poj3122 二分答案+贪心

My birthday is coming up and traditionally I’m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

#include<stdio.h>const double pi=3.1415926535897932;double sum;int cnt,n,f,T;double p[10005];inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int read(){    register int x=0,f=1;    register char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f*=-1;ch=getchar();}    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();    return f*x;}int main(){    T=read();    while(T--){       sum=0,n=read(),f=read();       f++;       for(register int i=1;i<=n;i++){          p[i]=read();          p[i]=p[i]*p[i];          sum+=p[i];       }       double lf=0.0,rg=sum/f,mid;       while(rg-lf>0.00001){             mid=(lf+rg)/2;             cnt=0;             for(register int i=1;i<=n;i++)                cnt+=(int)(p[i] / mid);             if(cnt>=f) lf=mid;             else rg=mid;       }       printf("%0.4lf\n",mid*pi);    }}