在只允许使用2位数字(4和7)的序列中查找第n个元素|集合2(log(n)方法)

考虑一系列仅由数字4和7组成的数字。该系列中的前几个数字是4、7、44、47、74、77、444，..等。给定数字n，我们需要找到第n个系列中的数字。
例子：

Input : n = 2
Output : 7

Input : n = 3
Output : 44

Input  : n = 5
Output : 74

Input  : n = 6
Output : 77

我们在下面的文章中讨论了O(n)解决方案。
在只允许两位数字(4和7)的序列中查找第n个元素
在这篇文章中，将讨论O(log n)解决方案，该解决方案基于以下数字模式。可以看到数字

""
              /      \
            4         7
          /   \     /   \ 
        44    47   74    77
       / \   / \   / \  / \

这个想法是从头开始填写所需的数字。我们知道可以观察到，如果n为奇数，则最后一位为4，如果n为偶数，则最后一位为7。填完最后一位数字后，我们移至树中的父节点。如果n为奇数，则父节点对应于(n-1 / 2。其他父节点对应于(n-2)/ 2。

C++

// C++ program to find n-th number containing
// only 4 and 7.
#include
using namespace std;
 
string findNthNo(int n)
{
    string res = "";
    while (n >= 1)
    {
        // If n is odd, append 4 and
        // move to parent
        if (n & 1)
        {
            res = res + "4";
            n = (n-1)/2;       
        }
 
        // If n is even, append 7 and
        // move to parent
        else
        {
            res = res + "7";
            n = (n-2)/2;     
        }
    }
 
   // Reverse res and return.
   reverse(res.begin(), res.end());
   return res;
}
 
// Driver code
int main()
{
    int n = 13;
    cout << findNthNo(n);
    return 0;
}

Java

// java program to find n-th number
// containing only 4 and 7.
public class GFG {
     
    static String findNthNo(int n)
    {
        String res = "";
        while (n >= 1)
        {
             
            // If n is odd, append
            // 4 and move to parent
            if ((n & 1) == 1)
            {
                res = res + "4";
                n = (n - 1) / 2;    
            }
     
            // If n is even, append
            // 7 and move to parent
            else
            {
                res = res + "7";
                n = (n - 2) / 2;    
            }
        }
     
        // Reverse res and return.
        StringBuilder sb =
            new StringBuilder(res);
        sb.reverse();
        return new String(sb);
    }
     
    // Driver code
    public static void main(String args[])
    {
        int n = 13;
     
        System.out.print( findNthNo(n) );
    }
}
 
// This code is contributed by Sam007

Python3

# Python3 program to find
# n-th number containing
# only 4 and 7.
def reverse(s):
    if len(s) == 0:
        return s
    else:
        return reverse(s[1:]) + s[0]
         
def findNthNo(n):
    res = "";
    while (n >= 1):
         
        # If n is odd, append
        # 4 and move to parent
        if (n & 1):
            res = res + "4";
            n = (int)((n - 1) / 2);
             
            # If n is even, append7
            # and move to parent
        else:
            res = res + "7";
            n = (int)((n - 2) / 2);
             
    # Reverse res
    # and return.
    return reverse(res);
 
# Driver code
n = 13;
print(findNthNo(n));
 
# This code is contributed
# by mits

C#

// C# program to find n-th number
// containing only 4 and 7.
using System;
class GFG {
 
static string findNthNo(int n)
{
    string res = "";
    while (n >= 1)
    {
         
        // If n is odd, append 4 and
        // move to parent
        if ((n & 1) == 1)
        {
            res = res + "4";
            n = (n - 1) / 2;    
        }
 
        // If n is even, append 7 and
        // move to parent
        else
        {
            res = res + "7";
            n = (n - 2) / 2;    
        }
    }
 
    // Reverse res and return.
    char[] arr = res.ToCharArray();
    Array.Reverse(arr);
    return new string(arr);
 
}
 
// Driver Code
public static void Main()
{
        int n = 13;
        Console.Write( findNthNo(n) );
}
}
 
// This code is contributed by Sam007

PHP

= 1)
    {
        // If n is odd, append
        // 4 and move to parent
        if ($n & 1)
        {
            $res = $res . "4";
            $n = (int)(($n - 1) / 2);
        }
 
        // If n is even, append
        // 7 and move to parent
        else
        {
            $res = $res . "7";
            $n = (int)(($n - 2) / 2);
        }
    }
     
// Reverse res
// and return.
return strrev($res);
}
 
// Driver code
$n = 13;
echo findNthNo($n);
 
// This code is contributed
// by mits
?>

Javascript

 
// javascript program to find n-th number
// containing only 4 and 7.
     
function findNthNo(n)
{
res = "";
while (n >= 1)
{
 
// If n is odd, append
// 4 and move to parent
if ((n & 1) == 1)
{
res = res + "4";
n = (n - 1) / 2;    
}
 
// If n is even, append
// 7 and move to parent
else
{
res = res + "7";
n = parseInt((n - 2) / 2);    
}
}
 
// Reverse res and return.
return res.split("").reverse().join("");
}
 
// Driver code
var n = 13;
 
document.write( findNthNo(n) );
 
// This code is contributed by 29AjayKumar
 

输出：

774