计算长度为N且仅包含0和1的二进制字符串的数量

给定一个整数N ，任务是计算长度N为仅0和1的二进制字符串的数量。
注意：由于计数可能非常大，请以10 ^ 9 + 7为模返回答案。

例子：

Input: 2
Output: 4
Explanation: The numbers are 00, 01, 11, 10. Hence the count is 4.

Input: 3
Output: 8
Explanation: The numbers are 000, 001, 011, 010, 111, 101, 110, 100. Hence the count is 8.

方法：使用置换和组合可以轻松解决该问题。在字符串的每个位置只能有两种可能性，即0或1。因此，长度为N的字符串中的0和1的置换总数由2 * 2 * 2 * …(N次)给出，即2 ^ N。答案可能非常大，因此返回以10 ^ 9 + 7为模的值。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long
#define mod (ll)(1e9 + 7)
  
// Iterative Function to calculate (x^y)%p in O(log y)
ll power(ll x, ll y, ll p)
{
    ll res = 1; // Initialize result
  
    x = x % p; // Update x if it is more than or
    // equal to p
  
    while (y > 0) {
  
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function to count the number of binary
// strings of length N having only 0's and 1's
ll findCount(ll N)
{
    int count = power(2, N, mod);
    return count;
}
  
// Driver code
int main()
{
    ll N = 25;
  
    cout << findCount(N);
  
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
  
class GFG 
{
  
static int mod = (int) (1e9 + 7);
  
// Iterative Function to calculate (x^y)%p in O(log y)
static int power(int x, int y, int p)
{
    int res = 1; // Initialize result
  
    x = x % p; // Update x if it is more than or
    // equal to p
  
    while (y > 0) 
    {
  
        // If y is odd, multiply x with result
        if ((y & 1)==1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function to count the number of binary
// strings of length N having only 0's and 1's
static int findCount(int N)
{
    int count = power(2, N, mod);
    return count;
}
  
// Driver code
public static void main(String[] args)
{
        int N = 25;
        System.out.println(findCount(N));
}
} 
  
/* This code contributed by PrinciRaj1992 */

Python3

# Python 3 implementation of the approach
mod = 1000000007
  
# Iterative Function to calculate (x^y)%p in O(log y)
def power(x, y, p):
    res = 1 # Initialize result
  
    x = x % p # Update x if it is more than or
              # equal to p
  
    while (y > 0):
          
        # If y is odd, multiply x with result
        if (y & 1):
            res = (res * x) % p
  
        # y must be even now
        y = y >> 1 # y = y/2
        x = (x * x) % p
  
    return res
  
# Function to count the number of binary
# strings of length N having only 0's and 1's
def findCount(N):
    count = power(2, N, mod)
    return count
  
# Driver code
if __name__ == '__main__':
    N = 25
    print(findCount(N))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach 
using System;
  
class GFG 
{ 
  
    static int mod = (int) (1e9 + 7); 
      
    // Iterative Function to calculate (x^y)%p in O(log y) 
    static int power(int x, int y, int p) 
    { 
        int res = 1; // Initialize result 
      
        x = x % p; // Update x if it is more than or 
        // equal to p 
      
        while (y > 0) 
        { 
      
            // If y is odd, multiply x with result 
            if ((y & 1) == 1) 
                res = (res * x) % p; 
      
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        } 
        return res; 
    } 
      
    // Function to count the number of binary 
    // strings of length N having only 0's and 1's 
    static int findCount(int N) 
    { 
        int count = power(2, N, mod); 
        return count; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
            int N = 25; 
            Console.WriteLine(findCount(N)); 
    } 
} 
  
// This code is contributed by Ryuga

PHP

 0)
    {
  
        // If y is odd, multiply x with result
        if ($y & 1)
            $res = ($res * $x) % $p;
  
        // y must be even now
        $y = $y >> 1; // y = y/2
        $x = ($x * $x) % $p;
    }
    return $res;
}
  
// Function to count the number of binary
// strings of length N having only 0's and 1's
function findCount($N)
{
    $count = power(2, $N);
    return $count;
}
  
// Driver code
$N = 25;
  
echo findCount($N);
      
// This code is contributed by Rajput-Ji
?>

输出：

33554432