1. Practice
2. Mathematics
3. Number Theory
4. Manasa loves Maths

6. Manasa loves Maths Hackerrank

Solution:

#include<bits/stdc++.h>
using namespace  std ;
typedef long long  ll;
typedef pair<ll, ll> Pll;
#define F first
#define S second
#define MP make_pair
#define PB push_back
#define rad(X) cout << (X) << endl
#define ASH(X) cout << (X) << " "
#define debug(x) cout << #x << " " << x << endl;
#define debug2(x,y) cout << #x << " " << x << " " << #y << " " << y << endl;
#define FOR(I,A,B) for(ll I = (A); I <= (B); I++)
#define cir(I,B,A) for(ll I = (A); I >=(B); I--)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define SORT(c) (sort(c.begin(),c.end()))
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define CASES int ___T; cin >> ___T; for(int cs=1;cs<=___T;cs++)
#define FAST()  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
ll fx[] = {1, 1, 0, -1, -1, -1, 0, 1};
ll fy[] = {0, 1, 1, 1, 0, -1, -1, -1};
vector<ll>a,b,c;
map<ll,ll>m1,m2,m3;

/************************************************/

void Execute()
{   ll n, nb, k, q, a, b, c, l, r, d, x, y, p, m;
    string s;
    cin >> s;
    l =  s.size();

    vector<ll>per;

    FOR(i,0,l-1)
    {   d = s[i]-'0';
        per.PB(d);
    }

    //SORT_UNIQUE(per);

    if(per.size()==1)
    {   if(per[0]%8==0)
            rad("YES");
        else
            rad("NO");
    }
    else if(per.size()<3)
    {   ll prothom_dig = per[0];
        ll ditio_dig  = per[1];
        ll res1 = (prothom_dig*10) + ditio_dig;
        ll res2 = (ditio_dig*10) + prothom_dig;
        if(res1%8 == 0 || res2%8 == 0)
            rad("YES");
        else
            rad("NO");
    }

    else
    {

        bool ok = false;

        FOR(i,0,per.size()-1)
        {   FOR(j,0,per.size()-1)
            {   FOR(k,0,per.size()-1)
                {   if(i!=j && j!=k && k!=i)
                    {   ll ans = (per[i]*10) + per[j];
                        ans = (ans*10)+per[k];
                        //debug(ans);
                        if(ans%8 == 0)
                        {   ok = true;
                            break;
                        }
                    }
                }
                if(ok == true )
                    break;
            }
            if(ok == true )
                break;

        }
        if(ok == true)
            rad("YES");
        else
            rad("NO");
    }



}


int32_t main()
{   FAST();
    CASES
    {   Execute();
    }
}