CF #277 (Div. 2) A. (query rule)

CF #277 (Div. 2) A. (query rule)
A. Calculating Functiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output question link: http://codeforces.com/contest/486/problem/A

For a positive integerNLet's define a functionF:

F(N)? = ?? -? 1? +? 2? -? 3? + ?..? +? (? -? 1)NN

Your task is to calculateF(N) For a given integerN.

Input

The single line contains the positive integerN(1? ≤?N? ≤? 1015 ).

Output

PrintF(N) In a single line.

Sample test (s) input

4

Output

2

Input

5

Output

-3

Note

F(4 )? = ?? -? 1? +? 2? -? 3? +? 4? =? 2

F(5 )? = ?? -? 1? +? 2? -? 3? +? 4? -? 5? = ?? -? 3

Solution:

Give you n, ask f (n ). This is a regular question. First of all, it is determined that the brute force solution cannot be solved. Because n is too large, the brute force will time out. At the same time, it cannot open such a large array for brute force.

Let's take a few simple notes and find that f (1) =-1, f (2) = 1, f (3) =-2, f (4) = 2, f (5) =-3, f (6) = 3 · this rule shows that there are two groups of numbers, if n/2 is an even number, the symbol is positive and the odd number is negative. And if n/2 is an odd number, we take an integer up, that is, n/2 + 1.

Finally, the median value is converted into long.

Complete code:

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                     using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;int main(){ #ifdef DoubleQ freopen("in.txt","r",stdin); #endif LL n; while(~scanf("%lld",&n)) { LL c = n / 2; if(n % 2 != 0) { c += 1; printf("-"); } printf("%lld\n" , c); }}