奇怪的数字

在数论中，奇怪的数是一个丰富但不半完美的自然数。换句话说，该数的真因数(除数包括 1 但不包括其本身)之和大于该数，但这些因数的任何子集都不与该数本身相加。
给定一个数字 N，任务是检查这个数字是否奇怪。

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, …

处理方法：检查数量是否充足。该方法已在此处讨论。检查完成后，检查数字是否为半完美。检查半完美数的方法已在此处讨论。
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 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 divisors
            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 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
            // divisors
            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 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 divisors
            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 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
            // divisors
            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

Javascript

 
// Javascript program to check if the
// number is weird or not
 
// code to find all the factors of
// the number excluding the number itself
function factors(n)
{
    // vector to store the factors
    var v = [];
    v.push(1);
 
    // note that this loop runs till sqrt(n)
    for (var i = 2; i <= Math.sqrt(n); i++) {
 
        // if the value of i is a factor
        if (n % i == 0) {
            v.push(i);
 
            // condition to check the
            // divisor is not the number itself
            if (n / i != i) {
                v.push(n / i);
            }
        }
    }
    // return the vector
    return v;
}
 
// Function to check if the number
// is abundant or not
function checkAbundant(n)
{
    var v = [];
 
    var sum = 0;
 
    // find the divisors using function
    v = factors(n);
 
    // sum all the factors
    for (var i = 0; i < v.length; 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
function checkSemiPerfect(n)
{
    var v = [];
 
    // find the divisors
    v = factors(n);
 
    // sorting the vector
    v.sort()
 
    var r = v.length;
 
    // subset to check if no is semiperfect
    var subset = Array.from(Array(r+1), ()=>Array(n+1));
 
    // initialising 1st column to true
    for (var i = 0; i <= r; i++)
        subset[i][0] = true;
 
    // initialing 1st row except zero position to 0
    for (var i = 1; i <= n; i++)
        subset[0][i] = false;
 
    // loop to find whether the number is semiperfect
    for (var i = 1; i <= r; i++) {
        for (var j = 1; j <= n; j++) {
 
            // calculation to check if the
            // number can be made by summation of divisors
            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
function checkweird(n)
{
    if (checkAbundant(n) == true &&
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
 
// Driver Code
var n = 70;
if (checkweird(n))
    document.write( "Weird Number");
else
    document.write( "Not Weird Number");
 
 

输出：

Weird Number

时间复杂度： O(N * 因子数)
辅助空间： O(N * 因子数)

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