C中整数形式的特定位数

Question

For example: if user input is 11234517 and wants to see the number of 1's in this input, output will be "number of 1's is 3. i hope you understand what i mean.

i am only able to count number of digits in an integer.

#include <stdio.h>
int main()
{
    int n, count = 0;

    printf("Enter an integer number:");
    scanf("%d",&n);
    while (n != 0)
    {
        n/=10;
        count++;
    }
    printf("Digits in your number: %d",count);
    return 0;
}

maybe arrays are the solution. Any help would be appreciated. thank you!

Answer 1

You don't need array. Try something like this:

int countDigits(int number, int digitToCount)
{
  // Store how many times given number occured
  int counter = 0;

  while(number != 0)
  {
   int tempDigit = number % 10;
   if(tempDigit == digitToCount)
     counter++;
   number = number/10;
  }

 return counter;
}

Answer 2

So, you've already found that you can convert 1234 to 123 (that is, remove the least significant digit) by using number / 10 .

If we wanted to acquire the least significant digit, we could use number % 10 . For 1234 , that would have the value of 4 .

Understanding this, we can then modify your code to take this into account:

int main() {
    int n, count = 0;

    printf("Enter an integer number:");
    scanf("%d",&n);

    while (n != 0) {
        if (n % 10 == 1)
            count++;
        n /= 10;
    }

    printf("Number of 1s in your number: %d", count);
    return 0;
}

Answer 3

You may want to use convert your int to a string like this :

char str[100];
sprintf(str, "%d", n);

Then, you can just iterate on str in order to find the occurrences of your digit.