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LCM för första n naturliga talen

Givet ett nummer n så att 1<= N <= 10^6 the Task is to Find the LCM of First n Natural Numbers. 

Exempel:  

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

Vi rekommenderar starkt att du klickar här och tränar på det innan du går vidare till lösningen.

Vi har diskuterat en enkel lösning i artikeln nedan. 
Det minsta talet är delbart med de första n talen
Ovanstående lösning fungerar bra för enkel ingång. Men om vi har flera ingångar är det en bra idé att använda Sil av Eratosthenes att lagra alla primära faktorer. Som vi vet om LCM(a b) = X så kommer vilken primfaktor som helst av a eller b också att vara primfaktorn för 'X'.  



  1. Initiera lcm-variabeln med 1
  2. Generera en sikt av Eratosthenes (boolvektor är Prime) med längden 10^6 (helst måste vara lika med antalet siffror i faktorial)
  3. För varje tal i boolvektorn isPrime om talet är primtal (isPrime[i] är sant) hitta det maximala antalet som är mindre än det givna talet och lika med potensen av primtal.
  4. Multiplicera sedan detta tal med variabeln lcm.
  5. Upprepa steg 3 och 4 tills primtal är mindre än det angivna talet.

Illustration:  

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

Nedan är implementeringen av ovanstående idé.  

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 ?>

Produktion 
420

Tidskomplexitet : O(n2) 
Hjälputrymme: På)

Ett annat tillvägagångssätt:

 Tanken är att om talet är mindre än 3 så returnera numret. Om talet är större än 2, hitta LCM för nn-1

  • Låt oss säga x=LCM(nn-1)
  • igen x=LCM(xn-2)
  • igen x=LCM(xn-3) ...
  • .
  • .
  • igen x=LCM(x1) ...

nu är resultatet x.

För att hitta LCM(ab) använder vi en funktion hcf(ab) som returnerar HCF av (ab)

Det vet vi LCM(ab)= (a*b)/HCF(ab)

Illustration:  

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

Nedan är implementeringen av ovanstående tillvägagångssätt

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>

Produktion 
420

Tidskomplexitet: O(n log n)
Hjälputrymme: O(1)