2014
02-27

# Buried memory

Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather not happened.
The world king Sconbin is not the exception.One day,Sconbin was sleeping,then swakened by one nightmare.It turned out that his love letters to Dufein were made public in his dream.These foolish letters might ruin his throne.Sconbin decided to destroy the letters by the military exercises’s opportunity.The missile is the best weapon.Considered the execution of the missile,Sconbin chose to use one missile with the minimum destruction.
Sconbin had writen N letters to Dufein, she buried these letters on different places.Sconbin got the places by difficult,he wants to know where is the best place launch the missile,and the smallest radius of the burst area. Let’s help Sconbin to get the award.

There are many test cases.Each case consists of a positive integer N(N<500,^V^,our great king might be a considerate lover) on a line followed by N lines giving the coordinates of N letters.Each coordinates have two numbers,x coordinate and y coordinate.N=0 is the end of the input file.

There are many test cases.Each case consists of a positive integer N(N<500,^V^,our great king might be a considerate lover) on a line followed by N lines giving the coordinates of N letters.Each coordinates have two numbers,x coordinate and y coordinate.N=0 is the end of the input file.

3
1.00 1.00
2.00 2.00
3.00 3.00
0

2.00 2.00 1.41

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const double eps=1e-8;
struct Point{
double x,y;
}p[505];
double dis(const Point &a,const Point &b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
Point circumcenter(const Point &a,const Point &b,const Point &c)
{ //返回三角形的外心
Point ret;
double a1=b.x-a.x,b1=b.y-a.y,c1=(a1*a1+b1*b1)/2;
double a2=c.x-a.x,b2=c.y-a.y,c2=(a2*a2+b2*b2)/2;
double d=a1*b2-a2*b1;
ret.x=a.x+(c1*b2-c2*b1)/d;
ret.y=a.y+(a1*c2-a2*c1)/d;
return ret;
}
void min_cover_circle(Point *p,int n,Point &c,double &r){ //c为圆心，r为半径
random_shuffle(p,p+n); //
c=p[0]; r=0;
for(int i=1;i<n;i++)
{
if(dis(p[i],c)>r+eps)  //第一个点
{
c=p[i]; r=0;
for(int j=0;j<i;j++)
if(dis(p[j],c)>r+eps) //第二个点
{
c.x=(p[i].x+p[j].x)/2;
c.y=(p[i].y+p[j].y)/2;
r=dis(p[j],c);
for(int k=0;k<j;k++)
if(dis(p[k],c)>r+eps) //第三个点
{//求外接圆圆心，三点必不共线
c=circumcenter(p[i],p[j],p[k]);
r=dis(p[i],c);
}
}
}
}
}
int main(){
int n;
Point c;
double r;
while(scanf("%d",&n)==1 && n){
for(int i=0;i<n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
min_cover_circle(p,n,c,r);
printf("%.2lf %.2lf %.2lf\n",c.x,c.y,r);
}
return 0;
}

1. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。

2. 第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。

3. I go through some of your put up and I uncovered a good deal of expertise from it. Many thanks for posting this sort of exciting posts

4. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.