Hello coders, today we are going to solve the Odd Pairs Codechef Solution whose Problem Code is ODDPAIRS.
Odd Pairs Codechef Solution
Problem
Given an integer NN, determine the number of pairs (A, B) such that:
- 1 ≤ A, B ≤ N;
- A + B is odd.
Input Format
- The first line of input will contain a single integer TT, denoting the number of test cases.
- Each test case consists of a single integer NN.
Output Format
For each test case, output the number of required pairs.
Constraints
Sample 1
Input
5 1 2 3 100 199
Output
0
2
4
5000
19800
Explanation:
Test case 1: There are no pairs satisfying the given conditions.
Test case 2: The pairs satisfying both conditions are: (1,2) and (2,1).
Test case 3: The pairs satisfying both conditions are: (1,2),(2,1),(2,3), and (3,2).
Odd Pairs Codechef Solution in CPP
#include <iostream>
using namespace std;
int main() {
int t,i;
cin>>t;
while(t--)
{
long int n,o=0,e=0,res;
cin>>n;
o=(n+1)/2;
e=n/2;
res=2*o*e;
cout<<res<<"\n";
}
return 0;
}
Odd Pairs Codechef Solution in JAVA
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t>0){
long n = sc.nextLong();
System.out.println((n/2)*(n-(n/2))*2);
t--;
}
}
}
Odd Pairs Codechef Solution in Python
# cook your dish here
#odd pairs 2 (1,2) (2,1) I.E N^2//2 METHODS ARE POSSIBLE
n = int(input())
while(n>0):
a = int(input())
print((a**2)//2)
n = n-1
Disclaimer: The above Problem (Odd Pairs) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.
