UVA - 12097 Pie

题意：有F+1个人来分N个圆派，每个人得到的必须是一整块派，而不是几块拼在一起，且面积要相同，求每个人最多能得到多大面积的派（不必是圆形）

思路：二分答案，因为派是不能拼起来的，所以一个半径r的派只能切出[pir^2/x]个派。向下取整，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 10005;
const double PI = acos(-1.0);

int n,f;
double A[MAXN];

bool ok(double area){
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += floor(A[i]/area);
    return sum >= f+1;
}

int main(){
    int t;
    scanf("%d",&t);
    while (t--){
        scanf("%d%d",&n,&f);
        double Max = -1;
        for (int i = 0; i < n; i++){
            int r;
            scanf("%d",&r);
            A[i] = PI*r*r;
            Max = max(Max,A[i]);
        }
        double L = 0,R = Max;
        while (R - L > 1e-5){
            double M = (L+R)/2;
            if (ok(M))
                L = M;
            else R = M;
        }
        printf("%.4lf\n",L);
    }
    return 0;
}