HDU 2876 Ellipse, again and again
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2876
Ellipse, again and againTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
Sample Output
Source
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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);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
xQ=sqrt((a*a*b*b*1.0)/(a*a*k1*k1+b*b));//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
yQ=k1*xQ;int flag= 0;//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
int w=a*a-b*b;//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
if((x0*x0)/(a*a)+(y0*y0)/(b*b)<1)//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
{
printf("In ellipse\n");//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
continue;
}
if(w < 0)//标记焦点所在轴
{
flag =1;
w=-w;
}
double F1,F2;//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
double c =sqrt(w*1.0);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
if(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));//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
F2 = 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);//FUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCKFUCK
double 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;
}
以下给出证明:
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");
else
printf("%d\n",a*a*b*b);
}
}
return 0;
}
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