OJ 1507 deletion problem

describe

Enter a high precision positive integer ｎ（<=240 position ）,

Remove any ｓ The remaining numbers after the first number will form a new integer in the original left-right order .

Programming for a given ｎ and ｓ, Look for a solution , Minimize the remaining number .

Input
first line ： The length is n The positive integer

The second line ： Number of numbers to delete s

Output
The minimum number remaining after deletion

sample input 1

178543
4
sample output 1

13

The title requires deleting some data to achieve the minimum number after deletion , That is to delete the maximum value in a continuous non descending order every time , Why not in descending order , Because it's possible There is only one data in the whole sequence , This is in descending order compared with the following , At this time, delete the first data in descending order , It's similar. In fact, this is an ascending order of data .
 

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string n;
    int a;
    while(cin>>n>>a)
    {
        while(a--)
        {
            int i=0;
            while(n[i]<=n[i+1])
                i++;
            n.erase(i,1);
        }
        int i=0;
        for(i;i<n.size();i++)
            if(n[i]!='0')
            break;
        for(i;i<n.size();i++)
            cout<<n[i];
        cout<<endl;
    }
    return 0;
}