Dados dos enteros N1 y N2 , la tarea es verificar si ambos pares son primos ormiston.
Los números primos de Ormiston son aquellos números que son primos y tienen los mismos dígitos en diferente orden.
Los primeros pares de primos de Ormiston son:
(1913, 1931), (18379, 18397), (19013, 19031), (25013, 25031) ……etc.
Ejemplos:
Entrada: N1 = 1913, N2 = 1931
Salida: SIEntrada: n1 = 5, n2 = 7
Salida: NO
Planteamiento : La idea es comprobar que tanto N1 como N2 son números primos y anagrama o no el uno del otro. Si los números son primos y anagramas entre sí, entonces el número es pares primos de Ormiston.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to
// check Ormiston prime
#include <iostream>
using namespace std;
// Function to check if the
// number is a prime or not
bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
const int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
void updateFreq(int n, int freq[])
{
// While there are digits
// left to process
while (n) {
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagrams of each other
bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int freqA[TEN] = { 0 };
int freqB[TEN] = { 0 };
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++) {
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Returns true if n1 and
// n2 are Ormiston primes
bool OrmistonPrime(int n1, int n2)
{
return (isPrime(n1) && isPrime(n2) &&
areAnagrams(n1, n2));
}
// Driver code
int main()
{
int n1 = 1913, n2 = 1931;
if (OrmistonPrime(n1, n2))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
Java
// Java implementation to
// check Ormiston prime
import java.util.*;
class GFG{
// Function to check if the
// number is a prime or not
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 ||
n % (i + 2) == 0)
return false;
return true;
}
static int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
static void updateFreq(int n, int freq[])
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagrams of each other
static boolean areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int freqA[] = new int[TEN];
int freqB[] = new int[TEN];
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++)
{
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Returns true if n1 and
// n2 are Ormiston primes
static boolean OrmistonPrime(int n1, int n2)
{
return (isPrime(n1) && isPrime(n2) &&
areAnagrams(n1, n2));
}
// Driver code
public static void main(String[] args)
{
int n1 = 1913, n2 = 1931;
if (OrmistonPrime(n1, n2))
System.out.print("YES" +"\n");
else
System.out.print("NO" +"\n");
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 implementation to
# check Ormiston prime
# Function to check if the
# number is a prime or not
def isPrime(n):
# Corner cases
if (n <= 1):
return False
if (n <= 3):
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return False
i = 5
while(i * i <= n):
if (n % i == 0 or n % (i + 2) == 0):
return False
i = i + 6
return True
TEN = 10
# Function to update the frequency array
# such that freq[i] stores the
# frequency of digit i in n
def updateFreq(n, freq):
# While there are digits
# left to process
while (n):
digit = n % TEN
# Update the frequency of
# the current digit
freq[digit] += 1
# Remove the last digit
n = n // TEN
# Function that returns true if a and b
# are anagrams of each other
def areAnagrams(a, b):
# To store the frequencies of
# the digits in a and b
freqA = [0] * TEN
freqB = [0] * TEN
# Update the frequency of
# the digits in a
updateFreq(a, freqA)
# Update the frequency of
# the digits in b
updateFreq(b, freqB)
# Match the frequencies of
# the common digits
for i in range(TEN):
# If frequency differs for any digit
# then the numbers are not
# anagrams of each other
if (freqA[i] != freqB[i]):
return False
return True
# Returns true if n1 and
# n2 are Ormiston primes
def OrmistonPrime(n1, n2):
return (isPrime(n1) and
isPrime(n2) and
areAnagrams(n1, n2))
# Driver code
n1, n2 = 1913, 1931
if (OrmistonPrime(n1, n2)):
print("YES")
else:
print("NO")
# This code is contributed by divyeshrabadiya07
C#
// C# implementation to
// check Ormiston prime
using System;
class GFG{
// Function to check if the
// number is a prime or not
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for(int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
static void updateFreq(int n, int []freq)
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagrams of each other
static bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int []freqA = new int[TEN];
int []freqB = new int[TEN];
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for(int i = 0; i < TEN; i++)
{
// If frequency differs for any
// digit then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Returns true if n1 and
// n2 are Ormiston primes
static bool OrmistonPrime(int n1, int n2)
{
return (isPrime(n1) && isPrime(n2) &&
areAnagrams(n1, n2));
}
// Driver code
public static void Main(String[] args)
{
int n1 = 1913, n2 = 1931;
if (OrmistonPrime(n1, n2))
Console.Write("YES" + "\n");
else
Console.Write("NO" + "\n");
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript implementation to
// check Ormiston prime
// Function to check if the
// number is a prime or not
function isPrime( n) {
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
const TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
function updateFreq( n, freq)
{
// While there are digits
// left to process
while (n > 0) {
let digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n = parseInt(n/TEN);
}
}
// Function that returns true if a and b
// are anagrams of each other
function areAnagrams( a ,b) {
// To store the frequencies of
// the digits in a and b
let freqA = Array(TEN).fill(0);
let freqB = Array(TEN).fill(0);
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for ( i = 0; i < TEN; i++) {
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Returns true if n1 and
// n2 are Ormiston primes
function OrmistonPrime( n1, n2)
{
return (isPrime(n1) && isPrime(n2) && areAnagrams(n1, n2));
}
// Driver code
let n1 = 1913, n2 = 1931;
if (OrmistonPrime(n1, n2))
document.write("YES" + "\n");
else
document.write("NO" + "\n");
// This code is contributed by todaysgaurav
</script>
Producción:
YES
Complejidad de tiempo: O(N)
Referencia: OEIS