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。。

 #include <stdio.h>
 #include <math.h>

 const double PI = acos(-);
 int r[], t, n, m;

 int main()
 {
     scanf("%d", &t);
     while(t--)
     {
         double low = , high = ;
         scanf("%d %d", &n, &m);
         for(int i = ; i < n; i++)
         {
             scanf("%d", &r[i]);
             high += PI * r[i] * r[i];
         }
         high /= m+;
         while(high - low > 1e-)
         {
             double mid = (high + low) / ;
             int cnt = ;
             for(int i = ; i < n; i++)
             {
                 cnt += PI * r[i] * r[i] / mid;
             }
             if(cnt >= m+)
             {
                 low = mid;
             }
             else
             {
                 high = mid;
             }
         }
         printf("%.4f\n", low);
     }
     return ;
 }