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

A. Calculating Function

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

Note

f(4) = - 1 + 2 - 3 + 4 = 2

f(5) = - 1 + 2 - 3 + 4 - 5 = - 3

题意：定义f(n)，求f(n)；

题解:奇偶关系

///

#include<bits/stdc++.h>

using namespace std ;

typedef long long ll;

#define mem(a) memset(a,0,sizeof(a))

#define pb push_back

#define meminf(a) memset(a,127,sizeof(a));

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){

        if(ch=='-')f=-;ch=getchar();

    }

    while(ch>=''&&ch<=''){

        x=x*+ch-'';ch=getchar();

    }return x*f;

}

//****************************************

#define maxn 1000+5

#define mod 1000000007

int main(){

    ll n=read();

    if(n%==){

        cout<<n/<<endl;

    }

    else {

        cout<<(n-)/-n<<endl;

    }

  return ;

}

代码

巴特西

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

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