A Pa kreisi saīsināms galvenais ir pirmskaitlis, kas dotajā bāzē (piemēram, 10) nesatur 0 un kas paliek pirmskaitlis, kad secīgi tiek noņemts pirmais (“kreisais”) cipars. Piemēram, 317 ir pa kreisi saīsināms pirmskaitlis, jo 317 17 un 7 ir pirmskaitļi. Kopā ir 4260 kreisi saīsināmi pirmskaitļi.
Uzdevums ir pārbaudīt, vai dotais skaitlis (N >0) ir pa kreisi saīsināms pirmskaitlis vai nav.
Piemēri:
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.
Ideja ir vispirms pārbaudīt, vai skaitlis satur 0 kā ciparu vai ne, un saskaitīt ciparu skaitu dotajā ciparā N. Ja tas satur 0, tad atgriezt false pretējā gadījumā ģenerē visus pirmskaitļus, kas ir mazāki vai vienādi ar doto skaitli N, izmantojot Eratostena siets. . Kad esam ģenerējuši visus šādus pirmskaitļus, mēs pārbaudām, vai skaitlis paliek galvenais, kad tiek secīgi noņemts pirmais (kreisais) cipars.
Zemāk ir aprakstīta iepriekš minētās pieejas īstenošana.
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>
Izvade
113 is left truncatable prime
Laika sarežģītība: O(N*N)
Palīgtelpa: O(N)
Saistīts raksts:
Labās puses saīsināmais galvenais
Atsauces: https://en.wikipedia.org/wiki/Truncatable_prime