how can i sum realy big number? [duplicate] - c

Question

This question already has answers here:
Digital Root in c
(2 answers)
Closed 11 months ago.
So you have to do:
11 = 1+1 = 2
3578 = 3+5+7+8 = 23 = 2+3 = 5
But the problem is that the number can be very large(consist of 10,000 digits)
But even with the easiest entrances it doesn't work:
Input : 11
Output: 2798 (and it always changes, but remains a 4-digit number)
Can someone explain why is this happening?
And how can I summarize each digit of a very large number?

Answer 1

You got that huge number becuase your program is adding the ASCII value of various characters.
Some improvements:
Don't use "%s", use "%<WIDTH>s", to avoid buffer-overflow
Use size_t to iterate through an array, instead of unsigned long long int
Instead of using bare return 0;, use return EXIT_SUCCESS;, which is defined in the header file stdlib.h.
always check whether scanf() input was successful or not
Don't check for '\n', because string from scanf() ends at both SPACES and NEWLINE.
adding +1 to array size for NULL terminating character
Use "%zu" instead of "%lld" for size_t
Final Code:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(void) {
char buffer[10001] = {0};
if(scanf("%10000s", buffer) != 1)
{
perror("bad input");
return EXIT_FAILURE;
}
size_t result = 0;
for(size_t i = 0; buffer[i]; i++) {
if(isdigit(buffer[i])){
result += buffer[i] - '0';
}
else {
perror("only digits are valid");
return EXIT_FAILURE;
}
}
printf("%zu\n", result);
return EXIT_SUCCESS;
}
Output:
1112
5
TRY IT ONLINE

Answer 2

You can do it without occupying memory
#include <ctype.h>
#include <stdio.h>
int main(void) {
int sum = 0;
for (;;) {
int ch = getchar();
if (!isdigit((unsigned char)ch)) break; // leave loop with ENTER, EOF, 'a', ...
sum += ch - '0';
}
printf("sum of digits is %d.\n", sum);
return 0;
}
Edit: see code running at ideone

Answer 3

Wiki Digital Root provides a shortcut for getting the final single digit.
Validate your input string has only numeric digits
Find the sum of all digits in ASCII form
Make use of congruence formula to get the result.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define MAX_NUM_LEN 10000
int digitRoot (int n) {
if (0 == n) return 0;
return (0 == (n % 9)) ? 9 : (n % 9);
}
int main () {
char str_num [MAX_NUM_LEN];
printf ("Finding Digital Root\nEnter a number : ");
if (NULL == fgets (str_num, sizeof (str_num), stdin)) {
perror ("Reading input string");
return 2;
}
int slen = strlen (str_num);
// remove new line if found
if ('\n' == str_num[slen - 1]) str_num[--slen] = '\0';
// validate input
int digitSum = 0;
for (int ni = 0; ni < slen; ++ni) {
if (!isdigit ((unsigned char) str_num[ni])) {
printf ("\nERROR: Invalid digit [%c]\n", str_num[ni]);
return 1;
}
digitSum += str_num[ni] - '0';
}
printf ("\nDigital Root is [%d]\n", digitRoot (digitSum));
return 0;
}