这题很好做，使用二分法就OK。
首先在读取所有派的半径后处理出所有派的面积，并且记录最大的那个派的面积。然后从 0 ~ maxsize 二分枚举一下，就能得到答案。
此外，这道题最后输出保留小数位数可以是 3, 4, 5，都可以。
附AC代码：
1: #include <stdio.h>
2: #include <math.h>
3: #include <iostream>
4: #include <cstdarg>
5: #include <algorithm>
6: #include <string.h>
7: #include <stdlib.h>
8: #include <string>
9: #include <list>
10: #include <vector>
11: #include <map>
12: #define LL long long
13: #define M(a) memset(a, 0, sizeof(a))
14: using namespace std;
15: void Clean(int count, ...)
16: {17: va_list arg_ptr;
18: va_start (arg_ptr, count);
19: for (int i = 0; i < count; i++)
20: M(va_arg(arg_ptr, int*));
21: va_end(arg_ptr);
22: }
23:
24: const double PI = acos(-1.0);
25: double buf[10009];
26: int n, f;
27:
28: bool Deal(double size)
29: {30: int res = 0;
31: for (int i = 1; i <= n; i++)
32: res += floor(buf[i] / size);
33: return (res >= (f + 1));
34: }
35:
36: int main()
37: {38: int T;
39: scanf("%d", &T);40: while (T--)
41: {42: double m = -1;
43: scanf("%d%d", &n, &f);44: for (int i = 1; i <= n; i++)
45: {46: int r;
47: scanf("%d", &r);48: buf[i] = PI * r * r;
49: m = max(m, buf[i]);
50: }
51: double tmp = 0.0;
52: while (m - tmp > 1e-5)
53: {54: double p = (m + tmp) / 2.0;
55: if (Deal(p)) tmp = p;
56: else m = p;
57: }
58: printf("%.5lf\n", tmp);59: }
60: return 0;
61: }