从给定的字符串中准确使用 P 辅音和 Q 元音可以组成的单词数
原文:https://www . geeksforgeeks . org/使用给定字符串中的精确 p 辅音和 q 元音可以生成的单词数/
给定一个字符串字符串和两个整数 P 和 Q 。任务是从给定的字符串中精确选择 P 辅音和 Q 元音，找到可以组成的单词总数。 举例:
输入: str = "geek "，P = 1，Q = 1 输出:8 “ge”、“ge”、“eg”、“ek”、“eg”、“ek”、 “ke”和“ke”是可能的单词。 输入: str = "crackathon "，P = 4，Q = 3 输出: 176400
方法:因为 P 辅音和 Q 元音必须从给定字符串中辅音和元音的原始计数中选择。因此，二项式系数可用于计算选择这些字符的组合，并且所选择的字符可使用其计数的因子自行排列。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
// Function to return the value of nCk
lli binomialCoeff(lli n, lli k)
{
if (k == 0 || k == n)
return 1;
return binomialCoeff(n - 1, k - 1)
+ binomialCoeff(n - 1, k);
}
// Function to return the factorial of n
lli fact(lli n)
{
if (n >= 1)
return n * fact(n - 1);
else
return 1;
}
// Function that returns true if ch is a vowel
bool isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i'
|| ch == 'o' || ch == 'u') {
return true;
}
return false;
}
// Function to return the number of words possible
lli countWords(string s, int p, int q)
{
// To store the count of vowels and
// consonanats in the given string
lli countc = 0, countv = 0;
for (int i = 0; i < s.length(); i++) {
// If current character is a vowel
if (isVowel(s[i]))
countv++;
else
countc++;
}
// Find the total possible words
lli a = binomialCoeff(countc, p);
lli b = binomialCoeff(countv, q);
lli c = fact(p + q);
lli ans = (a * b) * c;
return ans;
}
// Driver code
int main()
{
string s = "crackathon";
int p = 4, q = 3;
cout << countWords(s, p, q);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
class GFG
{
// Function to return the value of nCk
static long binomialCoeff(long n, long k)
{
if (k == 0 || k == n)
return 1;
return binomialCoeff(n - 1, k - 1) +
binomialCoeff(n - 1, k);
}
// Function to return the factorial of n
static long fact(long n)
{
if (n >= 1)
return n * fact(n - 1);
else
return 1;
}
// Function that returns true if ch is a vowel
static boolean isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' ||
ch == 'o' || ch == 'u')
{
return true;
}
return false;
}
// Function to return the number of words possible
static long countWords(String s, int p, int q)
{
// To store the count of vowels and
// consonanats in the given string
long countc = 0, countv = 0;
for (int i = 0; i < s.length(); i++)
{
// If current character is a vowel
if (isVowel(s.charAt(i)))
countv++;
else
countc++;
}
// Find the total possible words
long a = binomialCoeff(countc, p);
long b = binomialCoeff(countv, q);
long c = fact(p + q);
long ans = (a * b) * c;
return ans;
}
// Driver code
public static void main (String[] args)
{
String s = "crackathon";
int p = 4, q = 3;
System.out.println(countWords(s, p, q));
}
}
// This Code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to return the value of nCk
def binomialCoeff(n, k):
if (k == 0 or k == n):
return 1
return binomialCoeff(n - 1, k - 1) + \
binomialCoeff(n - 1, k)
# Function to return the factorial of n
def fact(n):
if (n >= 1):
return n * fact(n - 1)
else:
return 1
# Function that returns true if ch is a vowel
def isVowel(ch):
if (ch == 'a' or ch == 'e' or
ch == 'i' or ch == 'o' or ch == 'u'):
return True
return False
# Function to return the number of words possible
def countWords(s, p, q):
# To store the count of vowels and
# consonanats in the given string
countc = 0
countv = 0
for i in range(len(s)):
# If current character is a vowel
if (isVowel(s[i])):
countv += 1
else:
countc += 1
# Find the total possible words
a = binomialCoeff(countc, p)
b = binomialCoeff(countv, q)
c = fact(p + q)
ans = (a * b) * c
return ans
# Driver code
s = "crackathon"
p = 4
q = 3
print(countWords(s, p, q))
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the value of nCk
static long binomialCoeff(long n, long k)
{
if (k == 0 || k == n)
return 1;
return binomialCoeff(n - 1, k - 1) +
binomialCoeff(n - 1, k);
}
// Function to return the factorial of n
static long fact(long n)
{
if (n >= 1)
return n * fact(n - 1);
else
return 1;
}
// Function that returns true if ch is a vowel
static bool isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' ||
ch == 'o' || ch == 'u')
{
return true;
}
return false;
}
// Function to return the number of words possible
static long countWords(String s, int p, int q)
{
// To store the count of vowels and
// consonanats in the given string
long countc = 0, countv = 0;
for (int i = 0; i < s.Length; i++)
{
// If current character is a vowel
if (isVowel(s[i]))
countv++;
else
countc++;
}
// Find the total possible words
long a = binomialCoeff(countc, p);
long b = binomialCoeff(countv, q);
long c = fact(p + q);
long ans = (a * b) * c;
return ans;
}
// Driver code
public static void Main (String[] args)
{
String s = "crackathon";
int p = 4, q = 3;
Console.WriteLine(countWords(s, p, q));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the value of nCk
function binomialCoeff(n, k)
{
if (k == 0 || k == n)
return 1;
return binomialCoeff(n - 1, k - 1)
+ binomialCoeff(n - 1, k);
}
// Function to return the factorial of n
function fact(n)
{
if (n >= 1)
return n * fact(n - 1);
else
return 1;
}
// Function that returns true if ch is a vowel
function isVowel(ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i'
|| ch == 'o' || ch == 'u') {
return true;
}
return false;
}
// Function to return the number of words possible
function countWords(s, p, q)
{
// To store the count of vowels and
// consonanats in the given string
var countc = 0, countv = 0;
for (var i = 0; i < s.length; i++) {
// If current character is a vowel
if (isVowel(s[i]))
countv++;
else
countc++;
}
// Find the total possible words
var a = binomialCoeff(countc, p);
var b = binomialCoeff(countv, q);
var c = fact(p + q);
var ans = (a * b) * c;
return ans;
}
// Driver code
var s = "crackathon";
var p = 4, q = 3;
document.write(countWords(s, p, q));
</script>
Output:
176400
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