#practiceLinkDiv { näyttö: ei mitään !tärkeää; }Kun annetaan N kokonaislukuelementin taulukko arr[], tehtävänä on löytää tämän taulukon kaikkien osajoukkojen keskiarvon summa.
ydin java
Esimerkki:
Input : arr[] = [2 3 5]Recommended Practice Kaikkien osajoukkojen keskiarvon summa Kokeile sitä!
Output : 23.33
Explanation : Subsets with their average are
[2] average = 2/1 = 2
[3] average = 3/1 = 3
[5] average = 5/1 = 5
[2 3] average = (2+3)/2 = 2.5
[2 5] average = (2+5)/2 = 3.5
[3 5] average = (3+5)/2 = 4
[2 3 5] average = (2+3+5)/3 = 3.33
Sum of average of all subset is
2 + 3 + 5 + 2.5 + 3.5 + 4 + 3.33 = 23.33
Naivi lähestymistapa: Naiivi ratkaisu on iteroida läpi kaikki mahdolliset osajoukot saada an keskimäärin kaikista niistä ja lisää ne sitten yksitellen, mutta tämä vie eksponentiaalisesti aikaa ja on mahdotonta isommille taulukoille.
Voimme saada mallin ottamalla esimerkin
arr = [a0 a1 a2 a3]
sum of average =
a0/1 + a1/1 + a2/2 + a3/1 +
(a0+a1)/2 + (a0+a2)/2 + (a0+a3)/2 + (a1+a2)/2 +
(a1+a3)/2 + (a2+a3)/2 +
(a0+a1+a2)/3 + (a0+a2+a3)/3 + (a0+a1+a3)/3 +
(a1+a2+a3)/3 +
(a0+a1+a2+a3)/4
If S = (a0+a1+a2+a3) then above expression
can be rearranged as below
sum of average = (S)/1 + (3*S)/2 + (3*S)/3 + (S)/4
Kerroin osoittajilla voidaan selittää seuraavasti, oletetaan, että iteroitamme osajoukkoja, joissa on K elementtiä, jolloin nimittäjä on K ja osoittaja on r*S, missä 'r' tarkoittaa, kuinka monta kertaa tietty taulukkoelementti lisätään, kun iteroidaan samankokoisten osajoukkojen yli. Tarkastelemalla voimme nähdä, että r tulee olemaan nCr(N - 1 n - 1), koska yhden elementin summauksen jälkeen meidän on valittava (n - 1) elementtiä (N - 1) alkioista, jotta jokaisen elementin taajuus on nCr(N - 1 n - 1) samalla kun otetaan huomioon samankokoiset osajoukot, koska kaikki alkiot osallistuvat, ja tämä summauskertojen määrä on yhtä suuri. ilmaisua.
Alla olevassa koodissa nCr on toteutettu dynaamisella ohjelmointimenetelmällä voit lukea siitä lisää täältä
C++// C++ program to get sum of average of all subsets #include
// java program to get sum of // average of all subsets import java.io.*; class GFG { // Returns value of Binomial // Coefficient C(n k) static int nCr(int n int k) { int C[][] = new int[n + 1][k + 1]; int i j; // Calculate value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i k); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } return C[n][k]; } // method returns sum of average of all subsets static double resultOfAllSubsets(int arr[] int N) { // Initialize result double result = 0.0; // Find sum of elements int sum = 0; for (int i = 0; i < N; i++) sum += arr[i]; // looping once for all subset of same size for (int n = 1; n <= N; n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ result += (double)(sum * (nCr(N - 1 n - 1))) / n; return result; } // Driver code to test above methods public static void main(String[] args) { int arr[] = { 2 3 5 7 }; int N = arr.length; System.out.println(resultOfAllSubsets(arr N)); } } // This code is contributed by vt_m
// C# program to get sum of // average of all subsets using System; class GFG { // Returns value of Binomial // Coefficient C(n k) static int nCr(int n int k) { int[ ] C = new int[n + 1 k + 1]; int i j; // Calculate value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.Min(i k); j++) { // Base Cases if (j == 0 || j == i) C[i j] = 1; // Calculate value using // previously stored values else C[i j] = C[i - 1 j - 1] + C[i - 1 j]; } } return C[n k]; } // method returns sum of average // of all subsets static double resultOfAllSubsets(int[] arr int N) { // Initialize result double result = 0.0; // Find sum of elements int sum = 0; for (int i = 0; i < N; i++) sum += arr[i]; // looping once for all subset // of same size for (int n = 1; n <= N; n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ result += (double)(sum * (nCr(N - 1 n - 1))) / n; return result; } // Driver code to test above methods public static void Main() { int[] arr = { 2 3 5 7 }; int N = arr.Length; Console.WriteLine(resultOfAllSubsets(arr N)); } } // This code is contributed by Sam007
<script> // javascript program to get sum of // average of all subsets // Returns value of Binomial // Coefficient C(n k) function nCr(n k) { let C = new Array(n + 1); for (let i = 0; i <= n; i++) { C[i] = new Array(k + 1); for (let j = 0; j <= k; j++) { C[i][j] = 0; } } let i j; // Calculate value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i k); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } return C[n][k]; } // method returns sum of average of all subsets function resultOfAllSubsets(arr N) { // Initialize result let result = 0.0; // Find sum of elements let sum = 0; for (let i = 0; i < N; i++) sum += arr[i]; // looping once for all subset of same size for (let n = 1; n <= N; n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ result += (sum * (nCr(N - 1 n - 1))) / n; return result; } let arr = [ 2 3 5 7 ]; let N = arr.length; document.write(resultOfAllSubsets(arr N)); </script>
// PHP program to get sum // of average of all subsets // Returns value of Binomial // Coefficient C(n k) function nCr($n $k) { $C[$n + 1][$k + 1] = 0; $i; $j; // Calculate value of Binomial // Coefficient in bottom up manner for ($i = 0; $i <= $n; $i++) { for ($j = 0; $j <= min($i $k); $j++) { // Base Cases if ($j == 0 || $j == $i) $C[$i][$j] = 1; // Calculate value using // previously stored values else $C[$i][$j] = $C[$i - 1][$j - 1] + $C[$i - 1][$j]; } } return $C[$n][$k]; } // method returns sum of // average of all subsets function resultOfAllSubsets($arr $N) { // Initialize result $result = 0.0; // Find sum of elements $sum = 0; for ($i = 0; $i < $N; $i++) $sum += $arr[$i]; // looping once for all // subset of same size for ($n = 1; $n <= $N; $n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ $result += (($sum * (nCr($N - 1 $n - 1))) / $n); return $result; } // Driver Code $arr = array( 2 3 5 7 ); $N = sizeof($arr) / sizeof($arr[0]); echo resultOfAllSubsets($arr $N) ; // This code is contributed by nitin mittal. ?> # Python3 program to get sum # of average of all subsets # Returns value of Binomial # Coefficient C(n k) def nCr(n k): C = [[0 for i in range(k + 1)] for j in range(n + 1)] # Calculate value of Binomial # Coefficient in bottom up manner for i in range(n + 1): for j in range(min(i k) + 1): # Base Cases if (j == 0 or j == i): C[i][j] = 1 # Calculate value using # previously stored values else: C[i][j] = C[i-1][j-1] + C[i-1][j] return C[n][k] # Method returns sum of # average of all subsets def resultOfAllSubsets(arr N): result = 0.0 # Initialize result # Find sum of elements sum = 0 for i in range(N): sum += arr[i] # looping once for all subset of same size for n in range(1 N + 1): # each element occurs nCr(N-1 n-1) times while # considering subset of size n */ result += (sum * (nCr(N - 1 n - 1))) / n return result # Driver code arr = [2 3 5 7] N = len(arr) print(resultOfAllSubsets(arr N)) # This code is contributed by Anant Agarwal.
Lähtö
63.75
Aika monimutkaisuus: O(n3)
Aputila: O(n2)
Tehokas lähestymistapa: tilan optimointi O(1)
Yllä olevan lähestymistavan tilan monimutkaisuuden optimoimiseksi voimme käyttää tehokkaampaa lähestymistapaa, joka välttää koko matriisin tarpeen C[][] tallentaa binomikertoimia. Sen sijaan voimme käyttää yhdistelmäkaavaa laskeaksemme binomikertoimen suoraan tarvittaessa.
Käyttöönoton vaiheet:
- Toista taulukon elementit ja laske kaikkien elementtien summa.
- Toista jokaisen osajoukon kokoa 1:stä N:ään.
- Laske silmukan sisällä keskimäärin elementtien summasta kerrottuna osajoukon koon binomiaalikertoimella. Lisää tulokseen laskettu keskiarvo.
- Palauta lopputulos.
Toteutus:
C++#include
import java.util.Arrays; public class Main { // Method to calculate binomial coefficient C(n k) static int binomialCoeff(int n int k) { int res = 1; // Since C(n k) = C(n n-k) if (k > n - k) k = n - k; // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for (int i = 0; i < k; i++) { res *= (n - i); res /= (i + 1); } return res; } // Method to calculate the sum of the average of all subsets static double resultOfAllSubsets(int arr[] int N) { double result = 0.0; int sum = 0; // Calculate the sum of elements for (int i = 0; i < N; i++) sum += arr[i]; // Loop for each subset size for (int n = 1; n <= N; n++) result += (double) (sum * binomialCoeff(N - 1 n - 1)) / n; return result; } // Driver code to test the above methods public static void main(String[] args) { int arr[] = {2 3 5 7}; int N = arr.length; System.out.println(resultOfAllSubsets(arr N)); } }
using System; public class MainClass { // Method to calculate binomial coefficient C(n k) static int BinomialCoeff(int n int k) { int res = 1; // Since C(n k) = C(n n-k) if (k > n - k) k = n - k; // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for (int i = 0; i < k; i++) { res *= (n - i); res /= (i + 1); } return res; } // Method to calculate the sum of the average of all subsets static double ResultOfAllSubsets(int[] arr int N) { double result = 0.0; int sumVal = 0; // Calculate the sum of elements for (int i = 0; i < N; i++) sumVal += arr[i]; // Loop for each subset size for (int n = 1; n <= N; n++) result += (double)(sumVal * BinomialCoeff(N - 1 n - 1)) / n; return result; } // Driver code to test the above methods public static void Main() { int[] arr = { 2 3 5 7 }; int N = arr.Length; Console.WriteLine(ResultOfAllSubsets(arr N)); } }
// Function to calculate binomial coefficient C(n k) function binomialCoeff(n k) { let res = 1; // Since C(n k) = C(n n-k) if (k > n - k) k = n - k; // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for (let i = 0; i < k; i++) { res *= (n - i); res /= (i + 1); } return res; } // Function to calculate the sum of the average of all subsets function resultOfAllSubsets(arr) { let result = 0.0; let sum = arr.reduce((acc val) => acc + val 0); // Loop for each subset size for (let n = 1; n <= arr.length; n++) { result += (sum * binomialCoeff(arr.length - 1 n - 1)) / n; } return result; } const arr = [2 3 5 7]; console.log(resultOfAllSubsets(arr));
# Method to calculate binomial coefficient C(n k) def binomialCoeff(n k): res = 1 # Since C(n k) = C(n n-k) if k > n - k: k = n - k # Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for i in range(k): res *= (n - i) res //= (i + 1) return res # Method to calculate the sum of the average of all subsets def resultOfAllSubsets(arr N): result = 0.0 sum_val = 0 # Calculate the sum of elements for i in range(N): sum_val += arr[i] # Loop for each subset size for n in range(1 N + 1): result += (sum_val * binomialCoeff(N - 1 n - 1)) / n return result # Driver code to test the above methods arr = [2 3 5 7] N = len(arr) print(resultOfAllSubsets(arr N))
Lähtö
63.75 Aika monimutkaisuus: O(n^2)
Aputila: O(1)