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  • SDUT 2603 Rescue The Princess

    Rescue The Princess

    Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^

    题目描述

        Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue the princess, a prince who wanted to marry the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.

        Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x1,y1) and B(x2,y2). The third coordinate C(x3,y3) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x1,y1) and B(x2,y2), but he doesn’t know where the C(x3,y3) is. The prince need your help. Can you calculate the C(x3,y3) and tell him?

    输入

        The first line is an integer T(1 <= T <= 100) which is the number of test cases. T test cases follow. Each test case contains two coordinates A(x1,y1) and B(x2,y2), described by four floating-point numbers x1, y1, x2, y2 ( |x1|, |y1|, |x2|, |y2| <= 1000.0).
        Please notice that A(x1,y1) and B(x2,y2) and C(x3,y3) are in an anticlockwise direction from the equilateral triangle. And coordinates A(x1,y1) and B(x2,y2) are given by anticlockwise.

    输出

        For each test case, you should output the coordinate of C(x3,y3), the result should be rounded to 2 decimal places in a line.

    示例输入

    4
    -100.00 0.00 0.00 0.00
    0.00 0.00 0.00 100.00
    0.00 0.00 100.00 100.00
    1.00 0.00 1.866 0.50

    示例输出

    (-50.00,86.60)
    (-86.60,50.00)
    (-36.60,136.60)
    (1.00,1.00)

    提示

     

    来源

    2013年山东省第四届ACM大学生程序设计竞赛

    示例程序

     

    atan 和 atan2 都是求反正切函数,如:有两个点 point(x1,y1), 和 point(x2,y2);

    那么这两个点形成的斜率的角度计算方法分别是:

    float angle = atan( (y2-y1)/(x2-x1) );

    float angle = atan2( y2-y1, x2-x1 );

     atan 和 atan2 区别:

    1:参数的填写方式不同;

    2:atan2 的优点在于 如果 x2-x1等于0 依然可以计算,但是atan函数就会导致程序出错;

    结论: atan 和 atan2函数,建议用 atan2函数;

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    
    using namespace std;
    
    #define PI acos(-1.0)
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        int t;
        scanf("%d",&t);
        double a,b,a1,b1,a2,b2;
        while(t--){
            scanf("%lf%lf%lf%lf",&a1,&b1,&a2,&b2);
            double degree=atan2(b2-b1,a2-a1);
            double len=sqrt((b2-b1)*(b2-b1)+(a2-a1)*(a2-a1));
            a=a1+len*cos(degree+PI/3);
            b=b1+len*sin(degree+PI/3);
            printf("(%.2lf,%.2lf)
    ",a,b);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jackge/p/3142534.html
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