使用给定的N个字符，长度为M的非回文字符串的计数

给定两个正整数N和M ，任务是使用给定的N个不同字符来计算长度为M的非回文字符串的数量。
注意：每个不同的字符可以多次使用。

例子：

Input: N = 3, M = 2 
Output: 6 
Explanation: 
Since only 3 characters are given, those 3 characters can be used to form 3² different strings. Out of these, only 3 strings are palindromic. Hence, the remaining 6 strings are palindromic.

Input: N = 26, M = 5 
Output: 11863800 
 

方法：
请按照以下步骤解决问题：

• 使用给定的N个字符，长度为M的字符串总数为N ^M。
• 要使字符串成为回文，前半部分和后半部分应相等。对于M的偶数值，我们需要选择从给定的N个字符仅M / 2个字符。为奇数值，我们需要选择M /从给定的N个字符2 + 1字符。由于允许重复，因此长度为M的回文字符串总数为N ^{(M / 2 + M％2)} 。
• 非回文字符串的所需计数由以下公式给出：

NM - N(M/2 + M%2)

下面是上述方法的实现：

C++

// C++ Program to count
// non-palindromic strings
// of length M using N
// distinct characters
#include 
using namespace std;
 
 
// Iterative Function to calculate
// base^pow in O(log y)
unsigned long long power(
    unsigned long long base,
    unsigned long long pow)
{
    unsigned long long res = 1;
    while (pow > 0) {
        if (pow & 1)
            res = (res * base);
        base = (base * base);
        pow >>= 1;
    }
    return res;
}
 
// Function to return the
// count of non palindromic strings
unsigned long long countNonPalindromicString(
    unsigned long long n,
    unsigned long long m)
{
    // Count of strings using n
    // characters with
    // repetitions allowed
    unsigned long long total
        = power(n, m);
     
    // Count of palindromic strings
    unsigned long long palindrome
        = power(n, m / 2 + m % 2);
     
    // Count of non-palindromic strings
    unsigned long long count
        = total - palindrome;
 
    return  count;
}
int main()
{
 
    int n = 3, m = 5;
    cout<< countNonPalindromicString(n, m);
    return 0;
}

Java

// Java program to count non-palindromic
// strings of length M using N distinct
// characters
import java.util.*;
 
class GFG{
 
// Iterative Function to calculate
// base^pow in O(log y)
static long power(long base, long pow)
{
    long res = 1;
    while (pow > 0)
    {
        if ((pow & 1) == 1)
            res = (res * base);
        base = (base * base);
        pow >>= 1;
    }
    return res;
}
 
// Function to return the
// count of non palindromic strings
static long countNonPalindromicString(long n,
                                      long m)
{
     
    // Count of strings using n
    // characters with
    // repetitions allowed
    long total = power(n, m);
     
    // Count of palindromic strings
    long palindrome = power(n, m / 2 + m % 2);
     
    // Count of non-palindromic strings
    long count = total - palindrome;
 
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3, m = 5;
     
    System.out.println(
        countNonPalindromicString(n, m));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to count non-palindromic strings
# of length M using N distinct characters
 
# Iterative Function to calculate
# base^pow in O(log y)
def power(base, pwr):
 
    res = 1
    while(pwr > 0):
        if(pwr & 1):
            res = res * base
        base = base * base
        pwr >>= 1
 
    return res
 
# Function to return the count
# of non palindromic strings
def countNonPalindromicString(n, m):
 
    # Count of strings using n
    # characters with
    # repetitions allowed
    total = power(n, m)
 
    # Count of palindromic strings
    palindrome = power(n, m // 2 + m % 2)
 
    # Count of non-palindromic strings
    count = total - palindrome
 
    return count
 
# Driver code
if __name__ == '__main__':
 
    n = 3
    m = 5
     
    print(countNonPalindromicString(n, m))
 
# This code is contributed by Shivam Singh

C#

// C# program to count non-palindromic
// strings of length M using N distinct
// characters
using System;
 
class GFG{
 
// Iterative Function to calculate
// base^pow in O(log y)
static long power(long Base, long pow)
{
    long res = 1;
    while (pow > 0)
    {
        if ((pow & 1) == 1)
            res = (res * Base);
             
        Base = (Base * Base);
        pow >>= 1;
    }
    return res;
}
 
// Function to return the
// count of non palindromic strings
static long countNonPalindromicString(long n,
                                      long m)
{
     
    // Count of strings using n
    // characters with
    // repetitions allowed
    long total = power(n, m);
     
    // Count of palindromic strings
    long palindrome = power(n, m / 2 + m % 2);
     
    // Count of non-palindromic strings
    long count = total - palindrome;
 
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3, m = 5;
     
    Console.WriteLine(
        countNonPalindromicString(n, m));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript program to count non-palindromic
// strings of length M using N distinct
// characters
 
// Iterative Function to calculate
// base^pow in O(log y)
function power(base, pow)
{
    let res = 1;
    while (pow > 0)
    {
        if ((pow & 1) == 1)
            res = (res * base);
             
        base = (base * base);
        pow >>= 1;
    }
    return res;
}
 
// Function to return the
// count of non palindromic strings
function countNonPalindromicString(n, m)
{
     
    // Count of strings using n
    // characters with
    // repetitions allowed
    let total = power(n, m);
     
    // Count of palindromic strings
    let palindrome = power(n, m / 2 + m % 2);
     
    // Count of non-palindromic strings
    let count = total - palindrome;
 
    return count;
}
     
// Driver Code
let n = 3, m = 5;
 
document.write(countNonPalindromicString(n, m));
 
// This code is contributed by sanjoy_62
 

输出：

216

时间复杂度： O(log(N))