UVALive 3635 Pie(二分法)
简单的二分法应用,循环1000次精度就满足要求了。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<stack> #include<queue> #include<cctype> #include<sstream> using namespace std; #define pii pair<int,int> #define LL long long int const double eps=1e-10; const int INF=1000000000; const int maxn=10000+10; const double pi=acos(-1); int N,F,T; double r[maxn],maxr,s[maxn]; bool ok(double x) { int f=0; for(int i=0;i<N;i++) { f+=(int)(s[i]/x); if(f>=F) return true; } return false; } int main() { //freopen("in1.txt","r",stdin); //freopen("out.txt","w",stdout); scanf("%d",&T); while(T--) { scanf("%d%d",&N,&F); maxr=-1; F++; for(int i=0; i<N; i++) { scanf("%lf",&r[i]); maxr=max(r[i],maxr); s[i]=r[i]*r[i]*pi; } double L=0,R=maxr*maxr*pi; for(int i=0;i<1000;i++) { double m=(L+R)/2; if(ok(m)) L=m; else R=m; } printf("%.4lf\n",R); } //fclose(stdin); //fclose(stdout); return 0; }