Sum of digits not including the same digit twice - c

Question

(Purpose of the code is write in the title) my code work only if i put the same number once and in the end like - 123455 but if i write 12345566 is dosent work or 11234 it dosent wort to someone know why? i have been trying for a few days and i faild agine and agine.
while(num)
{
dig = num % 10 // dig is the digit in the number
num /= 10 // num is the number the user enter
while(num2) // num2 = num
{
num2 /= 10
dig2 = num2 % 10 // dig2 is is the one digit next to dig
num2 /= 10
if(dig2 == dig) // here I check if I got the same digit twice to
// not include him
{
dig2 = 0
dig = 0
}
}
sum = sum + dig + hold
}
printf("%d", sum)

Answer 1

Well your code's almost correct but there's are somethings wrong ,
When you declared num2 to be equal to num1 it was out of the loop so as soon as one full loop execution is done , num2 still remains to be less than zero or to be zero if it was a unsigned int, so according to the condition the second loop wont execute after its first run.
So mind adding ,
num2 = num1
inside your first loop .
Also your updating num2 twice which i think you wont need to do after the first change .
Full code which i tried
#include<stdio.h>
int main(void)
{
int num;
int num2;
int sum = 0;
int dig, dig2;
scanf_s("%d", &num);
while (num>0)
{
dig = num % 10; //dig is the digit in the number
num /= 10;
num2 = num;// num is the number the user enter
while (num2>0) // num2 = num
{
dig2 = num2 % 10; // dig2 is is the one digit next to dig
if(dig2 == dig) // here i check if i got the same digit twice to //not include him
{
dig = 0;
}
num2 /= 10;
}
sum = sum + dig ;
}
printf("%d", sum);
return 0;
}

Answer 2

O(n)
#include <stdio.h>
int main () {
unsigned int input = 0;
printf("Enter data : ");
scanf("%u", &input);
int sum = 0;
int dig = 0;
int check[10] = {0};
printf("Data input: %u\n", input);
while(input) {
dig = 0;
dig = input%10;
input = input/10;
if (check[dig] == 0) {
check[dig]++;
sum += dig;
}
}
printf("Sum of digits : %d\n", sum);
return 0;
}

Answer 3

My program uses a character array (string) for storing an integer. We convert its every character into an integer and I remove all integers repeated and I calculate the sume of integers without repetition
My code :
#include <stdio.h>
#include <stdlib.h>
int remove_occurences(int size,int arr[size])
{int s=0;
for (int i = 0; i < size; i++)
{
for(int p=i+1;p<size;p++)
{
if (arr[i]==arr[p])
{
for (int j = i+1; j < size ;j++)
{
arr[j-1] = arr[j];
}
size--;
p--;
}
}
}
for(int i=0;i<size;i++)
{
s=s+arr[i];
}
return s;
}
int main()
{
int i=0, sum=0,p=0;
char n[1000];
printf("Input an integer\n");
scanf("%s", n);
int T[strlen(n)];
while (n[i] != '\0')
{
T[p]=n[i] - '0'; // Converting character to integer and make it into the array
p++;i++;
}
printf("\nThe sum of digits of this number :%d\n",remove_occurences(strlen(n),T));
return 0;
}
Example :
Input : 1111111111 Output : 1
Input : 12345566 Output : 21
Or ,you can use this solution :
#include <stdio.h>
#include <stdbool.h>
bool Exist(int *,int,int);
int main ()
{
unsigned int n = 0;
int m=0,s=0;
printf("Add a number please :");
scanf("%u", &n);
int T[(int)floor(log10(abs(n)))+1];
// to calculate the number of digits use : (int)floor(log10(abs(n)))+1
printf("\nThe length of this number is : %d\n",(int)floor(log10(abs(n)))+1);
int p=0;
while(n!=0)
{
m=n%10;
if(Exist(T,p,m)==true)
{
T[p]=m;
p++;
}
n=n/10;
}
for(int i=0;i<p;i++)
{
s=s+T[i];
}
printf("\nSum of digits : %d\n", s);
return 0;
}
bool Exist(int *T,int k,int c)
{
for(int i=0;i<k;i++)
{
if(T[i]==c)
{
return false;
}
}
return true;
}