題目描述

? ??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?

輸入

? ? 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.

輸出

? ? 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.

示例輸入

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

示例輸出

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

提示

來源

2013年山東省第四屆ACM大學生程序設計競賽
分兩種：一種是A點B點的x坐標相同，二種：X坐標不相同。 #include<stdio.h> #include<math.h> int main() {double b[2],c[3],a,x[3],y[3],tx[2],ty[2],k,bb,edglen;int t;scanf("%d",&t);while(t--){scanf("%lf%lf%lf%lf",&x[0],&y[0],&x[1],&y[1]);if(x[0]==x[1]){edglen=sqrt(pow(x[0]-x[1],2.0)+pow(y[0]-y[1],2.0));tx[0]=x[0]+sqrt(3.0)/2*edglen;tx[1]=x[0]-sqrt(3.0)/2*edglen;if(y[1]>y[0]){ty[0]=y[0]+edglen/2;printf("(%.2lf,%.2lf)\n",tx[1],ty[0]);}else{ty[0]=y[1]+edglen/2;printf("(%.2lf,%.2lf)\n",tx[0],ty[0]);}continue;}c[0]=(x[0]*x[0]-x[1]*x[1]+y[0]*y[0]-y[1]*y[1])/(2*x[0]-2*x[1]);b[0]=-(y[0]-y[1])/(x[0]-x[1]);a=b[0]*b[0]+1; b[1]=2*(c[0]*b[0]-x[1]*b[0]-y[1]);c[1]=x[0]*x[0]-2*x[0]*x[1]+y[0]*y[0]-2*y[0]*y[1];c[2]=c[0]*c[0]-2*c[0]*x[1]-c[1];k=-b[0]; bb=y[0]-k*x[0];ty[0]=(-b[1]+sqrt(b[1]*b[1]-4*a*c[2]))/(2*a);tx[0]=c[0]+b[0]*ty[0];ty[1]=(-b[1]-sqrt(b[1]*b[1]-4*a*c[2]))/(2*a);tx[1]=c[0]+b[0]*ty[1];if(x[0]<x[1]){if(ty[0]-k*tx[0]-bb>0){x[2]=tx[0];y[2]=ty[0];}else{x[2]=tx[1];y[2]=ty[1];}}else if(x[0]>x[1]){if(ty[0]-k*tx[0]-bb<0){x[2]=tx[0];y[2]=ty[0];}else{x[2]=tx[1];y[2]=ty[1];}}printf("(%.2lf,%.2lf)\n",x[2],y[2]);}}

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