给定一个整数N ,任务是查找不带前导零的N位二进制数的计数。
例子:
Input: N = 2
Output: 2
10 and 11 are the only possible binary numbers.
Input: N = 4
Output: 8
方法:由于数字不能有前导零,因此必须将最左边的位设置为1 。现在,对于其余的N – 1位,有两种选择,它们可以设置为0或1 。因此,可能的数量为2 N – 1 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count
// of possible numbers
int count(int n)
{
return pow(2, n - 1);
}
// Driver code
int main()
{
int n = 4;
cout << count(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the count
// of possible numbers
static int count(int n)
{
return (int)Math.pow(2, n - 1);
}
// Driver code
public static void main (String[] args)
{
int n = 4;
System.out.println(count(n));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the approach
# Function to return the count
# of possible numbers
def count(n):
return pow(2, n - 1)
# Driver code
n = 4
print(count(n))
# This code is contributed by mohit kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of possible numbers
static int count(int n)
{
return (int)Math.Pow(2, n - 1);
}
// Driver code
public static void Main (String[] args)
{
int n = 4;
Console.WriteLine(count(n));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
8