可以在 K 步中减少到 0 的最大数 N

给定一个整数N 。执行以下任务：

• 编号已注明。
• 选自N前导数字选自N减去所得到的值存储回N.
• 再次记录N的新值。
• 这个过程一直持续到 N 变为 0。最后，最后记下0 。

例如，取N = 13 。在将 13 减少到 0 的过程中记下的数字将是：

• 13
• 13 – 1 = 12
• 12 – 1 = 11
• 11 – 1 = 10
• 10 – 1 = 9
• 9 – 9 = 0

给定一个整数K ，它是在将数字N减少到 0的过程中记下的数字总数，任务是找到起始数字N的最大可能值。

例子：

Input: k = 2 
Output: 9 
Explanation: Numbers noted = {9} 

• Subtract Leading digit: 9 – 9 = 0. Numbers noted = {9, 0} 
   

Input: k = 3 
Output: 10 
Explanation: Numbers noted = {10} 

• Subtract Leading digit: 10 – 1 = 9. Numbers noted = {10, 9}
• Subtract Leading digit: 9 – 9 = 0. Numbers noted = {10, 9, 0}

方法：想法是观察 N 的最大可能值将始终小于 10*K。因此，想法是在 K 到 10*K 的范围内应用二分搜索，并寻找可以在 K 步中减少到 0 的最大数。
要找到最大可能的数字：

1. 将左初始化为k ，右初始化为k*10 。
2. 将mid初始化为(left + right) / 2 。
3. 获取记下的数字的计数以将mid转换为 0并将其存储在len 中。
4. 遵循分而治之的方法：虽然len不等于k ：
   • 将 mid 更新为当前(left + right) / 2 。
   • 从mid开始时获取计数。
   • 如果计数大于中旬，中旬更新权更大。
   • 否则，更新left到mid 。
5. 现在， mid有一个值，这将导致写下k 个整数。
6. 要找到最大的这样的数字：
   • 虽然计数等于k ：
   • 如果当前计数不等于采用mid+1得到的计数，则中断
   • 否则，继续将mid增加 1。
7. 如果写下k 个整数， mid现在是最大可能的整数。

下面是上述方法的实现：

C++

// C++ program to implement above approach
 
#include 
using namespace std;
 
// Utility function to return the first
// digit of a number.
int firstDigit(int n)
{
    // Remove last digit from number
    // till only one digit is left
    while (n >= 10) {
        n /= 10;
    }
 
    // return the first digit
    return n;
}
 
// Utility function that returns the count of numbers
// written down when starting from n
int getCount(int n)
{
    int count = 1;
 
    while (n != 0) {
        int leadDigit = firstDigit(n);
        n -= leadDigit;
        count++;
    }
    return count;
}
 
// Function to find the largest number N which
// can be reduced to 0 in K steps
int getLargestNumber(int k)
{
    int left = k;
    int right = k * 10;
    int mid = (left + right) / 2;
 
    // Get the sequence length of the mid point
    int len = getCount(mid);
 
    // Until k sequence length is reached
    while (len != k) {
 
        // Update mid point
        mid = (left + right) / 2;
 
        // Get count of the new mid point
        len = getCount(mid);
 
        if (len > k) {
 
            // Update right to mid
            right = mid;
        }
        else {
 
            // Update left to mid
            left = mid;
        }
    }
 
    // Increment mid point by one while count
    // is equal to k to get the maximum value
    // of mid point
    while (len == k) {
 
        if (len != getCount(mid + 1)) {
            break;
        }
 
        mid++;
    }
 
    return (mid);
}
 
// Driver Code
int main()
{
    int k = 3;
 
    cout << getLargestNumber(k);
 
    return 0;
}

Java

// Java program to implement above approach
import java.io.*;
 
class GFG
{
     
// Utility function to return the first
// digit of a number.
static int firstDigit(int n)
{
    // Remove last digit from number
    // till only one digit is left
    while (n >= 10)
    {
        n /= 10;
    }
 
    // return the first digit
    return n;
}
 
// Utility function that returns the count of numbers
// written down when starting from n
static int getCount(int n)
{
    int count = 1;
 
    while (n != 0)
    {
        int leadDigit = firstDigit(n);
        n -= leadDigit;
        count++;
    }
    return count;
}
 
// Function to find the largest number N which
// can be reduced to 0 in K steps
static int getLargestNumber(int k)
{
    int left = k;
    int right = k * 10;
    int mid = (left + right) / 2;
 
    // Get the sequence length of the mid point
    int len = getCount(mid);
 
    // Until k sequence length is reached
    while (len != k)
    {
 
        // Update mid point
        mid = (left + right) / 2;
 
        // Get count of the new mid point
        len = getCount(mid);
 
        if (len > k)
        {
 
            // Update right to mid
            right = mid;
        }
        else
        {
 
            // Update left to mid
            left = mid;
        }
    }
 
    // Increment mid point by one while count
    // is equal to k to get the maximum value
    // of mid point
    while (len == k)
    {
 
        if (len != getCount(mid + 1))
        {
            break;
        }
 
        mid++;
    }
 
    return (mid);
}
 
    // Driver Code
    public static void main (String[] args)
    {
 
        int k = 3;
        System.out.println (getLargestNumber(k));
    }
}
 
// This code is contributed by jit_t

Python3

# Python3 program to implement above approach
 
# Utility function to return the first
# digit of a number.
def firstDigit(n) :
 
    # Remove last digit from number
    # till only one digit is left
    while (n >= 10) :
        n //= 10;
     
    # return the first digit
    return n;
 
# Utility function that returns
# the count of numbers written
# down when starting from n
def getCount(n) :
     
    count = 1;
 
    while (n != 0) :
        leadDigit = firstDigit(n);
        n -= leadDigit;
        count += 1;
     
    return count;
 
# Function to find the largest number N
# which can be reduced to 0 in K steps
def getLargestNumber(k) :
 
    left = k;
    right = k * 10;
    mid = (left + right) // 2;
 
    # Get the sequence length of the mid point
    length = getCount(mid);
 
    # Until k sequence length is reached
    while (length != k) :
 
        # Update mid point
        mid = (left + right) // 2;
 
        # Get count of the new mid point
        length = getCount(mid);
 
        if (length > k) :
 
            # Update right to mid
            right = mid;
         
        else :
 
            # Update left to mid
            left = mid;
         
    # Increment mid point by one while count
    # is equal to k to get the maximum value
    # of mid point
    while (length == k) :
 
        if (length != getCount(mid + 1)) :
            break;
         
        mid += 1;
     
    return mid;
 
# Driver Code
if __name__ == "__main__" :
 
    k = 3;
 
    print(getLargestNumber(k));
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Utility function to return the first
// digit of a number.
static int firstDigit(int n)
{
    // Remove last digit from number
    // till only one digit is left
    while (n >= 10)
    {
        n /= 10;
    }
 
    // return the first digit
    return n;
}
 
// Utility function that returns the count of numbers
// written down when starting from n
static int getCount(int n)
{
    int count = 1;
 
    while (n != 0)
    {
        int leadDigit = firstDigit(n);
        n -= leadDigit;
        count++;
    }
    return count;
}
 
// Function to find the largest number N which
// can be reduced to 0 in K steps
static int getLargestNumber(int k)
{
    int left = k;
    int right = k * 10;
    int mid = (left + right) / 2;
 
    // Get the sequence length of the mid point
    int len = getCount(mid);
 
    // Until k sequence length is reached
    while (len != k)
    {
 
        // Update mid point
        mid = (left + right) / 2;
 
        // Get count of the new mid point
        len = getCount(mid);
 
        if (len > k)
        {
 
            // Update right to mid
            right = mid;
        }
        else
        {
 
            // Update left to mid
            left = mid;
        }
    }
 
    // Increment mid point by one while count
    // is equal to k to get the maximum value
    // of mid point
    while (len == k)
    {
 
        if (len != getCount(mid + 1))
        {
            break;
        }
 
        mid++;
    }
 
    return (mid);
}
 
// Driver Code
public static void Main(String []args)
{
    int k = 3;
 
    Console.WriteLine( getLargestNumber(k));
}
}
 
// This code is contributed by Arnab Kundu

PHP

= 10)
    {
        $n = (int)($n / 10);
    }
 
    // return the first digit
    return $n;
}
 
// Utility function that returns the
// count of numbers written down when
// starting from n
function getCount($n)
{
    $count = 1;
 
    while ($n != 0)
    {
        $leadDigit = firstDigit($n);
        $n -= $leadDigit;
        $count++;
    }
    return $count;
}
 
// Function to find the largest number N
// which can be reduced to 0 in K steps
function getLargestNumber($k)
{
    $left = $k;
    $right = $k * 10;
    $mid = (int)(($left + $right) / 2);
 
    // Get the sequence length
    // of the mid point
    $len = getCount($mid);
 
    // Until k sequence length is reached
    while ($len != $k)
    {
 
        // Update mid point
        $mid = (int)(($left + $right) / 2);
 
        // Get count of the new mid point
        $len = getCount($mid);
 
        if ($len > $k)
        {
 
            // Update right to mid
            $right = $mid;
        }
        else
        {
 
            // Update left to mid
            $left = $mid;
        }
    }
 
    // Increment mid point by one while count
    // is equal to k to get the maximum value
    // of mid point
    while ($len == $k)
    {
        if ($len != getCount($mid + 1))
        {
            break;
        }
 
        $mid++;
    }
 
    return ($mid);
}
 
// Driver Code
$k = 3;
echo(getLargestNumber($k));
 
// This code is contributed by Code_Mech.
?>

Javascript

    // Javascript implementation of the approach
 
      // Utility function to return the first
    // digit of a number.
    function firstDigit(n)
    {
        // Remove last digit from number
        // till only one digit is left
        while (n >= 10)
        {
            n = parseInt(n / 10, 10);
        }
 
        // return the first digit
        return n;
    }
 
    // Utility function that returns the count of numbers
    // written down when starting from n
    function getCount(n)
    {
        let count = 1;
 
        while (n != 0)
        {
            let leadDigit = firstDigit(n);
            n -= leadDigit;
            count++;
        }
        return count;
    }
 
    // Function to find the largest number N which
    // can be reduced to 0 in K steps
    function getLargestNumber(k)
    {
        let left = k;
        let right = k * 10;
        let mid = parseInt((left + right) / 2, 10);
 
        // Get the sequence length of the mid point
        let len = getCount(mid);
 
        // Until k sequence length is reached
        while (len != k)
        {
 
            // Update mid point
            mid = parseInt((left + right) / 2, 10);
 
            // Get count of the new mid point
            len = getCount(mid);
 
            if (len > k)
            {
 
                // Update right to mid
                right = mid;
            }
            else
            {
 
                // Update left to mid
                left = mid;
            }
        }
 
        // Increment mid point by one while count
        // is equal to k to get the maximum value
        // of mid point
        while (len == k)
        {
 
            if (len != getCount(mid + 1))
            {
                break;
            }
 
            mid++;
        }
 
        return (mid);
    }
     
    let k = 3;
   
    document.write( getLargestNumber(k));
 

输出：

10