# CSES Problem Set Two Sets | CSES Solution

## CSES Problem Set Two Sets | CSES Solution

• Time limit: 1.00 s
•
• Memory limit: 512 MB
Your task is to divide the numbers $1,2,\dots ,n$ into two sets of equal sum.

Input

The only input line contains an integer $n$.

Output

Print "YES", if the division is possible, and "NO" otherwise.

After this, if the division is possible, print an example of how to create the sets. First, print the number of elements in the first set followed by the elements themselves in a separate line, and then, print the second set in a similar way.

Constraints
• $1\le n\le {10}^{6}$
Example 1

Input:
7

Output:
YES41 2 4 733 5 6

Example 2

Input:
6

Output:
NO

#include <bits/stdc++.h>using namespace std;#define int            long long int#define F              first#define S              second#define pb             push_back#define si             set <int>#define vi             vector <int>#define pii            pair <int, int>#define vpi            vector <pii>#define vpp            vector <pair<int, pii>>#define mii            map <int, int>#define mpi            map <pii, int>#define spi            set <pii>#define endl           "\n"#define sz(x)          ((int) x.size())#define all(p)         p.begin(), p.end()#define double         long double#define que_max        priority_queue <int>#define que_min        priority_queue <int, vi, greater<int>>#define bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)#define print(a)       for(auto x : a) cout << x << " "; cout << endl#define print1(a)      for(auto x : a) cout << x.F << " " << x.S << endl#define print2(a,x,y)  for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endlinline int power(int a, int b){  int x = 1;  while (b)  {    if (b & 1) x *= a;    a *= a;    b >>= 1;  }  return x;}template <typename Arg1>void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }template <typename Arg1, typename... Args>void __f (const char* names, Arg1&& arg1, Args&&... args){  const char* comma = strchr (names + 1, ',');  cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);}const int N = 200005;void solve() {int n;cin>>n;int total=(n*(n+1))/2;if(total%2)cout<<"NO"<<endl;else{    vi v(n,0);    for(int i=0; i<n; i++)v[i]=i+1;    set<int>s1,s2;    if(n%4==0){      int i=0;      int k=n/4;      for(i=0; i<k; i++)s1.insert(v[i]);      for(int i=k; i<3*k; i++)s2.insert(v[i]);      for(int i=3*k; i<n; i++)s1.insert(v[i]);      cout<<"YES"<<endl;      cout<<n/2<<endl;      print(s1);      cout<<n/2<<endl;      print(s2);     }    else if(n%4==3){      s1.insert(1);s1.insert(2);s2.insert(3);      if(n>3){        int k=(n-3)/4;        for(int i=3; i<k+3; i++)s1.insert(v[i]);        for(int i=(k+3); i<(3*k+3); i++)s2.insert(v[i]);        for(int i=(3*k+3); i<n; i++)s1.insert(v[i]);      }        cout<<"YES"<<endl;        cout<<(n/2)+1<<endl;        print(s1);        cout<<n/2<<endl;        print(s2);    }}}int32_t main(){  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);  clock_t z = clock();  int t = 1;//   cin >> t;  while (t--) solve();  return 0;}