HackerEarth Game of sequence problem solution

In this HackerEarth Game of sequence problem solution, Two players P and Q are playing a game of sequence on an array A of size n. In each move a player will choose two distinct numbers X and Y, both numbers should be present in the array. Now the player will pick all the positions where the number X is present and will replace the value there with the number Y.

The game stops if the array has only one distinct value. The player who is unable to make a move in his turn loses. You have to decide who will lose the game. The first move is played by player P.

HackerEarth Game of sequence problem solution.

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define MM 1000000009
#define reset(a) memset(a,0,sizeof(a))
#define rep(i,j,k) for(i=j;i<=k;++i)
#define per(i,j,k) for(i=j;i>=k;--i)
#define print(a,start,end) for(i=start;i<=end;++i) cout<<a[i];
#define endl "n"
#define inf 100000000000000
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
int main()
  {
    ios_base::sync_with_stdio(0);
    int t;
    cin >> t;
    while(t--)
      {
        int n;
        cin >> n;
        set<int> s;
        for(int i = 1; i <= n ; i++)
          {
            int x;
            cin >> x;
            s.insert(x);
          }
        if(s.size() % 2)
          {
            cout << "Pn";
          }
        else
            cout << "Qn";
      }
  }

Second solution

#include<bits/stdc++.h>
using namespace std;
int main()
{
  int t;
  cin>>t;
  assert(t>=1 && t<=10);
  while(t--)
  {
    int n;
    cin>>n;
    set<int>s;
    assert(n>=2 && n<=1e5);
    for(int i=0;i<n;i++)
    {
      int temp;
      cin>>temp;
      assert(temp>=1 && temp<=1e9);
      s.insert(temp);
    }
    if(s.size()&1)cout<<"Pn";
    else cout<<"Qn";
  }
  return 0;
}

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