Dada una array arr[] de tamaño N y un número entero K. La tarea es encontrar el número de dígitos K más grande divisible por todos los números de arr[].
Ejemplos:
Entrada: arr[] = {2, 3, 5}, K = 3
Salida: 990
Explicación: 990 es el mayor número de 3 dígitos divisible por 2, 3 y 5.Entrada: arr[] = {91, 93, 95}, K = 3
Salida: -1
Explicación: No hay un número de 3 dígitos divisible por todos los 91, 93 y 95.
Enfoque: La solución se basa en la idea similar a encontrar el número de dígitos K más grande divisible por X. Siga los pasos que se mencionan a continuación.
- Encuentre MCM de todos los números de la array arr[]
- Encuentra el múltiplo más grande de MCM que tiene K dígitos usando la fórmula:
MCM(arr) * ((10 K -1)/MCM(arr))
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// Base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to find LCM of the array
int findLCM(int arr[], int n, int idx)
{
// lcm(a, b) = (a*b / gcd(a, b))
if (idx == n - 1)
{
return arr[idx];
}
int a = arr[idx];
int b = findLCM(arr, n, idx + 1);
double gcd = __gcd(a, b);
return (a * (int)floor(b / gcd));
}
// Function to find the number
int findNum(int arr[], int n, int K)
{
int lcm = findLCM(arr, n, 0);
int ans = (int)floor((pow(10, K) - 1) / lcm);
ans = (ans) * lcm;
if (ans == 0)
return -1;
return ans;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 3;
cout << findNum(arr, n, K);
}
// This code is contributed by Samim Hossain Mondal.
Java
// Java code to implement above approach
class GFG
{
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// Base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to find LCM of the array
static int findLCM(int []arr, int idx)
{
// lcm(a, b) = (a*b / gcd(a, b))
if (idx == arr.length - 1)
{
return arr[idx];
}
int a = arr[idx];
int b = findLCM(arr, idx + 1);
double gcd = __gcd(a, b);
return (a * (int)Math.floor(b / gcd));
}
// Function to find the number
static int findNum(int []arr, int K)
{
int lcm = findLCM(arr, 0);
int ans = (int)Math.floor((Math.pow(10, K) - 1) / lcm);
ans = (ans) * lcm;
if (ans == 0)
return -1;
return ans;
}
// Driver code
public static void main(String []args)
{
int []arr = { 2, 3, 5 };
int K = 3;
System.out.println(findNum(arr, K));
}
}
// This code is contributed by 29AjayKumar
Python3
# python code to implement above approach import math # Function to find LCM of the array def findLCM(arr, idx): # lcm(a, b) = (a*b / gcd(a, b)) if (idx == len(arr) - 1): return arr[idx] a = arr[idx] b = findLCM(arr, idx + 1) return (a * b // math.gcd(a, b)) # Function to find the number def findNum(arr, K): lcm = findLCM(arr, 0) ans = (pow(10, K) - 1) // lcm ans = (ans)*lcm if (ans == 0): return -1 return ans # Driver code if __name__ == "__main__": arr = [2, 3, 5] K = 3 print(findNum(arr, K)) # This code is contributed by rakeshsahni
C#
// C# code to implement above approach
using System;
using System.Collections;
class GFG
{
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// Base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to find LCM of the array
static int findLCM(int []arr, int idx)
{
// lcm(a, b) = (a*b / gcd(a, b))
if (idx == arr.Length - 1)
{
return arr[idx];
}
int a = arr[idx];
int b = findLCM(arr, idx + 1);
double gcd = __gcd(a, b);
return (a * (int)Math.Floor(b / gcd));
}
// Function to find the number
static int findNum(int []arr, int K)
{
int lcm = findLCM(arr, 0);
int ans = (int)Math.Floor((Math.Pow(10, K) - 1) / lcm);
ans = (ans) * lcm;
if (ans == 0)
return -1;
return ans;
}
// Driver code
public static void Main()
{
int []arr = { 2, 3, 5 };
int K = 3;
Console.Write(findNum(arr, K));
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// JavaScript code to implement above approach
function __gcd(a, b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// Base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to find LCM of the array
function findLCM(arr, idx)
{
// lcm(a, b) = (a*b / gcd(a, b))
if (idx == arr.length - 1)
{
return arr[idx];
}
let a = arr[idx];
let b = findLCM(arr, idx + 1);
return Math.floor((a * b / __gcd(a, b)));
}
// Function to find the number
function findNum(arr, K)
{
let lcm = findLCM(arr, 0);
let ans = Math.floor((Math.pow(10, K) - 1) / lcm);
ans = (ans) * lcm;
if (ans == 0)
return -1;
return ans;
}
// Driver code
let arr = [ 2, 3, 5 ];
let K = 3;
document.write(findNum(arr, K));
// This code is contributed by Potta Lokesh
</script>
Producción
990
Complejidad de Tiempo: O(N*logD) donde D es el elemento máximo del arreglo
Espacio Auxiliar: O(1)