解题思路：二分
1 /************************************************************************** 2 user_id: SCNU20102200088 3 problem_id: uvalive 3635 4 problem_name: Pie 5 **************************************************************************/ 6 7 #include <algorithm> 8 #include <iostream> 9 #include <iterator> 10 #include <iomanip> 11 #include <sstream> 12 #include <fstream> 13 #include <cstring> 14 #include <cstdlib> 15 #include <climits> 16 #include <bitset> 17 #include <string> 18 #include <vector> 19 #include <cstdio> 20 #include <cctype> 21 #include <ctime> 22 #include <cmath> 23 #include <queue> 24 #include <stack> 25 #include <list> 26 #include <set> 27 #include <map> 28 using namespace std; 29 30 //线段树 31 #define lson l,m,rt<<1 32 #define rson m+1,r,rt<<1|1 33 34 //手工扩展栈 35 #pragma comment(linker,"/STACK:102400000,102400000") 36 37 const double EPS=1e-5; 38 const double PI=acos(-1.0); 39 const double E=2.7182818284590452353602874713526; //自然对数底数 40 const double R=0.5772156649015328606065120900824; //欧拉常数:(1+1/2+...+1/n)-ln(n) 41 42 const int x4[]={-1,0,1,0}; 43 const int y4[]={0,1,0,-1}; 44 const int x8[]={-1,-1,0,1,1,1,0,-1}; 45 const int y8[]={0,1,1,1,0,-1,-1,-1}; 46 47 typedef long long LL; 48 49 typedef int T; 50 T max(T a,T b){ return a>b? a:b; } 51 T min(T a,T b){ return a<b? a:b; } 52 T gcd(T a,T b){ return b==0? a:gcd(b,a%b); } 53 T lcm(T a,T b){ return a/gcd(a,b)*b; } 54 55 /////////////////////////////////////////////////////////////////////////// 56 //Add Code: 57 int n,f,r[10005]; 58 double s[10005]; 59 60 int cal(double x){ 61 int ret=0; 62 for(int i=1;i<=n;i++) ret+=floor(s[i]/x); 63 return ret; 64 } 65 66 double erfen(){ 67 double l=0,r=1e9; 68 while(r-l>EPS){ 69 double m=(l+r)/2; 70 if(f+1<=cal(m)) l=m; 71 else r=m; 72 } 73 return l; 74 } 75 /////////////////////////////////////////////////////////////////////////// 76 77 int main(){ 78 std::ios::sync_with_stdio(false); 79 //freopen("in.txt","r",stdin); 80 //freopen("out.txt","w",stdout); 81 /////////////////////////////////////////////////////////////////////// 82 //Add Code: 83 int Case,i; 84 scanf("%d",&Case); 85 while(Case--){ 86 scanf("%d%d",&n,&f); 87 for(i=1;i<=n;i++){ 88 scanf("%d",&r[i]); 89 s[i]=PI*r[i]*r[i]; 90 } 91 double ans=erfen(); 92 printf("%.4lf ",ans); 93 } 94 /////////////////////////////////////////////////////////////////////// 95 return 0; 96 } 97 98 /************************************************************************** 99 Testcase: 100 Input: 101 3 102 3 3 103 4 3 3 104 1 24 105 5 106 10 5 107 1 4 2 3 4 5 6 5 4 2 108 Output: 109 25.1327 110 3.1416 111 50.2655 112 **************************************************************************/