第n个数字，其数字为{0，1，2，3，4，5}

给定数字n，我们必须找到第n个数字，使得它的数字仅包含0、1、2、3、4或5。

例子 ：

Input: n = 6
Output: 5

Input:  n = 10
Output: 13

我们首先将0、1、2、3、4、5存储在一个数组中。我们可以看到下一个数字将是10、11、12、13、14、15，之后的数字将是20、21、23、24、25，依此类推。我们可以看到不断重复的模式。我们保存计算的结果，并将其用于进一步的计算。
接下来的6个数字是-
1 * 10 + 0 = 10
1 * 10 + 1 = 11
1 * 10 + 2 = 12
1 * 10 + 3 = 13
1 * 10 + 4 = 14
1 * 10 + 5 = 15
在那之后，接下来的6个数字将是-
2 * 10 + 0 = 20
2 * 10 + 1 = 21
2 * 10 + 2 = 22
2 * 10 + 3 = 23
2 * 10 + 4 = 24
2 * 10 + 5 = 25
我们使用这种模式来找到第n个数字。下面是完整的算法。

1) push 0 to 5 in ans vector
2) for i=0 to n/6
     for j=0 to 6  
          // this will be the case when first 
          // digit will be zero 
          if (ans[i]*10! = 0) 
              ans.push_back(ans[i]*10 + ans[j])
3) print ans[n-1]

CPP

// C++ program to find n-th number with digits
// in {0, 1, 2, 3, 4, 5}
#include 
using namespace std;
 
// Returns the N-th number with given digits
int findNth(int n)
{
    // vector to store results
    vector ans;
 
    // push first 6 numbers in the answer
    for (int i = 0; i < 6; i++)
        ans.push_back(i);
 
    // calculate further results
    for (int i = 0; i <= n / 6; i++)
        for (int j = 0; j < 6; j++)
            if ((ans[i] * 10) != 0)
                ans.push_back(ans[i]
                              * 10 + ans[j]);
 
    return ans[n - 1];
}
 
// Driver code
int main()
{
    int n = 10;
    cout << findNth(n);
    return 0;
}

Java

// Java program to find n-th number with digits
// in {0, 1, 2, 3, 4, 5}
import java.io.*;
import java.util.*;
 
class GFG
{
 
  // Returns the N-th number with given digits
  public static int findNth(int n)
  {
    // vector to store results
    ArrayList ans = new ArrayList();
 
    // push first 6 numbers in the answer
    for (int i = 0; i < 6; i++)
      ans.add(i);
 
    // calculate further results
    for (int i = 0; i <= n / 6; i++)
      for (int j = 0; j < 6; j++)
        if ((ans.get(i) * 10) != 0)
          ans.add(ans.get(i) * 10 + ans.get(j));
    return ans.get(n - 1);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int n = 10;
    int ans = findNth(n);
    System.out.println(ans);
  }
}
 
// This code is contributed by RohitOberoi.

---

输出

13

高效方法：

算法 ：

1. 首先将数字n转换为以6为底的数字。
2. 同时将转换后的值存储在数组中。
3. 以相反的顺序打印该数组。

下面是上述算法的实现：

C++

// CPP code to find nth number
// with digits 0, 1, 2, 3, 4, 5
#include 
using namespace std;
 
#define max 100000
 
// function to convert num to base 6
int baseconversion(int arr[], int num, int base)
 
{
    int i = 0, rem, j;
 
    if (num == 0) {
        return 0;
    }
 
    while (num > 0) {
        rem = num % base;
 
        arr[i++] = rem;
 
        num /= base;
    }
 
    return i;
}
 
// Driver code
int main()
{
 
    // initialize an array to 0
    int arr[max] = { 0 };
 
    int n = 10;
 
    // function calling to convert
    // number n to base 6
    int size = baseconversion(arr, n - 1, 6);
 
    // if size is zero then return zero
    if (size == 0)
 
        cout << size;
 
    for (int i = size - 1; i >= 0; i--) {
 
        cout << arr[i];
    }
 
    return 0;
}
 
// Code is contributed by Anivesh Tiwari.

Java

// Java code to find nth number
// with digits 0, 1, 2, 3, 4, 5
class GFG {
     
    static final int max = 100000;
     
    // function to convert num to base 6
    static int baseconversion(int arr[],
                          int num, int base)
    {
        int i = 0, rem, j;
     
        if (num == 0) {
            return 0;
        }
     
        while (num > 0) {
             
            rem = num % base;
            arr[i++] = rem;
            num /= base;
        }
     
        return i;
    }
     
    // Driver code
    public static void main (String[] args)
    {
         
        // initialize an array to 0
        int arr[] = new int[max];
     
        int n = 10;
     
        // function calling to convert
        // number n to base 6
        int size = baseconversion(arr, n - 1, 6);
     
        // if size is zero then return zero
        if (size == 0)
            System.out.print(size);
     
        for (int i = size - 1; i >= 0; i--) {
            System.out.print(arr[i]);
        }
    }
}
 
// This code is contributed by Anant Agarwal.

C＃

// C# code to find nth number
// with digits 0, 1, 2, 3, 4, 5
using System;
 
class GFG {
     
    static int max = 100000;
     
    // function to convert num to base 6
    static int baseconversion(int []arr,
                              int num, int bas)
    {
        int i = 0, rem;
     
        if (num == 0) {
            return 0;
        }
     
        while (num > 0) {
             
            rem = num % bas;
            arr[i++] = rem;
            num /= bas;
        }
     
        return i;
    }
     
    // Driver code
    public static void Main ()
    {
        // initialize an array to 0
        int []arr = new int[max];
     
        int n = 10;
     
        // function calling to convert
        // number n to base 6
        int size = baseconversion(arr, n - 1, 6);
     
        // if size is zero then return zero
        if (size == 0)
            Console.Write(size);
     
        for (int i = size - 1; i >= 0; i--) {
            Console.Write(arr[i]);
        }
    }
}
 
// This code is contributed by nitin mittal

输出：

13

另一种有效的方法：

算法：

1. 将数字N减1。
2. 2.将数字N转换为以6为底的数字。

下面是上述算法的实现：

C++

// CPP code to find nth number
// with digits 0, 1, 2, 3, 4, 5
 
#include 
using namespace std;
 
int ans(int n){
  // If the Number is less than 6 return the number as it is.
  if(n < 6){
    return n;
  }
  //Call the function again and again the get the desired result.
  //And convert the number to base 6.
  return n%6 + 10*(ans(n/6));
}
 
int getSpecialNumber(int N)
{
  //Decrease the Number by 1 and Call ans function
  // to convert N to base 6
  return ans(--N);
}
 
/*Example:-
Input: N = 17
Output: 24
 
Explaination:-
decrease 17 by 1
N = 16
call ans() on 16
 
ans():
    16%6 + 10*(ans(16/6))
        since 16/6 = 2 it is less than 6 the ans returns value as it is.
    4 + 10*(2)
    = 24
 
hence answer is 24.*/
 
int main()
{
  int N = 17;
  int answer = getSpecialNumber(N);
  cout<

输出

24

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