Dado un número N, la tarea es imprimir el máximo entre la suma y la multiplicación de los dígitos del número dado hasta que el número se reduzca a un solo dígito.
Nota: La suma y la multiplicación de dígitos se realizarán hasta que el número se reduzca a un solo dígito.
Tomemos un ejemplo donde N = 19,
19 se divide en 1+9=10 luego 10 se divide en 1+0=1. 1 es una suma de un solo dígito.
Además, 19 se divide en 1*9 = 9. 9 es una multiplicación de un solo dígito.
Entonces, la salida es 9, es decir, un máximo de 9 y 1.
Input: N = 631 Output: 8 Input: 110 Output: 2
Acercarse:
- Comprueba si un número es menor que 10, entonces la suma y el producto serán iguales. Entonces, devuelve ese número.
- Más,
- Encuentre la suma de dígitos repetidamente usando el Método 2 de Encontrar la suma de dígitos de un número hasta que la suma se convierta en un solo dígito .
- Y encuentre el producto de dígitos repetidamente usando el Método 1 de Encontrar la suma de dígitos de un número hasta que la suma se convierta en un solo dígito .
- Devuelve el máximo de ambos.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP implementation of above approach
#include<bits/stdc++.h>
using namespace std;
// Function to sum the digits until it
// becomes a single digit
long repeatedSum(long n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
// Function to product the digits until it
// becomes a single digit
long repeatedProduct(long n)
{
long prod = 1;
// Loop to do sum while
// sum is not less than
// or equal to 9
while (n > 0 || prod > 9) {
if (n == 0) {
n = prod;
prod = 1;
}
prod *= n % 10;
n /= 10;
}
return prod;
}
// Function to find the maximum among
// repeated sum and repeated product
long maxSumProduct(long N)
{
if (N < 10)
return N;
return max(repeatedSum(N), repeatedProduct(N));
}
// Driver code
int main()
{
long n = 631;
cout << maxSumProduct(n)<<endl;
return 0;
}
// This code is contributed by mits
Java
// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
// Function to sum the digits until it
// becomes a single digit
public static long repeatedSum(long n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
// Function to product the digits until it
// becomes a single digit
public static long repeatedProduct(long n)
{
long prod = 1;
// Loop to do sum while
// sum is not less than
// or equal to 9
while (n > 0 || prod > 9) {
if (n == 0) {
n = prod;
prod = 1;
}
prod *= n % 10;
n /= 10;
}
return prod;
}
// Function to find the maximum among
// repeated sum and repeated product
public static long maxSumProduct(long N)
{
if (N < 10)
return N;
return Math.max(repeatedSum(N), repeatedProduct(N));
}
// Driver code
public static void main(String[] args)
{
long n = 631;
System.out.println(maxSumProduct(n));
}
}
Python3
# Python 3 implementation of above approach # Function to sum the digits until # it becomes a single digit def repeatedSum(n): if (n == 0): return 0 return 9 if(n % 9 == 0) else (n % 9) # Function to product the digits # until it becomes a single digit def repeatedProduct(n): prod = 1 # Loop to do sum while # sum is not less than # or equal to 9 while (n > 0 or prod > 9) : if (n == 0) : n = prod prod = 1 prod *= n % 10 n //= 10 return prod # Function to find the maximum among # repeated sum and repeated product def maxSumProduct(N): if (N < 10): return N return max(repeatedSum(N), repeatedProduct(N)) # Driver code if __name__ == "__main__": n = 631 print(maxSumProduct(n)) # This code is contributed # by ChitraNayal
C#
// C# implementation of
// above approach
using System;
class GFG
{
// Function to sum the digits
// until it becomes a single digit
public static long repeatedSum(long n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ?
9 : (n % 9);
}
// Function to product the digits
// until it becomes a single digit
public static long repeatedProduct(long n)
{
long prod = 1;
// Loop to do sum while
// sum is not less than
// or equal to 9
while (n > 0 || prod > 9)
{
if (n == 0)
{
n = prod;
prod = 1;
}
prod *= n % 10;
n /= 10;
}
return prod;
}
// Function to find the maximum among
// repeated sum and repeated product
public static long maxSumProduct(long N)
{
if (N < 10)
return N;
return Math.Max(repeatedSum(N),
repeatedProduct(N));
}
// Driver code
public static void Main()
{
long n = 631;
Console.WriteLine(maxSumProduct(n));
}
}
// This code is contributed
// by inder_verma
Javascript
<script>
// javascript implementation of above approach
// Function to sum the digits until it
// becomes a single digit
function repeatedSum(n) {
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
// Function to product the digits until it
// becomes a single digit
function repeatedProduct(n) {
var prod = 1;
// Loop to do sum while
// sum is not less than
// or equal to 9
while (n > 0 || prod > 9) {
if (n == 0) {
n = prod;
prod = 1;
}
prod *= n % 10;
n = parseInt(n/10);
}
return prod;
}
// Function to find the maximum among
// repeated sum and repeated product
function maxSumProduct(N) {
if (N < 10)
return N;
return Math.max(repeatedSum(N), repeatedProduct(N));
}
// Driver code
var n = 631;
document.write(maxSumProduct(n));
// This code contributed by aashish1995
</script>
Producción:
8