Codeforces Round #277 (Div. 2)---A. Calculating Function （规律）

For a positive integer n let's define a function f:

f(n) = - 1 + 2 - 3 + .. + ( - 1)nn

Your task is to calculate f(n) for a given integer n.

Input

The single line contains the positive integer n (1 ≤ n ≤ 1015).

Output

Print f(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

解题思路：大水题一枚。直接找规律。n%2==0时，f = n/2; 否则，f = -(n+1)/2.

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int main()
{
//    #ifdef sxk
//        freopen("in.txt","r",stdin);
//    #endif
    long long n;
    while(scanf("%lld",&n)!=EOF)
    {
        if(n & 1) printf("%lld
", -(n+1)/2);
        else printf("%lld
", n/2);
    }
    return 0;
}