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(-1);
int r[10010], t, n, m;

int main()
{
    scanf("%d", &t);
    while(t--)
    {
        double low = 0, high = 0;
        scanf("%d %d", &n, &m);
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &r[i]);
            high += PI * r[i] * r[i];
        }
        high /= m+1;
        while(high - low > 1e-6)
        {
            double mid = (high + low) / 2;
            int cnt = 0;
            for(int i = 0; i < n; i++)
            {
                cnt += PI * r[i] * r[i] / mid;
            }
            if(cnt >= m+1)
            {
                low = mid;
            }
            else
            {
                high = mid;
            }
        }
        printf("%.4f
", low);
    }
    return 0;
}