CF#277 (Div. 2) A.（找规律） - c++编程基础

A. Calculating Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output 题目链接：http://codeforces.com/contest/486/problem/A

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

output

input

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：

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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); } }