MINOPER-Editorial

PROBLEM LINK:

PRACTICE
Editorialist: Avishake Maji

DIFFICULTY:

EASY

PREREQUISITES:

Number Theory

PROBLEM:

You are given a number X. You have to make this number divisible by 2. The operations you can perform is as follows:
⦁ Add 1 to X
⦁ Substract 1 from X
You can perform these operations zero or more no of times. You have to find the minimum number of operations required to make X divisible by 2.

Input:

t: Number of test cases
X: the given number

Output:

The minimum number of operations required to make X divisible by 2

Constraints:

1<=t<=1000
0<=X<=10^9

Sample Input:

2
4
0

###Sample Output:

0
0

Explanation:

Since 4 is divisible by 2 so the number of operation will be 0

Since 0 is divisible by 2 so the number of operation will be 0

EXPLANATION:

According to the question we are told to find the minimum of operations required to make a number by 2 using the given operations( either add the number by 1 or substract 1 from the number). If the number is already even i.e divisible by 2 then the minimum no of operations will be 0. If the odd then we can easily convert it to even either by adding 1 or substracting 1 from the number. So the minimum number of operations will be 1. So all we need to find is whether the number is even or odd and print 0 or 1 respectively.

SOLUTION:

#include <iostream> 
#include<ctime>
#include<map>
#define ll long long int
using namespace std; 
void solve(); 
int main() 
{ 
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL); 

	int t;
	 cin >> t; 
	while (t--) { 
		solve(); 
		// cout << "\n"; 
	} 

	return 0; 
} 
void solve() 
{ 
	ll n;
	cin>>n;
	if(n&1)
		cout<<"1";
	else
		cout<<"0";
	cout<<endl;
}