A_Permutation_Warm_Up.cpp

#include <bits/stdc++.h>
#include <algorithm>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")

#define nl '\n'
#define sp ' '
#define pi 2 * acos(0.0)

// Types of declarations /////////////////////////////////
#define ui unsigned int
#define us unsigned short
#define all(x) x.begin(), x.end()
#define ull unsigned long long
#define ll long long
#define ld long double
#define vstr vector<string>
#define vll vector<ll>
#define vi vector<int>
#define vvi vector<vector<int>>
#define vii vector<pair<int, int>>
#define pii pair<int, int>

// Odd Even /////////////////////////////////////////////
bool odd(ll num) { return ((num & 1) == 1); }
bool even(ll num) { return ((num & 1) == 0); }

//////////////////////////////////////////////////////// Prime

bool isPrime(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}

///////////////////////////////////////////////////////// LCM GCD
long long gcd(long long a, long long b)
{
    while (b != 0)
    {
        long long temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
long long lcm(long long a, long long b)
{
    return (a / gcd(a, b)) * b;
}
////////////////////////////////////////////////////////// SQR ROOT

long long sqrt(long long x)
{
    long long s = 0, e = 2e9, res = s;
    while (s <= e)
    {
        long long m = (s + e) / 2;
        if (m * m <= x)
            res = m, s = m + 1;
        else
            e = m - 1;
    }
    return res;
}
/*
check all edge cases
check for integer overflow
check for corner cases
check for constraints
check for time complexity
check for array bounds
check for negative values
*/
/*
    vi a(n);
    for(int i=0; i<n; i++){
        cin>>a[i];
    }
*/
#define cno cout << "NO\n"
#define cyes cout << "YES\n"
/*----------------------------------------------------------------------------*/

void solve()
{
    int n;
    cin >> n;

    int ans = (n * n) / 4 + 1;
    cout << ans << nl;
}

/*
For a permutation p
 of length n
∗
, we define the function:

f(p)=∑i=1n|pi−i|

You are given a number n
. You need to compute how many distinct values the function f(p)
 can take when considering all possible permutations of the numbers from 1
 to n
.

∗
A permutation of length n
 is an array consisting of n
 distinct integers from 1
 to n
 in arbitrary order. For example, [2,3,1,5,4]
 is a permutation, but [1,2,2]
 is not a permutation (2
 appears twice in the array), and [1,3,4]
 is also not a permutation (n=3
 but there is 4
 in the array).

Input
Each test contains multiple test cases. The first line contains the number of test cases t
 (1≤t≤100
). The description of the test cases follows.

The first line of each test case contains an integer n
 (1≤n≤500
) — the number of numbers in the permutations.

Output
For each test case, output a single integer — the number of distinct values of the function f(p)
 for the given length of permutations.

Example
InputCopy
5
2
3
8
15
43
OutputCopy
2
3
17
57
463
Note
Consider the first two examples of the input.

For n=2
, there are only 2
 permutations — [1,2]
 and [2,1]
. f([1,2])=|1−1|+|2−2|=0
, f([2,1])=|2−1|+|1−2|=1+1=2
. Thus, the function takes 2
 distinct values.

For n=3
, there are already 6
 permutations: [1,2,3]
, [1,3,2]
, [2,1,3]
, [2,3,1]
, [3,1,2]
, [3,2,1]
, the function values of which will be 0,2,2,4,4
, and 4
 respectively, meaning there are a total of 3
 values.



 */

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
        solve();
}