1 olacak şekilde bir n sayısı verildiğinde<= N <= 10^6 the Task is to Find the LCM of First n Natural Numbers.
Örnekler:
Input : n = 5 Output : 60 Input : n = 6 Output : 60 Input : n = 7 Output : 420
Çözüme geçmeden önce buraya tıklayıp pratik yapmanızı önemle tavsiye ederiz.
Aşağıdaki makalede basit bir çözümü tartıştık.
İlk n sayıya bölünebilen en küçük sayı
Yukarıdaki çözüm tek giriş için gayet iyi çalışıyor. Ancak birden fazla girdimiz varsa bunu kullanmak iyi bir fikirdir. Eratostenes Eleği tüm asal faktörleri saklamak için. Bildiğimiz gibi LCM(a b) = X ise a veya b'nin herhangi bir asal çarpanı aynı zamanda 'X'in de asal faktörü olacaktır.
- Lcm değişkenini 1 ile başlat
- 10^6 uzunluğunda bir Eratosthenes Eleği (bool vektörü isPrime) oluşturun (ideal olarak faktöriyeldeki basamak sayısına eşit olmalıdır)
- Şimdi bool vektörü isPrime'daki her sayı için, eğer sayı asalsa (isPrime[i] doğrudur), verilen sayıdan küçük ve asal sayının kuvvetine eşit olan maksimum sayıyı bulun.
- Daha sonra bu sayıyı lcm değişkeniyle çarpın.
- Asal verilen sayıdan küçük olana kadar 3. ve 4. adımları tekrarlayın.
İllüstrasyon:
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
Aşağıda yukarıdaki fikrin uygulanması yer almaktadır.
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 ?> Çıkış
420
Zaman Karmaşıklığı : Açık2)
Yardımcı Alan: Açık)
Başka Bir Yaklaşım:
Buradaki fikir, eğer sayı 3'ten küçükse sayıyı döndürmektir. Sayı 2'den büyükse nn-1'in LCM'sini bulun
- Diyelim ki x=LCM(nn-1)
- tekrar x=LCM(xn-2)
- tekrar x=LCM(xn-3) ...
- .
- .
- tekrar x=LCM(x1) ...
şimdi sonuç x'tir.
LCM(ab)'yi bulmak için (ab)'nin HCF'sini döndürecek hcf(ab) fonksiyonunu kullanırız.
Bunu biliyoruz LCM(ab)= (a*b)/HCF(ab)
İllüstrasyon:
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
Yukarıdaki yaklaşımın uygulanması aşağıdadır
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>
Çıkış
420
Zaman karmaşıklığı: O(n log n)
Yardımcı Alan: Ç(1)