HackerEarth Partition it! problem solution

YASH PAL, 31 July 202417 February 2026

In this HackerEarth Partition it! problem solution Tubby the doggu, likes sets which are closed. Your task is to help him/her solve a problem.

Now the task is that you are given a prime number P and an array A of N integers , in which the ith integer is denoted by Ai, 1<=i<=N, 1<=Ai<=P.

You have to partition the set of indices {1,2,3,4,…,N} into a partition of minimum length such that if Closure({Ai}) subset of Closure({Aj}) then i and j must be in different sets of the partition.

Here, i!=j , 1<=i,j<=N.

HackerEarth Partition it! problem solution

HackerEarth Partition it! problem solution.

#include <bits/stdc++.h>
#define ll long long
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define mk make_pair
#define pb push_back
#define eps 1e-12
#define MAXN 200009
using namespace std;
vector<ll> v;
vector<int> ans[MAXN],out_g[MAXN];
ll p;
inline ll fastexpo(ll base,ll expo,ll mod)
{
  ll result=1;
  while(expo)
  {
    if(expo%2==1)
    {
      result=result*base;
      result%=mod;
    }
    base=base*base;
    base%=mod;
    expo=expo/2ll;
  }
  return result;
}
inline void fact(ll n)
{
  for(ll i=1;i*i<=n;i++)
  {
    if(n%i==0)
    {
      v.pb(i);
      v.pb(n/i);
    }
  }
  sort(v.begin(),v.end());
}
inline ll find_order(ll x)
{
  for(int i=0;i<v.size();i++)
  {
    ll res=fastexpo(x,v[i],p);
    if(res==1)
    {
      return v[i];
    }
  }
  assert(false);
}
inline bool check_if_prime(ll n)
{
  if(n==1||n==0)
  {
    return false;
  }
  for(ll i=2;i*i<=n;i++)
  {
    if(n%i==0)
    {
      return false;
    }
  }
  return true;
}
int main()
{
  // ios_base::sync_with_stdio(false);
  // cin.tie(0);
  // cout.tie(0);
  int t;
  cin>>t;
  assert(t<=100);
  ll s=0;
  while(t--)
  {
    v.clear();
    ll n;
    cin>>p>>n;
    assert(check_if_prime(p));
    s=s+n;
    assert(p<=1000000000);
    assert(p>=2);
    assert(n<=1000);
    ll a[n+1];
    for(int i=1;i<=n;i++)
    {
      cin>>a[i];
      assert(a[i]<p);
      assert(a[i]>0);
    }
    fact(p-1);
    pli ord[n+1];
    for(int i=1;i<=n;i++)
    {
      ord[i]=mk(find_order(a[i]),i);
      assert(((p-1)%ord[i].first)==0);
    }
    sort(ord+1,ord+n+1);
    int indeg[n+1];
    memset(indeg,0,sizeof(indeg));
    for(int i=1;i<=n;i++)
    {
      for(int j=i+1;j<=n;j++)
      {
        if(ord[j].first%ord[i].first==0)
        {
          out_g[i].pb(j);
          indeg[j]++;
        }
      }
    }
    int dp[n+1],maxx=0;
    queue<int> q;
    for(int i=1;i<=n;i++)
    {
      if(indeg[i]==0)
      {
        maxx=max(maxx,1);
        dp[i]=1;
        q.push(i);
        ans[dp[i]].pb(i);
      }
    }
    while(!q.empty())
    {
      int x=q.front();
      q.pop();
      maxx=max(maxx,dp[x]);
      for(int i=0;i<out_g[x].size();i++)
      {
        indeg[out_g[x][i]]--;
        if(!indeg[out_g[x][i]])
        {
          q.push(out_g[x][i]);
          dp[out_g[x][i]]=dp[x]+1;
          ans[dp[out_g[x][i]]].pb(out_g[x][i]);
        }
      }
    }
    cout<<maxx<<"n";
    // for(int i=1;i<=maxx;i++)
    // {
    //  for(int j=0;j<ans[i].size();j++)
    //  {
    //    cout<<a[ord[ans[i][j]].second]<<" ";
    //  }
    //  cout<<"n";
    // }
    //clear
    for(int i=0;i<=n;i++)
    {
      out_g[i].clear();
      ans[i].clear();
    }
  }
  assert(s<=25000);
  //cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.n";
}

Second solution

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int A[1011];
int dp[1011];
ll bpow(ll x,ll n, ll mod) {
    ll ans = 1;
    while(n>0) {
        if(n&1) ans*=x;
        x*=x;
        ans%=mod;
        x%=mod;
        n/=2;
    }
    return ans;
}
int main()
{
    int T;
    cin >> T;
    while(T--) {
        int P,N;
        cin >> P >> N;
        vector<int>divs;
        int sz = sqrt(P-1);
        int num = P-1;

        for(int j=1;j<=sz;j++) {
            if(num%j==0) divs.push_back(j), divs.push_back(num/j);
        }
        sort(divs.begin(),divs.end());
        vector<int>periods;
        for(int i=0;i<N;i++) {
            assert(cin >> A[i]);
            assert(A[i]>=1 and A[i]<P);
            ll cur = 1;
            for(auto d:divs) {
                if(bpow(A[i],d,P)==1){
                    periods.push_back(d);
                    break;
                }
            }
        }
        int maxLen  = 0;
        sort(periods.begin(),periods.end());
        for(int i=N-1;i>=0;i--) {
            dp[i] = 1;
            for(int j=i+1;j<N;j++) {
                if(periods[j] % periods[i] == 0) dp[i] = max(dp[i], dp[j]+1);
            }
            maxLen = max(maxLen, dp[i]);
        }
        cout << maxLen << "n";
    }
}

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