LA3635 Pie (二分法)

Posted 2020-08-08

链接：http://vjudge.net/problem/33697

分析：二分答案。假设每个人得到派的最大面积是x，判断是否能满足F+1个人（因为派是不可以拼起来的，所以一个半径为r的派只能切出[πr²/x]个派，其他部分就浪费了）。

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 const double PI = acos(-1.0);
 7 const int maxn = 10000 + 5;
 8 
 9 int n, f;
10 double A[maxn];
11 
12 bool ok(double area) {
13     int sum = 0;
14     for (int i = 0; i < n; i++) sum += floor(A[i] / area);
15     return sum >= f + 1;
16 }
17 
18 int main() {
19     int T;
20     scanf("%d", &T);
21     while (T--) {
22         scanf("%d%d", &n, &f);
23         double maxa = -1;
24         for (int i = 0; i < n; i++) {
25             int r; scanf("%d", &r);
26             A[i] = PI * r * r;
27             maxa = max(maxa, A[i]);
28         }
29         double L = 0, R = maxa;
30         while (R - L > 1e-5) {
31             double M = (L + R) / 2;
32             if (ok(M)) L = M;
33             else R = M;
34         }
35         printf("%.4lf\n", L);
36     }
37     return 0;
38 }