可能的最小数量，使其具有 X 位并且通过旋转增长 Y 倍

给定两个正整数X和Y 。任务是找到最小可能的数字，使得该数字具有X个十进制数字，并且在将其最后一个数字移到第一个数字后增长Y倍

例子：

Input: X = 6, Y = 5 
Output: 142857
Explanation: Following are the operations performed to get the desired result.
Take last digit of 142857 i.e. 7 and put it forward at the beginning the number becomes 714285. 
Which is equal to 142857 * Y = 142857 * 5 = 714285.
Therefore, the number if 714285 which is minimum possible. 

Input: X = 1, Y = 2 
Output: -1
Explanation: The number that consists of a single digit cannot stay what it is when multiplied by 2 so no minimum number possible in this case.

方法：这个问题的解决方案是基于实现的。请按照以下步骤解决给定的问题。

• 声明一个向量num来存储形成的数字。
• 声明并初始化一个布尔变量possible = false ，以跟踪是否可以形成数字。
• 尝试通过迭代所有可能的数字来形成具有所需条件的X位数字。
• 如果数字可以在任何点形成，则设置可能 = true并从循环中中断。
• 最后，检查 if possible = true ：
  • 如果是，打印存储在 num 中的数字。
  • 否则不，找到这样的数字，所以返回-1 。
• 打印最终结果。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find minimum number
void findmin(int X, int Y)
{
    // Additional space required to form a number
    vector num;
 
    int a, b;
    bool possible = 0;
 
    // Forming a number
    for (int j = Y; j < 10; j++) {
        num.push_back(j);
        a = 0;
 
        // Forming X digit number
        for (int i = 0; i < X; i++) {
            b = a + num[i] * Y;
            a = b / 10;
            num.push_back(b - 10 * a);
        }
 
        if (num[X] == j && a == 0) {
            possible = 1;
            break;
        }
    }
 
    if (possible) {
 
        // Number can be formed
        for (int i = X - 1; i >= 0; i--) {
            cout << num[i];
        }
    }
 
    // There is no such number possible
    else
        cout << "-1";
}
 
// Driver code
int main()
{
    int X = 6;
    int Y = 4;
 
    // Function Call
    findmin(X, Y);
 
    return 0;
}

Java

// Java program for above approach
import java.util.ArrayList;
 
class GFG {
 
  // Function to find minimum number
  static void findmin(int X, int Y) {
 
    // Additional space required to form a number
    ArrayList num = new ArrayList();
 
    int a = 0, b = 0;
    boolean possible = false;
 
    // Forming a number
    for (int j = Y; j < 10; j++) {
      num.add(j);
      a = 0;
 
      // Forming X digit number
      for (int i = 0; i < X; i++) {
        b = a + (int) num.get(i) * Y;
        a = b / 10;
        num.add(b - 10 * a);
      }
 
      if ((int) num.get(X) == j && a == 0) {
        possible = true;
        break;
      }
    }
 
    if (possible) {
 
      // Number can be formed
      for (int i = X - 1; i >= 0; i--) {
        System.out.print(num.get(i));
      }
    }
 
    // There is no such number possible
    else
      System.out.print("-1");
  }
 
  // Driver code
  public static void main(String args[]) {
    int X = 6;
    int Y = 4;
 
    // Function Call
    findmin(X, Y);
 
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# python3 program for above approach
 
# Function to find minimum number
def findmin(X, Y):
 
    # Additional space required to form a number
    num = []
 
    a, b = 0, 0
    possible = 0
 
    # Forming a number
    for j in range(Y, 10):
        num.append(j)
        a = 0
 
        # Forming X digit number
        for i in range(0, X):
            b = a + num[i] * Y
            a = b // 10
            num.append(b - 10 * a)
 
        if (num[X] == j and a == 0):
            possible = 1
            break
 
    if (possible):
 
        # Number can be formed
        for i in range(X - 1, -1, -1):
            print(num[i], end="")
 
    # There is no such number possible
    else:
        print("-1")
 
# Driver code
if __name__ == "__main__":
 
    X = 6
    Y = 4
 
    # Function Call
    findmin(X, Y)
 
# This code is contributed by rakeshsahni

C#

// C# program for above approach
using System;
using System.Collections;
 
class GFG
{
   
  // Function to find minimum number
  static void findmin(int X, int Y)
  {
     
    // Additional space required to form a number
    ArrayList num = new ArrayList();
 
    int a = 0, b = 0;
    bool possible = false;
 
    // Forming a number
    for (int j = Y; j < 10; j++) {
      num.Add(j);
      a = 0;
 
      // Forming X digit number
      for (int i = 0; i < X; i++) {
        b = a + (int)num[i] * Y;
        a = b / 10;
        num.Add(b - 10 * a);
      }
 
      if ((int)num[X] == j && a == 0) {
        possible = true;
        break;
      }
    }
 
    if (possible) {
 
      // Number can be formed
      for (int i = X - 1; i >= 0; i--) {
        Console.Write(num[i]);
      }
    }
 
    // There is no such number possible
    else
      Console.Write("-1");
  }
 
  // Driver code
  public static void Main()
  {
    int X = 6;
    int Y = 4;
 
    // Function Call
    findmin(X, Y);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

      // JavaScript code for the above approach
 
      // Function to find minimum number
      function findmin(X, Y) {
          // Additional space required to form a number
          let num = [];
 
          let a, b;
          let possible = 0;
 
          // Forming a number
          for (let j = Y; j < 10; j++) {
              num.push(j);
              a = 0;
 
              // Forming X digit number
              for (let i = 0; i < X; i++) {
                  b = a + num[i] * Y;
                  a = Math.floor(b / 10);
                  num.push(b - 10 * a);
              }
 
              if (num[X] == j && a == 0) {
                  possible = 1;
                  break;
              }
          }
 
          if (possible) {
 
              // Number can be formed
              for (let i = X - 1; i >= 0; i--) {
                  document.write(num[i]);
              }
          }
 
          // There is no such number possible
          else
              document.write("-1")
      }
 
      // Driver code
 
      let X = 6;
      let Y = 4;
 
      // Function Call
      findmin(X, Y);
 
 
     // This code is contributed by Potta Lokesh
  

输出

102564

时间复杂度： O(X*Y)
辅助空间： O(X)