由至少一个 1 组成的给定长度的二进制字符串的计数

给定一个整数N ，任务是打印长度为 N 且至少为“1”的二进制字符串的数量。
例子：

Input: 2 
Output: 3 
Explanation: 
“01”, “10” and “11” are the possible strings
Input: 3 
Output: 7 
Expalnation: 
“001”, “011”, “010”, “100”, “101”, “110” and “111” are the possible strings 
 

方法：
我们可以观察到：

Only one string of length N does not contain any 1, the one filled with only 0’s. 
Since 2^N strings are possible of length N, the required answer is 2^N – 1. 
 

请按照以下步骤解决问题：

• 初始化 X = 1。
• 通过对 X 执行按位左移 N-1 次，最多计算2 ^N。
• 最后，打印 X – 1 作为所需答案。

下面是我们方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to return
// the count of strings
long count_strings(long n)
{
    int x = 1;
 
    // Calculate pow(2, n)
    for (int i = 1; i < n; i++) {
        x = (1 << x);
    }
 
    // Return pow(2, n) - 1
    return x - 1;
}
 
// Driver Code
int main()
{
    long n = 3;
 
    cout << count_strings(n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to return
// the count of Strings
static long count_Strings(long n)
{
    int x = 1;
 
    // Calculate Math.pow(2, n)
    for(int i = 1; i < n; i++)
    {
       x = (1 << x);
    }
 
    // Return Math.pow(2, n) - 1
    return x - 1;
}
 
// Driver Code
public static void main(String[] args)
{
    long n = 3;
 
    System.out.print(count_Strings(n));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to return
# the count of Strings
def count_Strings(n):
     
    x = 1;
 
    # Calculate pow(2, n)
    for i in range(1, n):
        x = (1 << x);
 
    # Return pow(2, n) - 1
    return x - 1;
 
# Driver Code
if __name__ == '__main__':
     
    n = 3;
 
    print(count_Strings(n));
 
# This code is contributed by Princi Singh

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return
// the count of Strings
static long count_Strings(long n)
{
    int x = 1;
 
    // Calculate Math.Pow(2, n)
    for(int i = 1; i < n; i++)
    {
       x = (1 << x);
    }
     
    // Return Math.Pow(2, n) - 1
    return x - 1;
}
 
// Driver Code
public static void Main(String[] args)
{
    long n = 3;
 
    Console.Write(count_Strings(n));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript program to implement
// the above approach
 
    // Function to return
    // the count of Strings
    function count_Strings(n)
    {
        var x = 1;
 
        // Calculate Math.pow(2, n)
        for (i = 1; i < n; i++) {
            x = (1 << x);
        }
 
        // Return Math.pow(2, n) - 1
        return x - 1;
    }
 
    // Driver Code
     
        var n = 3;
 
        document.write(count_Strings(n));
 
// This code is contributed by todaysgaurav
 

输出：

3

时间复杂度： O(N)
辅助空间： O(1)

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