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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.
The single line contains the positive integer n (1 ≤ n ≤ 1015).
Print f(n) in a single line.
4
2
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
///1085422276 #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=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){ if(ch==‘-‘)f=-1;ch=getchar(); } while(ch>=‘0‘&&ch<=‘9‘){ x=x*10+ch-‘0‘;ch=getchar(); }return x*f; } //**************************************** #define maxn 1000+5 #define mod 1000000007 int main(){ ll n=read(); if(n%2==0){ cout<<n/2<<endl; } else { cout<<(n-1)/2-n<<endl; } return 0; }
Codeforces Round #277 (Div. 2)A. Calculating Function 水
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原文地址：http://www.cnblogs.com/zxhl/p/4948729.html
