Contar formas de representar un número como suma de cuadrados perfectos

Posted on by Rudeus Greyrat

Dado un número entero N , la tarea es encontrar el número de formas de representar el número N como suma de cuadrados perfectos .

Ejemplos:

Entrada: N = 9
Salida: 4
Explicación:
Hay cuatro maneras de representar 9 como la suma de cuadrados perfectos:
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9
1 + 1 + 1 + 1 + 1 + 4 = 9
1 + 4 + 4 = 9
9 = 9

Entrada: N =8
Salida: 3

 

Enfoque ingenuo: la idea es almacenar todos los cuadrados perfectos menores o iguales a N en una array . El problema ahora se reduce a encontrar las formas de sumar a N usando elementos de array con repetición permitida , lo que puede resolverse usando recursividad . Siga los pasos a continuación para resolver el problema:

  • Almacene todos los cuadrados perfectos menores o iguales a N y en una array psquare[] .
  • Cree una función recursiva countWays(index, target) que tome dos parámetros index , (inicialmente N-1) y target (inicialmente N):
    • Manejar los casos base:
      • Si el destino es 0, devuelve 1.
      • Si el índice o el destino es menor que 0, devuelve 0.
    • De lo contrario, incluya el elemento psquare[index] en la suma restándolo del objetivo y llamando recursivamente al valor restante de target .
    • Excluya el elemento, psquare[index] de la suma moviéndose al siguiente índice y llamando recursivamente al mismo valor de target .
    • Devuelve la suma obtenida al incluir y excluir el elemento.
  • Imprime el valor de countWays(N-1, N) como resultado.

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Store perfect squares
// less than or equal to N
vector<int> psquare;
 
// Utility function to calculate perfect
// squares less than or equal to N
void calcPsquare(int N)
{
    for (int i = 1; i * i <= N; i++)
        psquare.push_back(i * i);
}
 
// Function to find the number
// of ways to represent a number
// as sum of perfect squares
int countWays(int index, int target)
{
    // Handle the base cases
    if (target == 0)
        return 1;
 
    if (index < 0 || target < 0)
        return 0;
 
    // Include the i-th index element
    int inc = countWays(
        index, target - psquare[index]);
 
    // Exclude the i-th index element
    int exc = countWays(index - 1, target);
 
    // Return the result
    return inc + exc;
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 9;
 
    // Precalculate perfect
    // squares <= N
    calcPsquare(N);
 
    // Function Call
    cout << countWays(psquare.size() - 1, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Store perfect squares
    // less than or equal to N
    static ArrayList<Integer> psquare = new ArrayList<>();
 
    // Utility function to calculate perfect
    // squares less than or equal to N
    static void calcPsquare(int N)
    {
        for (int i = 1; i * i <= N; i++)
            psquare.add(i * i);
    }
 
    // Function to find the number
    // of ways to represent a number
    // as sum of perfect squares
    static int countWays(int index, int target)
    {
        // Handle the base cases
        if (target == 0)
            return 1;
 
        if (index < 0 || target < 0)
            return 0;
 
        // Include the i-th index element
        int inc
            = countWays(index, target - psquare.get(index));
 
        // Exclude the i-th index element
        int exc = countWays(index - 1, target);
 
        // Return the result
        return inc + exc;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given Input
        int N = 9;
 
        // Precalculate perfect
        // squares <= N
        calcPsquare(N);
 
        // Function Call
        System.out.print(countWays(psquare.size() - 1, N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Store perfect squares
# less than or equal to N
psquare = []
 
# Utility function to calculate perfect
# squares less than or equal to N
def calcPsquare(N):
     
    for i in range(1, N):
        if i * i > N:
            break
         
        psquare.append(i * i)
 
# Function to find the number
# of ways to represent a number
# as sum of perfect squares
def countWays(index, target):
     
    # Handle the base cases
    if (target == 0):
        return 1
 
    if (index < 0 or target < 0):
        return 0
 
    # Include the i-th index element
    inc = countWays(index, target - psquare[index])
 
    # Exclude the i-th index element
    exc = countWays(index - 1, target)
 
    # Return the result
    return inc + exc
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    N = 9
 
    # Precalculate perfect
    # squares <= N
    calcPsquare(N)
 
    # Function Call
    print (countWays(len(psquare) - 1, N))
 
# This code is contributed by mohit kumar 29

C#

using System.IO;
using System;
using System.Collections;
 
class GFG {
    // Store perfect squares
    // less than or equal to N
    static ArrayList psquare = new ArrayList();
 
    // Utility function to calculate perfect
    // squares less than or equal to N
    static void calcPsquare(int N)
    {
        for (int i = 1; i * i <= N; i++)
            psquare.Add(i * i);
    }
 
    // Function to find the number
    // of ways to represent a number
    // as sum of perfect squares
    static int countWays(int index, int target)
    {
        // Handle the base cases
        if (target == 0)
            return 1;
 
        if (index < 0 || target < 0)
            return 0;
 
        // Include the i-th index element
        int inc = countWays(index,
                            target - (int)psquare[index]);
 
        // Exclude the i-th index element
        int exc = countWays(index - 1, target);
 
        // Return the result
        return inc + exc;
    }
 
    static void Main()
    {
       
        // Given Input
        int N = 9;
 
        // Precalculate perfect
        // squares <= N
        calcPsquare(N);
 
        // Function Call
        Console.WriteLine(countWays(psquare.Count - 1, N));
    }
}
 
// This code is contributed by abhinavjain194.

Javascript

<script>
 
// JavaScript program for the above approach
 
// Store perfect squares
// less than or equal to N
var psquare = []
 
// Utility function to calculate perfect
// squares less than or equal to N
function calcPsquare(N)
{
    var i;
    for (i = 1; i * i <= N; i++)
        psquare.push(i * i);
}
 
// Function to find the number
// of ways to represent a number
// as sum of perfect squares
function countWays(index, target)
{
    // Handle the base cases
    if (target == 0)
        return 1;
 
    if (index < 0 || target < 0)
        return 0;
 
    // Include the i-th index element
    var inc = countWays(
        index, target - psquare[index]);
 
    // Exclude the i-th index element
    var exc = countWays(index - 1, target);
 
    // Return the result
    return inc + exc;
}
 
// Driver Code
    // Given Input
    var N = 9;
 
    // Precalculate perfect
    // squares <= N
    calcPsquare(N);
 
    // Function Call
    document.write(countWays(psquare.length - 1, N));
 
</script>
Producción: 

4

 

Complejidad Temporal: O(2 K ), donde K es el número de cuadrados perfectos menores o iguales a N
Espacio Auxiliar: O(1)

Enfoque eficiente: este problema tiene subproblemas superpuestos y una propiedad de subestructura óptima . Para optimizar el enfoque anterior, la idea es usar programación dinámica al memorizar las llamadas recursivas anteriores usando una array 2D de tamaño K*N , donde K es el número de cuadrados perfectos menores o iguales que N.

A continuación se muestra la implementación del enfoque anterior: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Store perfect squares
// less than or equal to N
vector<int> psquare;
 
// Utility function to calculate
// perfect squares <= N
void calcPsquare(int N)
{
    for (int i = 1; i * i <= N; i++)
        psquare.push_back(i * i);
}
 
// DP array for memoization
vector<vector<int> > dp;
 
// Recursive function to count
// number of ways to represent
// a number as a sum of perfect squares
int countWaysUtil(int index, int target)
{
    // Handle base cases
    if (target == 0)
        return 1;
    if (index < 0 || target < 0)
        return 0;
 
    // If already computed, return the result
    if (dp[index][target] != -1)
        return dp[index][target];
 
    // Else, compute the result
    return dp[index][target]
           = countWaysUtil(
                 index, target - psquare[index])
 
             + countWaysUtil(
                   index - 1, target);
}
 
// Function to find the number of ways to
// represent a number as a sum of perfect squares
int countWays(int N)
{
    // Precalculate perfect squares less
    // than or equal to N
    calcPsquare(N);
 
    // Create dp array to memoize
    dp.resize(psquare.size() + 1,
              vector<int>(N + 1, -1));
 
    // Function call to fill dp array
    return countWaysUtil(psquare.size() - 1, N);
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 9;
 
    // Function Call
    cout << countWays(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
     
    // Store perfect squares
    // less than or equal to N
    private static ArrayList<Integer> psquare;
 
     
    // Utility function to calculate
    // perfect squares <= N
    private static void calcPsquare(int n) {
 
        for (int i = 1; i * i <= n; i++)
            psquare.add(i * i);
         
    }
     
    // DP array for memoization
    private static int[][] dp;
     
    // Recursive function to count
    // number of ways to represent
    // a number as a sum of perfect squares
    private static int countWaysUtil(int index, int target) {
        // Handle base cases
        if (target == 0)
            return 1;
        if (index < 0 || target < 0)
            return 0;
      
        // If already computed, return the result
        if (dp[index][target] != -1)
            return dp[index][target];
      
        // Else, compute the result
        return dp[index][target]
               = countWaysUtil(
                     index, target - psquare.get(index))
      
                 + countWaysUtil(
                       index - 1, target);
    }
     
    // Function to find the number of ways to
    // represent a number as a sum of perfect squares
    private static int countWays(int n) {
        // Precalculate perfect squares less
        // than or equal to N
        psquare = new ArrayList<Integer>();
        calcPsquare(n);
      
        // Create dp array to memoize
        dp = new int[psquare.size()+1][n+1];
        for(int i = 0; i<=psquare.size(); i++)Arrays.fill(dp[i], -1);
      
        // Function call to fill dp array
        return countWaysUtil(psquare.size() - 1, n);
    }
 
 
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given Input
        int N = 9;
 
        // Function Call
        System.out.print(countWays(N));
    }
     
}
 
// This code is contributed by Dheeraj Bhagchandani.

Python3

# Python3 program for the above approach
from math import sqrt
 
# Store perfect squares
# less than or equal to N
psquare = []
 
# DP array for memoization
dp = []
 
# Utility function to calculate
# perfect squares <= N
def calcPsquare(N):
     
    global psquare
    for i in range(1, int(sqrt(N)) + 1, 1):
        psquare.append(i * i)
 
# Recursive function to count
# number of ways to represent
# a number as a sum of perfect squares
def countWaysUtil(index, target):
     
    global dp
     
    # Handle base cases
    if (target == 0):
        return 1
    if (index < 0 or target < 0):
        return 0
 
    # If already computed, return the result
    if (dp[index][target] != -1):
        return dp[index][target]
 
    dp[index][target] = (countWaysUtil(
                               index, target - psquare[index]) +
                         countWaysUtil(index - 1, target))
 
    # Else, compute the result
    return dp[index][target]
 
# Function to find the number of ways to
# represent a number as a sum of perfect squares
def countWays(N):
     
    global dp
    global psquare
     
    # Precalculate perfect squares less
    # than or equal to N
    calcPsquare(N)
    temp = [-1 for i in range(N + 1)]
     
    # Create dp array to memoize
    dp = [temp for i in range(len(psquare) + 1)]
 
    # Function call to fill dp array
    return countWaysUtil(len(psquare) - 1, N) - 1
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    N = 9
     
    # Function Call
    print(countWays(N))
 
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Store perfect squares
// less than or equal to N
static List<int> psquare;
 
// Utility function to calculate
// perfect squares <= N
private static void calcPsquare(int n)
{
    for(int i = 1; i * i <= n; i++)
        psquare.Add(i * i);
}
 
// DP array for memoization
private static int[,]dp;
 
// Recursive function to count
// number of ways to represent
// a number as a sum of perfect squares
private static int countWaysUtil(int index,
                                 int target)
{
     
    // Handle base cases
    if (target == 0)
        return 1;
    if (index < 0 || target < 0)
        return 0;
  
    // If already computed, return the result
    if (dp[index, target] != -1)
        return dp[index, target];
  
    // Else, compute the result
    return dp[index, target] = countWaysUtil(index,
                                             target - psquare[index]) +
                               countWaysUtil(index - 1, target);
}
 
// Function to find the number of ways to
// represent a number as a sum of perfect squares
private static int countWays(int n)
{
     
    // Precalculate perfect squares less
    // than or equal to N
    psquare = new List<int>();
    calcPsquare(n);
  
    // Create dp array to memoize
    dp = new int[psquare.Count + 1, n + 1];
    for(int i = 0; i <= psquare.Count; i++)
    {
        for(int j = 0; j <= n; j++)
        {
            dp[i, j] = -1;
        }
         
        //Array.Fill(dp[i], -1);
    }
  
    // Function call to fill dp array
    return countWaysUtil(psquare.Count - 1, n);
}
 
// Driver Code
static void Main()
{
     
    // Given Input
    int N = 9;
 
    // Function Call
   Console.Write(countWays(N));
}
}
 
// This code is contributed by SoumikMondal

Javascript

<script>
 
// JavaScript program for the above approach
 
let psquare;
 
function calcPsquare(n)
{
     for (let i = 1; i * i <= n; i++)
            psquare.push(i * i);
}
 
 // DP array for memoization
let dp;
 
// Recursive function to count
    // number of ways to represent
    // a number as a sum of perfect squares
function countWaysUtil(index,target)
{
    // Handle base cases
        if (target == 0)
            return 1;
        if (index < 0 || target < 0)
            return 0;
       
        // If already computed, return the result
        if (dp[index][target] != -1)
            return dp[index][target];
       
        // Else, compute the result
        return dp[index][target]
               = countWaysUtil(
                     index, target - psquare[index])
       
                 + countWaysUtil(
                       index - 1, target);
}
 
// Function to find the number of ways to
    // represent a number as a sum of perfect squares
function countWays(n)
{
    // Precalculate perfect squares less
        // than or equal to N
        psquare = [];
        calcPsquare(n);
       
        // Create dp array to memoize
        dp = new Array(psquare.length+1);
        for(let i=0;i<psquare.length+1;i++)
        {
            dp[i]=new Array(n+1);
            for(let j=0;j<n+1;j++)
            {
                dp[i][j]=-1;
            }
        }
         
       
        // Function call to fill dp array
        return countWaysUtil(psquare.length - 1, n);
}
 
// Driver Code
// Given Input
let N = 9;
 
// Function Call
document.write(countWays(N));
 
 
// This code is contributed by patel2127
 
</script>
Producción

4

Complejidad Temporal: O(K*N), donde K es el número de cuadrados perfectos menores o iguales a N
Espacio Auxiliar: O(K*N)

Publicación traducida automáticamente

Artículo escrito por saikumarkudikala y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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