找出至少有一个字符重复的 M 个字符单词的个数
原文:https://www . geesforgeks . org/find-m 字符单词的计数-至少有一个字符重复/
给定两个整数 N 和 M ，任务是计算由给定的 N 个不同字符形成的 M 个字符长度的总字数，使得这些单词至少有一个字符重复一次以上。 举例:
输入: N = 3，M = 2 输出: 3 假设字符为{'a '，' b '，' c'} 所有可以用这些字符构成的 2 个长度字 均为“aa”、“ab”、“ac”、“ba”、“bb”、“bc”、“ca”、“cb”和“cc”。 在这些单词中，只有“aa”、“bb”和“cc”至少有一个字符重复了一次以上。 输入: N = 10，M = 5 输出: 69760
方法: 从 N 个字符中可能的 M 个字符单词的总数，总数= N M 。 无字符重复出现的 N 个字符中可能出现的 M 个字符单词的总数，no repeat =NPM。 因此，至少单个字符出现一次以上的总单词是total–no repeat，即NM–NPM。 以下是上述方法的实施:
C++
// C++ implementation for the above approach
#include <math.h>
#include <iostream>
using namespace std;
// Function to return the
// factorial of a number
int fact(int n)
{
if (n <= 1)
return 1;
return n * fact(n - 1);
}
// Function to return the value of nPr
int nPr(int n, int r)
{
return fact(n) / fact(n - r);
}
// Function to return the total number of
// M length words which have at least a
// single character repeated more than once
int countWords(int N, int M)
{
return pow(N, M) - nPr(N, M);
}
// Driver code
int main()
{
int N = 10, M = 5;
cout << (countWords(N, M));
return 0;
}
// This code is contributed by jit_t
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the
// factorial of a number
static int fact(int n)
{
if (n <= 1)
return 1;
return n * fact(n - 1);
}
// Function to return the value of nPr
static int nPr(int n, int r)
{
return fact(n) / fact(n - r);
}
// Function to return the total number of
// M length words which have at least a
// single character repeated more than once
static int countWords(int N, int M)
{
return (int)Math.pow(N, M) - nPr(N, M);
}
// Driver code
public static void main(String[] args)
{
int N = 10, M = 5;
System.out.print(countWords(N, M));
}
}
Python 3
# Python3 implementation for the above approach
# Function to return the
# factorial of a number
def fact(n):
if (n <= 1):
return 1;
return n * fact(n - 1);
# Function to return the value of nPr
def nPr(n, r):
return fact(n) // fact(n - r);
# Function to return the total number of
# M length words which have at least a
# single character repeated more than once
def countWords(N, M):
return pow(N, M) - nPr(N, M);
# Driver code
N = 10; M = 5;
print(countWords(N, M));
# This code is contributed by Code_Mech
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the
// factorial of a number
static int fact(int n)
{
if (n <= 1)
return 1;
return n * fact(n - 1);
}
// Function to return the value of nPr
static int nPr(int n, int r)
{
return fact(n) / fact(n - r);
}
// Function to return the total number of
// M length words which have at least a
// single character repeated more than once
static int countWords(int N, int M)
{
return (int)Math.Pow(N, M) - nPr(N, M);
}
// Driver code
static public void Main ()
{
int N = 10, M = 5;
Console.Write(countWords(N, M));
}
}
// This code is contributed by ajit.
java 描述语言
// javascript implementation of the approach
// Function to return the
// factorial of a number
function fact(n)
{
if (n <= 1)
return 1;
return n * fact(n - 1);
}
// Function to return the value of nPr
function nPr( n, r)
{
return fact(n) / fact(n - r);
}
// Function to return the total number of
// M length words which have at least a
// single character repeated more than once
function countWords( N, M)
{
return Math.pow(N, M) - nPr(N, M);
}
// Driver code
var N = 10 ;
var M = 5;
document.write(countWords(N, M));
// This code is contributed by bunnyram19.
Output:
69760
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