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Kairysis sutrumpinamas pagrindinis

A Kairėn sutrumpinamas pirmas yra pirminis skaičius, kuris tam tikroje bazėje (tarkime, 10) neturi 0 ir kuris išlieka pirminis, kai iš eilės pašalinamas pagrindinis ('kairysis') skaitmuo. Pavyzdžiui, 317 yra sutrumpinamas kairėje pusėje, nes 317 17 ir 7 yra pirminiai. Iš viso yra 4260 kairėje sutrumpinamų pirminių skaitmenų.
Užduotis – patikrinti, ar duotas skaičius (N >0) yra kairėje sutrumpinamas pirminis, ar ne.

Pavyzdžiai: 

  Input:   317   Output:   Yes   Input:   293   Output:   No 293 is not left-truncatable prime because numbers formed are 293 93 and 3. Here 293 and 3 are prime but 93 is not prime.

Idėja yra pirmiausia patikrinti, ar skaičiuje yra 0 kaip skaitmuo, ar ne, ir suskaičiuoti skaitmenų skaičių pateiktame skaičiuje N. Jei jame yra 0, tada grąžinkite false, kitaip generuokite visus pirminius skaičius, mažesnius arba lygius duotam skaičiui N, naudodami Eratosteno sietelis. . Sugeneravę visus tokius pirminius skaičius, patikriname, ar skaičius išlieka pirminis, kai iš eilės pašalinamas pagrindinis (kairysis) skaitmuo.



Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas.

C++
// Program to check whether a given number // is left-truncatable prime or not. #include   using namespace std; /* Function to calculate x raised to the power y */ int power(int x unsigned int y) {  if (y == 0)  return 1;  else if (y%2 == 0)  return power(x y/2)*power(x y/2);  else  return x*power(x y/2)*power(x y/2); } // Generate all prime numbers less than n. bool sieveOfEratosthenes(int n bool isPrime[]) {  // Initialize all entries of boolean array  // as true. A value in isPrime[i] will finally  // be false if i is Not a prime else true  // bool isPrime[n+1];  isPrime[0] = isPrime[1] = false;  for (int i=2; i<=n; i++)  isPrime[i] = true;  for (int p=2; p*p<=n; p++)  {  // If isPrime[p] is not changed then it is  // a prime  if (isPrime[p] == true)  {  // Update all multiples of p  for (int i=p*2; i<=n; i += p)  isPrime[i] = false;  }  } } // Returns true if n is right-truncatable else false bool leftTruPrime(int n) {  int temp = n cnt = 0 temp1;  // Counting number of digits in the  // input number and checking whether it  // contains 0 as digit or not.  while (temp)  {  cnt++; // counting number of digits.  temp1 = temp%10; // checking whether digit is 0 or not  if (temp1==0)  return false; // if digit is 0 return false.  temp = temp/10;  }  // Generating primes using Sieve  bool isPrime[n+1];  sieveOfEratosthenes(n isPrime);  // Checking whether the number remains prime  // when the leading ('left') digit is successively  // removed  for (int i=cnt; i>0; i--)  {  // Checking number by successively removing  // leading ('left') digit.  /* n=113 cnt=3  i=3 mod=1000 n%mod=113  i=2 mod=100 n%mod=13  i=3 mod=10 n%mod=3 */  int mod= power(10i);  if (!isPrime[n%mod]) // checking prime  return false; // if not prime return false  }  return true; // if remains prime return true } // Driver program int main() {  int n = 113;  if (leftTruPrime(n))  cout << n << ' is left truncatable prime' << endl;  else  cout << n << ' is not left truncatable prime' << endl;  return 0; } 
Java
// Program to check whether  // a given number is left // truncatable prime or not. import java.io.*; class GFG {    // Function to calculate x  // raised to the power y  static int power(int xint y)  {  if (y == 0)  return 1;  else if (y%2 == 0)  return power(x y/2)  *power(x y/2);  else  return x*power(x y/2)  *power(x y/2);  }    // Generate all prime   // numbers less than n.  static void sieveOfEratosthenes  (int n boolean isPrime[])  {    // Initialize all entries of boolean  // array as true. A value in isPrime[i]  // will finally be false if i is Not  // a prime else true bool isPrime[n+1];  isPrime[0] = isPrime[1] = false;  for (int i = 2; i <= n; i++)  isPrime[i] = true;    for (int p = 2; p*p <= n; p++)  {  // If isPrime[p] is not changed  // then it is a prime  if (isPrime[p] == true)  {    // Update all multiples of p  for (int i = p*2; i <= n; i += p)  isPrime[i] = false;  }  }  }    // Returns true if n is   // right-truncatable else false  static boolean leftTruPrime(int n)  {  int temp = n cnt = 0 temp1;    // Counting number of digits in the  // input number and checking whether  // it contains 0 as digit or not.  while (temp != 0)  {  // counting number of digits.  cnt++;     temp1 = temp%10;    // checking whether digit is  // 0 or not  if (temp1 == 0)  return false;  temp = temp/10;  }    // Generating primes using Sieve  boolean isPrime[] = new boolean[n+1];  sieveOfEratosthenes(n isPrime);    // Checking whether the number  // remains prime when the leading  // ('left') digit is successively removed  for (int i = cnt; i > 0; i--)  {  // Checking number by successively  // removing leading ('left') digit.  /* n=113 cnt=3  i=3 mod=1000 n%mod=113  i=2 mod=100 n%mod=13  i=3 mod=10 n%mod=3 */  int mod = power(10i);    if (!isPrime[n%mod])   return false;   }    // if remains prime return true  return true;   }    // Driver program  public static void main(String args[])  {  int n = 113;    if (leftTruPrime(n))  System.out.println  (n+' is left truncatable prime');  else  System.out.println  (n+' is not left truncatable prime');  } } /*This code is contributed by Nikita Tiwari.*/ 
Python3
# Python3 Program to  # check whether a  # given number is left # truncatable prime  # or not. # Function to calculate  # x raised to the power y  def power(x y) : if (y == 0) : return 1 elif (y % 2 == 0) : return(power(x y // 2) * power(x y // 2)) else : return(x * power(x y // 2) * power(x y // 2)) # Generate all prime  # numbers less than n. def sieveOfEratosthenes(n isPrime) : # Initialize all entries # of boolean array # as true. A value in # isPrime[i] will finally # be false if i is Not a # prime else true # bool isPrime[n+1]; isPrime[0] = isPrime[1] = False for i in range(2 n+1) : isPrime[i] = True p=2 while(p * p <= n) : # If isPrime[p] is not # changed then it is # a prime if (isPrime[p] == True) : # Update all multiples # of p i=p*2 while(i <= n) : isPrime[i] = False i = i + p p = p + 1 # Returns true if n is # right-truncatable  # else false def leftTruPrime(n) : temp = n cnt = 0 # Counting number of # digits in the input # number and checking # whether it contains # 0 as digit or not. while (temp != 0) : # counting number  # of digits. cnt=cnt + 1 # checking whether # digit is 0 or not temp1 = temp % 10; if (temp1 == 0) : # if digit is 0 # return false. return False temp = temp // 10 # Generating primes  # using Sieve isPrime = [None] * (n + 1) sieveOfEratosthenes(n isPrime) # Checking whether the # number remains prime # when the leading  # ('left') digit is # successively removed for i in range(cnt 0 -1) : # Checking number by  # successively removing # leading ('left') digit. # n=113 cnt=3 # i=3 mod=1000 n%mod=113 # i=2 mod=100 n%mod=13 # i=3 mod=10 n%mod=3  mod = power(10 i) # checking prime if (isPrime[n % mod] != True) : # if not prime # return false return False # if remains prime #  return true return True # Driver program n = 113 if (leftTruPrime(n)) : print(n 'is left truncatable prime') else : print(n 'is not left truncatable prime') # This code is contributed by Nikita Tiwari. 
C#
// Program to check whether  // a given number is left // truncatable prime or not. using System; class GFG {    // Function to calculate x  // raised to the power y  static int power(int x int y)  {  if (y == 0)  return 1;  else if (y%2 == 0)  return power(x y/2)  *power(x y/2);  else  return x*power(x y/2)  *power(x y/2);  }    // Generate all prime   // numbers less than n.  static void sieveOfEratosthenes  (int n bool []isPrime)  {  // Initialize all entries of boolean  // array as true. A value in isPrime[i]  // will finally be false if i is Not  // a prime else true bool isPrime[n+1];  isPrime[0] = isPrime[1] = false;  for (int i = 2; i <= n; i++)  isPrime[i] = true;    for (int p = 2; p * p <= n; p++)  {  // If isPrime[p] is not changed  // then it is a prime  if (isPrime[p] == true)  {  // Update all multiples of p  for (int i = p * 2; i <= n; i += p)  isPrime[i] = false;  }  }  }    // Returns true if n is   // right-truncatable else false  static bool leftTruPrime(int n)  {  int temp = n cnt = 0 temp1;    // Counting number of digits in the  // input number and checking whether  // it contains 0 as digit or not.  while (temp != 0)  {  // counting number of digits.  cnt++;     temp1 = temp%10;    // checking whether digit is  // 0 or not  if (temp1 == 0)  return false;  temp = temp/10;  }    // Generating primes using Sieve  bool []isPrime = new bool[n+1];  sieveOfEratosthenes(n isPrime);    // Checking whether the number  // remains prime when the leading  // ('left') digit is successively removed  for (int i = cnt; i > 0; i--)  {  // Checking number by successively  // removing leading ('left') digit.  /* n=113 cnt=3  i=3 mod=1000 n%mod=113  i=2 mod=100 n%mod=13  i=3 mod=10 n%mod=3 */  int mod = power(10 i);    if (!isPrime[n%mod])   return false;   }    // if remains prime return true  return true;   }    // Driver program  public static void Main()  {  int n = 113;    if (leftTruPrime(n))  Console.WriteLine  (n + ' is left truncatable prime');  else  Console.WriteLine  (n + ' is not left truncatable prime');  } }   //This code is contributed by Anant Agarwal. 
PHP
 // PHP Program to check whether a // given number is left-truncatable // prime or not. /* Function to calculate x raised to the power y */ function power($x $y) { if ($y == 0) return 1; else if ($y % 2 == 0) return power($x $y/2) * power($x $y/2); else return $x*power($x $y/2) * power($x $y/2); } // Generate all prime numbers // less than n. function sieveOfEratosthenes($n $l $isPrime) { // Initialize all entries of  // boolean array as true. A  // value in isPrime[i] will  // finally be false if i is  // Not a prime else true // bool isPrime[n+1]; $isPrime[0] = $isPrime[1] = -1; for ($i = 2; $i <= $n; $i++) $isPrime[$i] = true; for ( $p = 2; $p * $p <= $n; $p++) { // If isPrime[p] is not  // changed then it is // a prime if ($isPrime[$p] == true) { // Update all multiples // of p for ($i = $p * 2; $i <= $n; $i += $p) $isPrime[$i] = false; } } } // Returns true if n is  // right-truncatable else false function leftTruPrime($n) { $temp = $n; $cnt = 0; $temp1; // Counting number of digits in // the input number and checking // whether it contains 0 as digit // or not. while ($temp) { // counting number of digits. $cnt++; // checking whether digit is // 0 or not $temp1 = $temp % 10; if ($temp1 == 0) // if digit is 0 return // false. return -1; $temp = $temp / 10; } // Generating primes using Sieve $isPrime[$n + 1]; sieveOfEratosthenes($n $isPrime); // Checking whether the number // remains prime when the leading // ('left') digit is successively // removed for ($i = $cnt; $i > 0; $i--) { // Checking number by  // successively removing // leading ('left') digit. /* n=113 cnt=3  i=3 mod=1000 n%mod=113  i=2 mod=100 n%mod=13  i=3 mod=10 n%mod=3 */ $mod= power(10 $i); // checking prime if (!$isPrime[$n % $mod]) // if not prime return // false return -1; } // if remains prime return true return true; } // Driver program $n = 113; if (leftTruPrime($n)) echo $n ' is left truncatable' ' prime' 'n'; else echo $n ' is not left ' 'truncatable prime' 'n'; // This code is contributed by ajit ?> 
JavaScript
<script> // Javascript program to check whether  // a given number is left // truncatable prime or not.  function power(x y)  {  if (y == 0)  return 1;  else if (y%2 == 0)  return power(x Math.floor(y/2))  *power(x Math.floor(y/2));  else  return x*power(x Math.floor(y/2))  *power(x Math.floor(y/2));  }    // Generate all prime   // numbers less than n.  function sieveOfEratosthenes  (n isPrime)  {    // Initialize all entries of boolean  // array as true. A value in isPrime[i]  // will finally be false if i is Not  // a prime else true bool isPrime[n+1];  isPrime[0] = isPrime[1] = false;  for (let i = 2; i <= n; i++)  isPrime[i] = true;    for (let p = 2; p*p <= n; p++)  {  // If isPrime[p] is not changed  // then it is a prime  if (isPrime[p] == true)  {    // Update all multiples of p  for (let i = p*2; i <= n; i += p)  isPrime[i] = false;  }  }  }    // Returns true if n is   // right-truncatable else false  function leftTruPrime(n)  {  let temp = n cnt = 0 temp1;    // Counting number of digits in the  // input number and checking whether  // it contains 0 as digit or not.  while (temp != 0)  {  // counting number of digits.  cnt++;     temp1 = temp%10;    // checking whether digit is  // 0 or not  if (temp1 == 0)  return false;  temp = Math.floor(temp/10);  }    // Generating primes using Sieve  let isPrime = Array.from({length: n+1} (_ i) => 0);  sieveOfEratosthenes(n isPrime);    // Checking whether the number  // remains prime when the leading  // ('left') digit is successively removed  for (let i = cnt; i > 0; i--)  {  // Checking number by successively  // removing leading ('left') digit.  /* n=113 cnt=3  i=3 mod=1000 n%mod=113  i=2 mod=100 n%mod=13  i=3 mod=10 n%mod=3 */  let mod = power(10i);    if (!isPrime[n%mod])   return false;   }    // if remains prime return true  return true;   } // Driver Code  let n = 113;    if (leftTruPrime(n))  document.write  (n+' is left truncatable prime');  else  document.write  (n+' is not left truncatable prime'); // This code is contributed by sanjoy_62. </script> 

Išvestis
113 is left truncatable prime 

Laiko sudėtingumas: O (N*N) 
Pagalbinė erdvė: O(N)

Susijęs straipsnis:  
Dešinysis trumpinamasis pagrindinis
Nuorodos: https://en.wikipedia.org/wiki/Truncatable_prime

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