POJ 3122 Pie 二分枚举

题目:http://poj.org/problem?id=3122

这个题就好多了,没有恶心的精度问题,所以1A了。。

 1 #include <stdio.h>
 2 #include <math.h>
 3 
 4 const double PI = acos(-1);
 5 int r[10010], t, n, m;
 6 
 7 int main()
 8 {
 9     scanf("%d", &t);
10     while(t--)
11     {
12         double low = 0, high = 0;
13         scanf("%d %d", &n, &m);
14         for(int i = 0; i < n; i++)
15         {
16             scanf("%d", &r[i]);
17             high += PI * r[i] * r[i];
18         }
19         high /= m+1;
20         while(high - low > 1e-6)
21         {
22             double mid = (high + low) / 2;
23             int cnt = 0;
24             for(int i = 0; i < n; i++)
25             {
26                 cnt += PI * r[i] * r[i] / mid;
27             }
28             if(cnt >= m+1)
29             {
30                 low = mid;
31             }
32             else
33             {
34                 high = mid;
35             }
36         }
37         printf("%.4f\n", low);
38     }
39     return 0;
40 }
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posted @ 2013-10-25 20:01  Anti-Magic  阅读(147)  评论(0编辑  收藏  举报