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 (Clang 10.0.0) (judge verdict: Accepted)

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

int main(){
  char s[200001];
  scanf("%s",s);
  
  long long ans=0;
  
  int n[200001];
  for(int i=0;i<=200001;i++){
    if(s[i]<='0'){n[i]=0;}
    if(s[i]=='1'){n[i]=1;}
    if(s[i]=='2'){n[i]=2;}
    if(s[i]=='3'){n[i]=3;}
    if(s[i]=='4'){n[i]=4;}
    if(s[i]=='5'){n[i]=5;}
    if(s[i]=='6'){n[i]=6;}
    if(s[i]=='7'){n[i]=7;}
    if(s[i]=='8'){n[i]=8;}
    if(s[i]>='9'){n[i]=9;}
  }
  
  
  for(int i=0;i<=200001;i++){
    ans+=n[i];
  }
  
  if(ans%9==0){printf("Yes");return 0;}else{printf("No");return 0;}
}