I was wondering, how do I put in an index number for n elements. The program provided asks you to enter the numbers and it puts them in descending order, but I need an index number to be put with the values you enter for n to be the same for that exact number you put in.

Here's the code I have:

#include <iostream>
using namespace std;

int Partition(int low,int high,int arr[]);
void Selection_sort(int low,int high,int arr[]);
int main()
{
int *a,n,low,high,i;
cout<<"/**************************Selection Sort Algorithm Implementation*****************/";
cout<<"Enter number of elements: ";
cin>>n;

a=new int[n];

cout<<"enter the elements: "<<endl;
for(i=0;i<n;i++)
{
cin >> a[i];
}
//for(i=0;i<n;i++)
//a[i]=rand()%100;


cout<<"Element: \tInitial Order of elements: \tIndex"<<endl;
 for(i=0;i<n;i++)
            
  cout<<i+1<<"\t"<<a[i]<<endl;


high=n-1;
low=0;
Selection_sort(low,high,a);
cout<<"Element: \tFinal Array After Sorting: \tIndex:"<<endl;

  for(i=0;i<n;i++)
  cout<<i+1<<"\t"<<a[i]<<endl;

system("pause");
return 0;

}


int Partition(int low,int high,int arr[])
{ int i,high_vac,low_vac,pivot/*,itr*/;
   pivot=arr[low];
   while(high>low)
{ high_vac=arr[high];

  while(pivot>high_vac)
  {
    if(high<=low) break;
    high--;
    high_vac=arr[high];
  }

  arr[low]=high_vac;
  low_vac=arr[low];
  while(pivot<low_vac)
  {
    if(high<=low) break;
    low++;
    low_vac=arr[low];
  }
  arr[high]=low_vac;
}
  arr[low]=pivot;
   return low;
}

void Selection_sort(int low,int high,int arr[])
{
  int Piv_index,i;
  if(low<high)
  {
   Piv_index=Partition(low,high,arr);
   Selection_sort(low,Piv_index-1,arr);
   Selection_sort(Piv_index+1,high,arr);
  }
}

Thanks for your help

Member Avatar for iamthwee

Sounds like you need a parallel array or class/struct?

Yet another orphan post (see http://www.daniweb.com/forums/thread148725.html - 3 hours later). Alas, the post header does not bear a relation to the true problem (as in the 2nd thread too ;)).
Don't forget to mark this orphan as solved too when you will get an answer to the 2nd thread (better use one problem/one thread approach in future ;))...

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