As a general rule, we need to press one button per one character to display \(S\). The exception is 0, where 00 button is available.

If there is a portion of \(z\) consecutive 0s, it can be entered by pressing the button \(\lceil z/2 \rceil\) times. (\(\lceil x \rceil\) is \(x\) rounded up.)

Solution 1. press greedily from the beginning

By pressing the buttons as follows, one can minimize the number of operations required.

• If the uninputted part starts with two consecutive 0s, then press them at once.
• Otherwise, insert the first character.

This can be simulated using a loop.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  string s;
  cin >> s;
  int l=s.size();
  int p=0,res=0;
  while(p<l){
    res++;
    if((p+1)<l && s[p]=='0' && s[p+1]=='0'){p+=2;}
    else{p++;}
  }
  cout << res << "\n";
  return 0;
}

Solution 2. find consecutive 0s

One can (exhaustively) enumerate consecutive occurrences of 0, and for each run, count how many 0 there are, to find the answer.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  string s;
  cin >> s;
  int l=s.size();
  int z=0,res=0;
  for(auto &nx : s){
    if(nx=='0'){
      z++;
    }
    else{
      res+=(z+1)/2;
      z=0;
      res++;
    }
  }
  res+=(z+1)/2;
  cout << res << "\n";
  return 0;
}