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>

#define MAX_NUM 200000

int main(void){
    char number[MAX_NUM];
    int i;
    int dummy =0;
    int sum =0;

    //入力
    for(i=0;i<MAX_NUM;i++){
        scanf("%c",&number[i]);
        if((48 > number[i]) || (number[i] > 57)) break;
        dummy = (number[i]-48);
        sum = sum + dummy;
        //printf("number = %d, sum = %d",dummy,sum);
    }
    //9の倍数
    if(sum % 9 == 0){
        printf("Yes");
        return 0;
    }else{
        printf("No");
        return 0;
    }

}