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>

int main(void){

    int i;
    
    char a[200000];
    
    scanf("%s",a);
    
    int b[200000];
    
    for(i = 0;i < 200000; i++){
        b[i] = 0;
    
    }
    
    //printf("%s",a);
    
    for(i = 0;i < 200000; i++){
        b[i] = atoi(&a[i]);
    
    }
    
    long long sum = 0;
    
    for(i = 0;i < 200000; i++){
        sum += b[i];
    
    }

    
    if(sum%9 == 0){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    
    
    
}