不可更改的号码

给定数字N ，任务是检查N是否为不可触摸数字。如果N是不可触摸的数字，则打印“是”，否则打印“否” 。

Untouchable Number are numbers which are not the sum of the proper divisors of any number. 
 

例子：

Input: N = 5 
Output: Yes
Input: N = 20 
Output: No 
 

方法：这个想法是找到数字N的适当除数的总和，并检查总和是否等于N。如果sum不等于N ，则N是不可触摸的数字，然后打印“是”，否则打印“否” 。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate sum of
// all proper divisors of num
int divSum(int num)
{
    // Final result of summation
    // of divisors
    int result = 0;
 
    // Find all divisors of num
    for (int i = 2; i <= sqrt(num); i++) {
 
        // If 'i' is divisor of 'num'
        if (num % i == 0) {
 
            // If both divisors are same
            // then add it only once else
            // add both
            if (i == (num / i))
                result += i;
            else
                result += (i + num / i);
        }
    }
 
    // Add 1 to the result as
    // 1 is also a divisor
    return (result + 1);
}
 
// Function to check if N is a
// Untouchable Number
bool isUntouchable(int n)
{
    for (int i = 1; i <= 2 * n; i++) {
        if (divSum(i) == n)
            return false;
    }
    return true;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 52;
 
    // Function Call
    if (isUntouchable(n))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program for the above approach
class GFG{
 
// Function to calculate sum of
// all proper divisors of num
static int divSum(int num)
{
 
    // Final result of summation
    // of divisors
    int result = 0;
 
    // Find all divisors of num
    for(int i = 2; i <= Math.sqrt(num); i++)
    {
        
       // If 'i' is divisor of 'num'
       if (num % i == 0)
       {
            
           // If both divisors are same
           // then add it only once else
           // add both
           if (i == (num / i))
               result += i;
           else
               result += (i + num / i);
       }
    }
     
    // Add 1 to the result as
    // 1 is also a divisor
    return (result + 1);
}
 
// Function to check if N is a
// Untouchable Number
static boolean isUntouchable(int n)
{
    for(int i = 1; i <= 2 * n; i++)
    {
       if (divSum(i) == n)
           return false;
    }
    return true;
}
 
// Driver code
public static void main(String[] args)
{
 
    // Given Number N
    int n = 52;
 
    // Function Call
    if (isUntouchable(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
import math;
 
# Function to calculate sum of
# all proper divisors of num
def divSum(num):
 
    # Final result of summation
    # of divisors
    result = 0;
 
    # Find all divisors of num
    for i in range(2, int(math.sqrt(num))):
 
        # If 'i' is divisor of 'num'
        if (num % i == 0):
 
            # If both divisors are same
            # then add it only once else
            # add both
            if (i == (num // i)):
                result += i;
            else:
                result += (i + num // i);
         
    # Add 1 to the result as
    # 1 is also a divisor
    return (result + 1);
 
# Function to check if N is a
# Untouchable Number
def isUntouchable(n):
 
    for i in range(1, 2 * n):
        if (divSum(i) == n):
            return False;
     
    return True;
 
# Driver Code
 
# Given Number N
N = 52;
 
# Function Call
if (isUntouchable(N)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to calculate sum of
// all proper divisors of num
static int divSum(int num)
{
 
    // Final result of summation
    // of divisors
    int result = 0;
 
    // Find all divisors of num
    for(int i = 2; i <= Math.Sqrt(num); i++)
    {
        
       // If 'i' is divisor of 'num'
       if (num % i == 0)
       {
            
           // If both divisors are same
           // then add it only once else
           // add both
           if (i == (num / i))
               result += i;
           else
               result += (i + num / i);
       }
    }
     
    // Add 1 to the result as
    // 1 is also a divisor
    return (result + 1);
}
 
// Function to check if N is a
// Untouchable Number
static bool isUntouchable(int n)
{
    for(int i = 1; i <= 2 * n; i++)
    {
       if (divSum(i) == n)
           return false;
    }
    return true;
}
 
// Driver code
public static void Main()
{
 
    // Given Number N
    int n = 52;
 
    // Function Call
    if (isUntouchable(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Code_Mech

Javascript

// JavaScript program for the above approach
 
// Function to calculate sum of
// all proper divisors of num
function divSum(num)
{
 
    // Final result of summation
    // of divisors
    let result = 0;
 
    // Find all divisors of num
    for (let i = 2; i <= Math.sqrt(num); i++)
    {
 
        // If 'i' is divisor of 'num'
        if (num % i == 0)
        {
 
            // If both divisors are same
            // then add it only once else
            // add both
            if (i == (num / i))
                result += i;
            else
                result += (i + num / i);
        }
    }
 
    // Add 1 to the result as
    // 1 is also a divisor
    return (result + 1);
}
 
// Function to check if N is a
// Untouchable Number
function isUntouchable(n)
{
    for (let i = 1; i <= 2 * n; i++)
    {
        if (divSum(i) == n)
            return false;
    }
    return true;
}
 
// Driver Code
 
// Given Number n
let n = 52;
 
// Function Call
if (isUntouchable(n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by blalverma92

输出：

Yes

时间复杂度： O(sqrt(N))