hdu 6154 CaoHaha's staff

CaoHaha's staff

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 487    Accepted Submission(s): 286

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

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题意：

一笔可以画一条长为1的边或者一条根号二的对角线，问围成一个面积是n 的图形最少需要几条边

题解：

找规律画图

这题可以先把问题转化为已知画m笔，最大可以画出多大的面积

当m%4==0，画sqrt（2）的边，可以达成最大的面积

当m%4==1，就把[m/4]*4围成的图像的 最长边往外推sqrt（2）/2，比原来增加一个梯形的面积

当m%4==2，就把[m/4]*4围成的图像的 最长边往外推sqrt（2）长度，增加一个小矩形面积

当m%4==3，在m%4==2的基础上实行m%4==1一样的操作

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int t;
int n,sd;

int sumS(int m)  //m是周长，推出公式就可以了
{
    int d=m/;
    if (m%==) return *d*d;
    if (m%==) return *d*d+d-;
    if (m%==) return *d*d+*d;
    if (m%==) return *d*d+*d;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sd=(int)sqrt(n/2.0)*;  //最小的周长
        while()
        {
            if (sumS(sd)>=n) {printf("%d\n",sd); break;}
            sd++;
        }
    }
  return ;
}