# PROBLEM LINK:

* Author and Editorialist:* Pankaj Sharma

*Jatin Kashyap*

**Tester:**# DIFFICULTY:

CAKEWALK.

# PREREQUISITES:

Basics of Maths.

# PROBLEM:

You are given an integer **N**.

Your task is to find **any** two integers **A,B** such that 1\leq A , B \leq 2*N and difference of **A** and **B** exactly equals to **N** i.e. A - B = N.

# QUICK EXPLANATION:

## Check

Answer is A = N + 1 and B = 1

# EXPLANATION:

Let’s try to brute force i.e. generate all pairs of (A, B) such that between 1 \leq A, B, \leq 2*N and run two loops to check whether A - B == N or not.

## Code

```
#include<iostream>
using namespace std;
int main()
{
int t;
// Input number of test cases
cin >> t;
while (t--) {
int N;
// Input N
cin >> N;
for (int A = 1; A <= 2 * N; A++)
{
bool foundAnswer = false;
for (int B = 1; B <= 2 * N; B++)
{
if (A - B == N) {
cout << A << " " << B << "\n";
foundAnswer = true;
break;
}
}
if (foundAnswer == true) {
break;
}
}
}
return 0;
}
```

Above code will result in TLE i.e. Time Limit Exceeded error as N can be 10^9 and T can be 10^5 and 1 second = 10^8 operations but we are performing around 10^{14} operations as N * T = 10^9 * 10^5 = 10^{14}

So we need to think about something different.

We if closely observe

A - B = N can be written as A = N + B

Now A \leq 2*N

So , N + B \leq 2 * N

So, B \leq N

Hence B lies between [1, N] and A lies between [N + 1, 2 * N].

Minimum value of N can be 1.

So B = 1 and A = N + 1 will be a valid answer.

Also A = 2*N and B = N is also a valid answer.

But A = 2 and B = N + 2 fails on case when N = 1 as B becomes greater than 2 * N

# Time Complexity:

The time complexity is O(1) per test case.

# SOLUTIONS:

## Setter's Solution

```
#include<iostream>
using namespace std;
int main()
{
int t;
// Input number of test cases
cin >> t;
while (t--) {
int N;
// Input N
cin >> N;
int A = N + 1, B = 1;
// Print A,B in newline
cout << A << " " << B << "\n";
}
return 0;
}
```

## Tester's Solution

```
import java.util.*;
class test{
public static void main (String args[]){
Scanner sc=new Scanner(System.in);
for(int i=sc.nextInt();i>0;i--)
System.out.println((sc.nextInt()+1)+" 1");
}
}
```

## Editorialist's Solution

```
#include<iostream>
using namespace std;
int main()
{
int t;
// Input number of test cases
cin >> t;
while (t--) {
int N;
// Input N
cin >> N;
int A = N + 1, B = 1;
// Print A,B in newline
cout << A << " " << B << "\n";
}
return 0;
}
```

For doubts, please leave them in the comment section, I’ll address them.