给定整数N ，任务是计算长度为N的仅具有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。
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_GangwarC#
// 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 RyugaPHP
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
?>Javascript
输出：
33554432