（C）代码无法在所有需要的条件下工作

Question

#include <stdio.h>
#include <stdlib.h>

#define TEN 10 
int main ()
{
 int number = 0;
 int digit = 0;
 int last_digit = 0;
 int digit_sum = 0;
 int i = 0;
while (i == 0)
{
 printf("Please Enter A Positive Number! \n"); //explaining
 scanf("%d",&number);
 if (number > 0)
{
    i++;
}

}

    while (number > 0)
    {
    digit = number % TEN; //breaking number into digits
    number /= TEN;

    if (last_digit != digit) //comparing digits
    {
       last_digit = digit;
       digit_sum += digit;
    }


    }


    printf("The Sum Of The Digits Is : %d",digit_sum);
    return 0;

}

---

the code will divide the number into digits and check if there are duped digits, in case there are, only one of them will be calculated for exmple: 3211 3+2+1, but my problem is thats the code wont work with numbers like 31211 Im thankful for any kind of help.

Answer 1

The code doesn't work because there is no guarantee that duplicate's will appear consecutive manner. your code handles that not the other ways. That's why it fails.

A simple solution would be to consider a 10 element array where you will keep count of which element appeared and which didn't.

The idea is to map the digits to the array indices of the 10 element array. Intialized with 0 .

...
int digitAppeared[10]={0};
while (number)
{
    digit = number % TEN; //breaking number into digits
    number /= TEN;
    digit_sum  += (1 - digitAppeared[digit]) * digit;
    digitAppeared[digit] = 1;
}
...

To give you a clear idea this line basically checks whether the element appeared or not and as per the result it will add the digit.

If digit D appeared then digitAppeared[D]=1 and if it didn't then digitAppeared[D]=0 .

We will add it to digitsum if it appears first time. That's why the (1-digitAppeared[D]) will tell us whether to add it or not.

    digit_sum  += (1 - digitAppeared[digit]) * digit;

Answer 2

Convert the number to string use itoa() . sort it and then walk through it , looking for unique number and do your calculation

Answer 3

You can mark which digits were already added by using setting logical flags as represented by done in program below:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define TEN 10 
int main ()
{
 int number = 0;
 int digit = 0;
 int last_digit = 0;
 int digit_sum = 0;
 int i = 0;
 int done[10];
while (i == 0)
{
 printf("Please Enter A Positive Number! \n"); //explaining
 scanf("%d",&number);
 if (number > 0)
{
    i++;
}

}
    memset(done,0,sizeof(done));
    while (number > 0)
    {
    digit = number % TEN; //breaking number into digits
    number /= TEN;
      if(done[digit] == 0)
      {
       digit_sum += digit;
       done[digit] = 1;
      }
    }


    printf("The Sum Of The Digits Is : %d",digit_sum);
    return 0;
}