由至少一个元音和一个辅音组成的N个大小的字符串的数量

给定一个代表字符串长度的整数N ，任务是计算长度为N的可能字符串的数量，该长度仅由一个元音和一个辅音组成。
注意：由于输出可以以模1000000007为大字体
例子：

Input: N = 2 
Output: 210 
Explanation: 
There are 5 vowels and 21 consonants in English alphabets. 
So for vowel ‘a’ we can have 42 strings of the form ‘ab’, ‘ba’, ‘ac’, ‘ca’, ‘ad’, ‘da’ and so on. 
For the other 4 vowels, the same process repeats, and we get a total of 210 such strings.
Input: N = 3 
Output: 8190 

方法：
为了解决上述问题，我们需要忽略仅包含元音(允许至少一个辅音)和仅包含辅音(允许至少一个元音)的字符串。因此，所需的答案是：

All N length strings possible – (N length strings consisting of only vowels + N length strings consisting of only consonants) = 26 ^N – (5 ^N + 21 ^N) 
 

下面是上述方法的实现：

C++

// C++ program to count all
// possible strings of length N
// consisting of atleast one
// vowel and one consonant
#include 
using namespace std;
 
const unsigned long long mod = 1e9 + 7;
 
// Function to return base^exponent
unsigned long long expo(
    unsigned long long base,
    unsigned long long exponent)
{
 
    unsigned long long ans = 1;
 
    while (exponent != 0) {
        if ((exponent & 1) == 1) {
            ans = ans * base;
            ans = ans % mod;
        }
 
        base = base * base;
        base %= mod;
        exponent >>= 1;
    }
 
    return ans % mod;
}
 
// Function to count all possible strings
unsigned long long findCount(
    unsigned long long N)
{
    // All possible strings of length N
    unsigned long long ans
        = (expo(26, N)
 
           // vowels only
           - expo(5, N)
 
           // consonants only
           - expo(21, N))
 
          % mod;
 
    ans += mod;
    ans %= mod;
 
    // Return the
    // final result
    return ans;
}
 
// Driver Program
int main()
{
    unsigned long long N = 3;
    cout << findCount(N);
 
    return 0;
}

Java

// Java program to count all
// possible Strings of length N
// consisting of atleast one
// vowel and one consonant
class GFG{
 
static int mod = (int) (1e9 + 7);
 
// Function to return base^exponent
static int expo(int base, int exponent)
{
    int ans = 1;
 
    while (exponent != 0)
    {
        if ((exponent & 1) == 1)
        {
            ans = ans * base;
            ans = ans % mod;
        }
        base = base * base;
        base %= mod;
        exponent >>= 1;
    }
    return ans % mod;
}
 
// Function to count all possible Strings
static int findCount(int N)
{
     
    // All possible Strings of length N
    int ans = (expo(26, N) -
               
               // Vowels only
               expo(5, N) -
                
               // Consonants only
               expo(21, N))% mod;
    ans += mod;
    ans %= mod;
 
    // Return the
    // final result
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
    System.out.print(findCount(N));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to count all
# possible strings of length N
# consisting of atleast one
# vowel and one consonant
mod = 1e9 + 7
 
# Function to return base^exponent
def expo(base, exponent):
    ans = 1
    while (exponent != 0):
        if ((exponent & 1) == 1):
            ans = ans * base
            ans = ans % mod
 
        base = base * base
        base %= mod
        exponent >>= 1
 
    return ans % mod
 
# Function to count all
# possible strings
def findCount(N):
 
    # All possible strings
    # of length N
    ans = ((expo(26, N) -
             
            # vowels only
            expo(5, N) -
 
            # consonants only
            expo(21, N)) %
            mod)
 
    ans += mod
    ans %= mod
 
    # Return the
    # final result
    return ans
 
# Driver Program
if __name__ == "__main__":
    N = 3
    print (int(findCount(N)))
 
# This code is contributed by Chitranayal

C#

// C# program to count all possible Strings
// of length N consisting of atleast one
// vowel and one consonant
using System;
 
class GFG{
 
static int mod = (int)(1e9 + 7);
 
// Function to return base^exponent
static int expo(int Base, int exponent)
{
    int ans = 1;
 
    while (exponent != 0)
    {
        if ((exponent & 1) == 1)
        {
            ans = ans * Base;
            ans = ans % mod;
        }
        Base = Base * Base;
        Base %= mod;
        exponent >>= 1;
    }
    return ans % mod;
}
 
// Function to count all possible Strings
static int findCount(int N)
{
     
    // All possible Strings of length N
    int ans = (expo(26, N) -
                
               // Vowels only
               expo(5, N) -
                
               // Consonants only
               expo(21, N)) % mod;
    ans += mod;
    ans %= mod;
 
    // Return the
    // readonly result
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 3;
     
    Console.Write(findCount(N));
}
}
 
// This code is contributed by Rajput-Ji

输出：

8190