geometry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 340 Accepted Submission(s): 256
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|.
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
1
2 1
Sample Output
4
in 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：给你一点p的坐标（x,y）现在一条直线过p点分别与X，Y正半轴交于点A，B，求PA*PB最小值
题解：
如图：PA=AC/sinθ, PB=OC/sinθ 所以PA*PB=(AC*OC)/(sinθ*sinθ) 因为AC=PC*tanθ PC=x，OC=y所以 PA*PB=(x*y)/(sinθ*cosθ)
所以当sinθ*cosθ最大时结果最小，此时θ等于45度所以PA*PB=x*y*2
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 1010
#define mod 10007
using namespace std;
int main()
{
int n,m,j,i,s,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
printf("%d\n",n*m*2);
}
return 0;
}