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…

A. 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…

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) f…

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…

Problem 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) …

#include<iostream> #include<cstring> #include<cstdio> /* 题意:计算f(n) = -1 + 2 -3 +4.....+(-1)^n *n的值 思路:偶数和 - 奇数和(或者用等差数列计算化简得到结果) */ #include<algorithm> #define N 10000 using namespace std; int main(){ long long n; cin>>n; ==)…

题目链接:www.codeforces.com/problemset/problem/486/A题意:求表达式f(n)的值.(f(n)的表述见题目)C++代码: #include <iostream> using namespace std; long long f(long long n) { == ) ; else - n; } int main() { long long n; cin >> n; cout << f(n) << endl; ; } C…

A. 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…

A. Calculating Function 水题,分奇数偶数处理一下就好了 #include<stdio.h> #include<iostream> using namespace std; int main() { long long n;scanf("%lld",&n); ==) printf("%lld\n",(n-1LL)/2LL - n); else printf("%lld\n",n/2LL); }…

题目地址:http://codeforces.com/contest/486 A题.Calculating Function 奇偶性判断,简单推导公式. #include<cstdio> #include<iostream> using namespace std; int main() { long long n; cin>>n; ==) { cout<<(-)*((n-)/+)+n<<endl; } else cout<<((n-…