思路：

這道題就是簡單的數學題，只不過被n維嚇住了而已，只需要按照二維的思路處理n維即可。

程式碼：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const int N = 105;
const int M = 2005;
int o[N];//圓點座標
double d[M];//點到圓點距離
double rd[M];//點的切線距離
int p[M][N];//點的座標 
 

int n,m;
double r;
double ans[M];
inline double sqr(double x){
    return x * x;
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin>>n>>m;
    cin>>r;
    for(int i = 0;i<n;i++){
        cin>>o[i];
    }
    for(int i = 0;i<m;i++){
        double s = 0;
        
 
for(int j = 0;j<n;j++){
            cin>>p[i][j];
            s += sqr(p[i][j] - o[j]);
        }
        d[i] = sqrt(s);//求出點到圓心距離 
        rd[i] = sqrt(s - sqr(r));//求出切線距離 
    }
    for(int i = 0;i<m;i++){
        for(int j = 0;j<i;j++){
            double s = 0;
            for(int k = 0;k<n;k++){
                s 
 
+= sqr(p[i][k] - p[j][k]);
            }
            double c = sqrt(s);
            double a = d[i];
            double b = d[j];
            double t = (a + b + c) / 2;
            double area = sqrt(t * (t - a) * (t - b) * (t - c));
            double h = 2 * area / c;
            if(h >= r || sqr(a) >= sqr(b) + s || sqr(b) >= sqr(a) + s){
                ans[i] += c, ans[j] += c;
                continue;
            }
            double angle = acos((sqr(a) + sqr(b) - sqr(c)) / (2 * a * b));
            double angle1 = acos( r / a);
            double angle2 = acos( r / b);
            double res = (angle - angle1 - angle2) * r + rd[i] + rd[j];
            ans[i] += res, ans[j] += res;
        }
    }
    for(int i = 0;i<m;i++){
        printf("%.13f\n", ans[i]);
    }
}