思路：就是二分派的体积，但是要处理一下高精度。

#include
     
      #include
      
       #include
       
        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\n", R / 10000.0);    }}