I need help figuring out how to nest this loop

Question

I am attempting a Hacker Rank problem for my Programming II class. I can either get the program to print the correct values but out of order, or it prints out the wrong values in the right order

I've tried nesting the loops inside one another in different ways to change the order, or to keep the first input variable value so it can be read by the second loop. So far I haven't gotten the desired result.

int main() 
{
int n;

for(int k = 0; k < 2; ++k)
{
    cin >> n;
    if(n == 1)
    {
        cout << "one\n";
    }
    else if(n == 2)
    {
        cout << "two\n";
    }
    else if(n == 3)
    {
        cout << "three\n";
    }
    else if(n == 4)
    {
        cout << "four\n";
    }
    else if(n == 5)
    {
        cout << "five\n";
    }
    else if(n == 6)
    {
        cout << "six\n";
    }
    else if(n == 7)
    {
        cout << "seven\n";
    }
    else if(n == 8)
    {
        cout << "eight\n";
    }
    else if (n >= 9)
    {
        cout << "nine\n";
    }
}
for(int j = 0; j < 2; j++)
{
    if(n % 2 == 0)
    {
        cout << "even\n";
    }
    else
    {
        cout << "odd\n";
    }
}


return 0;
}

Correct result with 8 and 11 used

eight
nine
even
odd

My result

eight
nine
odd
odd

Answer 1

The problem with your code is that after 8 is accepted as input, it is not being saved any where since it is replaced by 11 as input. Hence after the end of the first for loop, n is still 11 and hence you get odd twice. Use arrays instead.

#include<iostream>

int main() 
{
int n;
int a[2],counter=0;
for(int k = 0; k < 2; ++k)
{
    std::cin >> n;
    a[counter++] = n;
    if(n == 1)
    {
        std::cout << "one\n";
    }
    else if(n == 2)
    {
        std::cout << "two\n";
    }
    else if(n == 3)
    {
        std::cout << "three\n";
    }
    else if(n == 4)
    {
        std::cout << "four\n";
    }
    else if(n == 5)
    {
        std::cout << "five\n";
    }
    else if(n == 6)
    {
        std::cout << "six\n";
    }
    else if(n == 7)
    {
        std::cout << "seven\n";
    }
    else if(n == 8)
    {
        std::cout << "eight\n";
    }
    else if (n >= 9)
    {
        std::cout << "nine\n";
    }
}
for(int k = 0;k<counter;k++){
    if(a[k] % 2 == 0)
    {
        std::cout << "even\n";
    }
    else
    {
        std::cout << "odd\n";
    }
}


return 0;
}

Answer 2

This loop:

for(int j = 0; j < 2; j++)
{
    if(n % 2 == 0)
    {
        cout << "even\n";
    }
    else
    {
        cout << "odd\n";
    }
}

is outside of your first loop.

And you don't really want a loop at all, just an if for even or odd.

Answer 3

In the second loop you aren't checking both values, but just the last one(when you assign to n using std::cin the second time you are overwriting the first value). To solve this you could declare n as an array using the syntax int n[2]; and then iterate through it in your loops writing n[k] or n[j] (in this specific program, remember that n, j and k are arbitrary names). You don't really need nested loops in this case but they can easily be achieved and are often useful.The syntax is as follows:

for(int i = 0; i < 23; i++)
{
    /*bar*/
    for(int j = 0; j < 42; j++)
    {
        /*foo*/
    }
}

in this case bar is executed 23 times and foo is executed 23*42 time

When a program has different behaviour based on a variable's value, and you only need to check for equality but against many different values it is advisable to use the switch-case block instead of if and else if.

So if I personally had to write the program that you're writing I would do it like this:

#include <iostream>
int main()
{
using namespace std;

int n[2];
for(int k = 0; k < 2; ++k)
{
    cin >> n[k];
    switch(n[k])
    {
    case 1:
        cout << "one\n";
        break;
    case 2:
        cout << "two\n";
        break;
    case 3:
        cout << "three\n";
        break;
    case 4:
        cout << "four\n";
                break;
    case 5:
        cout << "five\n";
        break;
    case 6:
        cout << "six\n";
        break;
    case 7:
        cout << "seven\n";
        break;
    case 8:
        cout << "eight\n";
        break;
    default:
        if(n[k]>=9)
            cout << "nine\n";
        break;
    }
}
for(int j = 0; j < 2; j++)
    if(n[j] % 2 == 0)
        cout << "even\n";
    else
        cout << "odd\n";


return 0;
}

Answer 4

You could do this in one loop:

static char const * const number_names[] = 
{
  "zero", "one", "two", "three", "four",
  "five", "six", "seven", "eight", "nine"
};  

for (int k = 0; k < 2; ++k)
{
    std::cin >> k;
    if ((k >= 0) && (k < 10))
    {
        std::cout << number_names[k] << "\n";
    }
    else
    {
        std::cout << "number not within range\n";
    }
    if ((k%2) == 0)
    {
        std::cout << "even";
    }
    else
    {
        std::cout << "odd";
    }
    std::cout << "\n";
}

You can use the number as an index into an array of number-names. No need for switch or if-else-if ladders. Usually, online judges don't care about how the assignment is implemented. They have other criteria.

The even/odd test is included in the same loop. This eliminates the need to remember numbers (i.e. store them into an array).

The "\n" is common to both the even and odd printing, so it is factored out. (Sorry, but a habit from the olden days).

The array is declared as static because it is private and to indicate there is only one instance.

The array contains pointers that don't change, e.g. constant.
The array contains pointers that point to constant data.

An array of string literals is chosen so it can be directly accessed from the data area of the program, without going through an intermediate data structure like std::string.