给定一个字符串str，任务是计算所有str的子串它们是回文并且其长度为素数。
例子：
Input: str = “geeksforgeeks”
Output: 2
“ee” and “ee” are the only valid sub-strings
Input: str = “abccc”
Output: 3
方法：使用埃拉托色尼的筛，找到所有的素数，直到str的长度，因为这是海峡的子字符串可以有最大长度。现在从最小素数开始，即j = 2直到j≤len(str) 。如果j为质数，则计算其长度= j的str的所有回文子串。最后打印总数。
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 sub-string
// starting at i and ending at j in str is a palindrome
bool isPalindrome(string str, int i, int j)
{
while (i < j) {
if (str[i] != str[j])
return false;
i++;
j--;
}
return true;
}
// Function to count all palindromic substring
// whose lwngth is a prime number
int countPrimePalindrome(string str, int len)
{
bool prime[len + 1];
memset(prime, true, sizeof(prime));
// 0 and 1 are non-primes
prime[0] = prime[1] = false;
for (int p = 2; p * p <= len; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p]) {
// 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 <= len; i += p)
prime[i] = false;
}
}
// To store the required number of sub-strings
int count = 0;
// Starting from the smallest prime till
// the largest length of the sub-string possible
for (int j = 2; j <= len; j++) {
// If j is prime
if (prime[j]) {
// Check all the sub-strings of length j
for (int i = 0; i + j - 1 < len; i++) {
// If current sub-string is a palindrome
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
// Driver Code
int main()
{
string s = "geeksforgeeks";
int len = s.length();
cout << countPrimePalindrome(s, len);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GfG
{
// Function that returns true if
// sub-string starting at i and
// ending at j in str is a palindrome
static boolean isPalindrome(String str, int i, int j)
{
while (i < j)
{
if (str.charAt(i) != str.charAt(j))
return false;
i++;
j--;
}
return true;
}
// Function to count all palindromic substring
// whose lwngth is a prime number
static int countPrimePalindrome(String str, int len)
{
boolean[] prime = new boolean[len + 1];
Arrays.fill(prime, true);
// 0 and 1 are non-primes
prime[0] = prime[1] = false;
for (int p = 2; p * p <= len; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// 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 <= len; i += p)
prime[i] = false;
}
}
// To store the required number of sub-strings
int count = 0;
// Starting from the smallest prime till
// the largest length of the sub-string possible
for (int j = 2; j <= len; j++)
{
// If j is prime
if (prime[j])
{
// Check all the sub-strings of length j
for (int i = 0; i + j - 1 < len; i++)
{
// If current sub-string is a palindrome
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
// Driver code
public static void main(String []args)
{
String s = "geeksforgeeks";
int len = s.length();
System.out.println(countPrimePalindrome(s, len));
}
}
// This code is contributed by Rituraj JainPython3
# Python3 implementation of the approach
import math as mt
# Function that returns True if sub-str1ing
# starting at i and ending at j in str1
# is a palindrome
def isPalindrome(str1, i, j):
while (i < j):
if (str1[i] != str1[j]):
return False
i += 1
j -= 1
return True
# Function to count all palindromic substr1ing
# whose lwngth is a prime number
def countPrimePalindrome(str1, Len):
prime = [True for i in range(Len + 1)]
# 0 and 1 are non-primes
prime[0], prime[1] = False, False
for p in range(2, mt.ceil(mt.sqrt(Len + 1))):
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# 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, Len + 1, p):
prime[i] = False
# To store the required number
# of sub-str1ings
count = 0
# Starting from the smallest prime
# till the largest Length of the
# sub-str1ing possible
for j in range(2, Len + 1):
# If j is prime
if (prime[j]):
# Check all the sub-str1ings of
# Length j
for i in range(Len + 1 - j):
# If current sub-str1ing is a palindrome
if (isPalindrome(str1, i, i + j - 1)):
count += 1
return count
# Driver Code
s = "geeksforgeeks"
Len = len(s)
print( countPrimePalindrome(s, Len))
# This code is contributed by
# Mohit kumar 29C#
// C# implementation of the approach
using System;
class GfG
{
// Function that returns true if
// sub-string starting at i and
// ending at j in str is a palindrome
static bool isPalindrome(string str,
int i, int j)
{
while (i < j)
{
if (str[i] != str[j])
return false;
i++;
j--;
}
return true;
}
// Function to count all palindromic
// substring whose lwngth is a prime number
static int countPrimePalindrome(string str,
int len)
{
bool[] prime = new bool[len + 1];
Array.Fill(prime, true);
// 0 and 1 are non-primes
prime[0] = prime[1] = false;
for (int p = 2; p * p <= len; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// 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 <= len; i += p)
prime[i] = false;
}
}
// To store the required number
// of sub-strings
int count = 0;
// Starting from the smallest prime
// till the largest length of the
// sub-string possible
for (int j = 2; j <= len; j++)
{
// If j is prime
if (prime[j])
{
// Check all the sub-strings of
// length j
for (int i = 0;
i + j - 1 < len; i++)
{
// If current sub-string is a
// palindrome
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
// Driver code
public static void Main()
{
string s = "geeksforgeeks";
int len = s.Length;
Console.WriteLine(countPrimePalindrome(s, len));
}
}
// This code is contributed by Code_MechPHP
输出：
2
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