杭电5605 geometry

geometry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 205    Accepted Submission(s): 155

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.

 

Source

BestCoder Round #68 (div.2)

 

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题意就是过一点p的直线与x轴y轴正方向分别交于A,B，求pA于pB的乘积 的最小值。

设一个斜率k可以得到Ａ，Ｂ两个点，然后用基本不等式去解，可以解得ｋ衡等于－１的，长度即为2*x*y；

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int p,m,n;int main(){scanf("%d",&p);while(p--){scanf("%d%d",&m,&n);printf("%d\n",2*m*n);}}