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: Accepted)

#include<stdio.h>
int main()
{
  int i,sum,a;
  char n[200000];
  
  scanf("%s", &n);
  i=0;
  sum=0;
  while(1) {
  if (n[i] != '\0'){
    a=n[i]-'0';
    //printf("a=");
    //printf("%d\n",a);
    
    sum=sum+a;
    //printf("SUM=");
  	//printf("%d\n",sum);
    i++;
  } else {
    break;
  }
  }
   //printf("%d",sum);
  if (sum % 9 == 0) {
    printf("Yes");
  } else {
    printf("No");
  }
  return 0;
}