CODEWITHSUNDEEP
c++sorting
Aidan
Aidan

Reputation: 3

Sorting output numbers without an array

So the main task is to find all numbers divisible by 7 between 0 - 100 then sort them in descending order without using an array. I'm just starting c++ and one of my first lab tasks was this, however when I finished it I was told that I shouldn't have used an array. I'm now curious as to how to do so otherwise. The code here only finds the numbers divisible by 7 and naturally displays them in an ascending sort.

I'm unsure how I would sort them without storing the value in an array then changing switching the values that way.

#include <iostream>

using namespace std;

int main() {

for( int i = 0; i <= 100; i++){

    if(i%7 == 0){

        //Display every integer divisible by 7
        cout << i << endl;

    }

}


return 0;

}

Upvotes: 0

Views: 322

Answers (5)

Anders Cedronius
Anders Cedronius

Reputation: 2076

Bling it like its 2020 .. (You will get great grades ;-) )

#include <iostream>
using namespace std;
int main() {
    int i=101;
    while(--i)i%7?cout:cout<<i<<endl;
}

Upvotes: 0

NiVeR
NiVeR

Reputation: 9786

Just reverse the for loop:

for( int i = 100; i >= 7; i--){ //There is no integer lower than 7 that is divisible by 7
    if(i%7 == 0){
        cout << i << endl;

    }    
}

Upvotes: 1

Little Boy Blue
Little Boy Blue

Reputation: 311

What you can do is start at max and go down from there; as follows:

  for(int i = 100; i >= 0; i--){
    if(i % 7 == 0){
        cout << i << endl;
    }
  }

Upvotes: 0

Scott Hunter
Scott Hunter

Reputation: 49803

Change your loop to go in descending order; then you won't have to sort anything.

Upvotes: 0

gt6989b
gt6989b

Reputation: 4203

One approach is to find the lagest number divisible by 7 (here, 98) and just continue removing 7 to it until you run across the lowest boundary.

Upvotes: 2

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