SPOJ BLMIRINA 计算几何

最新推荐文章于 2018-01-01 13:28:40 发布

black_miracle

最新推荐文章于 2018-01-01 13:28:40 发布

阅读量615

点赞数

CC 4.0 BY-SA版权

分类专栏：每日一题计算几何

本文链接：https://blog.youkuaiyun.com/black_miracle/article/details/60591085

每日一题同时被 2 个专栏收录

51 篇文章

订阅专栏

计算几何

9 篇文章

订阅专栏

本文介绍了一个几何问题，即从原点出发的弓箭如何击中以(x,y)为中心，半径为r的圆上的最远点。通过二分查找算法确定最佳射击位置，并提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Archery Training

#geometry

Mirana is an archer with superpower. Every arrow she shoots will get stronger the further it travels. Mirana is currently on the way to warzone.

Since the road is still a long way, Mirana remembers about when she's still in training. In each of her training, Mirana stands on the (0,0) point in a cartesian scale. From that point, she must shoot a circle centered in (x,y) with radius r. Everything happens in z=0.

To maximize the arrow's power, Mirana must shoot the furthest point of the enemy possible. Her arrow travels at the speed of light and will instantly stops the moment it touches the target. On the target, determine the coordinate point that Mirana has to shoot to get maximum power. If multiple coordinate exists, choose the one with the lower x value.

Input

First line is T, number of training (T < 100000). Next T lines each contains 3 space separeted integers x, y, and r for each training (1 < r < x,y < 1000)

Output

For each training, output a line containing the coordinate of the arrow's destination separated by space. Output until 6 digit after decimal point.

Example

Input:
3
1 1 1
2 2 1
4 5 2

Output:
0.000000 1.000000
1.088562 2.411438
2.126155 5.699076

题意：从原点发射一颗子弹，当子弹遇到障碍物的时候就会停下。
障碍物为一个圆心为(x, y), 半径为r的圆。
问子弹停下后，距离原点最远的位置坐标是多少，多个时答案输出x坐标最小的那个。

题解：最远的点，那肯定是切线。

我们可以二分x来确定r

判断条件是当前(x,y)与r的距离来二分x

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<deque>
#include<map>
#include<cmath>
using namespace std;
typedef long long ll;
int main(){  
    int t;
    scanf("%d",&t);
    while(t--){
    	double x,y,d;
    	scanf("%lf%lf%lf",&x,&y,&d);
        double l=0,r=x;
        double dis=x*x+y*y-d*d;
        for(int i=1;i<=100;i++){
            double m=(l+r)/2;
            if((m-x)*(m-x)+(sqrt(dis-m*m)-y)*(sqrt(dis-m*m)-y)>d*d)l=m;
            else r=m;
        }
        printf("%.6f %.6f\n",l,sqrt(dis-l*l));
    }
    return 0;  
}