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

Calculating Function

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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：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", -(n+1)/2);

        else printf("%lld\n", n/2);

    }

    return 0;

}

巴特西

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

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