BZOJ 1600 建造栅栏

O(N)分成1,2与3,4两部分搞一搞。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,ans=0,lim;
int ask(int x)
{
    return min(lim,x-1)-max(x-lim,1)+1;
}
int main()
{
    scanf("%d",&n);lim=n/2-(!(n&1));
    for (int i=2;i<=n-2;i++)
        ans+=ask(i)*ask(n-i);
    printf("%d
",ans);
    return 0;
}