傳送門

題意：在平面中，有一個圓，有一個點，問能在這個圓中圍出最大的圓的圓心坐標和半徑。要求這個最大圓不包含這個點。

思路：比較基礎的計算幾何，要分三種情況，第一種就是這個點在圓外的情況。第二種是點在圓內。第三種是這個點和圓心重合。

ac代碼

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include 
 
<set>
#include <queue>
#include <list>
#include <iterator>
#include <cmath>
using namespace std;

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define
 
 pb push_back
#define pq priority_queue

#define Pll pair<ll,ll>
#define Pii pair<int,int>

#define fi first
#define se second

#define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double espp = 0.0000000001;

/*-----------------show time----------------
 
*/
double R,x,y,x2,y2;
double dis(double a,double b,double x,double y)
{
    return sqrt((a-x)*(a-x) + (b-y)*(b-y));
}
int main(){
    scanf("%lf%lf%lf%lf%lf", &R, &x, &y, &x2, &y2);
    double d = dis(x,y,x2,y2);
    
    if(d > R ||abs(d-R) < espp)
    {
        printf("%lf %lf %lf\n",x,y,R);
    }
    else
    {
        double r1 = (R + d)/2;  
        double dx = 1.0*(x-x2)*(x-x2) + 1.0*(y-y2)*(y-y2);
        dx = sqrt(dx);  
        double ans1,ans2;
        if(dx!=0)
        {
             ans1 = x2 + (x-x2)/dx*r1;
             ans2 = y2 + (y-y2)/dx*r1;
        }
        else 
        {            
            ans1 = x;
            ans2 = y+R/2; 
        }
        printf("%.9lf %.9lf %.9lf\n",ans1,ans2,r1);
    }
    
    
    return 0;
}

Codeforces Round #465 &935C. Fifa and Fafa計算幾何