units) —> (100 portions of 1 unit)

Input: N = 65 
Output: 7 
Explanation: 
(65 units) —> (1 portion of 32 units, 1 portion of 33 units) —> (3 portions of 16 units, 1 portion of 17 units) —> (7 portions of 8 units, 1 portions of 9 units) —> (15 portions of 4 units, 1 portion of 5 units) —> (31 portions of 2 units, 1 portions of 3 units) —> (63 portions of 1 unit, 1 portion of 2 units) —> (65 portions of 1 unit) 
  

方法：
由于我们可以在特定的时间切割一次棒的任何部分，因此我们需要在每次切割后将其最大化。因此，我们将在第一次切割时将木棍切成尽可能长的两个部分。在下一个实例中，在下一次切割中将两个获得的部分进一步切割成两个最长的部分。重复此步骤，直到获得N个单位。

Illustration: 
N = 100 
1st Cut: (50) + (50) 
2nd Cut: (25) + (25) + (25) + (25) 
3rd Cut: (12) + (13) + (12) + (13) + (12) + (13) + (12) + (13) 
4th Cut: (6) + (6) + (6) + (7) + (6) + (6) + (6) + (7) + (6) + (6) + (6) + (7) + (6) + (6) + (6) + (7) 
5th Cut: (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) + (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) + (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) + (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) 
6th Cut: 28 portions of 1 unit, 36 portions of 2 units 
7th Cut: 100 portions of 1 unit 
 

因此，将N长棒分成1个单位所需的最短时间为ceil(log ₂ N) 。

下面是上述方法的实现：

C++

// C++ program to find minimum
// time required to split a
// stick of N length into
// unit pieces
 
#include 
using namespace std;
 
// Function to return the
// minimum time required
// to split stick of N into
// length into unit pieces
int min_time_to_cut(int N)
{
    if (N == 0)
        return 0;
    // Return the minimum
    // unit of time required
    return ceil(log2(N));
}
// Driver Code
int main()
{
    int N = 100;
    cout << min_time_to_cut(N);
    return 0;
}

Java

// Java program to find minimum
// time required to split a
// stick of N length into
// unit pieces
import java.lang.*;
 
class GFG{
     
// Function to return the
// minimum time required
// to split stick of N into
// length into unit pieces
static int min_time_to_cut(int N)
{
    if (N == 0)
        return 0;
         
    // Return the minimum
    // unit of time required
    return (int)Math.ceil(Math.log(N) /
                          Math.log(2));
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 100;
    System.out.print(min_time_to_cut(N));
}
}
 
// This code is contributed by rock_cool

Python3

# Python3 program to find minimum
# time required to split a stick
# of N length into unit pieces
import math
 
# Function to return the
# minimum time required
# to split stick of N into
# length into unit pieces
def min_time_to_cut(N):
     
    if (N == 0):
        return 0
         
    # Return the minimum
    # unit of time required
    return int(math.log2(N)) + 1
 
# Driver Code
N = 100
 
print(min_time_to_cut(N))
 
# This code is contributed by Vishal Maurya

C#

// C# program to find minimum
// time required to split a
// stick of N length into
// unit pieces
using System;
class GFG{
     
// Function to return the
// minimum time required
// to split stick of N into
// length into unit pieces
static int min_time_to_cut(int N)
{
    if (N == 0)
        return 0;
         
    // Return the minimum
    // unit of time required
    return (int)Math.Ceiling(Math.Log(N) /
                             Math.Log(2));
}
 
// Driver Code
public static void Main()
{
    int N = 100;
    Console.Write(min_time_to_cut(N));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript program to find minimum
// time required to split a stick of
// N length into unit pieces
 
// Function to return the
// minimum time required
// to split stick of N into
// length into unit pieces
function min_time_to_cut(N)
{
    if (N == 0)
        return 0;
         
    // Return the minimum
    // unit of time required
    return Math.ceil(Math.log(N) /
                     Math.log(2));
}
 
// Driver code
let N = 100;
 
document.write(min_time_to_cut(N));
 
// This code is contributed by divyesh072019
 

输出：

7

时间复杂度： O(1)
辅助空间： O(1)