标签：$$ using const print 方程组 for i+1 表示 形式
给你一个n维球体上的n+1个点，让你求这个n维球体的球心。数据保证球心是唯一的。
将球心设出来为$(x_1, x_2, \cdots, x_n)$，设半径为$r$。设球上一点为$(y_1, y_2, \cdots, y_n)$，根据n维空间内两点之间距离公式得$\sum_{i=1}^n (y_i-x_i)^2=r^2$。
设每个点表示为$(a_{i,1}, a_{i,2}, \cdots, a_{i,n})$。可以得到方程组：
$$\begin{cases}
\sum_{i=1}^n (a_{1,i}-x_i)^2=r^2\\
\sum_{i=1}^n (a_{2,i}-x_i)^2=r^2\\
\cdots\\
\sum_{i=1}^n (a_{n+1,i}-x_i)^2=r^2\\
\end{cases}
$$
将第$i$个方程与第$i+1$个方程相减得：
$$\begin{cases}
\sum_{i=1}^n (a_{1,i}^2-a_{2,i}^2-2x_i(a_{1,i}-a_{2,i}))=0\\
\sum_{i=1}^n (a_{2,i}^2-a_{3,i}^2-2x_i(a_{2,i}-a_{3,i}))=0\\
\cdots\\
\sum_{i=1}^n (a_{n,i}^2-a_{n+1,i}^2-2x_i(a_{n,i}-a_{n+1,i}))=0\\
\end{cases}
$$
再变成标准形式：
$$\begin{cases}
\sum_{i=1}^n (2x_i(a_{1,i}-a_{2,i}))=\sum_{i=1}^n a_{1,i}^2-a_{2,i}^2\\
\sum_{i=1}^n (2x_i(a_{2,i}-a_{3,i}))=\sum_{i=1}^n a_{2,i}^2-a_{3,i}^2\\
\cdots\\
\sum_{i=1}^n (2x_i(a_{n,i}-a_{n+1,i}))=\sum_{i=1}^n a_{n,i}^2-a_{n+1,i}^2\\
\end{cases}
$$
n个未知数n个方程，直接高斯消元。因为题目保证有解所以不需要做过多的判断。
#include <iostream> #include <cstdio> using namespace std; const int N = 11; int n; double a[N][N], A[N][N], B[N]; void Gauss() { for (int i = 1; i <= n; i++) { int mx = i; for (int j = i + 1; j <= n; j++) { if (A[j][i] > A[mx][i]) { mx = j; } } for (int j = i; j <= n; j++) { swap(A[i][j], A[mx][j]); } swap(B[i], B[mx]); if (!A[i][i]) continue; for (int j = 1; j <= n; j++) { if (i == j) continue; double Rate = A[j][i] / A[i][i]; for (int k = i; k <= n; k++) { A[j][k] -= A[i][k] * Rate; } B[j] -= B[i] * Rate; } } } int main() { scanf("%d", &n); for (int i = 1; i <= n + 1; i++) { for (int j = 1; j <= n; j++) { scanf("%lf", &a[i][j]); } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { A[i][j] = 2.0 * (a[i][j] - a[i + 1][j]); B[i] += a[i][j] * a[i][j] - a[i + 1][j] * a[i + 1][j]; } } Gauss(); for (int i = 1; i <= n; i++) printf("%.3lf ", B[i] / A[i][i]); return 0; }
标签：$$ using const print 方程组 for i+1 表示 形式
原文地址：https://www.cnblogs.com/zcr-blog/p/13155348.html
