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 <bits/stdc++.h>
using namespace std;

int main() {
  string n;
  cin >> n;
  int sum=0;

  for(int i=0; i<n.length(); ++i) {
    if (n[i] == '0') sum += 0;
    else if (n[i] == '1') sum += 1;
    else if (n[i] == '2') sum += 2;
    else if (n[i] == '3') sum += 3;
    else if (n[i] == '4') sum += 4;
    else if (n[i] == '5') sum += 5;
    else if (n[i] == '6') sum += 6;
    else if (n[i] == '7') sum += 7;
    else if (n[i] == '8') sum += 8;
    else if (n[i] == '9') sum += 9;
  }

  if (sum % 9 ==0) cout << "Yes" << endl;
  else cout << "No" << endl;
  return 0;
}