Circles Inside a Square（几何题）

题目描述

You have 8 circles of equal size and you want to pack them inside a square. You want to minimize the size of the square. The following figure illustrates the minimum way of packing 8 circles inside a square:

Given the radius, r, find the area of the minimum square into which 8 circles of that radius can be packed.

输入

There is one input line, it consists of a positive real number (between 0.001 and 1000, inclusive) denoting the radius, r.

输出

Print the area of the minimum square where 8 circles of radius r can be packed. Print 5 digits after the decimal. Your output is considered correct if it is within ±0.00001 of the judge’s output.

样例输入

0.1

样例输出

0.34383

题意：
将八个圆放到一个正方形中，给出圆的半径求出正方形的面积
相信很多人都会去推公式小声说我也去推了公式 ，后来采用了偷工减料的方法，根据给出的数据推出可能会成立的结论，没想到还真过了[手动滑稽]

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
/**bool prime(int n)
{
    if(n==1||n==1)
        return false;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0) return false;
    }
    return true;
}
int num[maxn],n,cnt,judge[maxn];
void shai()
{
    for(int i=2;i<=maxn;i++)
    {
        if(!judge[i])
            num[++cnt]=i;
        for(int j=1;j<=cnt&&num[j]*i<=maxn;j++)
        {
            judge[i*num[j]]=1;
            if(i%num[j]==0)
                break;
        }
    }
}**/
start{
    ///
    double r;
    scanf("%lf",&r);
    double ans=0;
    ans+=r*4;
    printf("%f
",r);
    ans+=(1.8637*r);
    printf("%f
",ans);
    printf("%.7f
",ans*ans);
	end;
}

解释一下这里的1.8367的来源：
r==0.1时的结果0.34383进行开方，减去四倍的半径就是0.18637然后目前得到的这个数就是比较难球求的中间部分的长度，它必然和半径有关，而且这个关系肯定是成特定比例的因此呢，中间难求的部分和2r的关系通过这个关系来解决0.18637/（2r）等于1.8637
上面给出的代码有测试输出