怪异的数字

在数论中，一个怪异的数字是一个自然数，它很丰富，但不是半完美的。换句话说，该数字的适当除数(除数本身包括1，但不包括自身)的总和大于该数字，但是这些除数的子集都不等于该数字本身。

给定数字N，任务是检查数字是否怪异。

例子：

Input: 40
Output: The number is not weird
1+4+5+10+20=40, hence it is not weird.

Input: 70
Output: The number is Weird
The smallest weird number is 70. Its proper divisors are 1, 2, 5, 7, 10, 14, and 35; these sum to 74, but no subset of these sums to 70.

The number 12, for example, is abundant but not weird, because the proper divisors of 12 are 1, 2, 3, 4, and 6, which sum to 16; but 2+4+6 = 12. The first few weird numbers are 70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110, 12530, 12670, 13370, 13510, ..

方法：检查数量是否充足。该方法已在此处讨论。完成检查后，检查数字是否为半完美。这里讨论了检查半完美数字的方法。

下面是上述方法的实现：

C++

// C++ program to check if the 
// number is weird 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 abundant or not 
bool checkAbundant(int n)
{
    vector v;
  
    int sum = 0;
  
    // find the divisors using function
    v = factors(n);
  
    // sum all the factors
    for (int i = 0; i < v.size(); i++) {
        sum += v[i];
    }
  
    // check for abundant or not
    if (sum > n)
        return true;
    else
        return false;
}
  
// Function to check if the
// number is semi-perfecr or not
bool checkSemiPerfect(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;
  
    // initialing 1st row except zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
  
    // loop to find whther 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;
}
  
// Function to check for 
// weird or not 
bool checkweird(int n)
{
    if (checkAbundant(n) == true && 
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
  
// Driver Code
int main()
{
    int n = 70;
  
    if (checkweird(n))
        cout << "Weird Number";
    else
        cout << "Not Weird Number";
    return 0;
}

Java

// Java program to check if  
// the number is weird or not
import java.util.*;
class GFG
{ 
// code to find all the 
// factors of the number 
// excluding the number itself
static ArrayList factors(int n)
{
    // ArrayList to store
    // the factors
    ArrayList v = new ArrayList();
    v.add(1);
  
    // note that this loop 
    // runs till 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 ArrayList
    return v;
}
  
// Function to check if the 
// number is abundant or not 
static boolean checkAbundant(int n)
{
    ArrayList v;
  
    int sum = 0;
  
    // find the divisors
    // using function
    v = factors(n);
  
    // sum all the factors
    for (int i = 0; i < v.size(); i++) 
    {
        sum += v.get(i);
    }
  
    // check for abundant
    // or not
    if (sum > n)
        return true;
    else
        return false;
}
  
// Function to check if the
// number is semi-perfecr or not
static boolean checkSemiPerfect(int n)
{
    ArrayList v;
  
    // find the divisors
    v = factors(n);
  
    // sorting the ArrayList
    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;
  
    // initialing 1st row except 
    // zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
  
    // loop to find whther 
    // 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.get(i - 1))
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] || 
                               subset[i - 1][j - 
                                v.get(i - 1)];
            }
        }
    }
  
    // if not possible to make 
    // the number by any 
    // combination of divisors
    if ((subset[r][n]) == false)
        return false;
    else
        return true;
}
  
// Function to check 
// for weird or not 
static boolean checkweird(int n)
{
    if (checkAbundant(n) == true && 
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
  
// Driver Code
public static void main(String args[])
{
    int n = 70;
  
    if (checkweird(n))
        System.out.println("Weird Number");
    else
        System.out.println("Not Weird Number");
}
}
  
// This code is contributed
// by Arnab Kundu

Python3

# Python 3 program to check if the 
# number is weird or not
from math import sqrt
  
# 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)
    for i in range(2, int(sqrt(n)) + 1, 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 (int(n / i) != i):
                v.append(int(n / i))
  
    # return the vector
    return v
  
# Function to check if the number 
# is abundant or not 
def checkAbundant(n):
    sum = 0
  
    # find the divisors using function
    v = factors(n)
  
    # sum all the factors
    for i in range(len(v)):
        sum += v[i]
  
    # check for abundant or not
    if (sum > n):
        return True
    else:
        return False
  
# Function to check if the
# number is semi-perfecr or not
def checkSemiPerfect(n):
      
    # find the divisors
    v = factors(n)
  
    # sorting the vector
    v.sort(reverse = False)
    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
  
    # initialing 1st row except zero position to 0
    for i in range(1, n + 1):
        subset[0][i] = False
  
    # loop to find whther 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
  
# Function to check for 
# weird or not 
def checkweird(n):
    if (checkAbundant(n) == True and 
        checkSemiPerfect(n) == False):
        return True
    else:
        return False
  
# Driver Code
if __name__ == '__main__':
    n = 70
  
    if (checkweird(n)):
        print("Weird Number")
    else:
        print("Not Weird Number")
          
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to check if 
// the number is weird 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)
{
    // List to store
    // the factors
    List v = new List();
    v.Add(1);
  
    // note that this loop 
    // runs till 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 List
    return v;
}
  
// Function to check if the 
// number is abundant or not 
static Boolean checkAbundant(int n)
{
    List v;
  
    int sum = 0;
  
    // find the divisors
    // using function
    v = factors(n);
  
    // sum all the factors
    for (int i = 0; i < v.Count; i++) 
    {
        sum += v[i];
    }
  
    // check for abundant
    // or not
    if (sum > n)
        return true;
    else
        return false;
}
  
// Function to check if the
// number is semi-perfecr or not
static Boolean checkSemiPerfect(int n)
{
    List v;
  
    // find the divisors
    v = factors(n);
  
    // sorting the List
    v.Sort();
  
    int r = v.Count;
  
    // 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;
  
    // initialing 1st row except 
    // zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0,i] = false;
  
    // loop to find whther 
    // 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;
}
  
// Function to check 
// for weird or not 
static Boolean checkweird(int n)
{
    if (checkAbundant(n) == true && 
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
  
// Driver Code
public static void Main(String []args)
{
    int n = 70;
  
    if (checkweird(n))
        Console.WriteLine("Weird Number");
    else
        Console.WriteLine("Not Weird Number");
}
}
  
// This code is contributed by Princi Singh

输出：

Weird Number

时间复杂度： O(N *因素数)
辅助空间： O(N *因子数量)