Make a program for the output of all three-digit numbers, the amount of numbers of which is equal to this natural number. In solving the problem, it is necessary to use a cycle with a detonation

# include "PCH.H"
#Include & lt; iostream & gt;
Using Namespace STD;
INT MAIN ()
{
  INT Z, Y, X, N, I;
  SETLOCALE (LC_ALL, "RUS");
  COUT & LT; & LT; "\ N. Natural N:";
  CIN & GT; & GT; n;
  z = y = i = 0;
  x = 1;
  do.
  {
    if (n = x + y + z) cout & lt; & lt; "\ NKVV X, Y, Z" & LT; & LT; x * 100 + y * 10 + z;
    Z ++;
    if (z = 10) y ++, z = 0;
    if (y = 10) x ++, y = 0;
    COUT & LT; & LT; "\ NKVV z" & lt; & lt; z;
    COUT & LT; & LT; "\ NKVV Y" & lt; & lt; y;
    COUT & LT; & LT; "\ NKVV X" & lt; & lt; x;
  }
  While (x & lt; = 10);
}

Answer 1

You can not bother not so much:

size_t digitsum (Size_t Number)
{
  Size_t Sum {0};
  While (Number & gt; 0)
  {
     SUM + = NUMBER% 10;
     Number / = 10;
  }
  Return Sum;
}
INT MAIN ()
{
  Size_t Requested_sum {0};
  STD :: CIN & GT; & GT; Requested_sum;
  Size_t Number {100};
  do.
  {
    If (Digitsum (Number) == Requested_SUM) {
      STD :: COUT & LT; & LT; Number & lt; & lt; STD :: ENDL;
    }
    Number ++;
  } While (Number & LT; 1000);
  Return 0;
}

And yes, as surely noticed in the comments, do not be confused = and == 🙂

Answer 2

int x = 0, y = 0, n, k;
if (CIN & GT; & gt; n) {// if the specified number
  do {
    CIN & GT; & GT; k;
    x = k / 100;
    if (x & amp; & amp;! (x / 10)) {// If a three-digit number is introduced
      y = k / 10; // Second digit
      IF (x + y% 10 + k% 10 == n)
        COUT & LT; & LT; "Number" & lt; & lt; k & lt; & lt; "Satisfies the requirement \ n";
    }
  } While (CIN); // While the number has been introduced
}