在只允许两位数字(4和7)的序列中查找第n个元素

考虑仅由数字4和7组成的一系列数字。该系列中的前几个数字是4、7、44、47、74、44744等。等。给定一个数字n，我们需要在其中找到第n个数字。该系列。
例子：

Input : n = 2
Output : 7

Input : n = 3
Output : 44

Input  : n = 5
Output : 74

Input  : n = 6
Output : 77

这个想法是基于这样的事实，即最后一位的值是交替排列的。例如，如果第i个数字的最后一位为4，则第(i-1)和第(i + 1)个数字的最后一位必须为7。
我们创建一个大小为(n + 1)的数组，然后将其推入4和7(这两个始终是序列的前两个元素)。有关更多元素，请检查
1)如果我很奇怪，
arr [i] = arr [i / 2] * 10 + 4;
2)如果是偶数，
arr [i] = arr [(i / 2)-1] * 10 + 7;
最后返回arr [n]。

C++

// C++ program to find n-th number in a series
// made of digits 4 and 7
#include 
using namespace std;
 
// Return n-th number in series made of 4 and 7
int printNthElement(int n)
{
    // create an array of size (n+1)
    int arr[n+1];
    arr[1] = 4;
    arr[2] = 7;
 
    for (int i=3; i<=n; i++)
    {
        // If i is odd
        if (i%2 != 0)
            arr[i] = arr[i/2]*10 + 4;
        else
            arr[i] = arr[(i/2)-1]*10 + 7;
    }
    return arr[n];
}
 
// Driver code
int main()
{
    int n = 6;
    cout << printNthElement(n);
    return 0;
}

Java

// Java program to find n-th number in a series
// made of digits 4 and 7
 
class FindNth
{
    // Return n-th number in series made of 4 and 7
    static int printNthElement(int n)
    {
        // create an array of size (n+1)
        int arr[] = new int[n+1];
        arr[1] = 4;
        arr[2] = 7;
      
        for (int i=3; i<=n; i++)
        {
            // If i is odd
            if (i%2 != 0)
                arr[i] = arr[i/2]*10 + 4;
            else
                arr[i] = arr[(i/2)-1]*10 + 7;
        }
        return arr[n];
    }   
     
    // main function
    public static void main (String[] args)
    {
        int n = 6;
        System.out.println(printNthElement(n));
    }
}

Python3

# Python3 program to find n-th number
# in a series made of digits 4 and 7
 
# Return n-th number in series made
# of 4 and 7
def printNthElement(n) :
     
    # create an array of size (n + 1)
    arr =[0] * (n + 1);
    arr[1] = 4
    arr[2] = 7
 
    for i in range(3, n + 1) :
        # If i is odd
        if (i % 2 != 0) :
            arr[i] = arr[i // 2] * 10 + 4
        else :
            arr[i] = arr[(i // 2) - 1] * 10 + 7
     
    return arr[n]
     
# Driver code
n = 6
print(printNthElement(n))
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to find n-th number in a series
// made of digits 4 and 7
using System;
 
class GFG
{
    // Return n-th number in series made of 4 and 7
    static int printNthElement(int n)
    {
        // create an array of size (n+1)
        int []arr = new int[n+1];
        arr[1] = 4;
        arr[2] = 7;
     
        for (int i = 3; i <= n; i++)
        {
            // If i is odd
            if (i % 2 != 0)
                arr[i] = arr[i / 2] * 10 + 4;
            else
                arr[i] = arr[(i / 2) - 1] * 10 + 7;
        }
        return arr[n];
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 6;
        Console.Write(printNthElement(n));
    }
}
 
// This code is contributed by vt_m.

Javascript

// javascript program to find n-th number in a series
// made of digits 4 and 7
 
 
    // Return n-th number in series made of 4 and 7
    function printNthElement(n) {
        // create an array of size (n+1)
        var arr = Array(n + 1).fill(0);
        arr[1] = 4;
        arr[2] = 7;
 
        for (var i = 3; i <= n; i++) {
            // If i is odd
            if (i % 2 != 0)
                arr[i] = arr[i / 2] * 10 + 4;
            else
                arr[i] = arr[(i / 2) - 1] * 10 + 7;
        }
        return arr[n];
    }
 
    // main function
     
        var n = 6;
        document.write(printNthElement(n));
 
// This code is contributed by Princi Singh

输出：

77