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Programmingoneonone
Programmingoneonone

Learn everything about programming

CODECHEF Different Medians Problem Solution (DIFFMED)

YASH PAL, 31 July 2024

In this CodeChef different medians problem solution Chef is given an integer N. Chef wants to construct a permutation , it must hold that median(Pi) != median(P(i+1))

codechef different medians problem solution

CODECHEF Different Medians Problem Solution (DIFFMED)

//Divyansh Vinayak
#include<bits/stdc++.h>
using namespace std;
#define gc getchar_unlocked
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define si(x)	scanf("%d",&x)
#define sl(x)	scanf("%lld",&x)
#define ss(s)	scanf("%s",s)
#define pi(x)	printf("%d",x)
#define pl(x)	printf("%lld",x)
#define ps(s)	printf("%s",s)
#define pnl()   printf("n")
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pl>		vpl;
typedef vector<vi>		vvi;
typedef vector<vl>		vvl;
int mpow(int base, int exp); 
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 18 * 103, M = N;
#define top(a,b) (ll)((a+b-1)/b)
//=======================
vi g[N];
void matrix(ll n, ll m){
  char s[101][101];
    while(cin>>n>>m){
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>s[i][j];
            }
        }
}
}

int fact(ll n);
 

 
// Returns factorial of n
int fact(ll n)
{
    ll res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

int main() {
  ll t;
  cin >> t;
  while(t--){
    ll n;
    cin >> n;
    vi a;
    for(int i = 0; i < n; i++){
        a.push_back(i+1);
    }
    ll mid;
    if(n%2 == 1)
        mid = (n - 1)/2;
    else
        mid = (n/2) - 1;
    ll right = mid + 1;
    ll left = mid - 1;
    while(right != n - 1){
        cout << a[right] << " ";
        cout << a[left] << " ";
        right++;
        left--;
    }
    if(n%2 == 1){
        cout << a[right] << " ";
        cout << a[left] << " ";
    }
    else
        cout << a[right] << " ";
    cout << a[mid] << endl;
  }
  return 0;
}      


int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % mod;
    base = ((ll)base * base) % mod;
    exp >>= 1;
  }
  return result;
}

void ipgraph(int n, int m){
	int i, u, v;
	while(m--){
		cin>>u>>v;
		g[u-1].pb(v-1);
		g[v-1].pb(u-1);
	}
}

void dfs(int u, int par){
	for(int v:g[u]){
		if (v == par) continue;
		dfs(v, u);
	}
}
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