POJ 3122 （集训比赛2B_B题）解题报告

#include <cstdio>
#include <cmath>

int N =0;
int F=0;
double pim =0;
double sum=0;
const int MAXN =10000+10;
double r[MAXN];
double pie[MAXN];

const double PI =acos(-1.0);
const double eps=1e-6;

int jud(double mid){
    int count=0;
    for(int i=0;i<N;i++){
        count+=(int)(pie[i]/mid);
    }
    if(count>=(F+1))    return 1;    
    else return  0;
}
double proce(){
    double left =pim/(F+1);
    double right=sum/(F+1);
    double mid= (left+right)/2;
    while(right-left>eps){
        mid= (left+right)/2;
        if (jud(mid)){
            left = mid;
        }else{
            right =mid;
        }
    }

    return mid;
}

int main(void){
    int T=0;
    scanf("%d",&T);

    for(int i=0;i<T;i++){
        pim=0;
        sum=0;
        scanf("%d %d",&N,&F);
        for(int j=0;j<N;j++){
            scanf("%lf",&r[j]);
            pie[j]=PI*r[j]*r[j];
            if(pie[j]>pim)    pim=pie[j];
            sum+=pie[j];
        }
        double ans = proce();
        printf("%.4f\n",ans);
    }
    
        
    return 0;
}