C ++如何计算用户输入的数字中的偶数位数？

Question

In this problem I'm trying to find the number of even digits in a user input number. I am unable to use arrays. The program is a menu based program where this is option 1. I think I have the rest of the code working well so far, it is just the math for this part that I am getting hung up on. Any help would be greatly appreciated.

    int main() {

    int choice; // user input for menu options

    const int evenDig = 1, // constants for menu choices
                fact = 2,
                quit = 3;

    cout << "Welcome to playing with numbers!\n";

    do
    {
        showMenu (); // displays menu
        cin >> choice;

        // call getValidUserInputPosNumGT0, passing the value in choice
        // as an argument
        getValidUserInputPosNumGT0(choice);

            if (choice == evenDig) // option 1
            {
                int num; // user entered number

                cout << "Enter a positive number greater than zero...";
                cin >> num;
                numEvenDigits(num);
            }


    }while(choice != quit);

    return 0;
}

void showMenu ()
{
    cout << "1) Count the even digits in a number\n"
        << "2) Compute the factorial of a number\n"
        << "3) Quit\n"
        << "Select an option (1..3).. ";
}

void getValidUserInputPosNumGT0 (int choice) // validation function
{
    while ( choice < 1 || choice > 3) // input validation loop
    {
        cout << "Select an option (1..3).\n";
        cin >> choice;
    }
}

int numEvenDigits (int num)
{
    int even = 0; // int that will be returned as number of even numbers

    if ( num > 0)
    {
        int rem = num % 10;
        if (rem % 2 == 0)
            even++;

        cout << "numEvenDigits("<<num<<") = "<< even;
        cout << endl;
    }
    else
        cout << "Please enter a positive nonzero number." << endl;
}

Thank you!

Answer 1

Since there are multiple ways to solve what you are trying to acomplish, instead of providing a solution, I will try to give a couple hints on the math and thought process to help you figure out your own solution.

As Sam Varshavchik suggests, describe the problem and the solution and see what you have done in code and how it matches up.

• your approach using the moduluo function is correct:
  • number mod 10 = "right most digit"
  • digit mod 2, if odd=1 , if even=0

So you have sucessfully done this in your code once

    int rem = num % 10;
    if (rem % 2 == 0)
        even++;

What about the rest of the numbers? --> you are missing a loop somewhere

Take for exmple the number 123456

• First round num = 123456 , mod 10 = 6, mod 2 = 0 --> even++
• Next round?
  • repeat but with 12345

How to get to 12345 will be something I'll let you figure out.

• subtractions, divisions, truncation to integers, etc...

Then you will be missing how to define the end of the loop.

Cheers!

Answer 2

You Need to change some of your code

      void getValidUserInputPosNumGT0 (int choice);

To

      void getValidUserInputPosNumGT0 (int &choice);

Because If you want to modify choice you must pass it by reference not by value.

And

int numEvenDigits (int num)
{
    int even = 0; // int that will be returned as number of even numbers

    if ( num > 0)
    {
        int rem = num % 10;
        if (rem % 2 == 0)
            even++;

        cout << "numEvenDigits("<<num<<") = "<< even;
        cout << endl;
    }
    else
        cout << "Please enter a positive nonzero number." << endl;
}

To

int numEvenDigits (int num)
{
    int even = 0;
    if ( num > 0)
    {
       for (int val =num;val>0;val/=10)
       {
           if (val % 2 == 0)
           even++;
       }
        cout << "numEvenDigits("<<num<<") = "<< even;
        cout << endl;
    }
    else
       cout << "Please enter a positive nonzero number." << endl;

    return even; 
} 

Or, This is similar to your code

int numEvenDigits (int num)
{
    int even = 0,val=num,rem;
    if ( num > 0)
    {
       while(val>0)
       {
           rem=val%10;
           if(rem%2==0)
               even++;
           val/=10;
        }
        cout << "numEvenDigits("<<num<<") = "<< even;
        cout << endl;
    }
    else
       cout << "Please enter a positive nonzero number." << endl;

     return even;
}

Answer 3

Just simply do this to find even digits in a number 1.Your code to find even digits must be written in a loop 2. Decrease number after each iteration to move the next digit. Remember we count the digits from right to left.

Int number, x, even=0;
Cin>> number;
While(num>=0)
{
    X=num%10;
     If(x%2==0)
       Even++;
     Num=num/10;// decrease number
}