HDOJ 5605 geometry



geometry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 65    Accepted Submission(s): 49

Problem Description

There is a point P at coordinate (x,y).
A line goes through the point, and intersects with the postive part of X,Y axes at point A,B.
Please calculate the minimum possible value of |PA|∗|PB|.

 

Input

the first line contains a positive integer T,means the numbers of the test cases.

the next T lines there are two positive integers X,Y,means the coordinates of P.

T=500,0<X,Y≤10000.

 

Output

T lines,each line contains a number,means the answer to each test case.

 

Sample Input

12 1

 

Sample Output

4in the sample $P(2,1)$,we make the line $y=-x+3$,which intersects the positive axis of $X,Y$ at (3,0),(0,3).$|PA|=\sqrt{2},|PB|=2\sqrt{2},|PA|*|PB|=4$,the answer is checked to be the best answer.

 

题意：直接给中文题面链接吧：点击打开链接

题解：我是凭直觉判断出k=1时是正解的，化了两组数，就直接写了。

贴出BC上给的证明吧：

只用计算2*x*y就行了，我写的啰嗦，代码如下：

#include<cstdio>#include<cstring>#include<cmath>int main(){int x,y,b,t,x1,x2,y1,y2;scanf("%d",&t);while(t--){scanf("%d%d",&x,&y);b=y-x;y2=b; x2=0;y1=0; x1=-b;double ans=sqrt((x1-x)*(x1-x)*1.0+(y1-y)*(y1-y)*1.0)*sqrt((x2-x)*(x2-x)*1.0+(y2-y)*(y2-y)*1.0);        printf("%.0f\n",ans); }return 0;}