Shulou(Shulou.com)06/01 Report--

This article introduces the knowledge of "how to keep the maximum number in C++". In the operation of actual cases, many people will encounter such a dilemma. Then let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

Topic description

Given a decimal positive integer number, choose to remove part of the number from it, and the number you want to retain will make up the largest positive integer.

Enter description

Enter as two lines, the first line being a positive integer number,1 ≤ length (number) ≤ 50000. The second line is the number of numbers you want to remove cnt 1 ≤ cnt

Output description

Output the retained results.

Example

Input

325 1

Output

thirty-five

Analysis

The number remaining is the largest after several digits are removed from an integer. The key to this question lies in how we choose the number to be removed, that is, what is the standard for removing the number. At first, Xiaobai thought that it would be OK to get rid of the smallest digits, but this idea was not correct. For example, 3120, if you remove 0, it is 312, but if you remove 1, it is 320, which is larger than 312, so we can decide that it is not right to simply remove the minimum number.

The right idea is to get rid of the number that is "smaller than the next bit", and if there is no number smaller than the last digit (that is, each digit of the whole number is an incremental number), it makes more sense to use the list container because you need to divide the number, and list allows insertion at the front end, which further supports the method of getting each digit by dividing by more than 10 digits. One thing to pay attention to here is that you can only take out one number at a time, and then you need to look for the first number that is smaller than the last one.

Code

# include

# include

Using namespace std

Int main ()

{

Int number, cnt

While (cin > > number > > cnt)

{

List Lnumber

While (numbered _ blank _ 0)

{

Lnumber.push_front (number 10)

Number = number / 10

}

Lnumber.push_front (number)

For (int I = 0; I < cnt; iTunes +)

{

For (auto m = Lnumber.cbegin (); m! = Lnumber.cend (); masks +)

{

If (masks = (--Lnumber.cend ()

{

M = Lnumber.erase (m)

Break

}

Auto n = m

If (* m)

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