【JZOJ6345】ZYB建围墙

description

---

analysis

• 打表找规律，自认为样例给的提示很明显

• 容易想到最优方案是让家庭尽量先围成一个正六边形，剩下的在最外层绕一个圈

• 手推一波可以知道，如果正六边形有\(n\)层，剩下来\(n-1\)答案一样

• 其实每隔\(n-1,n,n,n,n,n+1\)答案都一样，中间间隔\(1\)

• 然后模拟

---

node

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

const ll answer[7]={0,6,8,9,10,11,12};
ll n;

inline ll query(ll n)
{
	if (n<7)return answer[n];
	ll tot=1,tmp=0;
	for (;n>=tmp+6;++tot)tmp+=6,n-=tmp;
	ll ans=tmp+6;
	if (n<=1)return ans;
	if (n+1<=tot)return ans+1;
	n-=tot,++ans;
	return n==0?ans:ans+1+(n-1)/tot;
}
int main()
{
	//freopen("T1.in","r",stdin);
	freopen("wall.in","r",stdin);
	freopen("wall.out","w",stdout);
	//fo(i,1,37)printf("%lld:%lld\n",i,query(i));
	scanf("%lld",&n),printf("%lld\n",query(n));
	return 0;
}