HDU5605——数学题(三角函数的应用)

来源：互联网发布：mac qq五笔拼音输入法编辑：程序博客网时间：2024/05/29 10:58

题目描述：

geometry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 356 Accepted Submission(s): 269

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：

在平面直角坐标系上有一个点P, 他的坐标是(x,y). 有一条直线y=kx+b经过了P, 且分别交x,y正半轴于A,B. 求∣PA∣∗∣PB∣的最小值.

解析：

这道题是典型的数学题，由于题设给出的条件比较少，我们可以想到用三角函数求解

假设直线与x轴负半轴的夹角为a（0，π/2），那么则有

PA = x/cosa,PB = y/sina;

则PA*PB = xy/(sina*cosa) = 2xy/sin2a;

而在此区间上sin2a的值域为(0,1];

所以PA*PB的最小值为2xy

完整代码实现：

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

总结：这道题在比赛的时候，因为看错了题目，还误打误撞的AC了，后面发现其实自己是想错了，这种高中的几何题也要注意，不能学过的就忘了，也要注意与三角函数之间的联系，会推导一些基本的公式。

如有错误，还请指正，O(∩_∩)O谢谢

0 0