Breaking The Summation Formula (Part 1)

 Q. f(n) =  - 1 + 2 - 3 + .. + ( - 1) ⁿ n . Given n, find out f(n)

Approach(1)- Bruteforce:
1. Calculation sum=n*(n+1)/2
2. loop[i=1,i=n : i+=2] odd+=i
3.ans=sum-2*odd

Code:

1. #include <bits/stdc++.h>
2. using namespace std;
3. int main(){
4.  
6. long long x;
7. cin>>x;
8. long long p=(x*(x+1))/2;
9. long long bad=0;
10. for(long long i=1;i<=x;i+=2)
11. bad+=i;
13. cout<<p-2*bad<<endl;
15. }

Approach(2)-Greedy:
Basic:
s=1+2+3+4+....+n
Formula: sum=n*(n+1)/2= (n/2) + (n+1).2
here n/2 denotes the summation of all even numbers and (n+1)/2 denotes the summation of all odd numbers.

• 1.if n is even answer=n/2

        explanation:
        n=6
       -1+2-3+4-5+6 =(2+4+6)-(1+3+5)= 12-9=3

• if n is odd answer=(n-1)/(2-n)=-(n+1)/2

         explanation:
         n=5
        -1+2-3+4-5 =(2+4)-(1+3+5)=6-9=-3
        We got -1 in the formula because of (-1)^n. here n is odd. that's why we'll
         get -1 for one time always in the case of odd numbers.

Code:

1. #include <bits/stdc++.h>
2. using namespace std;
3. int main(){
5. long long x,y,n;
6. cin>>n;
7. if(n%2==0)cout<<n/2<<endl;
8. else cout<<-(n+1)/2<<endl;
10. }

Test-Case Observation:

Problem:https://codeforces.com/contest/486/problem/A