可被 M 整除的第 K 个最大的 N 位数

给定三个正整数N 、 K和M 。任务是找到可被M整除的第K个最大的N位数。

注意： K 将是这样一个整数，即始终存在可被M整除的第 K个最大的N位数。

例子

Input: N = 2, K = 2, M = 2
Output: 96
Explanation: The 2nd largest 2 digit number divisible by 2 is 96.

Input: N = 9, K = 6, M = 4
Output: 999999976

方法：问题是基于数学的。给定三个数字N 、 K和M 。需要找到可被M整除的第K个最大的N位数。要获得可被M整除的最大N位数，首先需要找到最大的N位数（例如P ），即N乘以9 。
现在可被M整除的最大N位数是(P – (P%M)) 。
因此，从该值中减去 (K-1) 次 M以得到可被M整除的N位数的第 K个最大值。
下面给出了获得可被M整除的第 K个最大N位数的条件和数学表达式。

Let P be the largest N digit number.
Then the largest N digit number divisible by M is: (P – (P % M)).
Now the Kth largest N digit number divisible by M is: [(P – (P % M)) – ((K – 1) * M)]

编程需要懂一点英语

下面是根据上述公式的代码。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find Kth N
// digit number divisible by M
int findAnswer(int N, int K, int M)
{
    int i;
    long long int r = 0;
 
    // Loop to calculate the largest
    // N digit number.
    for (i = 1; i <= N; i++) {
        r = r * 10 + 9;
    }
 
    // Kth largest N digit number
    // divisible by M.
    long long int u = r - (r % M)
        - M * (K - 1);
 
    return u;
}
 
// Driver Code
int main()
{
    int N = 9;
    int K = 6;
    int M = 4;
 
    cout << findAnswer(N, K, M);
    return 0;
}

Java

// Java program for above approach
import java.util.*;
 
class GFG{
 
  // Function to find Kth N
  // digit number divisible by M
  static int findAnswer(int N, int K, int M)
  {
    int i;
    int r = 0;
 
    // Loop to calculate the largest
    // N digit number.
    for (i = 1; i <= N; i++) {
      r = r * 10 + 9;
    }
 
    // Kth largest N digit number
    // divisible by M.
    int u = r - (r % M)
      - M * (K - 1);
 
    return u;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 9;
    int K = 6;
    int M = 4;
 
    System.out.print(findAnswer(N, K, M));
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
 
# Function to find Kth N
# digit number divisible by M
def findAnswer(N, K, M):
    i = None
    r = 0;
 
    # Loop to calculate the largest
    # N digit number.
    for i in range(1, N + 1):
        r = r * 10 + 9;
 
    # Kth largest N digit number
    # divisible by M.
    u = r - (r % M) - M * (K - 1);
 
    return u;
 
# Driver Code
N = 9;
K = 6;
M = 4;
 
print(findAnswer(N, K, M));
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for above approach
using System;
class GFG
{
 
  // Function to find Kth N
  // digit number divisible by M
  static int findAnswer(int N, int K, int M)
  {
    long r = 0;
 
    // Loop to calculate the largest
    // N digit number.
    for (int i = 1; i <= N; i++) {
      r = r * 10 + 9;
    }
 
    // Kth largest N digit number
    // divisible by M.
    long u = r - (r % M)
      - M * (K - 1);
 
    return (int)u;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 9;
    int K = 6;
    int M = 4;
 
    Console.Write(findAnswer(N, K, M));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

999999976

时间复杂度： O(MaxDigit)，其中 maxDigit 是最大的 N 位数。
辅助空间： O(1)