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

A. Calculating Function

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/486/problem/A

Description

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 Input

4

Sample Output

2

HINT

题意
f(n) =  - 1 + 2 - 3 + .. + ( - 1)^n

给你n,然后让你求f(n)

题解:

分奇偶啦,奇数就是(n-1)/2-n,偶数就是n/2

代码

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
    long long n;scanf("%lld",&n);
    if(n%2==1) printf("%lld
",(n-1LL)/2LL - n);
    else printf("%lld
",n/2LL);
}
原文地址:https://www.cnblogs.com/qscqesze/p/4908638.html