计算可以使用给定运算将N转换为1的数字

给定正整数N(N≥2) ，任务是对范围[2，N]中的整数X进行计数，以使X可以使用以下操作将N转换为1：

• 如果N可被X整除，则将N的值更新为N / X。
• 否则，将N的值更新为N – X。

例子：

Input: N = 6 
Output: 3 
Explanation: 
The following integers can be used to convert N to 1: 
X = 2 => N = 6 -> N = 3 -> N = 1 
X = 5 => N = 6 -> N = 1 
X = 6 => N = 6 -> N = 1

Input: N = 48 
Output: 4

天真的方法：这个问题的天真的方法是遍历从2到N的所有整数，并对可以将N转换为1的整数进行计数。

下面是上述方法的实现：

C++

// C++ program to count the numbers
// which can convert N to 1
// using the given operation
 
#include 
using namespace std;
 
// Function to count the numbers
// which can convert N to 1
// using the given operation
int countValues(int n)
{
 
    int answer = 0;
 
    // Iterate through all the integers
    for (int i = 2; i <= n; i++) {
        int k = n;
 
        // Check if N can be converted
        // to 1
        while (k >= i) {
            if (k % i == 0)
                k /= i;
            else
                k -= i;
        }
 
        // Incrementing the count if it can
        // be converted
        if (k == 1)
            answer++;
    }
    return answer;
}
 
// Driver code
int main()
{
    int N = 6;
 
    cout << countValues(N);
 
    return 0;
}

Java

// Java program to count the numbers
// which can convert N to 1
// using the given operation
import java.io.*;
import java.util.*;
class GFG{
     
// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int n)
{
    int answer = 0;
 
    // Iterate through all the integers
    for (int i = 2; i <= n; i++)
    {
        int k = n;
 
        // Check if N can be converted
        // to 1
        while (k >= i)
        {
            if (k % i == 0)
                k /= i;
            else
                k -= i;
        }
 
        // Incrementing the count if it can
        // be converted
        if (k == 1)
            answer++;
    }
    return answer;
}
 
// Driver code
public static void main(String args[])
{
    int N = 6;
 
    System.out.print(countValues(N));
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 program to count the numbers
# which can convert N to 1
# using the given operation
 
# Function to count the numbers
# which can convert N to 1
# using the given operation
def countValues(n):
    answer = 0
 
    # Iterate through all the integers
    for i in range(2, n + 1, 1):
        k = n
 
        # Check if N can be converted
        # to 1
        while (k >= i):
            if (k % i == 0):
                k //= i
            else:
                k -= i
 
        # Incrementing the count if it can
        # be converted
        if (k == 1):
            answer += 1
    return answer
 
# Driver code
if __name__ == '__main__':
     
    N = 6
    print(countValues(N))
 
# This code is contributed by Samarth

C#

// C# program to count the numbers
// which can convert N to 1
// using the given operation
using System;
class GFG{
     
// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int n)
{
    int answer = 0;
 
    // Iterate through all the integers
    for (int i = 2; i <= n; i++)
    {
        int k = n;
 
        // Check if N can be converted
        // to 1
        while (k >= i)
        {
            if (k % i == 0)
                k /= i;
            else
                k -= i;
        }
 
        // Incrementing the count if it can
        // be converted
        if (k == 1)
            answer++;
    }
    return answer;
}
 
// Driver code
public static void Main()
{
    int N = 6;
 
    Console.Write(countValues(N));
}
}
 
// This code is contributed by Nidhi_biet

Javascript

 
    // Javascript program to count the numbers
    // which can convert N to 1
    // using the given operation
     
    // Function to count the numbers
    // which can convert N to 1
    // using the given operation
    function countValues(n)
    {
 
        let answer = 0;
 
        // Iterate through all the integers
        for (let i = 2; i <= n; i++) {
            let k = n;
 
            // Check if N can be converted
            // to 1
            while (k >= i) {
                if (k % i == 0)
                    k /= i;
                else
                    k -= i;
            }
 
            // Incrementing the count if it can
            // be converted
            if (k == 1)
                answer++;
        }
        return answer;
    }
     
    let N = 6;
  
    document.write(countValues(N));
     

输出：

3

时间复杂度： O(N) ，其中N是给定的数字。

高效的方法：想法是观察到，如果最初N不能被N整除，那么在整个减法过程中都将进行减法运算，因为每次减法N仍然不能被N整除。当N≤X时，这些运算也会停止， N的最终值将等于N mod X。

因此，对于从2到N的所有数字，只有两种可能的情况：

1. 没有除法运算：对于所有这些数字，最终值将等于N modX。只有在N mod X = 1时，N才最终变为1。显然，对于X = N – 1，并且N的所有除数– 1，N mod X = 1成立。
2. 除法运算不止一次发生：除法运算仅对N上的除数发生。对于N的每个除数，说d，执行除法直到N mod d！=0。如果最终N mod d = 1，则将其包括在否则就没有答案(使用从案例1派生的属性)。

下面是上述方法的实现：

C++

// C++ program to count the numbers
// which can convert N to 1
// using the given operation
 
#include 
using namespace std;
 
// Function to count the numbers
// which can convert N to 1
// using the given operation
int countValues(int N)
{
 
    vector div;
 
    // Store all the divisors of N
    for (int i = 2; i * i <= N; i++) {
 
        // If i is a divisor
        if (N % i == 0) {
            div.push_back(i);
 
            // If i is not equal to N / i
            if (N != i * i) {
                div.push_back(N / i);
            }
        }
    }
 
    int answer = 0;
 
    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for (int i = 1; i * i <= N - 1; i++) {
 
        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0) {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }
 
    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    for (auto d : div) {
        int K = N;
        while (K % d == 0)
            K /= d;
        if ((K - 1) % d == 0)
            answer++;
    }
    return answer;
}
 
// Driver code
int main()
{
    int N = 6;
 
    cout << countValues(N);
 
    return 0;
}

Java

// Java program to count the numbers
// which can convert N to 1
// using the given operation
import java.util.*;
 
class GFG{
 
// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int N)
{
    Vector div = new Vector<>();
 
    // Store all the divisors of N
    for(int i = 2; i * i <= N; i++)
    {
         
        // If i is a divisor
        if (N % i == 0)
        {
            div.add(i);
 
            // If i is not equal to N / i
            if (N != i * i)
            {
                div.add(N / i);
            }
        }
    }
 
    int answer = 0;
 
    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for(int i = 1; i * i <= N - 1; i++)
    {
         
        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0)
        {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }
 
    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    for(int d : div)
    {
        int K = N;
        while (K % d == 0)
            K /= d;
             
        if ((K - 1) % d == 0)
            answer++;
    }
    return answer;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 6;
 
    System.out.print(countValues(N));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program to count the numbers
# which can convert N to 1
# using the given operation
 
# Function to count the numbers
# which can convert N to 1
# using the given operation
def countValues(N):
     
    div = []
    i = 2
     
    # Store all the divisors of N
    while ((i * i) <= N):
         
        # If i is a divisor
        if (N % i == 0):
            div.append(i)
  
            # If i is not equal to N / i
            if (N != i * i):
                div.append(N // i)
                 
        i += 1 
         
    answer = 0
    i = 1
     
    # Iterate through all the divisors of
    # N - 1 and count them in answer
    while((i * i) <= N - 1):
  
        # Check if N - 1 is a divisor
        # or not
        if ((N - 1) % i == 0):
            if (i * i == N - 1):
                answer += 1
            else:
                answer += 2
                 
        i += 1
  
    # Iterate through all divisors and check
    # for N mod d = 1 or (N-1) mod d = 0
    for d in div:
        K = N
         
        while (K % d == 0):
            K //= d
        if ((K - 1) % d == 0):
            answer += 1
     
    return answer
 
# Driver code
if __name__=="__main__":
     
    N = 6
  
    print(countValues(N))
 
# This code is contributed by rutvik_56

C＃

// C# program to count the numbers
// which can convert N to 1
// using the given operation
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int N)
{
    List div = new List();
 
    // Store all the divisors of N
    for(int i = 2; i * i <= N; i++)
    {
         
        // If i is a divisor
        if (N % i == 0)
        {
            div.Add(i);
 
            // If i is not equal to N / i
            if (N != i * i)
            {
                div.Add(N / i);
            }
        }
    }
 
    int answer = 0;
 
    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for(int i = 1; i * i <= N - 1; i++)
    {
         
        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0)
        {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }
 
    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    foreach(int d in div)
    {
        int K = N;
        while (K % d == 0)
            K /= d;
             
        if ((K - 1) % d == 0)
            answer++;
    }
    return answer;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 6;
 
    Console.Write(countValues(N));
}
}
 
// This code is contributed by Amit Katiyar

输出：

3

时间复杂度：

$O(\sqrt{N})$

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