通过除以一个数字的因子或者从数组中删除第一个出现的数字来将一个数字转换成另一个数字
原文:https://www . geeksforgeeks . org/将一个数字转换为另一个数字除以其因子或从数组中移除第一个出现的数字/
给定两个正整数 A 、 B 和一个仅由数字【0-9】组成的数组D【】，任务是检查是否有可能通过重复除以数组D【】中存在的任何因子或通过删除数组中存在的任何数字的第一次出现来将 A 减少到 B
示例:
输入: A = 5643，B = 81，D[] = {3，8，1} 输出:是 说明: 运算 1: 将 A (= 5643)除以 3，则 A 的值变为 1881。 操作 2: 从 A(= 1881)中去掉第一个出现的 8，那么 A 的值就变成了 181。 操作 3: 从 A(= 181)中去掉第一个出现的 1，那么 A 的值变成 81。
输入: A = 82，B = 2，D[] = {8，2 } T3】输出:是
方法:通过使用队列对值 A 执行所有可能的操作，并检查在任何步骤中 A 的值是否修改为 B ，可以解决给定的问题。按照以下步骤解决问题:
- 初始化一个队列，说 Q ，初始推 A 给它。
- 初始化一个 HashMap ，说 M 存储队列中存在的元素 Q 并初始化一个变量 ans 为“否”存储需要的结果。
- 迭代至 Q 不为空，执行以下步骤:
- 完成上述步骤后，打印和的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a digit x is
// present in the number N or not
int isPresent(int n, int x)
{
// Convert N to string
string num = to_string(n);
// Traverse the string num
for (int i = 0; i < num.size();
i++) {
// Return first occurrence
// of the digit x
if ((num[i] - '0') == x)
return i;
}
return -1;
}
// Function to remove the character
// at a given index from the number
int removeDigit(int n, int index)
{
// Convert N to string
string num = to_string(n);
// Store the resultant string
string ans = "";
// Traverse the string num
for (int i = 0;
i < num.size(); i++) {
if (i != index)
ans += num[i];
}
// If the number becomes empty
// after deletion, then return -1
if (ans == "" || (ans.size() == 1
&& ans[0] == '0'))
return -1;
// Return the number
int x = stoi(ans);
return x;
}
// Function to check if A can be
// reduced to B by performing the
// operations any number of times
bool reduceNtoX(int a, int b,
int d[], int n)
{
// Create a queue
queue<int> q;
// Push A into the queue
q.push(a);
// Hashmap to check if the element
// is present in the Queue or not
unordered_map<int, bool> visited;
// Set A as visited
visited[a] = true;
// Iterate while the queue is not empty
while (!q.empty()) {
// Store the front value of the
// queue and pop it from it
int top = q.front();
q.pop();
if (top < 0)
continue;
// If top is equal to B,
// then return true
if (top == b)
return true;
// Traverse the array, D[]
for (int i = 0; i < n; i++) {
// Divide top by D[i] if
// it is possible and
// push the result in q
if (d[i] != 0 && top % d[i] == 0
&& !visited[top / d[i]]) {
q.push(top / d[i]);
visited[top / d[i]] = true;
}
// If D[i] is present at the top
int index = isPresent(top, d[i]);
if (index != -1) {
// Remove the first occurrence
// of D[i] from the top and
// store the new number
int newElement
= removeDigit(top, index);
// Push newElement into the queue q
if (newElement != -1
&& (!visited[newElement])) {
q.push(newElement);
visited[newElement] = true;
}
}
}
}
// Return false if A can
// not be reduced to B
return false;
}
// Driver Code
int main()
{
int A = 5643, B = 81;
int D[] = { 3, 8, 1 };
int N = sizeof(D) / sizeof(D[0]);
if (reduceNtoX(A, B, D, N))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to check if a digit x is
// present in the number N or not
static int isPresent(int n, int x)
{
// Convert N to string
String num = String.valueOf(n);
// Traverse the string num
for (int i = 0; i < num.length();
i++) {
// Return first occurrence
// of the digit x
if ((num.charAt(i) - '0') == x)
return i;
}
return -1;
}
// Function to remove the character
// at a given index from the number
static int removeDigit(int n, int index)
{
// Convert N to string
String num = String.valueOf(n);
// Store the resultant string
String ans = "";
// Traverse the string num
for (int i = 0;
i < num.length(); i++)
{
if (i != index)
ans += num.charAt(i);
}
// If the number becomes empty
// after deletion, then return -1
if (ans == "" || (ans.length() == 1
&& ans.charAt(0) == '0'))
return -1;
// Return the number
int x = Integer.valueOf(ans);
return x;
}
// Function to check if A can be
// reduced to B by performing the
// operations any number of times
static boolean reduceNtoX(int a, int b,
int d[], int n)
{
// Create a queue
Queue<Integer> q=new LinkedList<>();
// Push A into the queue
q.add(a);
// Hashmap to check if the element
// is present in the Queue or not
Map<Integer, Boolean> visited= new HashMap<>();
// Set A as visited
visited.put(a,true);
// Iterate while the queue is not empty
while (!q.isEmpty()) {
// Store the front value of the
// queue and pop it from it
int top = q.peek();
q.poll();
if (top < 0)
continue;
// If top is equal to B,
// then return true
if (top == b)
return true;
// Traverse the array, D[]
for (int i = 0; i < n; i++) {
// Divide top by D[i] if
// it is possible and
// push the result in q
if (d[i] != 0 && top % d[i] == 0
&& !visited.getOrDefault(top / d[i], false)) {
q.add(top / d[i]);
visited.put(top / d[i], true);
}
// If D[i] is present at the top
int index = isPresent(top, d[i]);
if (index != -1) {
// Remove the first occurrence
// of D[i] from the top and
// store the new number
int newElement
= removeDigit(top, index);
// Push newElement into the queue q
if (newElement != -1
&& (!visited.getOrDefault(newElement,false))) {
q.add(newElement);
visited.put(newElement, true);
}
}
}
}
// Return false if A can
// not be reduced to B
return false;
}
// Driver code
public static void main (String[] args)
{
// Given inputs
int A = 5643, B = 81;
int D[] = { 3, 8, 1 };
int N = D.length;
if (reduceNtoX(A, B, D, N))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by offbeat
Python 3
# Python3 program for the above approach
from collections import deque
# Function to check if a digit x is
# present in the number N or not
def isPresent(n, x):
# Convert N to string
num = str(n)
# Traverse the num
for i in range(len(num)):
# Return first occurrence
# of the digit x
if ((ord(num[i]) - ord('0')) == x):
return i
return -1
# Function to remove the character
# at a given index from the number
def removeDigit(n, index):
# Convert N to string
num = str(n)
# Store the resultant string
ans = ""
# Traverse the num
for i in range(len(num)):
if (i != index):
ans += num[i]
# If the number becomes empty
# after deletion, then return -1
if (ans == "" or (len(ans) == 1 and
ans[0] == '0')):
return -1
# Return the number
x = int(ans)
return x
# Function to check if A can be
# reduced to B by performing the
# operations any number of times
def reduceNtoX(a, b, d, n):
# Create a queue
q = deque()
# Push A into the queue
q.append(a)
# Hashmap to check if the element
# is present in the Queue or not
visited = {}
# Set A as visited
visited[a] = True
# Iterate while the queue is not empty
while (len(q) > 0):
# Store the front value of the
# queue and pop it from it
top = q.popleft()
if (top < 0):
continue
# If top is equal to B,
# then return true
if (top == b):
return True
# Traverse the array, D[]
for i in range(n):
# Divide top by D[i] if
# it is possible and
# push the result in q
if (d[i] != 0 and top % d[i] == 0 and
(top // d[i] not in visited)):
q.append(top // d[i])
visited[top // d[i]] = True
# If D[i] is present at the top
index = isPresent(top, d[i])
if (index != -1):
# Remove the first occurrence
# of D[i] from the top and
# store the new number
newElement = removeDigit(top, index)
# Push newElement into the queue q
if (newElement != -1 and
(newElement not in visited)):
q.append(newElement)
visited[newElement] = True
# Return false if A can
# not be reduced to B
return False
# Driver Code
if __name__ == '__main__':
A, B = 5643, 81
D = [ 3, 8, 1 ]
N = len(D)
if (reduceNtoX(A, B, D, N)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check if a digit x is
// present in the number N or not
static int isPresent(int n, int x)
{
// Convert N to string
string num = n.ToString();
// Traverse the string num
for (int i = 0; i < num.Length;
i++) {
// Return first occurrence
// of the digit x
if (((int)num[i] - 97) == x)
return i;
}
return -1;
}
// Function to remove the character
// at a given index from the number
static int removeDigit(int n, int index)
{
// Convert N to string
string num = n.ToString();
// Store the resultant string
string ans = "";
// Traverse the string num
for (int i = 0;
i < num.Length; i++) {
if (i != index)
ans += num[i];
}
// If the number becomes empty
// after deletion, then return -1
if (ans == "" || (ans.Length == 1
&& ans[0] == '0'))
return -1;
// Return the number
int x = Int32.Parse(ans);;
return x;
}
// Function to check if A can be
// reduced to B by performing the
// operations any number of times
static bool reduceNtoX(int a, int b,
int []d, int n)
{
// Create a queue
Queue<int> q = new Queue<int>();
// Push A into the queue
q.Enqueue(a);
// Hashmap to check if the element
// is present in the Queue or not
Dictionary<int,bool> visited = new Dictionary<int,bool>();
// Set A as visited
visited[a] = true;
// Iterate while the queue is not empty
while (q.Count>0) {
// Store the front value of the
// queue and pop it from it
int top = q.Peek();
q.Dequeue();
if (top < 0)
continue;
// If top is equal to B,
// then return true
if (top == b)
return true;
// Traverse the array, D[]
for (int i = 0; i < n; i++) {
// Divide top by D[i] if
// it is possible and
// push the result in q
if (d[i] != 0 && top % d[i] == 0 &&
visited.ContainsKey(top / d[i]) && visited[top / d[i]]==false) {
q.Enqueue(top / d[i]);
visited[top / d[i]] = true;
}
// If D[i] is present at the top
int index = isPresent(top, d[i]);
if (index != -1) {
// Remove the first occurrence
// of D[i] from the top and
// store the new number
int newElement = removeDigit(top, index);
// Push newElement into the queue q
if (newElement != -1 && (visited.ContainsKey(newElement) && visited[newElement]==false)) {
q.Enqueue(newElement);
visited[newElement] = true;
}
}
}
}
// Return false if A can
// not be reduced to B
return true;
}
// Driver Code
public static void Main()
{
int A = 5643, B = 81;
int []D = { 3, 8, 1 };
int N = D.Length;
if (reduceNtoX(A, B, D, N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by SURENDRA_GANGWAR.
java 描述语言
<script>
// Javascript program for the above approach
// Function to check if a digit x is
// present in the number N or not
function isPresent(n,x)
{
// Convert N to string
let num = (n).toString();
// Traverse the string num
for (let i = 0; i < num.length;
i++) {
// Return first occurrence
// of the digit x
if ((num[i].charCodeAt(0) - '0'.charCodeAt(0)) == x)
return i;
}
return -1;
}
// Function to remove the character
// at a given index from the number
function removeDigit(n,index)
{
// Convert N to string
let num = (n).toString();
// Store the resultant string
let ans = "";
// Traverse the string num
for (let i = 0;
i < num.length; i++)
{
if (i != index)
ans += num[i];
}
// If the number becomes empty
// after deletion, then return -1
if (ans == "" || (ans.length == 1
&& ans[0] == '0'))
return -1;
// Return the number
let x = parseInt(ans);
return x;
}
// Function to check if A can be
// reduced to B by performing the
// operations any number of times
function reduceNtoX(a,b,d,n)
{
// Create a queue
let q=[];
// Push A into the queue
q.push(a);
// Hashmap to check if the element
// is present in the Queue or not
let visited= new Map();
// Set A as visited
visited.set(a,true);
// Iterate while the queue is not empty
while (q.length!=0) {
// Store the front value of the
// queue and pop it from it
let top = q.shift();
if (top < 0)
continue;
// If top is equal to B,
// then return true
if (top == b)
return true;
// Traverse the array, D[]
for (let i = 0; i < n; i++) {
// Divide top by D[i] if
// it is possible and
// push the result in q
if(!visited.has(top / d[i]))
visited.set(top / d[i],false);
if (d[i] != 0 && top % d[i] == 0
&& !visited.get(top / d[i])) {
q.push(top / d[i]);
visited.set(top / d[i], true);
}
// If D[i] is present at the top
let index = isPresent(top, d[i]);
if (index != -1) {
// Remove the first occurrence
// of D[i] from the top and
// store the new number
let newElement
= removeDigit(top, index);
if(!visited.has(newElement))
visited.set(newElement,false);
// Push newElement into the queue q
if (newElement != -1
&& (!visited.get(newElement))) {
q.push(newElement);
visited.set(newElement, true);
}
}
}
}
// Return false if A can
// not be reduced to B
return false;
}
// Driver code
// Given inputs
let A = 5643, B = 81;
let D = [ 3, 8, 1 ];
let N = D.length;
if (reduceNtoX(A, B, D, N))
document.write("Yes");
else
document.write("No");
// This code is contributed by unknown2108
</script>
Output:
Yes
时间复杂度:O(2N) 辅助空间: O(2 N )
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