HDOJ 5605-geometry【数学】

客官°小女子只卖身不卖艺 2022-08-19 15:14 127阅读 0赞

# geometry #

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

Total Submission(s): 262 Accepted Submission(s): 195

Problem Description

There is a point  P![Image 1][] at coordinate  (x,y)![Image 1][].  
A line goes through the point, and intersects with the postive part of  X,Y![Image 1][] axes at point  A,B![Image 1][].  
Please calculate the minimum possible value of  |PA|∗|PB|![Image 1][].

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![Image 1][],  0<X,Y≤10000![Image 1][].

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) ][BestCoder Round _68 _div.2_]

解题思路：

题目大意就是 在平面直角坐标系上有一个点 PP , 他的坐标是 (x, y)(x,y) . 有一条直线 y = kx + by=kx\+b 经过了 PP , 且分别交 x, yx,y 正半轴于 A, BA,B . 求 |PA| \* |PB|∣PA∣∗∣PB∣ 的最小值.

利用数学化简将∣PA∣∗∣PB∣化简成只含x和y的表达式。只要利用数学中的均值不等式。

![Center][]

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

[Image 1]: 
[BestCoder Round _68 _div.2_]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2368+%28div.2%29&source=1&searchmode=source
[Center]: /images/20220731/7f7fe0705fe04187b11b97ebb001980d.png