Nos dan un número entero N. Necesitamos escribir un programa para encontrar el número entero menos positivo X formado por solo los dígitos 9 y 0, de modo que X sea un múltiplo de N.
Nota : Se supone que el valor de X no excederá de 10 6 .
Ejemplos:
Input : N = 5 Output : X = 90 Explanation: 90 is the smallest number made up of 9's and 0's which is divisible by 5. Input : N = 7 Output : X = 9009 Explanation: 9009 is smallest number made up of 9's and 0's which is divisible by 7.
La idea para resolver este problema es generar y almacenar todos los números que se pueden formar usando los dígitos 0 y 9. Luego encuentre el número más pequeño entre estos números generados que sea divisible por N.
Usaremos el método de generación de números binarios para generar todos los números que se pueden formar usando los dígitos 0 y 9.
A continuación se muestra la implementación de la idea anterior:
C++
// CPP program to find smallest multiple of a
// given number made of digits 0 and 9 only
#include <bits/stdc++.h>
using namespace std;
// Maximum number of numbers made of 0 and 9
#define MAX_COUNT 10000
// vector to store all numbers that can be formed
// using digits 0 and 9 and are less than 10^5
vector<string> vec;
/* Preprocessing function to generate all possible
numbers formed by 0 and 9 */
void generateNumbersUtil()
{
// Create an empty queue of strings
queue<string> q;
// enqueue the first number
q.push("9");
// This loops is like BFS of a tree with 9 as root
// 0 as left child and 9 as right child and so on
for (int count = MAX_COUNT; count > 0; count--)
{
string s1 = q.front();
q.pop();
// storing the front of queue in the vector
vec.push_back(s1);
string s2 = s1;
// Append "0" to s1 and enqueue it
q.push(s1.append("0"));
// Append "9" to s2 and enqueue it. Note that
// s2 contains the previous front
q.push(s2.append("9"));
}
}
// function to find smallest number made up of only
// digits 9’s and 0’s, which is a multiple of n.
string findSmallestMultiple(int n)
{
// traverse the vector to find the smallest
// multiple of n
for (int i = 0; i < vec.size(); i++)
// stoi() is used for string to int conversion
if (stoi(vec[i])%n == 0)
return vec[i];
}
// Driver Code
int main()
{
generateNumbersUtil();
int n = 7;
cout << findSmallestMultiple(n);
return 0;
}
Java
// Java program to find smallest
// multiple of a given number
// made of digits 0 and 9 only
import java.util.*;
class GFG
{
// Maximum number of
// numbers made of 0 and 9
static int MAX_COUNT = 10000;
// vector to store all numbers
// that can be formed using
// digits 0 and 9 and are
// less than 10^5
static List<String> vec = new LinkedList<String>();
/* Preprocessing function
to generate all possible
numbers formed by 0 and 9 */
static void generateNumbersUtil()
{
// Create an empty
// queue of Strings
Queue<String> q = new LinkedList<String>();
// enqueue the
// first number
q.add("9");
// This loops is like BFS of
// a tree with 9 as root
// 0 as left child and 9 as
// right child and so on
for (int count = MAX_COUNT;
count > 0; count--)
{
String s1 = q.peek();
q.remove();
// storing the Peek of
// queue in the vector
vec.add(s1);
String s2 = s1;
// Append "0" to s1
// and enqueue it
q.add(s1 + "0");
// Append "9" to s2 and
// enqueue it. Note that
// s2 contains the previous Peek
q.add(s2 + "9");
}
}
// function to find smallest
// number made up of only
// digits 9's and 0's, which
// is a multiple of n.
static String findSmallestMultiple(int n)
{
// traverse the vector
// to find the smallest
// multiple of n
for (int i = 0; i < vec.size(); i++) // stoi() is used for
// String to int conversion
{
if (Integer.parseInt(vec.get(i)) % n == 0)
{
return vec.get(i);
}
}
return "";
}
// Driver Code
public static void main(String[] args)
{
generateNumbersUtil();
int n = 7;
System.out.println(findSmallestMultiple(n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program to find smallest multiple of
# a given number made of digits 0 and 9 only
from queue import Queue
# Preprocessing function to generate
# all possible numbers formed by 0 and 9
def generateNumbersUtil():
global vec
# Create an empty queue of strings
q = Queue()
# enqueue the first number
q.put("9")
# This loops is like BFS of a tree
# with 9 as root, 0 as left child
# and 9 as right child and so on
for count in range(MAX_COUNT, -1, -1):
s1 = q.queue[0]
q.get()
# storing the front of queue
# in the vector
vec.append(s1)
s2 = s1
# Append "0" to s1 and enqueue it
s1 += "0"
q.put(s1)
# Append "9" to s2 and enqueue it. Note
# that s2 contains the previous front
s2 += "9"
q.put(s2)
# function to find smallest number made
# up of only digits 9’s and 0’s, which
# is a multiple of n.
def findSmallestMultiple(n):
global vec
# traverse the vector to find
# the smallest multiple of n
for i in range(len(vec)):
# int is used for string to
# conversion
if (int(vec[i]) % n == 0):
return vec[i]
# Driver Code
# Maximum number of numbers
# made of 0 and 9
MAX_COUNT = 10000
# stack to store all numbers that
# can be formed using digits 0 and
# 9 and are less than 10^5
vec = []
generateNumbersUtil()
n = 7
print(findSmallestMultiple(n))
# This code is contributed by PranchalK
C#
// C# program to find smallest
// multiple of a given number
// made of digits 0 and 9 only
using System;
using System.Collections.Generic;
class GFG
{
// Maximum number of
// numbers made of 0 and 9
static int MAX_COUNT = 10000;
// vector to store all numbers
// that can be formed using
// digits 0 and 9 and are
// less than 10^5
static List<string> vec = new List<string>();
/* Preprocessing function
to generate all possible
numbers formed by 0 and 9 */
static void generateNumbersUtil()
{
// Create an empty
// queue of strings
Queue<string> q = new Queue<string>();
// enqueue the
// first number
q.Enqueue("9");
// This loops is like BFS of
// a tree with 9 as root
// 0 as left child and 9 as
// right child and so on
for (int count = MAX_COUNT;
count > 0; count--)
{
string s1 = q.Peek();
q.Dequeue();
// storing the Peek of
// queue in the vector
vec.Add(s1);
string s2 = s1;
// Append "0" to s1
// and enqueue it
q.Enqueue(s1 + "0");
// Append "9" to s2 and
// enqueue it. Note that
// s2 contains the previous Peek
q.Enqueue(s2 + "9");
}
}
// function to find smallest
// number made up of only
// digits 9’s and 0’s, which
// is a multiple of n.
static string findSmallestMultiple(int n)
{
// traverse the vector
// to find the smallest
// multiple of n
for (int i = 0; i < vec.Count; i++)
// stoi() is used for
// string to int conversion
if (int.Parse(vec[i]) % n == 0)
return vec[i];
return "";
}
// Driver Code
static void Main()
{
generateNumbersUtil();
int n = 7;
Console.Write(findSmallestMultiple(n));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
Producción:
9009
Complejidad de tiempo : O(n), donde n representa el entero dado.
Espacio auxiliar : O(10000), no se requiere espacio adicional, por lo que es una constante.
Publicación traducida automáticamente
Artículo escrito por Ravi_Maurya y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA