HDU6154——找规律

传送门：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): 809    Accepted Submission(s): 463

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：输入需要拼成的面积大小s，求出最少需要多少条边可以拼成大于等于面积s的图形，每条边长度为1或者根号2

思路：下面这张图是从其他大神博客中借鉴的，这样看起来更加形象。在当时，理解题意上没有注意到可以大于所给面积这一个很致命的错误。其次思维过于局限没有想到图形的扩展，死想规律然后想到死了。。。

从图形入手，其实画法是有规律的，就是以4边5边这两种做扩展，比如6边就是在4边的中间加两条边，7边是在5边的基础上加两条对角线

可以根据4为周期(也可以分成奇偶)推出公式，然后二分找答案

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
typedef long long ll;
using namespace std;
int a[200005]={0,0,0,0};
int main(){
    int j,n,T;
    ll i;///会超int
    for(i=4;i<=100005;i++){
        if(i%4==0) a[i]=(i*i)/8;
        if(i%4==1) a[i]=a[i-1]+i/4-0.5;
        if(i%4==2) a[i]=(i/4)*(i/4+1)*2;
        if(i%4==3) a[i]=a[i-1]+i/4+0.5;
    }
    scanf("%d",&T);
    while(T--){
        ll s,ans;
        scanf("%lld",&s);
        ans=lower_bound(a,a+100000,s)-a;
        printf("%lld\n",ans);
    }
    return 0;
}