HDU 6154 CaoHaha's staff（2017中国大学生程序设计竞赛 - 网络选拔赛）

题目代号：HDU 6154

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6154

CaoHaha's staff

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 777    Accepted Submission(s): 438

Problem Description

"You shall not pass!"
After shouted out that,the Force Staff appered in CaoHaha's hand.
As we all know,the Force Staff is a staff with infinity power.If you can use it skillful,it may help you to do whatever you want.
But now,his new owner,CaoHaha,is a sorcerers apprentice.He can only use that staff to send things to other place.
Today,Dreamwyy come to CaoHaha.Requesting him send a toy to his new girl friend.It was so far that Dreamwyy can only resort to CaoHaha.
The first step to send something is draw a Magic array on a Magic place.The magic place looks like a coordinate system,and each time you can draw a segments either on cell sides or on cell diagonals.In additional,you need 1 minutes to draw a segments.
If you want to send something ,you need to draw a Magic array which is not smaller than the that.You can make it any deformation,so what really matters is the size of the object.
CaoHaha want to help dreamwyy but his time is valuable(to learn to be just like you),so he want to draw least segments.However,because of his bad math,he needs your help.

 

Input

The first line contains one integer T(T<=300).The number of toys.
Then T lines each contains one intetger S.The size of the toy(N<=1e9).

 

Output

Out put T integer in each line ,the least time CaoHaha can send the toy.

 

Sample Input

5
1
2
3
4
5

 

Sample Output

4
4
6
6
7

 

Source

[2017中国大学生程序设计竞赛 - 网络选拔赛]( http://acm.hdu.edu.cn/search.php?field=problem&key=2017%D6%D0%B9%FA%B4%F3%D1%A7%C9%FA%B3%CC%D0%F2%C9%E8%BC%C6%BE%BA%C8%FC+-+%CD%F8%C2%E7%D1%A1%B0%CE%C8%FC&source=1&searchmode=source)

题目大意：有一个巫师可以在一个带有xOy坐标系的魔术场地上画传送阵，他画的传送阵面积要大于或等于需要传送物品的面积，他每分钟只能画一条边，这条边能是每个单位方块的边或者对角线（即每次能画单位长度为1的边或者\(\sqrt2\)的对角线）现在他希望用最小的时间画出满足需要的面积的传送阵，他需要你的帮助。

解题思路：使面积最大化，即尽量使用对角线进行绘制传送阵。

现在需要讨论的是对于面积的特殊情况，现在根据下图来看一看：

很显然，在原来的基础上只增加了一条边，但是面积却增加了1.5，也就是说当面积为9的时候，只需要在8条边的基础上增加一条边即可使覆盖面积达到10.5，那么同理，当某条边的长度很长的时候那么在原先的基础上只需要增加一条边，面积却扩大了很多，并且这是最优的画法（只需增加一笔即可扩展出一个梯形）

并且当有两个连续边各增加一个梯形后，可以将两条小边扩展成直角如下图所示，



同理，当一个正方形每条边增加一笔即可将整个正方形的边长增加\(\sqrt2\)这是显然的，再看看下图



这样是最优的画法，相同的边时使面积最大化。

下面放上AC代码：

# include <bits/stdc++.h>
using namespace std;
# define IOS ios::sync_with_stdio(false);
typedef unsigned long long ULL;
typedef long long LL;
///coding...................................

int main()
{
    IOS
#ifdef FLAG
    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
#endif /// FLAG
    LL n,t,a[10]={0,4,4,6,6,7,8,8,8};
    cin>>t;
    while(t--) {
        cin>>n;
        if(n<=8)cout<<a[n]<<endl;
        else {
            LL l=0;
            while(2*l*l<=n)++l;--l;
            if(n==2*l*l)cout<<4*l<<endl;
            else if(n<=2*l*l+l-0.5)cout<<4*l+1<<endl;
            else if(n<=2*l*l+2*(l-0.5)+1)cout<<4*l+2<<endl;
            else if(n<=2*l*l+3*(l-0.5)+2)cout<<4*l+3<<endl;
            else cout<<4*l+4<<endl;
        }

    }
    return 0;
}