1015.Smallest_Integer_Divisible_by_K.cpp

/*
    written by Pankaj Kumar.
    country:-INDIA
    Institute: National Institute of Technology, Uttarakhand
*/
typedef long long ll ;
typedef vector<int> vl;
typedef vector<vector<int>> vvl;
#define speed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
/*  Abbrevations  */
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
// loops
#define forin(arr,n) for(ll i=0;i<n;i++) cin>>arr[i];
// Some print
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
// sort
#define all(V) (V).begin(),(V).end()
#define srt(V) sort(all(V))
#define srtGreat(V) sort(all(V),greater<ll>())
// some extra
#define printv(v) for(ll i=0;i<ll(v.size());i++){cout<<v[i]<<" ";} cout<<endl;
#define precision(x) cout<<fixed<<setprecision(x);
#define sz(V) ll(V.size())


/* ascii value
A=65,Z=90,a=97,z=122
*/

/* Some syntax 
//Syntax to create a min heap for priority queue
//priority_queue <int, vector<int>, greater<int>>pq;
*/
ll ppow(ll n, ll m, ll mod){
    if(m==0) return 1;
    ll tmp=ppow(n, m/2, mod);
    tmp=tmp*tmp%mod;
    return m%2 ? tmp*n%mod: tmp;
}
namespace mod_operations{
    ll modInv(ll n, ll mod){
        return ppow(n,mod-2, mod);
    }
    ll modAdd(ll n, ll m, ll mod){
        n=(n%mod+mod)%mod; m=(m%mod+mod)%mod;
        return (n+m)%mod;
    }
    ll modMul(ll n, ll m, ll mod){
        n=(n%mod+mod)%mod; m=(m%mod+mod)%mod;
        return n*m %mod;
    }
    ll modSub(ll n, ll m, ll mod){
        n=(n%mod+mod)%mod; m=(m%mod+mod)%mod;
        return modAdd(n,-m, mod);
    }
    ll modDiv(ll n, ll m, ll mod){
        return modMul(n, modInv(m, mod), mod);
    }
}
using namespace mod_operations;


/*  --------------------MAIN PROGRAM----------------------------*/
// to run ctrl+b
const ll INF=1e18;
const ll mod1=1e9+7;
const ll mod2=998244353;
// Techniques :
// divide into cases, brute force, pattern finding
// sort, greedy, binary search, two pointer
// transform into graph

// Experience :
// Cp is nothing but only observation and mathematics.


//Add main code here

class Solution {
public:
    int smallestRepunitDivByK(int k) {
        map<int,int> m;
        int remainder=1;
        m[1]++;
        int length=1;
        while(1){
            if(remainder%k==0){
                return length;
            }
            else{
                length++;
                remainder=(remainder*10+1)%k;
            }
            if(m[remainder]){
                return -1;
            }
            m[remainder]++;
        }
        return -1;
    }
};


/* -----------------END OF PROGRAM --------------------*/
/*
* stuff you should look before submission
* constraint and time limit
* int overflow
* special test case (n=0||n=1||n=2)
* don't get stuck on one approach if you get wrong answer
*/