在序列中找到不是给定数字倍数的第 N 个数字

给定四个整数A 、 N 、 L和R ，任务是在从L到R的连续整数序列中找到第N个不是A的倍数。假设序列至少包含N 个不能被A整除的数字，并且整数A总是大于 1。

例子：

Input: A = 2, N = 3, L = 1, R = 10 
Output: 5 
Explanation: 
The sequence is 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Here 5 is the third number which is not a multiple of 2 in the sequence.

Input: A = 3, N = 6, L = 4, R = 20 
Output: 11 
Explanation : 
11 is the 6th number which is not a multiple of 3 in the sequence. 

朴素的方法：朴素的方法是在循环中遍历范围[L, R]以找到第N^个数字。步骤是：

1. 将非倍数和当前数的计数初始化为 0。
2. 迭代范围[L, R]直到非倍数的计数不等于N 。
3. 如果当前数字不能被 A 整除，则将非倍数的计数加 1。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find Nth number not a
// multiple of A in the range [L, R]
void count_no (int A, int N, int L, int R)
{
     
    // To store the count
    int count = 0;
    int i = 0;
 
    // To check all the nos in range
    for(i = L; i < R + 1; i++)
    {
    if (i % A != 0)
        count += 1;
         
    if (count == N)
        break;
    }
    cout << i;
}
 
// Driver code
int main()
{
     
    // Given values of A, N, L, R
    int A = 3, N = 6, L = 4, R = 20;
     
    // Function Call
    count_no (A, N, L, R);
    return 0;
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
import java.util.*;
import java.io.*;
 
class GFG{
     
// Function to find Nth number not a
// multiple of A in the range [L, R]
static void count_no (int A, int N,
                      int L, int R)
{
 
    // To store the count
    int count = 0;
    int i = 0;
 
    // To check all the nos in range
    for(i = L; i < R + 1; i++)
    {
        if (i % A != 0)
            count += 1;
     
        if (count == N)
            break;
    }
    System.out.println(i);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given values of A, N, L, R
    int A = 3, N = 6, L = 4, R = 20;
 
    // Function call
    count_no (A, N, L, R);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to find Nth number not a
# multiple of A in the range [L, R]
def count_no (A, N, L, R):
 
    # To store the count
    count = 0
 
    # To check all the nos in range
    for i in range ( L, R + 1 ):
        if ( i % A != 0 ):
            count += 1
 
        if ( count == N ):
            break
    print ( i )
     
# Given values of A, N, L, R
A, N, L, R = 3, 6, 4, 20
 
# Function Call
count_no (A, N, L, R)

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find Nth number not a
// multiple of A in the range [L, R]
static void count_no (int A, int N,
                      int L, int R)
{
     
    // To store the count
    int count = 0;
    int i = 0;
 
    // To check all the nos in range
    for(i = L; i < R + 1; i++)
    {
        if (i % A != 0)
            count += 1;
     
        if (count == N)
            break;
    }
    Console.WriteLine(i);
}
 
// Driver code
public static void Main()
{
     
    // Given values of A, N, L, R
    int A = 3, N = 6, L = 4, R = 20;
 
    // Function call
    count_no (A, N, L, R);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// Javascript Program to implement
// the above approach
 
// Function to find Nth number not a
// multiple of A in the range [L, R]
function count_no (A, N, L, R)
{
  
    // To store the count
    let count = 0;
    let i = 0;
  
    // To check all the nos in range
    for(i = L; i < R + 1; i++)
    {
        if (i % A != 0)
            count += 1;
      
        if (count == N)
            break;
    }
   document.write(i);
}
 
// Driver Code
 
    // Given values of A, N, L, R
    let A = 3, N = 6, L = 4, R = 20;
  
    // Function call
    count_no (A, N, L, R);
     
    // This code is contributed by chinmoy1997pal.

输出：

11

时间复杂度： O(R – L)
辅助空间： O(1)

有效的方法：
关键的观察是在范围[1, A – 1]中有A – 1 个不能被A整除的数字。同样，在[A + 1, 2 * A – 1], [2 * A + 1, 3 * A – 1]等范围内有A – 1 个不能被A整除的数字。
在这个观察的帮助下，不能被A整除的^第N^个数字将是：

$N + \left \lfloor \frac{N - 1}{A - 1} \right \rfloor$

要找到范围 [ L, R ] 中的值，我们需要将原点从 ‘0’ 移到 ‘L – 1’，因此我们可以说该范围内不能被 A 整除的^第N^个数字将是：

$(L - 1) + \left \lfloor \frac{N - 1}{A - 1} \right \rfloor$

然而，有一个边缘情况，当( L – 1 ) + N + floor( ( N – 1 ) / ( A – 1 ) )本身的值是‘A’ 的倍数时，在这种情况下是第 N 个数字将 ：

$(L - 1) + \left \lfloor \frac{N - 1}{A - 1} \right \rfloor + 1$

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find Nth number
// not a multiple of A in range [L, R]
void countNo(int A, int N, int L, int R)
{
     
    // Calculate the Nth no
    int ans = L - 1 + N + floor((N - 1) /
                                (A - 1));
     
    // Check for the edge case
    if (ans % A == 0)
    {
        ans = ans + 1;
    }
    cout << ans << endl;
}
 
// Driver Code
int main()
{
     
    // Input parameters
    int A = 5, N = 10, L = 4, R = 20;
     
    // Function Call
    countNo(A, N, L, R);
     
    return 0;
}
 
// This code is contributed by avanitrachhadiya2155

Java

// Java program for the above approach
import java.io.*;
 
class GFG
{
 
  // Function to find Nth number
  // not a multiple of A in range [L, R]
  static void countNo(int A, int N, int L, int R)
  {
 
    // Calculate the Nth no
    int ans = L - 1 + N + (int)Math.floor((N - 1) / (A - 1));
 
    // Check for the edge case
    if (ans % A == 0)
    {
      ans = ans + 1;
    }
    System.out.println(ans);   
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Input parameters
    int A = 5, N = 10, L = 4, R = 20;
 
    // Function Call
    countNo(A, N, L, R);   
  }
}
 
//  This code is contributed by rag2127

蟒蛇3

# Python3 program for the above approach
import math
 
# Function to find Nth number
# not a multiple of A in range [L, R]
def countNo (A, N, L, R):
 
    # Calculate the Nth no
    ans = L - 1 + N \
          + math.floor( ( N - 1 ) / ( A - 1 ) )
     
    # Check for the edge case
    if ans % A == 0:
        ans = ans + 1;
    print(ans)
 
# Input parameters
A, N, L, R = 5, 10, 4, 20
 
# Function Call
countNo(A, N, L, R)

C＃

// C# program for the above approach
using System;
class GFG {
 
  // Function to find Nth number
  // not a multiple of A in range [L, R]
  static void countNo(int A, int N, int L, int R)
  {
 
    // Calculate the Nth no
    int ans = L - 1 + N + ((N - 1) / (A - 1));
 
    // Check for the edge case
    if (ans % A == 0)
    {
      ans = ans + 1;
    }
    Console.WriteLine(ans);
  }
 
  // Driver code
  static void Main()
  {
 
    // Input parameters
    int A = 5, N = 10, L = 4, R = 20;
 
    // Function Call
    countNo(A, N, L, R);
  }
}
 
// This code is contributed by divyesh072019.

Javascript

 
// Javascript program for
// the above approach
 
// Function to find Nth number
// not a multiple of A in range [L, R]
function countNo(A, N, L, R)
{
 
    // Calculate the Nth no
    var ans = L - 1 + N +
      Math.floor((N - 1) / (A - 1));
     
    // Check for the edge case
    if (ans % A == 0)
    {
        ans = ans + 1;
    }
    document.write(ans);   
}
 
// Driver code
 
// Input parameters
var A = 5, N = 10, L = 4, R = 20;
 
// Function Call
countNo(A, N, L, R); 
 
// This code is contributed by Khushboogoyal499
    

输出：

16

时间复杂度： O(1)
辅助空间： O(1)