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LCM של מספרים טבעיים n הראשונים

ניתן מספר n כך ש-1<= N <= 10^6 the Task is to Find the LCM of First n Natural Numbers. 

דוגמאות:  

Input : n = 5 Output : 60 Input : n = 6 Output : 60 Input : n = 7 Output : 420 

אנו ממליצים בחום ללחוץ כאן ולתרגל זאת לפני שתמשיך לפתרון.

דנו בפתרון פשוט במאמר שלהלן. 
המספר הקטן ביותר המתחלק ב-n המספרים הראשונים
הפתרון לעיל עובד מצוין עבור קלט בודד. אבל אם יש לנו מספר כניסות, זה רעיון טוב להשתמש מסננת של ארוטוסטנס לאחסן את כל הגורמים הראשוניים. כפי שאנו יודעים אם LCM(a b) = X אז כל גורם ראשוני של a או b יהיה גם הגורם הראשוני של 'X'.  



  1. אתחול משתנה lcm עם 1
  2. צור מסננת של Eratosthenes (וקטור bool isPrime) באורך 10^6 (באופן אידיאלי חייב להיות שווה למספר הספרות בפקטוריאלי)
  3. כעת עבור כל מספר ב-bool vector isPrime אם המספר הוא ראשוני (isPrime[i] נכון) מצא את המספר המקסימלי הקטן מהמספר הנתון ושווה בחזקת הראשוני.
  4. לאחר מכן תכפיל את המספר הזה עם משתנה lcm.
  5. חזור על שלב 3 ו-4 עד שהראשון קטן מהמספר הנתון.

אִיוּר:  

For example if n = 10 8 will be the first number which is equal to 2^3 then 9 which is equal to 3^2 then 5 which is equal to 5^1 then 7 which is equal to 7^1 Finally we multiply those numbers 8*9*5*7 = 2520

להלן יישום הרעיון לעיל.  

C++
// C++ program to find LCM of First N Natural Numbers. #include    #define MAX 100000 using namespace std; vector<bool> isPrime (MAX true); // utility function for sieve of sieve of Eratosthenes void sieve() {    for (int i = 2; i * i <= MAX; i++)  {  if (isPrime[i] == true)  for (int j = i*i; j<= MAX; j+=i)  isPrime[j] = false;  } } // Function to find LCM of first n Natural Numbers long long LCM(int n) {  long long lcm = 1;  int i=2;   while(i<=n) {  if(isPrime[i]){  int pp = i;  while (pp * i <= n)  pp = pp * i;  lcm *= pp;  }  i++;  }  return lcm; } // Driver code int main() {  // build sieve  sieve();  int N = 7;  // Function call  cout << LCM(N);  return 0; } 
Java
// Java program to find LCM of First N Natural Numbers. import java.util.*; class GFG  {  static int MAX = 100000;  // array to store all prime less than and equal to 10^6  static ArrayList<Integer> primes  = new ArrayList<Integer>();  // utility function for sieve of sieve of Eratosthenes  static void sieve()  {  boolean[] isComposite = new boolean[MAX + 1];  for (int i = 2; i * i <= MAX; i++)   {  if (isComposite[i] == false)  for (int j = 2; j * i <= MAX; j++)  isComposite[i * j] = true;  }  // Store all prime numbers in vector primes[]  for (int i = 2; i <= MAX; i++)  if (isComposite[i] == false)  primes.add(i);  }  // Function to find LCM of first n Natural Numbers  static long LCM(int n)  {  long lcm = 1;  for (int i = 0;  i < primes.size() && primes.get(i) <= n;   i++)   {  // Find the highest power of prime primes[i]  // that is less than or equal to n  int pp = primes.get(i);  while (pp * primes.get(i) <= n)  pp = pp * primes.get(i);  // multiply lcm with highest power of prime[i]  lcm *= pp;  lcm %= 1000000007;  }  return lcm;  }  // Driver code  public static void main(String[] args)  {  sieve();  int N = 7;    // Function call  System.out.println(LCM(N));  } } // This code is contributed by mits 
Python3
# Python3 program to find LCM of # First N Natural Numbers. MAX = 100000 # array to store all prime less # than and equal to 10^6 primes = [] # utility function for # sieve of Eratosthenes def sieve(): isComposite = [False]*(MAX+1) i = 2 while (i * i <= MAX): if (isComposite[i] == False): j = 2 while (j * i <= MAX): isComposite[i * j] = True j += 1 i += 1 # Store all prime numbers in # vector primes[] for i in range(2 MAX+1): if (isComposite[i] == False): primes.append(i) # Function to find LCM of # first n Natural Numbers def LCM(n): lcm = 1 i = 0 while (i < len(primes) and primes[i] <= n): # Find the highest power of prime # primes[i] that is less than or # equal to n pp = primes[i] while (pp * primes[i] <= n): pp = pp * primes[i] # multiply lcm with highest # power of prime[i] lcm *= pp lcm %= 1000000007 i += 1 return lcm # Driver code sieve() N = 7 # Function call print(LCM(N)) # This code is contributed by mits 
C#
// C# program to find LCM of First N // Natural Numbers. using System.Collections; using System; class GFG {  static int MAX = 100000;  // array to store all prime less than  // and equal to 10^6  static ArrayList primes = new ArrayList();  // utility function for sieve of  // sieve of Eratosthenes  static void sieve()  {  bool[] isComposite = new bool[MAX + 1];  for (int i = 2; i * i <= MAX; i++)   {  if (isComposite[i] == false)  for (int j = 2; j * i <= MAX; j++)  isComposite[i * j] = true;  }  // Store all prime numbers in vector primes[]  for (int i = 2; i <= MAX; i++)  if (isComposite[i] == false)  primes.Add(i);  }  // Function to find LCM of first  // n Natural Numbers  static long LCM(int n)  {  long lcm = 1;  for (int i = 0;  i < primes.Count && (int)primes[i] <= n;   i++)   {  // Find the highest power of prime primes[i]  // that is less than or equal to n  int pp = (int)primes[i];  while (pp * (int)primes[i] <= n)  pp = pp * (int)primes[i];  // multiply lcm with highest power of prime[i]  lcm *= pp;  lcm %= 1000000007;  }  return lcm;  }  // Driver code  public static void Main()  {  sieve();  int N = 7;    // Function call  Console.WriteLine(LCM(N));  } } // This code is contributed by mits 
JavaScript
<script>  // Javascript program to find LCM of First N  // Natural Numbers.    let MAX = 100000;    // array to store all prime less than  // and equal to 10^6  let primes = [];    // utility function for sieve of  // sieve of Eratosthenes  function sieve()  {  let isComposite = new Array(MAX + 1);  isComposite.fill(false);  for (let i = 2; i * i <= MAX; i++)  {  if (isComposite[i] == false)  for (let j = 2; j * i <= MAX; j++)  isComposite[i * j] = true;  }    // Store all prime numbers in vector primes[]  for (let i = 2; i <= MAX; i++)  if (isComposite[i] == false)  primes.push(i);  }    // Function to find LCM of first  // n Natural Numbers  function LCM(n)  {  let lcm = 1;  for (let i = 0;  i < primes.length && primes[i] <= n;  i++)  {  // Find the highest power of prime primes[i]  // that is less than or equal to n  let pp = primes[i];  while (pp * primes[i] <= n)  pp = pp * primes[i];    // multiply lcm with highest power of prime[i]  lcm *= pp;  lcm %= 1000000007;  }  return lcm;  }    sieve();  let N = 7;  // Function call  document.write(LCM(N)); // This code is contributed by decode2207. </script> 
PHP
 // PHP program to find LCM of // First N Natural Numbers. $MAX = 100000; // array to store all prime less // than and equal to 10^6 $primes = array(); // utility function for // sieve of Eratosthenes function sieve() { global $MAX $primes; $isComposite = array_fill(0 $MAX false); for ($i = 2; $i * $i <= $MAX; $i++) { if ($isComposite[$i] == false) for ($j = 2; $j * $i <= $MAX; $j++) $isComposite[$i * $j] = true; } // Store all prime numbers in // vector primes[] for ($i = 2; $i <= $MAX; $i++) if ($isComposite[$i] == false) array_push($primes $i); } // Function to find LCM of  // first n Natural Numbers function LCM($n) { global $MAX $primes; $lcm = 1; for ($i = 0; $i < count($primes) && $primes[$i] <= $n; $i++) { // Find the highest power of prime  // primes[i] that is less than or  // equal to n $pp = $primes[$i]; while ($pp * $primes[$i] <= $n) $pp = $pp * $primes[$i]; // multiply lcm with highest // power of prime[i] $lcm *= $pp; $lcm %= 1000000007; } return $lcm; } // Driver code sieve(); $N = 7; // Function call echo LCM($N); // This code is contributed by mits ?> 

תְפוּקָה
420

מורכבות זמן : O(n2
מרחב עזר: עַל)

גישה נוספת:

 הרעיון הוא שאם המספר קטן מ-3 אז החזר מספר. אם המספר גדול מ-2 אז מצא את LCM של nn-1

  • נניח x=LCM(nn-1)
  • שוב x=LCM(xn-2)
  • שוב x=LCM(xn-3) ...
  • .
  • .
  • שוב x=LCM(x1) ...

כעת התוצאה היא x.

למציאת LCM(ab) אנו משתמשים בפונקציה hcf(ab) שתחזיר HCF של (ab)

אנחנו יודעים את זה LCM(ab)= (a*b)/HCF(ab)

אִיוּר: 

For example if n = 7 function call lcm(76) now lets say a=7  b=6 Now  b!= 1 Hence a=lcm(76) = 42 and b=6-1=5 function call lcm(425) a=lcm(425) = 210 and b=5-1=4 function call lcm(2104) a=lcm(2104) = 420 and b=4-1=3 function call lcm(4203) a=lcm(4203) = 420 and b=3-1=2 function call lcm(4202) a=lcm(4202) = 420 and b=2-1=1 Now b=1 Hence return a=420

להלן יישום הגישה לעיל

C++
// C++ program to find LCM of First N Natural Numbers. #include    using namespace std; // to calculate hcf int hcf(int a int b) {  if (b == 0)  return a;  return hcf(b a % b); } int findlcm(int aint b) {  if (b == 1)    // lcm(ab)=(a*b)/hcf(ab)  return a;     // assign a=lcm of nn-1  a = (a * b) / hcf(a b);     // b=b-1  b -= 1;   return findlcm(a b); } // Driver code int main() {  int n = 7;  if (n < 3)  cout << n; // base case  else    // Function call  // pass nn-1 in function to find LCM of first n natural  // number  cout << findlcm(n n - 1);    return 0; } // contributed by ajaykr00kj 
Java
// Java program to find LCM of First N Natural Numbers public class Main {  // to calculate hcf  static int hcf(int a int b)  {  if (b == 0)  return a;  return hcf(b a % b);  }  static int findlcm(int aint b)  {  if (b == 1)  // lcm(ab)=(a*b)/hcf(ab)  return a;   // assign a=lcm of nn-1  a = (a * b) / hcf(a b);   // b=b-1  b -= 1;   return findlcm(a b);  }  // Driver code.  public static void main(String[] args)   {  int n = 7;  if (n < 3)  System.out.print(n); // base case  else  // Function call  // pass nn-1 in function to find LCM of first n natural  // number  System.out.print(findlcm(n n - 1));  } } // This code is contributed by divyeshrabadiya07. 
Python3
# Python3 program to find LCM  # of First N Natural Numbers. # To calculate hcf def hcf(a b): if (b == 0): return a return hcf(b a % b) def findlcm(a b): if (b == 1): # lcm(ab)=(a*b)//hcf(ab) return a # Assign a=lcm of nn-1 a = (a * b) // hcf(a b) # b=b-1 b -= 1 return findlcm(a b) # Driver code n = 7 if (n < 3): print(n) else: # Function call # pass nn-1 in function # to find LCM of first n  # natural number print(findlcm(n n - 1)) # This code is contributed by Shubham_Singh 
C#
// C# program to find LCM of First N Natural Numbers. using System; class GFG {  // to calculate hcf  static int hcf(int a int b)  {  if (b == 0)  return a;  return hcf(b a % b);  }  static int findlcm(int aint b)  {  if (b == 1)  // lcm(ab)=(a*b)/hcf(ab)  return a;   // assign a=lcm of nn-1  a = (a * b) / hcf(a b);   // b=b-1  b -= 1;   return findlcm(a b);  }  // Driver code  static void Main() {  int n = 7;  if (n < 3)  Console.Write(n); // base case  else  // Function call  // pass nn-1 in function to find LCM of first n natural  // number  Console.Write(findlcm(n n - 1));  } } // This code is contributed by divyesh072019. 
JavaScript
<script>  // Javascript program to find LCM of First N Natural Numbers.    // to calculate hcf  function hcf(a b)  {  if (b == 0)  return a;  return hcf(b a % b);  }  function findlcm(ab)  {  if (b == 1)  // lcm(ab)=(a*b)/hcf(ab)  return a;  // assign a=lcm of nn-1  a = (a * b) / hcf(a b);  // b=b-1  b -= 1;  return findlcm(a b);  }    let n = 7;  if (n < 3)  document.write(n); // base case  else    // Function call  // pass nn-1 in function to find LCM of first n natural  // number  document.write(findlcm(n n - 1));   </script> 

תְפוּקָה
420

מורכבות הזמן: O(n log n)
מרחב עזר: O(1)