 First - Editorial

#1

Author: Nishchith Shetty

Tester: Karan Sheth

Editorialist: Chirag Shetty

Easy

Simple Math

PROBLEM:

Given an integer n you have to print result adding all even nos till n and subtracting all odd nos tll n.

EXPLANATION:

Since n is very large adding even nos one by one and subtracting odd nos one by one will give you TLE.

There are 2 simple cases.

1. If n is even (say 10) then answer is n/2 (i.e 5)

If n is even then we can consider first n/2 odd numbers and
and first n/2 even numbers excluding 0 as 0 won’t add up to the sum.

For eg if n=10,
Odd nos - 1 3 5 7 9
Even nos - 2 4 6 8 10 (excluding 0)
Let k = n/2 (i.e 5)
Using sum of Arithmetic Progression Formula :
S = (k/2)*(2a+(k-1)*d)

Sum of even numbers (a=2,d=2)
S1 = (k/2)(4+(k-1)2)
S1 = (k/2)
(2k+2)
S1 = k
k+k

Sum of odd numbers (a=1,d=2)
S2 = (k/2)(2+(k-1)2)
S2 = (k/2)
(2k)
S2 = k
k

S1-S2 = k = n/2

1. If n is odd (say 9) then answer is -(n+1)/2 (i.e -5)

If we take n+1 then that is even and we just need to subtract the last even number i.e n+1 from above formula.
i.e (n+1)/2-(n+1) = -(n+1)/2

ALTERNATIVE SOLUTION

Observations skills AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.

Tester’s solution can be found here.

#2

Actually the solution is pretty simple . using some maths and properties of arithematic progressions i found out that if the number entered by the user is even then we just have to output the number entered by user divided by two . for example : if user enters 8 output 8/2 i.e. 4.
the A.P. formed will be 2,4,…n . and 1,3,…n-1 .find the sum of both the A.P.s and perform the given operation . you will get answer n/2

on the other hand if number entered by user is odd then you have to check if the number is 1 or greater than 1 . if one you have to print 1 and if greater than 1 print -(n+1)/2.
the A.P. formed will be 2,4,…n-1 . and 1,3,… .find the sum of both the A.P.s and perform the given operation . you will get answer -(n+1)/2