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: Time Limit Exceeded)

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

int main() {
    char n[200001];
    long long int sum=0;
    int i;
    char buf[2]={'\0'};
    char buf2[2]={'\0'};
    scanf("%s", n);
    for (i=0; i<strlen(n)/2;i++) {
        buf[0]=n[i];
        buf2[0]=n[strlen(n)-1-i];
        sum+=atoi(buf)+atoi(buf2);
    }
    if (strlen(n)%2==1) {
        buf[0]=n[strlen(n)/2];
        sum+=atoi(buf);
    }
    //printf("%d\n",sum);
    if (sum%9==0) printf("Yes\n");
    else printf("No\n");
    return 0;
}