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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>   
using namespace std;
using namespace __gnu_pbds;
typedef long long ll; 
#define int long long
typedef unsigned long long lu;
typedef vector<ll> v;
typedef vector<vector<ll> > vv;
typedef vector<string>  vs;
typedef vector<pair<ll,ll>> vpr;
typedef vector<bool>vb;
typedef vector<double>vd;
typedef long double ld;
#define f(i,n) for(ll i = 0; i < n; i++)
#define ff(i,n) for(ll i=1;i<=n;i++)
#define pb push_back
#define mp make_pair
#define endl "\n"
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define amax(x,y) if(y>x) x=y;
#define amin(x,y) if(y<x)x=y;
#define bg(x) x.begin()
#define sz(x) (ll)x.size()
#define in(x,n) for(ll i=0;i<n;i++)cin>>x[i]
#define out(x,n) for(ll i=0;i<n;i++)cout<<x[i]<<" " 
#define mxt(a) *(max_element(a.begin(),a.end()))
#define mnt(a) *(min_element(a.begin(),a.end())
#define tc ll t;cin>>t;while(t--)
typedef pair<ll,ll> pi;
#define yes cout<<"Yes\n";
#define no cout<<"No\n";
#define yesno(f) if(f) yes else no
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
const ld PI = 2 * acos(0.0);
ll cel(ll x1,ll y1){if((x1%y1)==0)return x1/y1;else return x1/y1+1;}
ll power(ll a,ll b,ll m)
{
    if(b==0)
        return 1;
    ll d=power(a,b/2,m);
    d=(d*d)%m;
    if(b&1)
        d=(d*a)%m;
    return d;
}
const ll mod=1e9+7;
int MOD(int a)
{
    if(a<0)
        a+=mod;
    if(a>=mod)
        a%=mod;
    return a;
}
// set_name.find_by_order(k) It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time
// set_name.order_of_key(k) It returns to the number of items that are strictly smaller than our item k in O(logn) time.
/*string operations :
str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
str.substr (x) : returns a substring str[x],... end of string
str.find(qtr) : returns the first occurenece of qtr in str */
int32_t main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    int mod=0;
    string s;cin>>s;
    for(auto it:s)
    {
    	int d=it-'0';
    	mod+=d;
    	mod%=9;
    }
    yesno(mod==0);

    return 0;
}