仅由数字 0 和 9 组成的给定数字的最小倍数
给定一个整数 N。 我们需要编写一个程序来找出仅由数字 9 和 0 组成的最小正整数 X，使得 X 是 N 的倍数。
注：假设 X 的值不超过 10 6 。
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例子：
Input : N = 5
Output : X = 90
Exaplanation: 90 is the smallest number made up
of 9's and 0's which is divisible by 5.
Input : N = 7
Output : X = 9009
Exaplanation: 9009 is smallest number made up
of 9's and 0's which is divisible by 7.
解决这个问题的想法是生成并存储所有可以用数字 0 和 9 组成的数字。然后在这些生成的数字中找到能被 N 整除的最小数字。
我们将使用生成二进制数的方法生成所有可以使用数字 0 和 9 组成的数字。
下面是上述想法的实现：
C++
// CPP program to find smallest multiple of a
// given number made of digits 0 and 9 only
#include
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 vec;
/* Preprocessing function to generate all possible
numbers formed by 0 and 9 */
void generateNumbersUtil()
{
// Create an empty queue of strings
queue q;
// enque 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 vec = new LinkedList();
/* Preprocessing function
to generate all possible
numbers formed by 0 and 9 */
static void generateNumbersUtil()
{
// Create an empty
// queue of Strings
Queue q = new LinkedList();
// enque 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()
# enque 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 PranchalKC#
// 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 vec = new List();
/* Preprocessing function
to generate all possible
numbers formed by 0 and 9 */
static void generateNumbersUtil()
{
// Create an empty
// queue of strings
Queue q = new Queue();
// enque 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) 输出：
9009
时间复杂度：O(n)