# AVGOF3 - Editorial

Setter: Jeevan Jyot Singh
Tester: Manan Grover
Editorialist: Lavish Gupta

Cakewalk

None

# PROBLEM:

It is Chef’s birthday. You know that Chef’s favourite number is X. You also know that Chef loves averages. Therefore you decide it’s best to gift Chef 3 integers A_1, A_2, A_3, such that:

• The mean of A_1, A_2 and A_3 is X.
• 1 \le A_1, A_2, A_3 \le 1000.
• A_1, A_2 and A_3 are distinct.

Output any suitable A_1, A_2 and A_3 which you could gift to Chef.

As a reminder, the mean of three numbers P, Q, R is defined as: mean(P, Q, R) = \dfrac{P + Q + R}{3}.

For example, mean(2, 3, 5) = \frac{2 + 3 + 5}{3} = \frac{10}{3} = 3.33\bar{3}, mean(2, 2, 5) = \frac{2 + 2 + 5}{3} = \frac{9}{3} = 3.

# EXPLANATION:

There are several possible solutions to this problem. Here are two of them:

Based on Total Sum

Given that the average is X, the total sum of three elements will be S = 3 \cdot X.
Now, given 2 \leq X \leq 100, we will have 6 \leq S \leq 300.
Because we need to choose 3 distinct positive integers, let’s choose A_1 = 1, A_2 = 2. Now, A_3 = S - 3, so 3 \leq A_3 \leq 297, which satisfies all the constraints.

Based on Average

We want to make the average of three elements to X. One way to achieve this is to symmetrically distribute elements around X.
So, we can have one of the elements as X itself. Now, we have two remaining elements which we want to assign symmetrically, so we can have one of the element as X-1 and other as X+1.
This assignment satisfies all the constraints.

# TIME COMPLEXITY:

O(1) for each test case.

# SOLUTION:

Setter's Solution
#include <bits/stdc++.h>
using namespace std;

#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

int32_t main()
{
IOS;
int T; cin >> T;
while(T--)
{
int n; cin >> n;
cout << n-1 << ' ' << n << ' ' << n+1 << '\n';
}
return 0;
}

Tester's Solution
#include <bits/stdc++.h>
using namespace std;
int main(){
ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
int t;
cin>>t;
while(t--){
int x;
cin>>x;
cout<<1<<" "<<2<<" "<<3 * x - 3<<"\n";
}
return 0;
}

Editorialist's Solution
#include<bits/stdc++.h>
#define ll long long
using namespace std ;

const ll z = 1000000007 ;

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

cout << n-1 << ' ' << n << ' ' << n+1 << endl ;
return ;
}

int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("inputf.txt" , "r" , stdin) ;
freopen("outputf.txt" , "w" , stdout) ;
freopen("error.txt" , "w" , stderr) ;
#endif

int t;
cin >> t ;
while(t--)
solve() ;

return 0;
}

1 Like
#include <iostream>
using namespace std;

int main() {
// your code goes here
int n;
cin>>n;
while(n--){
int a;
cin>>a;

int news=a*2;
int new1=news-1;
cout<<a<<" "<<new1<<" "<<1<<endl;

}
return 0;
}