HDU 2876 Ellipse, again and again（数学题）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2876

 

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Ellipse, again and again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 537    Accepted Submission(s): 200

Problem Description

There is an ellipse in the plane, its formula is  . We denote the focuses by F1 and F2. There is a point P in the plane. Draw a segment to associate the origin and P, which intersect the ellipse at point Q. Then draw a tangent of the ellipse which passes Q. Denote the distance from the center of the ellipse to the tangent by d. Calculate the value of  .

 

Input

The first line contains a positive integer n that indicates number of test cases.
And each test case contains a line with four integers. The value of parameters of the ellipse a, b(0<|a|,|b|<=100),and the coordinates x, y of P(|x|<=100,|y|<=100) are given successively.

 

Output

For each test case, output one line. If the given point P lies inside the given ellipse, print "In ellipse" otherwise print the value of d*d*QF1*QF2 rounded to the nearest integer.

 

Sample Input

11 1 1 1

 

Sample Output

1

 

Source

2009 Multi-University Training Contest 8 - Host by BJNU

 

Recommend

gaojie

  求距离！

#include <stdio.h>#include <math.h>int main(){double xQ,yQ;double k1;int a,b,x0,y0,T;while(~scanf("%d",&T))//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK{while(T--)//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK{scanf("%d%d%d%d",&a,&b,&x0,&y0);k1=y0/(x0*1.0);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKxQ=sqrt((a*a*b*b*1.0)/(a*a*k1*k1+b*b));//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKyQ=k1*xQ;int flag= 0;//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKint w=a*a-b*b;//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKif((x0*x0)/(a*a)+(y0*y0)/(b*b)<1)//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK{printf("In ellipse\n");//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKcontinue;}if(w < 0)//标记焦点所在轴{flag =1;w=-w;}double F1,F2;//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKdouble c =sqrt(w*1.0);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKif(flag == 0){ F1 = sqrt((xQ+c)*(xQ+c)+(yQ*yQ));//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK F2 = sqrt((xQ-c)*(xQ-c)+yQ*yQ);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK}else{ F1 = sqrt(xQ*xQ+(yQ+c)*(yQ+c));//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKF2 = sqrt(xQ*xQ+(yQ-c)*(yQ-c));//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK}double t = (sqrt)((xQ*xQ*b*b*b*b)*1.0+(yQ*yQ*a*a*a*a)*1.0);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKdouble d =a*a*b*b/(t*1.0);double D=d*d*F1*F2;//化简后D==a*a*b*b FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK D = a*a*b*b;printf("%.0lf\n",D);}}return 0;}

下面给出证明：

至此D=d*d*F1*F2可化简为D=a*a*b*b

所以给出简短代码：

#include<cstdio>int main(){int a,b,x0,y0;int T;while(~scanf("%d",&T)){while(T--){scanf("%d%d%d%d",&a,&b,&x0,&y0);if(x0*x0/(a*a)+y0*y0/(b*b)<1)printf("In ellipse\n");elseprintf("%d\n",a*a*b*b);}}return 0;}