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.
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).$|PA|=sqrt{2},|PB|=2sqrt{2},|PA|*|PB|=4$,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:
![](https://images2015.cnblogs.com/blog/733602/201601/733602-20160130224154224-1217847127.png)
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*m*2); } return 0; }