Skip to content
Programmingoneonone
Programmingoneonone
  • Home
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
  • Work with US
Programmingoneonone
Programmingoneonone

HackerEarth Partition it! problem solution

YASH PAL, 31 July 2024
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";
}
}
coding problems solutions

Post navigation

Previous post
Next post

Related website

The Computer Science

Pages

  • About US
  • Contact US
  • Privacy Policy

Programing Practice

  • C Programs
  • java Programs

HackerRank Solutions

  • C
  • C++
  • Java
  • Python
  • Algorithm

Other

  • Leetcode Solutions
  • Interview Preparation

Programming Tutorials

  • DSA
  • C

CS Subjects

  • Digital Communication
  • Human Values
  • Internet Of Things
  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2025 Programmingoneonone | WordPress Theme by SuperbThemes