在只允许数字4和7形成的序列中查找给定术语的位置

有一系列数字只有数字4和7，并且数字以升序排列。系列的前几个数字是4、7、44、47、74、77、444等。给定数字N ，任务是找到该数字在给定系列中的位置。
例子：

Input: N = 4 
Output: 1 
Explanation: 
The first number in the series is 4
Input: N = 777 
Output: 14 
Explanation: 
The 14th number in the series is 777 
 

方法：可以使用以下观察结果解决此问题：

• 以下是给定系列中观察到的模式。这些数字可以看成是：

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

• 如我们所见，模式的2的幂增加。因此，想法是从最低有效数字开始迭代数字的数字，并将数字的位置更新为：
  1. 如果当前数字= 7，则添加$2^{height + 1}$到该位置。
  2. 如果当前数字= 4，则添加$2^{height}$到该位置。
• 完成上述操作后，打印最终位置。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the position
// of the number N
void findPosition(int n)
{
    int i = 0;
 
    // To store the position of N
    int pos = 0;
 
    // Iterate through all digit of N
    while (n > 0) {
 
        // If current digit is 7
        if (n % 10 == 7) {
            pos = pos + pow(2, i + 1);
        }
 
        // If current digit is 4
        else {
            pos = pos + pow(2, i);
        }
 
        i++;
        n = n / 10;
    }
 
    // Print the final position
    cout << pos;
}
 
// Driver Code
int main()
{
    // Given number of the series
    int N = 777;
 
    // Function Call
    findPosition(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the position
// of the number N
static void findPosition(int n)
{
    int i = 0;
 
    // To store the position of N
    int pos = 0;
 
    // Iterate through all digit of N
    while (n > 0)
    {
 
        // If current digit is 7
        if (n % 10 == 7)
        {
            pos = pos + (int)Math.pow(2, i + 1);
        }
 
        // If current digit is 4
        else
        {
            pos = pos + (int)Math.pow(2, i);
        }
 
        i++;
        n = n / 10;
    }
 
    // Print the final position
    System.out.print(pos);
}
 
// Driver Code
public static void main(String[] args)
{
    // Given number of the series
    int N = 777;
 
    // Function Call
    findPosition(N);
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 program for the above approach
 
# Function to find the position
# of the number N
def findPosition(n):
     
    i = 0
 
    # To store the position of N
    pos = 0
 
    # Iterate through all digit of N
    while (n > 0):
 
        # If current digit is 7
        if (n % 10 == 7):
            pos = pos + pow(2, i + 1)
 
        # If current digit is 4
        else:
            pos = pos + pow(2, i)
             
        i += 1
        n = n // 10
 
    # Print the final position
    print(pos)
 
# Driver Code
if __name__ == '__main__':
     
    # Given number of the series
    N = 777
 
    # Function Call
    findPosition(N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find the position
// of the number N
static void findPosition(int n)
{
    int i = 0;
 
    // To store the position of N
    int pos = 0;
 
    // Iterate through all digit of N
    while (n > 0)
    {
 
        // If current digit is 7
        if (n % 10 == 7)
        {
            pos = pos + (int)Math.Pow(2, i + 1);
        }
 
        // If current digit is 4
        else
        {
            pos = pos + (int)Math.Pow(2, i);
        }
 
        i++;
        n = n / 10;
    }
 
    // Print the final position
    Console.Write(pos);
}
 
// Driver Code
public static void Main()
{
    // Given number of the series
    int N = 777;
 
    // Function Call
    findPosition(N);
}
}
 
// This code is contributed by Code_Mech

Javascript

// javascript program for the above approach
 
    // Function to find the position
    // of the number N
    function findPosition(n)
    {
        var i = 0;
 
        // To store the position of N
        var pos = 0;
 
        // Iterate through all digit of N
        while (n > 0) {
 
            // If current digit is 7
            if (n % 10 == 7) {
                pos = pos + parseInt( Math.pow(2, i + 1));
            }
 
            // If current digit is 4
            else {
                pos = pos + parseInt( Math.pow(2, i));
            }
 
            i++;
            n = parseInt(n / 10);
        }
 
        // Prvar the final position
        document.write(pos);
    }
 
    // Driver Code
     
    // Given number of the series
    var N = 777;
 
    // Function Call
    findPosition(N);
 
// This code is contributed by Princi Singh

输出：

14

时间复杂度： O(log ₁₀ N)