Tcefrep号码

Tcefrep编号为N的数字，使得reverse(n)= N的适当除数之和。

6, 498906, 20671542, 41673714…. 
 

检查N是否为Tcefrep号码

给定数字N ，任务是检查N是否为Tcefrep数字。如果N是Tcefrep号码，则打印“是”，否则打印“否” 。
例子：

Input: N = 498906 
Output: Yes 
Explanation: 
proper divisors of 498906 are 1, 2, 3, 6, 9, 18, 27, 54, 
9239, 18478, 27717, 55434, 83151, 166302, 249453, 
which sum to 609894, the reverse of 498906
Input: N = 120 
Output: No 
 

方法：

1. 我们将找到N的适当除数之和
2. 我们会发现N的反面
3. 然后，我们将检查是否N个适当除数的总和等于逆转的N或不是，如果相等，则打印“是”其他打印“否”。

下面是上述方法的实现：

C++

// C++ implementation to check if N
// is a Tcefrep number
#include 
using namespace std;
 
// Iterative function to
// reverse digits of num
int reverse(int num)
{
    int rev_num = 0;
    while(num > 0)
    {
        rev_num = rev_num*10 + num%10;
        num = num/10;
    }
    return rev_num;
}
 
// Function to calculate sum of
// all proper divisors
// num --> given natural number
int properDivSum(int num)
{
    // Final result of summation of divisors
    int result = 0;
   
    // find all divisors which divides '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);
}
 
bool isTcefrep(int n)
{
    return properDivSum(n) == reverse(n);
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 6;
 
    // Function Call
    if (isTcefrep(N))
        cout << "Yes";
    else
        cout << "No";
    return 0; 
     
}

Java

// Java program for above approach
class GFG{
 
// Iterative function to
// reverse digits of num
static int reverse(int num)
{
    int rev_num = 0;
    while(num > 0)
    {
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}
 
// Function to calculate sum of
// all proper divisors
// num --> given natural number
static int properDivSum(int num)
{
    // Final result of summation of divisors
    int result = 0;
     
    // find all divisors which divides '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);
}
 
static boolean isTcefrep(int n)
{
    return properDivSum(n) == reverse(n);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 6;
 
    // Function Call
    if (isTcefrep(N))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 implementation to check if N
# is a Tcefrep number
import math
 
# Iterative function to
# reverse digits of num
def reverse(num):
    rev_num = 0
    while(num > 0):
        rev_num = rev_num * 10 + num % 10
        num = num // 10
 
    return rev_num
 
# Function to calculate sum of
# all proper divisors
# num --> given natural number
def properDivSum(num):
     
    # Final result of summation of divisors
    result = 0
 
    # find all divisors which divides 'num'
    for i in range(2, (int)(math.sqrt(num)) + 1):
         
        # 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)
 
def isTcefrep(n):
    return properDivSum(n) == reverse(n);
 
# Driver Code
 
# Given Number N
N = 6
 
# Function Call
if(isTcefrep(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Sanjit Prasad

C#

// C# program for above approach
using System;
class GFG{
 
// Iterative function to
// reverse digits of num
static int reverse(int num)
{
    int rev_num = 0;
    while(num > 0)
    {
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}
 
// Function to calculate sum of
// all proper divisors
// num --> given natural number
static int properDivSum(int num)
{
    // Final result of summation of divisors
    int result = 0;
     
    // find all divisors which divides '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);
}
 
static bool isTcefrep(int n)
{
    return properDivSum(n) == reverse(n);
}
 
// Driver Code
public static void Main()
{
    int N = 6;
 
    // Function Call
    if (isTcefrep(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Nidhi_Biet

Javascript

 
// Javascript program for above approach
 
 
    // Iterative function to
    // reverse digits of num
    function reverse( num) {
        let rev_num = 0;
        while (num > 0) {
            rev_num = rev_num * 10 + num % 10;
            num = parseInt(num / 10);
        }
        return rev_num;
    }
 
    // Function to calculate sum of
    // all proper divisors
    // num --> given natural number
    function properDivSum( num) {
        // Final result of summation of divisors
        let result = 0;
 
        // find all divisors which divides 'num'
        for ( 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 isTcefrep( n) {
        return properDivSum(n) == reverse(n);
    }
 
    // Driver Code
      
        let N = 6;
 
        // Function Call
        if (isTcefrep(N))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by gauravrajput1
 

输出：

Yes

时间复杂度： O(n ^ 2)
参考：http://www.numbersaplenty.com/set/tcefrep_number/