2016京东笔试编程题

2016京东笔试编程题

题意如下：

The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered.

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x,  y,  r, where (x, y) are the coordinates of the stadium's center ( -  103 ≤ x,  y ≤ 103), and r (1 ≤ r  ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line.

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

解题思路：

其实很容易列出两个二元二次方程。可以直接通过求解该方程。只是比较复杂而已。可以参见如下程序：

//
//  main.cpp
//  hiho_110
//
//  Created by chenalong on 8/10/16.
//  Copyright (c) 2016 chenalong-Arc_Point_2014_hunan. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <vector>
#include <map>
#include <cmath>
#include <set>
#include <queue>

using namespace std;

#define eps 1e-6

struct node
{
    double val,x_val,x_2_val;
};

// 分别表示 常数项，一次项，二次项系数
void init_node(node& tmp)
{
    tmp.val = 0;
    tmp.x_2_val = 0;
    tmp.x_val = 0;
}

// 把 lamba(x - constant) ^ 2 相关系数加入到统计量中去
void add_val(node& tmp,int lamba,int constant)
{
    tmp.val += 1.0 * lamba * constant * constant;
    tmp.x_val += 1.0 * -lamba * 2 * constant;
    tmp.x_2_val += lamba;
}

// 每个系数都乘以一个常数
void multi_val(node& tmp, int constant)
{
    tmp.val *= constant;
    tmp.x_2_val *= constant;
    tmp.x_val *= constant;
}

// 两个方程相减，就是相对应的系数相减
void sub_val(node& tmp1,node& tmp2)
{
    tmp1.val -= tmp2.val;
    tmp1.x_val -= tmp2.x_val;
    tmp1.x_2_val -= tmp2.x_2_val;
}

// 程序的输入，表示圆心坐标和半径大小
int x[4],y[4],r[4];

//解二元一次方程
bool solve_quadratic(double a,double b,double c,double& answer1,double& answer2)
{
    if(fabs(a) < eps)
    {
        answer2 = answer1 = -c / b;
        return true;
    }
    else
    {
        double sq = b * b - 4 * a * c;
        if(sq < -eps)
            return false;
        answer1 = (-b + sqrt(sq)) / 2 / a;
        answer2 = (-b - sqrt(sq)) / 2 / a;
        return true;
    }
}

// 计算一个数的平方
double square(double a)
{
    return a * a;
}

// 判断得到的解是否正确
void verify(double x1,double y1)
{
    for(int i= 1;i<=3;i++)
    {
        cout << sqrt(square(x1 - x[i]) + square(y1 - y[i])) << endl;
    }
}


// 判断输出那个点
void output(double x1,double y1,double x2,double y2)
{
    if(square(x1 - x[1]) + square(y1 - y[1]) < square(x2-x[1]) + square(y2-x[1]))
    {
        printf("%.5lf %.5lf\n",x1,y1);
        //verify(x1,y1);
    }
    else
    {
        printf("%.5lf %.5lf\n",x2,y2);
        //verify(x2, y2);
    }
}

// 寻找满足条件的点
void solve(node& a1,node& b1,node& a2, node& b2)
{
    // 两个解
    double x1,y1,x2,y2;
    //二次方程的系数
    double a,b,c;
    //标志是否有解
    bool flag = true;
    
    if(fabs(a2.x_val) <eps)
    {
        y1 = y2 = (a2.val + b2.val) * 1.0/ (-b2.x_val);
        
        a = a1.x_2_val;
        b = a1.x_val;
        c = 0.0 + a1.val + b1.x_2_val * y1 * y1 + b1.x_val * y1 + b1.val;
        
        flag = solve_quadratic(a, b, c, x1, x2);
    }
    else if(fabs(b2.x_val) <eps)
    {
        x1 = x2 = (a2.val + b2.val) * 1.0/ (-a2.x_val);
        
        a = b1.x_2_val;
        b = b1.x_val;
        c = 0.0 + b1.val + a1.x_2_val * x1 * x1 + a1.x_val * x1 + a1.val;
        
        flag = solve_quadratic(a, b, c, y1, y2);
    }
    else
    {
        double constant = a2.val + b2.val;
        double coff1 = -constant / b2.x_val;
        double coff2 = -a2.x_val / b2.x_val;
        
        a = 0.0 + a1.x_2_val + b1.x_2_val * coff2 * coff2;
        b = 0.0 + a1.x_val + b1.x_2_val * (2 * coff1 * coff2) + b1.x_val * coff2;
        c = 0.0 + a1.val + b1.val + b1.x_2_val *(coff1 * coff1) + b1.x_val * coff1;
        
        flag = solve_quadratic(a, b, c, x1, x2);
        
        y1 = coff1 + coff2 * x1;
        y2 = coff1 + coff2 * x2;
    }
    
    if(flag)
    {
        output(x1,y1,x2,y2);
    }
}

int main()
{
    
    for(int i = 1;i<=3;i++)
        cin >> x[i] >> y[i] >> r[i];
    
    for(int i = 1;i<4;i++)
        r[i] *= r[i];

    node a1,b1,a2,b2;
    init_node(a1);
    init_node(a2);
    init_node(b1);
    init_node(b2);
    
    // 根据距离关系构建两个二元二次方程
    add_val(a1,r[2],x[1]);
    add_val(a1,-r[1],x[2]);
    
    add_val(b1,r[2],y[1]);
    add_val(b1,-r[1],y[2]);
    
    add_val(a2,r[3],x[1]);
    add_val(a2,-r[1],x[3]);
    
    add_val(b2,r[3],y[1]);
    add_val(b2,-r[1],y[3]);
    
    //根据不同条件消除二次项，把 y 用 x 来表示
    if(r[3] == r[1])
    {
        solve(a1,b1,a2,b2);
    }
    else if(r[2] == r[1])
    {
        solve(a2,b2,a1,b1);
    }
    else
    {
        // 两个方程相减消除平方项
        multi_val(a1, r[3] - r[1]);
        multi_val(b1, r[3] - r[1]);
        
        multi_val(a2, r[2] - r[1]);
        multi_val(b2, r[2] - r[1]);
        
        sub_val(a1,a2);
        sub_val(b1,b2);
        
        solve(a2,b2,a1,b1);
    }
    
    return 0;
}