Running For Life
One day,poor Bob had a dream.In his dream,he was being chased by a mystery man.Bob was so afraid that he ran so fast than ever.However,Bob knows that as long as his shortest distance with the mystery man is larger than the safe distance,then he will be safe.
Suppose that Bob and the mystery man were in a perfect 2D plane.Bob’s initial location was (Xb,Yb),and the mystery man’s initial location was (Xm,Ym).Also,Bob’s speed was given by (Vxb,Vyb),and the mystery man’s speed was (Vxm,Vym),and they both keep their speed constant during the whole dream!
Luckily,Bob also know about his safe distance,which was sd.However,Bob forgot whether he was safe at last,so he asks you,the smart ACMer,for help.
Input
There are about 1000 test cases.Each case contains a line which describe nine integer:Xb,Yb,Xm,Ym,Vxb,Vyb,Vxm,Vym,sd,(-1e6<=Xb,Yb,Xm,Ym,Vxb,Vyb,Vxm,Vym<=1e6,0<=sd<=1e6)
Output
If Bob was safe at last,please output “Lucky Bob!”,else please output “Poor Bob!”(without quotations).
Sample Input
6 2 7 3 7 0 5 1 0
4 3 1 0 2 0 4 4 6
Sample Output
Lucky Bob!
Poor Bob!
思路：
由Bob和man的初始位置和常速度的方向可以在直角坐标系中，找出他们关于时间的位置坐标，构建距离的方程式并分类求最短距离，与安全距离比较即可
注意：
１、代码中算出来的最短距离没有开方，相应的安全距离也要平方，且数据类型为ｌｏｎｇ ｌｏｎｇ类
２、位置，速度的数据范围为－１ｅ６到１ｅ６，需要用ｌｏｎｇ ｌｏｎｇ类型存储ａ，ｂ，ｃ，且先除后乘
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
LL xb,yb,xm,ym,vxb,vyb,vxm,vym,sd;
while(cin>>xb>>yb>>xm>>ym>>vxb>>vyb>>vxm>>vym>>sd)
{
LL x=xm-xb;
LL y=ym-yb;
LL vx=vxm-vxb;
LL vy=vym-vyb;
LL a=vx*vx+vy*vy;
LL b=2*x*vx+2*y*vy;
LL c=x*x+y*y;
sd=sd*sd; //最短距离没有开方，安全距离需要平方
bool res;
if(a==0) //a,b,c分别为0时的分类
if(b>=0) if(c>sd) res=true;
else res=false;
else res=false;
else if(b>=0)
if(c>sd) res=true;
else res=false;
else
{
double s=c-(b/4.0/a)*b; //最好先除后乘
//cout<<"AAA"<<s<<endl;
if(s>sd) res=true;
else res=false;
}
if(res==true) cout<<"Lucky Bob!"<<endl;
else cout<<"Poor Bob!"<<endl;
}
return 0;
}