Also in the task you need to put a space between the same characters, but it was possible to implement it.
Already 2 weeks I am trying to decide, about 10 hours spent during this time, but everything does not succeed. I decided to ask on the forums, and here came here.

Answer 1, Authority 100%

Somehow so

std :: string st = "abbcccabbccc";
std :: unique_copy (st.begin (), st.end (), std :: ostreambuf_terator & lt; char & gt; (std :: cout));
STD :: COUT & LT; & LT; '\ n';

Answer 2

ABYX working option. I can offer my own, it may be more understandable for you. If something is incomprehensible, ask questions 🙂

# include & lt; iostream & gt;
Using Namespace STD;
INT MAIN () {
  STRING S, RESULT; // S - Initial Row, Result - Finite
  char prev; // Last non-repeating symbol we met
  CIN & GT; & GT; s;
  Prev = S [0];
  result + = s [0];
  For (int i = 1; i & lt; s.size (); i ++) {
    if (s [i]! = prev) {
      prev = s [i];
      result + = s [i];
    }
  }
  COUT & LT; & LT; result;
  Return 0;
}

Answer 3

In the end, I decided 🙂

#include & lt; iostream & gt;
Using Namespace STD;
INT MAIN ()
{
  int n, m;
  String ST, STD;
  CIN & GT; & GT; ST;
  For (n = 0; N & LT; st.Size (); n ++) {
    IF (ST [N] == ST [N + 1]) {
      ST.ERASE (N, 1);
    }
  }
  STD = ST;
  For (n = 0; N & LT; st.Size (); n ++) {
    For (m = n + 1; m & lt; std.size () + 1; m ++) {
      IF (ST [N] == STD [M]) {
        STD.ERASE (M, 1);
      }
    }
  }
  COUT & LT; & LT; STD;
}

Answer 4

The function using iterators removes repeating characters from a string at the input and returns the result

string deleteduplicates (String Str) {
  For (String :: ITERATOR IT1 = STR.BEGIN (); it1! = str.end (); ++ IT1) {
    For (String :: Iterator It2 = it1 + 1; it2! = str.end ();) {
      if (* IT1 == * IT2) {
        STR.ERASE (IT2);
      }
      ELSE {
        ++ IT2;
      }
    }
  }
  Return str;
}

Answer 5

Removes from the line a repeating symbol I think to understand

for (int i = 0; i & lt; k.size ();) {
    if (used [k [i]])
      a.ERASE (I, I + 1);
    ELSE.
    {
      Used [k [i]] = 1;
      ++ I;
  }