HDU 5605 geometry

geometry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 516 Accepted Submission(s): 399



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

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.

Source

BestCoder Round #68 (div.2)

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题意：给你一个坐标p（x,y),一条直线穿过该点，分别与x轴y轴交于一点a,b,问|pa|*|pb|的最小值；

思路：貌似求最值得都需要用到基本不等式。设该直线的斜率为k,用k表示出与x轴y轴的交点坐标，再利用基本不等式。最后会发现其实最小值就是k为-1，最小值为2xy..

#include<stdio.h>
int main()
{
int t,x,y;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&x,&y);
printf("%d\n",2*x*y);
}
return 0;
}