zoj3203(三分)

最新推荐文章于 2019-04-07 14:02:11 发布

原创最新推荐文章于 2019-04-07 14:02:11 发布 · 495 阅读

0 ·

CC 4.0 BY-SA版权

二分专栏收录该内容

5 篇文章

订阅专栏

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3203

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
double H,h,D;
double cal(double x){
    return ( h*D - H*x )/(D- x) + x;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
       scanf("%lf %lf%lf",&H,&h,&D);
       double mid , midmid;
       double low = 0 ,high = D * (h/H),esp =1e-6;
       while(high - low >= esp){
           mid = (low+ high) / 2;
           midmid =(high + mid) / 2;
           doublecmid = cal(mid);
           doublecmidmid = cal(midmid);
           if(cmid> cmidmid)
              high = midmid;
           else
              low = mid;
       }
       printf("%.3lf\n",cal(low));
      //printf("%.3lf\n",cal(D-sqrt((H-h)*D)));直接求导求出极值点可证明
    }
    return 0;
}