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📜  半完美数

📅  最后修改于: 2021-05-06 08:38:45             🧑  作者: Mango

在数论中，半完美数或伪完美数是自然数n，它等于其所有或某些适当除数的和。等于其所有适当除数之和的半完美数是一个理想数。
给定一个数字，任务是检查该数字是否为半完美数字。
例子：

方法：将所有数量因子存储在数据结构(向量或数组)中。按升序对它们进行排序。一旦存储了因子，就可以使用动态编程来检查是否有任何组合形式为N。该问题变得类似于“子集总和问题”。我们可以使用相同的方法并检查数字是否为半完美数字。
下面是上述方法的实现。

C++
// C++ program to check if the number
// is semi-perfect or not
#include 
using namespace std;
 
// code to find all the factors of
// the number excluding the number itself
vector factors(int n)
{
    // vector to store the factors
    vector v;
    v.push_back(1);
 
    // note that this loop runs till sqrt(n)
    for (int i = 2; i <= sqrt(n); i++) {
 
        // if the value of i is a factor
        if (n % i == 0) {
            v.push_back(i);
 
            // condition to check the
            // divisor is not the number itself
            if (n / i != i) {
                v.push_back(n / i);
            }
        }
    }
    // return the vector
    return v;
}
 
// Function to check if the
// number is semi-perfecr or not
bool check(int n)
{
    vector v;
 
    // find the divisors
    v = factors(n);
 
    // sorting the vector
    sort(v.begin(), v.end());
 
    int r = v.size();
 
    // subset to check if no is semiperfect
    bool subset[r + 1][n + 1];
 
    // initialising 1st column to true
    for (int i = 0; i <= r; i++)
        subset[i][0] = true;
 
    // initializing 1st row except zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
 
    // loop to find whether the number is semiperfect
    for (int i = 1; i <= r; i++) {
        for (int j = 1; j <= n; j++) {
 
            // calculation to check if the
            // number can be made by summation of diviors
            if (j < v[i - 1])
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] ||
                               subset[i - 1][j - v[i - 1]];
            }
        }
    }
 
    // if not possible to make the
    // number by any combination of divisors
    if ((subset[r][n]) == 0)
        return false;
    else
        return true;
}
 
// driver code to check if possible
int main()
{
    int n = 40;
    if (check(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java
// Java program to check
// if the number is
// semi-perfect or not
import java.util.*;
class GFG{
 
// Code to find all the factors of
// the number excluding the number itself
static Vector factors(int n)
{
  // vector to store the factors
  Vector v = new Vector<>();
  v.add(1);
 
  // note that this loop runs till Math.sqrt(n)
  for (int i = 2; i <= Math.sqrt(n); i++)
  {
    // if the value of i is a factor
    if (n % i == 0)
    {
      v.add(i);
 
      // condition to check the
      // divisor is not the number itself
      if (n / i != i)
      {
        v.add(n / i);
      }
    }
  }
   
  // return the vector
  return v;
}
 
// Function to check if the
// number is semi-perfecr or not
static boolean check(int n)
{
  Vector v = new Vector<>();
 
  // find the divisors
  v = factors(n);
 
  // sorting the vector
  Collections.sort(v);
 
  int r = v.size();
 
  // subset to check if no
  // is semiperfect
  boolean [][]subset =
          new boolean[r + 1][n + 1];
 
  // initialising 1st column to true
  for (int i = 0; i <= r; i++)
    subset[i][0] = true;
 
  // initializing 1st row except
  // zero position to 0
  for (int i = 1; i <= n; i++)
    subset[0][i] = false;
 
  // loop to find whether the
  // number is semiperfect
  for (int i = 1; i <= r; i++)
  {
    for (int j = 1; j <= n; j++)
    {
      // calculation to check if the
      // number can be made by
      // summation of diviors
      if (j < v.elementAt(i - 1))
        subset[i][j] = subset[i - 1][j];
      else
      {
        subset[i][j] = subset[i - 1][j] ||
                       subset[i - 1][j - v.elementAt(i - 1)];
      }
    }
  }
 
  // If not possible to make the
  // number by any combination of divisors
  if ((subset[r][n]) == false)
    return false;
  else
    return true;
}
 
// Driver code to check if possible
public static void main(String[] args)
{
  int n = 40;
  if (check(n))
    System.out.print("Yes");
  else
    System.out.print("No");
}
}
 
// This code is contributed by gauravrajput1

Python3
# Python3 program to check if the number
# is semi-perfect or not
import math
 
# code to find all the factors of
# the number excluding the number itself
def factors( n):
 
    # vector to store the factors
    v = []
    v.append(1)
 
    # note that this loop runs till sqrt(n)
    sqt = int(math.sqrt(n))
    for i in range(2, sqt + 1):
 
        # if the value of i is a factor
        if (n % i == 0):
            v.append(i)
 
            # condition to check the
            # divisor is not the number itself
            if (n // i != i):
                v.append(n // i)
                 
    # return the vector
    return v
 
# Function to check if the
# number is semi-perfecr or not
def check( n):
 
    v = []
 
    # find the divisors
    v = factors(n)
 
    # sorting the vector
    v.sort()
 
    r = len(v)
 
    # subset to check if no is semiperfect
    subset = [[ 0 for i in range(n + 1)]
                  for j in range(r + 1)]
 
    # initialising 1st column to true
    for i in range(r + 1):
        subset[i][0] = True
 
    # initializing 1st row except zero position to 0
    for i in range(1, n + 1):
        subset[0][i] = False
 
    # loop to find whether the number is semiperfect
    for i in range(1, r + 1):
        for j in range(1, n + 1):
 
            # calculation to check if the
            # number can be made by summation of diviors
            if (j < v[i - 1]):
                subset[i][j] = subset[i - 1][j];
            else:
                subset[i][j] = (subset[i - 1][j] or
                                subset[i - 1][j - v[i - 1]])
 
    # if not possible to make the
    # number by any combination of divisors
    if ((subset[r][n]) == 0):
        return False
    else:
        return True
 
# Driver Code
if __name__ == "__main__":
    n = 40
    if (check(n)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by ChitraNayal

C#
// C# program to check
// if the number is
// semi-perfect or not
using System;
using System.Collections.Generic;
class GFG{
 
// Code to find all the
// factors of the number
// excluding the number itself
static List factors(int n)
{
  // vector to store the factors
  List v = new List();
  v.Add(1);
 
  // note that this loop runs
  // till Math.Sqrt(n)
  for (int i = 2;
           i <= Math.Sqrt(n); i++)
  {
    // if the value of i is
    // a factor
    if (n % i == 0)
    {
      v.Add(i);
 
      // condition to check the
      // divisor is not the number
      // itself
      if (n / i != i)
      {
        v.Add(n / i);
      }
    }
  }
 
// return the vector
return v;
}
 
// Function to check if the
// number is semi-perfecr or not
static bool check(int n)
{
  List v = new List();
 
  // find the divisors
  v = factors(n);
 
  // sorting the vector
  v.Sort();
 
  int r = v.Count;
 
  // subset to check if no
  // is semiperfect
  bool [,]subset = new bool[r + 1,
                            n + 1];
 
  // initialising 1st column
  // to true
  for (int i = 0; i <= r; i++)
    subset[i, 0] = true;
 
  // initializing 1st row except
  // zero position to 0
  for (int i = 1; i <= n; i++)
    subset[0, i] = false;
 
  // loop to find whether the
  // number is semiperfect
  for (int i = 1; i <= r; i++)
  {
    for (int j = 1; j <= n; j++)
    {
      // calculation to check if the
      // number can be made by
      // summation of diviors
      if (j < v[i - 1])
        subset[i, j] = subset[i - 1, j];
      else
      {
        subset[i, j] = subset[i - 1, j] ||
        subset[i - 1, (j - v[i - 1])];
      }
    }
  }
 
  // If not possible to make
  // the number by any combination
  // of divisors
  if ((subset[r, n]) == false)
    return false;
  else
    return true;
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 40;
   
  if (check(n))
    Console.Write("Yes");
  else
    Console.Write("No");
}
}
 
// This code is contributed by Princi Singh

输出： 
Yes

时间复杂度： O(因子数量* N)
辅助空间： O(因数* N)