Números de cruce

El número de unión es un número N si se puede escribir como K + SumOfDIgits(k) para al menos dos K.
 

101, 103, 105, 107, 109, 111, 113, 115…. 
 

Comprobar si un número N es un número de cruce

Dado un número N , la tarea es comprobar si N es un número de unión o no. Si N es un número de cruce, escriba «Sí» , de lo contrario, escriba «No» .
Ejemplos: 
 

Entrada: N = 1106 
Salida: Sí 
Explicación: 
1106 = 1093 + (1+0+9+3) = 1102 + (1+1+0+2)
Entrada: N = 11 
Salida: No 
 

Enfoque: dado que el número de unión es un número N si se puede escribir como K + SumOfDIgits(k) para al menos dos K. Entonces, para todos los números del 1 al N en un ciclo, verificaremos si i + sumOfdigitsof(i) es igual a N o no. Si es igual a N, escriba «Sí» , de lo contrario, escriba «No» .
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation for the
// above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// sum Of digits of N
int sum(int n)
{
 
    // To store sum of N and
    // sumOfdigitsof(N)
    int sum = 0;
 
    while (n != 0) {
        // extracting digit
        int r = n % 10;
        sum = sum + r;
        n = n / 10;
    }
 
    return sum;
}
 
// Function to check Junction numbers
bool isJunction(int n)
{
    // To store count of ways n can be
    // represented as i + SOD(i)
    int count = 0;
    for (int i = 1; i <= n; i++) {
        if (i + sum(i) == n)
            count++;
    }
 
    return count >= 2;
}
 
// Driver Code
int main()
{
    int N = 111;
 
    // Function Call
    if (isJunction(N))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

// Java program for above approach
class GFG{
 
// Function to find the
// sum Of digits of N
static int sum(int n)
{
 
    // To store sum of N and
    // sumOfdigitsof(N)
    int sum = 0;
 
    while (n != 0)
    {
        // extracting digit
        int r = n % 10;
        sum = sum + r;
        n = n / 10;
    }
    return sum;
}
 
// Function to check Junction numbers
static boolean isJunction(int n)
{
    // To store count of ways n can be
    // represented as i + SOD(i)
    int count = 0;
    for (int i = 1; i <= n; i++)
    {
        if (i + sum(i) == n)
            count++;
    }
    return count >= 2;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 111;
 
    // Function Call
    if (isJunction(N))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Pratima Pandey

# Python3 program for the above approach
import math
 
# Function to find the
# sum Of digits of N
def sum1(n):
 
    # To store sum of N and
    # sumOfdigitsof(N)
    sum1 = 0
 
    while (n != 0):
         
        # extracting digit
        r = n % 10
        sum1 = sum1 + r
        n = n // 10
     
    return sum1
 
# Function to check Junction numbers
def isJunction(n):
 
    # To store count of ways n can be
    # represented as i + SOD(i)
    count = 0
    for i in range(1, n + 1):
        if (i + sum1(i) == n):
            count = count + 1
 
    return count >= 2
 
# Driver Code
if __name__=='__main__':
 
    # Given Number
    n = 111
 
    # Function Call
    if (isJunction(n) == 1):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by rock_cool

// C# program for above approach
using System;
class GFG{
 
// Function to find the
// sum Of digits of N
static int sum(int n)
{
 
    // To store sum of N and
    // sumOfdigitsof(N)
    int sum = 0;
 
    while (n != 0)
    {
        // extracting digit
        int r = n % 10;
        sum = sum + r;
        n = n / 10;
    }
    return sum;
}
 
// Function to check Junction numbers
static bool isJunction(int n)
{
    // To store count of ways n can be
    // represented as i + SOD(i)
    int count = 0;
    for (int i = 1; i <= n; i++)
    {
        if (i + sum(i) == n)
            count++;
    }
    return count >= 2;
}
 
// Driver Code
public static void Main()
{
    int N = 111;
 
    // Function Call
    if (isJunction(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Nidhi Biet

<script>
 
// Javascript program for above approach
 
 
    // Function to find the
    // sum Of digits of N
    function sum( n) {
 
        // To store sum of N and
        // sumOfdigitsof(N)
        let sum = 0;
 
        while (n != 0) {
            // extracting digit
            let r = n % 10;
            sum = sum + r;
            n = parseInt(n / 10);
        }
        return sum;
    }
 
    // Function to check Junction numbers
    function isJunction( n) {
        // To store count of ways n can be
        // represented as i + SOD(i)
        let count = 0;
        for ( i = 1; i <= n; i++) {
            if (i + sum(i) == n)
                count++;
        }
        return count >= 2;
    }
 
    // Driver Code
      
        let N = 111;
 
        // Function Call
        if (isJunction(N))
            document.write("Yes");
        else
            document.write("No");
 
 
// This code contributed by Rajput-Ji
 
</script>

Yes

Complejidad de tiempo: O(n) 

Complejidad espacial: O(1)
Referencia : http://www.numbersaplenty.com/set/junction_number/