Atsižvelgiant į surūšiuotą skirtingų teigiamų sveikųjų skaičių masyvą, spausdinami visi trynukai, sudarantys geometrinę progresiją su integraliu bendruoju santykiu.
Geometrinė progresija yra skaičių seka, kurioje kiekvienas narys po pirmojo randamas padauginus ankstesnįjį iš fiksuoto ne nulio skaičiaus, vadinamo bendruoju santykiu. Pavyzdžiui, seka 2 6 18 54... yra geometrinė progresija su bendru santykiu 3.
Pavyzdžiai:
Input: arr = [1 2 6 10 18 54] Output: 2 6 18 6 18 54 Input: arr = [2 8 10 15 16 30 32 64] Output: 2 8 32 8 16 32 16 32 64 Input: arr = [ 1 2 6 18 36 54] Output: 2 6 18 1 6 36 6 18 54
Idėja yra pradėti nuo antrojo elemento ir pritvirtinti kiekvieną elementą kaip vidurinį elementą ir ieškoti kitų dviejų elementų triplete (vienas mažesnis ir vienas didesnis). Kad elementas arr[j] būtų geometrinės progresijos vidurys, turi egzistuoti tokie elementai arr[i] ir arr[k], kad
arr[j] / arr[i] = r and arr[k] / arr[j] = r where r is an positive integer and 0 <= i < j and j < k <= n - 1
Žemiau yra aukščiau pateiktos idėjos įgyvendinimas
C++// C++ program to find if there exist three elements in // Geometric Progression or not #include
// Java program to find if there exist three elements in // Geometric Progression or not import java.util.*; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets(int arr[] int n) { // One by fix every element as middle element for (int j = 1; j < n - 1; j++) { // Initialize i and k for the current j int i = j - 1 k = j + 1; // Find all i and k such that (i j k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet System.out.println(arr[i] +' ' + arr[j] + ' ' + arr[k]); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code public static void main(String[] args) { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int arr[] = {1 2 4 16}; // int arr[] = {1 2 3 6 18 22}; int n = arr.length; findGeometricTriplets(arr n); } } // This code is contributed by Rajput-Ji
# Python 3 program to find if # there exist three elements in # Geometric Progression or not # The function prints three elements # in GP if exists. # Assumption: arr[0..n-1] is sorted. def findGeometricTriplets(arr n): # One by fix every element # as middle element for j in range(1 n - 1): # Initialize i and k for # the current j i = j - 1 k = j + 1 # Find all i and k such that # (i j k) forms a triplet of GP while (i >= 0 and k <= n - 1): # if arr[j]/arr[i] = r and # arr[k]/arr[j] = r and r # is an integer (i j k) forms # Geometric Progression while (arr[j] % arr[i] == 0 and arr[k] % arr[j] == 0 and arr[j] // arr[i] == arr[k] // arr[j]): # print the triplet print( arr[i] ' ' arr[j] ' ' arr[k]) # Since the array is sorted and # elements are distinct. k += 1 i -= 1 # if arr[j] is multiple of arr[i] # and arr[k] is multiple of arr[j] # then arr[j] / arr[i] != arr[k] / arr[j]. # We compare their values to # move to next k or previous i. if(arr[j] % arr[i] == 0 and arr[k] % arr[j] == 0): if(arr[j] // arr[i] < arr[k] // arr[j]): i -= 1 else: k += 1 # else if arr[j] is multiple of # arr[i] then try next k. Else # try previous i. elif (arr[j] % arr[i] == 0): k += 1 else: i -= 1 # Driver code if __name__ =='__main__': arr = [1 2 4 16] n = len(arr) findGeometricTriplets(arr n) # This code is contributed # by ChitraNayal
// C# program to find if there exist three elements // in Geometric Progression or not using System; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets(int []arr int n) { // One by fix every element as middle element for (int j = 1; j < n - 1; j++) { // Initialize i and k for the current j int i = j - 1 k = j + 1; // Find all i and k such that (i j k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet Console.WriteLine(arr[i] +' ' + arr[j] + ' ' + arr[k]); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code static public void Main () { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int []arr = {1 2 4 16}; // int arr[] = {1 2 3 6 18 22}; int n = arr.Length; findGeometricTriplets(arr n); } } // This code is contributed by ajit.
<script> // Javascript program to find if there exist three elements in // Geometric Progression or not // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. function findGeometricTriplets(arrn) { // One by fix every element as middle element for (let j = 1; j < n - 1; j++) { // Initialize i and k for the current j let i = j - 1 k = j + 1; // Find all i and k such that (i j k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet document.write(arr[i] +' ' + arr[j] + ' ' + arr[k]+'
'); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; let arr = [1 2 4 16]; // int arr[] = {1 2 3 6 18 22}; let n = arr.length; findGeometricTriplets(arr n); // This code is contributed by avanitrachhadiya2155 </script>
Išvestis
1 2 4 1 4 16
Laiko sudėtingumas aukščiau pateiktas sprendimas yra O (n2), kaip ir kiekvienam j, randame i ir k tiesiniu laiku.
Pagalbinė erdvė: O(1) nes nenaudojome jokios papildomos vietos.