Distinct numbers in array - c++

Question

I have no idea what to do. Please help me with code or tell me what textbook to look up or something; I need code to finish this program and I would love an explanation of what I'm looking at..
#include<iostream>
using namespace std;
int main()
{
short num[100], size, //declare an array of type short that has 100 elements
unique[100], number, // declare a second array to help solve the problem; number counts the number of unique values
k; // loop control variable; may need other variables
cout<<"enter the number of values to store in the array\n";
cin>>size;
cout<<”enter the “<<size<<” values to be used as data for this program\n”;
for(k=0; k<size; k++)
cin>>num[k];
// print the contents of the array
cout<<"\nthere are "<<size<<" values in the array\n";
for(k=0; k<size; k++)
cout<<num[k]<<’ ‘; // there is one space between each number in the display
cout<<endl; // cursor moved to next output line
cout<<"the program will count the number of different (distinct) values found in the array\n";
//************************************************************
//Put the code here that counts the number of unique values stored in the
//array num. The variable number will contain the count.
//************************************************************
cout<<endl<<number<<" unique values were found in the "<<size<<" element array\n";
// pause the program to see the results
system("pause");
//return 0;
}
I have to do one of these two things and I don't know what they mean?
Algorithm – unique array is used to help find the solution, used to avoid counting any value more than one time
Set number to 0 - this represents the number of distinct values in the data set; also used as a subscript in the unique array
Loop from 0 to size by one, proceeding through successive elements of the data (num) array
Store value of current array element in non-array variable (SV)
Set event_flag to 0
Loop from 0 to number by one, proceeding through successive elements of unique array
If SV is equal to current element of unique array
Set event_flag to 1
Break (stop) inner loop
End of inner loop
If event_flag is equal to 0 (value not found in unique array and not previously counted)
Store SV in element number of unique array
Increment the value of number
End of outer loop
Solution – the variable number contains the count of distinct values in the data array
Alternate Algorithm
Algorithm that does not use the event_flag (loop control variable can be used to determine if event occurred)
Algorithm – unique array is used to help find the solution, used to avoid counting any value more than one time
Set number to 0 - this represents the number of distinct values in the data set; also used as a subscript in the unique array
Loop from 0 to size by one, proceeding through successive elements of the data (num) array
Store value of current array element in non-array variable (SV)
Loop from 0 to number by one, proceeding through successive elements of unique array
If SV is equal to current element of unique array
Break (stop) inner loop
End of inner loop
If loop control variable of inner loop is equal to value of number (SV not found in unique array and not previously counted)
Store SV in element number of unique array
Increment the value of number
End of outer loop
Solution – the variable number contains the count of distinct values in the data array
I put this in mine:
//************************************************************
//Put the code here that counts the number of unique values stored in the array num. The variable number will contain the count.
for(k=0; k<size; k++)
num=SV;
event_flag=0;
for(k=1; k<number; k++)
if(SV=unique)
return true;
return false;
//************************************************************
It's not working, obviously.

Answer 1

This is my code, it seems to work
//************************************************************
//Put the code here that counts the number of unique values
//stored in the array num. The variable number will contain the count.
number = 0;
for (k = 0; k < size; ++k)
{
short sv = num[k];
short event_flag = 0;
for (int i = 0; i < number; ++i)
{
if (sv == unique[i])
{
event_flag = 1;
break;
}
}
if (event_flag == 0)
{
unique[number] = sv;
++number;
}
}
For the alternative ,
his is my code, it seems to work
//************************************************************
//Put the code here that counts the number of unique values
//stored in the array num. The variable number will contain the count.
number = 0;
for (k = 0; k < size; ++k)
{
short sv = num[k];
int i;
for (i = 0; i < number; ++i)
if (sv == unique[i])
break;
if (number == i)
{
unique[number] = sv;
++number;
}
}

Answer 2

You are roughly asked to do the following:
#include <iostream>
using namespace std;
int main()
{
// This is the given array.
int given_array[5] = { 1, 1, 2, 2, 3 };
// This is the array where unique values will be stored.
int unique_array[5];
// This index is used to keep track of the size
// (different from capacity) of unique_array.
int unique_index = 0;
// This is used to determine whether we can
// insert an element into unique_array or not.
bool can_insert;
// This loop traverses given_array.
for (int i = 0; i < 5; ++i)
{
// Initially assume that we can insert elements
// into unique_array, unless told otherwise.
can_insert = true;
// This loop traverses unique_array.
for (int j = 0; j < unique_index; ++j)
{
// If the element is already in unique_array,
// then don't insert it again.
if (unique_array[j] == given_array[i])
{
can_insert = false;
break;
}
}
// This is the actual inserting.
if (can_insert)
{
unique_array[unique_index] = given_array[i];
unique_index++;
}
}
// Tell us how many elements are unique.
cout << unique_index;
return 0;
}

Answer 3

Try out this one...
You can insert cout statements wherever required.
#include <iostream>
using namespace std;
int main()
{
int Input[100], Unique[100], InSize, UniLength = 0;
cin >> InSize;
for (int ii = 0 ; ii < InSize ; ii++ )
{
cin >> Input[ii];
}
Unique[0] = Input[0];
UniLength++;
bool IsUnique;
for ( int ii = 1 ; ii < InSize ; ii++ )
{
IsUnique=true;
for (int jj = 0 ; jj < UniLength ; jj++ )
{
if ( Input[ii] == Unique[jj] )
{
IsUnique=false;
break;
}
}
if ( IsUnique )
{
Unique[UniLength] = Input[ii];
UniLength++;
}
}
cout<<"We've "<<UniLength<<" Unique elements and We're printing them"<<endl;
for ( int jj = 0 ; jj < UniLength ; jj++ )
{
cout << Unique[jj] << " ";
}
}
I hope this is what you were looking for.....
Have a nice day.

Answer 4

Here is my approach. Hope so it will helpful.
#include<iostream>
#include<algorithm>
int main(){
int arr[] = {3, 2, 3, 4, 1, 5, 5, 5};
int len = sizeof(arr) / sizeof(*arr); // Finding length of array
std::sort(arr, arr+len);
int unique_elements = std::unique(arr, arr+len) - arr;
std::cout << unique_elements << '\n'; // The output will 5
return 0;
}