UVALive - 3635 - Pie（二分）

题意：有F + 1（1 <= F <= 10000）个人分N（1 <= N <= 10000）个圆形派，每个人得到的派面积相同，且必须是一整块（不能够两个甚至多个派拼在一起），求每个人最多能得到多大面积的派。（误差最多到0.001）

因为答案是小数类型的，并且N高达10000，故不可暴力枚举。

可以二分枚举最大面积，然后检查是否切出来派的总个数大于等于F + 1。

（判相等时不可直接判相等，需要加精度控制）

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {-1, 1, 0, 0};
const ll MOD = 1e9 + 7;
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;  

int T, n, f;
double a[MAXT];  

bool judge(double area){
    int sum = 0;
    for(int i = 0; i < n; ++i)  sum += int(a[i] / area);
    return sum >= f;
}  

int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &f);
        for(int i = 0; i < n; ++i){
            scanf("%lf", a + i);
            a[i] = a[i] * a[i] * pi;
        }
        ++f;
        double l = 0.0, r = *max_element(a, a + n);
        while(l + EPS < r){
            double mid = (l + r) / 2;
            if(judge(mid))  l = mid;
            else  r = mid;
        }
        printf("%.4lf\n", l);
    }
    return 0;
}