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
题意
有一个大小为 n 的玩具，我们需要画出面积不小于它的多边形（只能沿着格子边缘或者对角线），每个边缘或者对角线为一步，问最少需要多少步。
思路
当步数为偶数时，显然我们需要找一个最接近正方形的矩形（长宽差值最小）才可以使得总面积最大，此时所有边都覆盖在对角线上。
假设当前步数为 i ，则该矩形面积为 2.0 * (i/4) * (i/2 - (i/4)) 。
而当步数为奇数时，我们需要在比它小的一个偶数步的基础上，将其最长边拆开，增加一条边使其面积增大 (d * 2 - 1) * 0.5 。
根据步数打表以后二分输出答案即可。
AC 代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
typedef __int64 LL;
const int mod = 1e9+7;
double ans[maxn];
void init()
{
double lastL,lastR;
for(int i=4; i<maxn; i++)
{
if(i&1)
{
double d = max(lastL,lastR);
ans[i] = ans[i-1] + (d * 2 - 1) * 0.5;
}
else
{
lastL = i/4;
lastR = i/2 - lastL;
ans[i] = 2.0 * lastL * lastR;
}
}
}
int main()
{
ios::sync_with_stdio(false);
init();
int T;
cin>>T;
while(T--)
{
LL n;
cin>>n;
cout<<lower_bound(ans,ans+maxn,n)-ans<<endl;
}
return 0;
}
