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>
int powerten(int n){
    int num=1;
    for(int i=0;i<n;i++){
        num*=10;
    }
    return num;
}
int main(){
    int N=0,i=1,sum=0;;
    scanf("%d",&N);
    while(1){
        if(N<powerten(i))break;
        else i++;
    }//Nはi桁
    sum+=N%10;
    if(i==1){
        if(sum%9==0)printf("Yes");
        else printf("No");
    }
    else{
        for(int j=1;j<i;j++){
            sum+=N/powerten(j);
        }
        if(sum%9==0)printf("Yes");
        else printf("No");
    }
    return 0;
}