CF#277 (Div. 2) A.（找规律）

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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：

给你n，求f(n) 。这是一道规律题，首先确定不能暴力求解，因为n实在太大，暴力必超时。同时也开不了那么大的数组来暴力。

提笔简单算了几个，发现f(1) = -1 , f(2) = 1, f(3) = -2, f(4) = 2, f(5) = -3, f(6) = 3··········这样规律就看出来了，两个数一组，n/2如果为偶数那么符号为正，奇数符号为负。并且如果n/2为奇数的话，我们向上取整，即为n / 2 + 1

最后，中间值什么的都开成long long。

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>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);    }}

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