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📜  无回文子字符串的长度为N的字符串数

📅  最后修改于: 2021-06-25 13:04:27             🧑  作者: Mango

给定两个正整数N,M 。任务是在大小为M的字母集下找到长度为N的字符串数,以使回文大小不大于1。
例子:

Input : N = 2, M = 3
Output : 6
In this case, set of alphabet are 3, say {A, B, C}
All possible string of length 2, using 3 letters are:
{AA, AB, AC, BA, BB, BC, CA, CB, CC}
Out of these {AA, BB, CC} contain palindromic substring, 
so our answer will be 
8 - 2 = 6.

Input : N = 2, M = 2
Output : 2
Out of {AA, BB, AB, BA}, only {AB, BA} contain 
non-palindromic substrings.

首先,观察,一个字符串不包含任何回文子,如果该字符串不具有长度2和3的任何回文串,因为更长的长度的所有的回文字符串包含的2或长度中的至少一个回文子串3,基本上在中心。
因此,以下是正确的:

  • 有M种方法选择字符串的第一个符号。
  • 然后,有(M – 1)种方法选择字符串的第二个符号。基本上,它不应该等于第一个。
  • 然后,有(M – 2)种方法来选择下一个符号。基本上,它不应与之前的符号(不相等)重合。

知道了这一点,我们可以通过以下方式评估答案:

  • 如果N = 1,则答案将是M。
  • 如果N = 2,则答案为M *(M – 1)。
  • 如果N> = 3,则M *(M – 1)*(M – 2) N-2

下面是上述想法的实现:

C++
// CPP program to count number of strings of
// size m such that no substring is palindrome.
#include 
using namespace std;
 
// Return the count of strings with
// no palindromic substring.
int numofstring(int n, int m)
{   
    if (n == 1)
        return m;
 
    if (n == 2)
        return m * (m - 1);
 
    return m * (m - 1) * pow(m - 2, n - 2);
}
 
// Driven Program
int main()
{   
    int n = 2, m = 3;
    cout << numofstring(n, m) << endl;
    return 0;
}


Java
// Java program to count number of strings of
// size m such that no substring is palindrome.
import java.io.*;
 
class GFG {
     
    // Return the count of strings with
    // no palindromic substring.
    static int numofstring(int n, int m)
    {
        if (n == 1)
            return m;
     
        if (n == 2)
            return m * (m - 1);
     
        return m * (m - 1) * (int)Math.pow(m - 2, n - 2);
    }
     
    // Driven Program
    public static void main (String[] args)
    {
        int n = 2, m = 3;
        System.out.println(numofstring(n, m));
    }
}
 
// This code is contributed by ajit.


Python3
# Python3 program to count number of strings of
# size m such that no substring is palindrome
 
# Return the count of strings with
# no palindromic substring.
def numofstring(n, m):
    if n == 1:
        return m
 
    if n == 2:
        return m * (m - 1)
 
    return m * (m - 1) * pow(m - 2, n - 2)
 
# Driven Program
n = 2
m = 3
print (numofstring(n, m))
 
# This code is contributed
# by Shreyanshi Arun.


C#
// C# program to count number of strings of
// size m such that no substring is palindrome.
using System;
 
class GFG {
     
    // Return the count of strings with
    // no palindromic substring.
    static int numofstring(int n, int m)
    {
        if (n == 1)
            return m;
     
        if (n == 2)
            return m * (m - 1);
     
        return m * (m - 1) * (int)Math.Pow(m - 2,
                                           n - 2);
    }
     
    // Driver Code
    public static void Main ()
    {
        int n = 2, m = 3;
        Console.Write(numofstring(n, m));
    }
}
 
// This code is contributed by Nitin Mittal.


PHP


Javascript


输出

6