思路:對n個數進行規律判斷,容易找到奇數個和偶數個所對應的規律
#include<iostream>
using namespace std;
int main()
{
long long int n,ans;
scanf("%lld",&n);
if(n%2==0)
{
ans=n/2;
}
else
{
ans=n/2-(n);
}
printf("%lld\n",ans);
return 0;
}