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;
double pi=3.14159265358979323846264338;
long long sum;
long long ma;
long long n,k,st,x,t;
string s;
long long pr(long long a, long long b){
	long long k=1;
	if(b==0)return 1;
	
	for(int i=0;i<b;i++)k=k*10;
	return k;
}
long long arr[200001];
bool che(long long a){
	long long cur=0;
	for(int i=0;i<n;i++){
		cur+=(arr[i]+a-1)/(a)-1;
	}
	if(cur<=k)return true;
	return false;
}
int main(){
	cin>>s;
	n=s.length();
	k=0;
	for(int i=0;i<n;i++){
		k+=(int)(s[i]-'0');
	}
	if(k%9==0)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
	return 0;
}