检查给定的字符串是否为元音素数
给定一个由小写英文字母组成的字符串str ，任务是检查该字符串是否是元音素数。如果字符串中的所有元音仅出现在素数索引处，则称该字符串为素数。
例子：
Input: str = “geeksforgeeks”
Output: No
str[1] = ‘e’ is a vowel but 1 is not prime.
Input: str = “bcae”
Output: Yes
All the vowels are at prime indices i.e. 2 and 3.
方法：使用埃拉托色尼筛法找出所有小于 N 的素数，以便检查字符串的每个索引是否为素数。现在，如果有一些非素数索引使得该位置的字符是元音，那么字符串不是元音素数，否则它是。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true if c is a vowel
bool isVowel(char c)
{
if (c == 'a' || c == 'e' || c == 'i'
|| c == 'o' || c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
bool isVowelPrime(string str, int n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
bool prime[n];
memset(prime, true, sizeof(prime));
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p < n; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (int i = 0; i < n; i++) {
// If current character is vowel
// and the index is not prime
if (isVowel(str[i]) && !prime[i])
return false;
}
return true;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int n = str.length();
if (isVowelPrime(str, n))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true
// if c is a vowel
static boolean isVowel(char c)
{
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' ||
c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
static boolean isVowelPrime(String str, int n)
{
// Create a boolean array "prime[0..n]" and
// initialize all entries it as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean []prime = new boolean[n];
Arrays.fill(prime, true);
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p < n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (int i = 0; i < n; i++)
{
// If current character is vowel
// and the index is not prime
if (isVowel(str.charAt(i)) && !prime[i])
return false;
}
return true;
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeks";
int n = str.length();
if (isVowelPrime(str, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Princi SinghPython3
# Python3 implementation of the approach
# Function that returns true if c is a vowel
def isVowel(c):
if (c == 'a' or c == 'e' or
c == 'i' or c == 'o' or
c == 'u'):
return True
return False
# Function that returns true if
# all the vowels in the given string
# are only at prime indices
def isVowelPrime(Str, n):
# Create a boolean array "prime[0..n]"
# and initialize all entries in it as true.
# A value in prime[i] will finally be false
# if i is Not a prime, else true.
prime = [True for i in range(n)]
# 0 and 1 are not prime
prime[0] = False
prime[1] = False
for p in range(2, n):
if p * p > n:
break
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p greater than or
# equal to the square of it
# numbers which are multiple of p and are
# less than p^2 are already been marked.
for i in range(2 * p, n, p):
prime[i] = False
# For every character of the given String
for i in range(n):
# If current character is vowel
# and the index is not prime
if (isVowel(Str[i]) and
prime[i] == False):
return False
return True
# Driver code
Str= "geeksforgeeks";
n = len(Str)
if (isVowelPrime(Str, n)):
print("Yes")
else:
print("No")
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true
// if c is a vowel
static Boolean isVowel(char c)
{
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' ||
c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
static Boolean isVowelPrime(String str, int n)
{
// Create a boolean array "prime[0..n]" and
// initialize all entries it as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
Boolean []prime = new Boolean[n];
for(int i = 0; i < n; i++)
prime[i] = true;
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p < n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p greater than
// or equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (int i = 0; i < n; i++)
{
// If current character is vowel
// and the index is not prime
if (isVowel(str[i]) && !prime[i])
return false;
}
return true;
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
int n = str.Length;
if (isVowelPrime(str, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Princi SinghJavascript
输出：
No