Question

program in c++ using array that determine odd and even numbers from input numbers.

sample run:
enter 10 int: 1 2 3 4 5 6 7 8 9 10
odd: 1 3 5 7 9
even; 2 4 6 8 10

Answer 1

I will explain. In C++ you should have something called a Module or Mod also known as the symbol %. Using your C++ knowledge(which you should have) you should be able to do this.

if(i % 2 == 0) //even

else if (i% 2 == 1) //odd

It's that simple.

Answer 2

i is an integer, so for example 2 % 2 = 0 is even and 3 % 2 = 1 is odd.

Answer 3

basically the % will be a remainder #. So 3/2 is 1 remainder 1. 26 % 2 is 13 remainder 0, and 29 % 2 is 14 remainder 1. Make sense?

Answer 4

yep i got it. i already made a program. but idk if it is correct lol.

Answer 5

#include "stdafx.h"
#include <iostream>

using namespace std;

const int ARRAY_SIZE = 10;
int item[ARRAY_SIZE];
int counter;



int _tmain(int argc, _TCHAR* argv[])
{
cout<<"Please enter 10 no.: ";
for(counter = 0;counter<10;counter++)
{
cin>>item[counter];
}

cout<<endl<<endl;

cout<<"The even no. are: ";
for (counter = 0;counter<10;counter++)
{
if(item[counter] %2==0)
{
cout<< item[counter]<< " ";
}
}

cout<<endl<<endl;

cout<<"The odd no. are: ";
for (counter = 0;counter<10;counter++)
{
if(item[counter] % 2!=0)
{
cout<<item[counter] << " ";
}
}



system("pause");
return 0;

Answer 6

only way to know is to try it out ;)

Answer 7

also you want 1 for loop. having 2 is redundant....

Answer 8

for (counter = 0;counter<10;counter++)
{
if(item[counter] % 2 == 0
{
//Even
}
else If //etc
{
//odd
}