Arrays in C++

I am not very good at programming so thanks to all who help in advance. I can only utilize basic code that involves while, for, and do loops. in this case i think for loops are the most efficient.

I need to have a vector given 20 integers, smallest to largest, and then remove the items in the vector in the fewest possible steps by using the following process:
An element X in the data set can be removed along with X – 1 or X + 1 if they are present in the set. After each deletion, print the values remaining in the data set.

heres my code so far (only have the input) Any ideas?

Code:

---

#include <stdio.h>

#define numdata 20

void getdata(int[]);
void riddata(int[]);
int main ()
{
  int data[numdata]; //array to hold data to sort

  getdata(data);
  riddata(data);
  return(0);
}

void getdata(int x[])
{
  int lcv; //loop control variable get data

  for(lcv = 0; lcv < numdata; lcv++)
  {
    printf("Enter data #%d: ", lcv+1);
    scanf("%d", &x[lcv]);
  }
}

void riddata(int x[])
{
  for
}

---

Quote:

---

Originally Posted by stinky123

I am not very good at programming so thanks to all who help in advance. I can only utilize basic code that involves while, for, and do loops. in this case i think for loops are the most efficient.

---

Concerning efficiency, there isn't a measurable difference in performance between while for or do, you choose one out of convenience, and that depends on what kind of loop it is. The for loop would be the right choice here, because it's a counting loop and for groups all three parts of a counting loop together.

Quote:

---

I need to have a vector given 20 integers, smallest to largest, and then remove the items in the vector in the fewest possible steps by using the following process:
An element X in the data set can be removed along with X – 1 or X + 1 if they are present in the set. After each deletion, print the values remaining in the data set.

---

Well the smallest X will be encountered first if they are in fact arranged smallest to largest. If the position of X is p then if *(p+1) == X+1 AND *(p+2) == X+2, then you can call erase(p, p+3); on the vector and proceed from there. If the test fails, you'll have to make the next X, *(p+1), the new X.

Of course, you aren't using a vector object, yet. You're using an array. With arrays, you can't really use erase. Instead, I would copy over all the Xs involved, where the above test fails, into a new array.

Why are you using scanf and printf in C++?