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\n",r); ans+=(1.8637*r); printf("%f\n",ans); printf("%.7f\n",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
2020.7.25更
代码正解：
#include<bits/stdc++.h> using namespace std; int main() { double a，b; cin>>a; b=(2*sqrt(2)+2*sqrt(3)+2)*a/sqrt(2); b=b*b; printf("%.5f",b); return 0; }
