# PROBLEM LINK:

Contest Division 1

Contest Division 2

Contest Division 3

Contest Division 4

Setter: Jeevan Jyot Singh

Tester: Nishank Suresh, Satyam

Editorialist: Devendra Singh

# DIFFICULTY:

Simple

# PREREQUISITES:

The mean of an array B of size M is defined as: \texttt{mean}(B) = \dfrac{\sum_{i = 1}^{M} B_i}{M}.

# PROBLEM:

You are given two integers N and X. Output an array A of length N such that:

- -1000 \le A_i \le 1000 for all 1 \le i \le N.
- All A_i are
**distinct**. - \texttt{mean}(A) = X.

If there are multiple answers, print any. It is guaranteed that under the given constraints at least one array satisfying the given conditions exists.

As a reminder, the mean of an array B of size M is defined as: \texttt{mean}(B) = \dfrac{\sum_{i = 1}^{M} B_i}{M}.

For example, \texttt{mean}([3, 1, 4, 8]) = \frac{3 + 1 + 4 + 8}{4} = \frac{16}{4} = 4.

# EXPLANATION:

The problem can be divided into two cases:

\textbf{Case 1}: N is even. To achieve an average of X , the sum of all the values of array A must be N\cdot X. We simply create N/2 pairs such that their sum is 2\cdot X each. Total sum of these pairs would be 2\cdot X\cdot N/2 = N\cdot X

Therefore the average of these N values is (N\cdot X)/N = X . One possible way to create these pairs is to pair values around X i.e X-1\: and\: X+1 ,X-2\: and\: X+2 and so on.

To print the answer for this case run a loop from i=1 to i=N/2 and on each iteration print two values X-i and X+i.

\textbf{Case 1}: N is odd. To achieve an average of X , the sum of all the values of array A must be N\cdot X. We simply create (N-1)/2 pairs such that their sum is 2\cdot X each and append X to the end of the array. Total sum of these pairs would be 2\cdot X\cdot (N-1)/2 +X= N\cdot X - X +X= N\cdot X

Therefore the average of these N values is (N\cdot X)/N = X . One possible way to create these pairs is to pair values around X i.e X-1\: and\: X+1 ,X-2\: and\: X+2 and so on.

To print the answer for this case run a loop from i=1 to i=(N-1)/2 and on each iteration print two values X-i and X+i. Then print X in the end.

The value of A_i never exceeds 600 and never drops below -500 in this approach which satisfies the given constraints.

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

## Setter's solution

```
#ifdef WTSH
#include <wtsh.h>
#else
#include <bits/stdc++.h>
using namespace std;
#define dbg(...)
#endif
#define int long long
#define endl "\n"
#define sz(w) (int)(w.size())
using pii = pair<int, int>;
const long long INF = 1e18;
const int N = 1e6 + 5;
void solve()
{
int n, x; cin >> n >> x;
vector<int> a;
for(int i = 1; i <= n / 2; i++)
a.push_back(x - i), a.push_back(x + i);
if(n % 2)
a.push_back(x);
for(int x: a)
cout << x << " ";
cout << endl;
}
int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int T; cin >> T;
for(int tc = 1; tc <= T; tc++)
{
// cout << "Case #" << tc << ": ";
solve();
}
return 0;
}
```

## Tester-1's Solution(Python)

```
for _ in range(int(input())):
n, x = map(int, input().split())
for i in range(n//2):
print(x-i-1, x+i+1, end = ' ')
if n%2 == 1:
print(x)
else:
print('')
```

## Tester-2's Solution

```
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif
/*
------------------------Input Checker----------------------------------
*/
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
int MAX=100000;
void solve(){
int n=readIntSp(1,1000);
int x=readIntLn(0,100);
if(n&1){
cout<<x<<" "; n--;
}
int l=-5,r=2*x+5;
for(int i=1;i<=n;i+=2){
cout<<l--<<" "<<r++<<" ";
}
cout<<"\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
int test_cases=readIntLn(1,100);
while(test_cases--){
solve();
}
assert(getchar()==-1);
return 0;
}
```

## 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>
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); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
int n, x;
cin >> n >> x;
if (n & 1)
{
cout << x << ' ';
for (int i = 1; i <= (n / 2); i++)
cout << x - i << ' ' << x + i << ' ';
cout << '\n';
}
else
{
for (int i = 1; i <= (n / 2); i++)
cout << x - i << ' ' << x + 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();
}
```