# DIVBYI-Editorial

Setter: Satyam
Tester: Lavish Gupta, Abhinav sharma, Takuki Kurokawa
Editorialist: Devendra Singh

1184

None

# PROBLEM:

You are given an integer N.

Construct a permutation P of length N such that

• For all i (1 \leq i \leq N-1), i divides abs(P_{i+1}-P_i).

Recall that a permutation of length N is an array where every integer from 1 to N occurs exactly once.

It can be proven that for the given constraints at least one such P always exists.

# EXPLANATION:

Let us look at the maximum difference needed (and is possible) for any two adjacent values in the permutation. It is equal to N-1 (needed to be divisible by index i=N-1) and since P only consists of the values from 1 to N. It is possible only when the last two values are 1,N or N,1.

Let us take the case when the last two values are 1,N i.e P_N=N and P_{N-1}=1. Now for index N-2 we need the absolute difference of P_{N-2} and P_{N-1} to be divisible by N-2 which is again the maximum possible difference that can be created from the remaining values (as N can’t be used anymore to create differences) which are 1 and N-2. Assign P_{N-2}=N-1. Now for index N-3 we need the absolute difference of P_{N-3} and P_{N-2} to be divisible by N-3 which is again the maximum possible difference that can be created from the remaining values (as N and 1 can’t be used anymore to create differences) which are 2 and N-1. Therefore P_{N-3}=2. Using same analysis keep assigning the values till all values of the permutation are exhausted.

This sequence can be generated in linear time: Let R=N and L=1. Keep iterating from the index i=N to 1 assign P_i=R and P_{i-1}=L(if\: i-1\geq 1), Decrease R by 1 and increase L by 1. The resulting sequence is same as described above and it satisfies the condition needed in the problem. Similar analysis can be carried out for the sequence when P_N=1 and P_{N-1}=N.

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

Setter's solution
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long
const ll INF_MUL=1e13;
#define pb push_back
#define mp make_pair
#define nline "\n"
#define f first
#define s second
#define pll pair<ll,ll>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vvvl vector<vector<vector<ll>>>
#define all(v) v.begin(),v.end()
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif
void _print(ll x){cerr<<x;}
void _print(string x){cerr<<x;}
template<class T,class V> void _print(pair<T,V> p) {cerr<<"{"; _print(p.first);cerr<<","; _print(p.second);cerr<<"}";}
template<class T>void _print(vector<T> v) {cerr<<" [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T>void _print(set<T> v) {cerr<<" [ "; for (T i:v){_print(i); cerr<<" ";}cerr<<"]";}
template<class T>void _print(multiset<T> v) {cerr<< " [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T,class V>void _print(map<T, V> v) {cerr<<" [ "; for(auto i:v) {_print(i);cerr<<" ";} cerr<<"]";}
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using muloset=tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
//--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
const ll MOD=1e9+7;
const ll MAX=500500;
void solve(){
ll n; cin>>n;
ll l=1,r=n;
vector<ll> a(n+5,0);
for(ll i=n;i>=1;i--){
if((i&1)==(n&1)){
a[i]=r--;
}
else{
a[i]=l++;
}
}
for(ll i=1;i<=n;i++){
cout<<a[i]<<" \n"[i==n];
}
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("error.txt", "w", stderr);
#endif
ll test_cases=1;
cin>>test_cases;
while(test_cases--){
solve();
}
cout<<fixed<<setprecision(15);
cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n";
}

Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
void sol(void)
{
int n;
cin >> n;
int arr[n], l = 1, r = n;
for (int i = n - 1; i >= 0; i -= 2)
{
arr[i] = l++;
if (i - 1 >= 0)
arr[i - 1] = r--;
}
for (int i = 0; i < n; i++)
cout << arr[i] << ' ';
cout << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}