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 <stdlib.h>
#include <string.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */
#define MAXSIZE 200001
int main(int argc, char *argv[]) {
//	char res[4];
	char N[MAXSIZE];
	scanf("%s", N);
	
	int i = 0;
	int sum = 0;
	for (i = 0; i < strlen(N); ++i)
	{
		sum = sum + (N[i] - '0');
	}
	if (N[0] - '0' == 0 && strlen(N) == 1) 
	{
		printf("%s", "Yes");
//		return;
	}
	
	if (sum % 9 == 0)
	{
		printf("%s", "Yes");
		
//		return;
	}
	else 
	{
		printf("%s", "No");
	}
	
	
	return 0;
}