Answer to Question #290801 in C++ for Young

Question

Listing 102

by CodeChum Admin

We've already made listing an array the easy way, but how about listing and printing the array in reverse order?

Make a program that will input an integer and then using loops, add items on an array/list one by one for the same number of times as that of the first inputted integer. Then, print out the array/list in ascending order. After doing that, print the array again on the next line in reverse order, that is, starting from the last item on the array/list down to the first one, each in separated lines.

Input

The first line contains the size of the array.

The next lines contains the items of the array (integers).

5
1
64
32
2
11

Output

The first line contains the elements of the array printed in ascending order.

The second line contains the elements of the array printed in reversed order.

1·64·32·2·11
11·2·32·64·1

Answer 1

#include<conio.h>
#include<stdio.h>
#include<stdlib.h>
#include <iostream>
using namespace std;


void swap(int *p,int *q) 
{
   unsigned int t;   
   t=*p; 
   *p=*q; 
   *q=t;
}


void sort(int a[],int n) 
{ 
   int i,j;
   int temp;


   for(i = 0;i < n-1;i++) 
   {
      for(j = 0;j < n-i-1;j++) 
	  {
         if(a[j] > a[j+1])	swap(&a[j],&a[j+1]);
      }
   }
    cout<<"\nAscending Order: ";
    for(i = 0;i < n;i++) cout<<a[i]<<" - ";
    cout<<"\nDescending Order: ";
	for(i = n-1;i >=0;i--) cout<<a[i]<<" - ";


}


int main()
{
	int Count;
	cout<<"\n\tEnter Count of Numbers: "; cin>>Count;
	int Nums[Count],n=0;
	
	for(n=0;n<Count;n++)
	{
		cout<<"\n\tEnter Number-"<<n+1<<": "; cin>>Nums[n];
	}
	
	sort(&Nums[0],Count);
	return(0);
}