Problem:
Score : 200 points

Problem Statement

An integer N is a multiple of 9 if and only if the sum of the digits in the decimal representation of N is a multiple of 9.

Determine whether N is a multiple of 9.

Constraints

0 \leq N < 10^{200000}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is a multiple of 9, print Yes; otherwise, print No.

Sample Input 1

123456789

Sample Output 1

Yes

The sum of these digits is 1+2+3+4+5+6+7+8+9=45, which is a multiple of 9, so 123456789 is a multiple of 9.

Sample Input 2

0

Sample Output 2

Yes

Sample Input 3

31415926535897932384626433832795028841971693993751058209749445923078164062862089986280

Sample Output 3

No

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int main(void){
  char n[200000] ;
  printf("input N > \n") ;
  fflush( stdout ) ;
  fscanf( stdin, "%s", n ) ;
  // printf( "%s\n", n ) ;
  int l = strlen(n) ;
  // printf( "%d\n", l ) ;
  int S = 0 ;
  int i, k ;
  for( i = 0 ; i < l ; i++ ) {
  	  k = n[i] - '0' ;
  	  if( 0 <= k && k<= 9 ) { 
  	  	  S += k ;
  	  } else {
  	  	  printf("This is not a integer.\n") ;
  	  	  return 1 ;
  	  }
  }
  // printf( "%d\n", S ) ;
  if( S % 9 == 0 ) {
  	  printf("Yes\n") ;
  } else {
  	  printf("No\n") ;
  }
  return 0 ;
}