派 二分水题

思路：就是二分派的体积，但是要处理一下高精度。

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
#define IP acos(-1.0)
const int maxn = 1e4 + 10;
#define ll long long
#define max(a, b) (a)>(b)?(a):(b)
int t, n, f;
ll a[maxn];
ll r, maxx;
bool ok(ll x){
    int ans = 0;
    for (int i = 0; i < n; ++i)
    {
        ans += a[i] / x;
        if (f + 1 <= ans)return 1;
    }
    return 0;
}

int main(){
    scanf("%d", &t);
    while (t--){
        maxx = 0;
        scanf("%d%d", &n, &f);
        for (int i = 0; i < n; ++i)
        {
            scanf("%lld", &r);    a[i] = IP*r*r * 10000;    maxx = max(maxx, IP*r*r*10000);
        }
        ll L = 0, R = maxx;
        while (R  >  L+1){
    //        cout << L << " " << R << endl;
            ll mid = (R + L) / 2.0;
            if (ok(mid))L = mid;  else R = mid;
        }
        printf("%.4lf
", R / 10000.0);
    }
}