📜  在序列中查找第N个数字,该数字不是给定数字的倍数

📅  最后修改于: 2021-05-08 17:55:13             🧑  作者: Mango

给定四个整数ANLR ,任务是在从LR的连续整数序列中找到第N个数字,该整数不是A的倍数。假定该序列至少包含N个不能被A整除的数字,并且整数A始终大于1。

例子:

天真的方法:天真的方法是循环遍历[L,R]范围以找到第N数字。这些步骤是:

  1. 将非整数和当前数字的计数初始化为0。
  2. [L,R]范围内迭代直到非整数的计数不等于N为止。
  3. 如果当前数字不能被A整除,则将非整数的计数增加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


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


Python3
# 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


输出:
11

时间复杂度: O(R – L)
辅助空间: O(1)

高效方法:
关键的观察结果是,在范围[1,A – 1]中,有A – 1个数字不能被A整除。同样,在范围[A + 1、2 * A – 1],[2 * A + 1、3 * A – 1]中,存在不能被A整除的A – 1个数字,依此类推。
借助此观察,无法被A整除的N个数将是:

要找到[ L,R ]范围内的值,我们需要将原点从’0’转换为’L – 1’,因此可以说,在该范围内被A整除的N数字将是:

但是,有一个极端的情况,当(L – 1)+ N + floor((N – 1)/(A – 1))的值本身是‘A’的倍数时,在这种情况下为第N个数将 :

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

Python3

# 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.

Java脚本


输出:
16

时间复杂度: O(1)
辅助空间: O(1)