题意：给出一个点到12个点的距离，求这个点的坐标。（坐标维数：11）
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4835
——>>因为一定有解，所以转化为线性方程组后直接高斯消元。（训练时我们队FA～激动）
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>using namespace std;const int maxn = 13;
typedef double Matrix[maxn][maxn];void gause_elimination(Matrix A, int n)
{int i, j, k, r;for(i = 0; i < n; i++){r = i;for(j = i+1; j < n; j++)if(fabs(A[j][i]) > fabs(A[r][i]))r = j;if(r != i) for(j = 0; j <= n; j++) swap(A[r][j], A[i][j]);for(k = i+1; k < n; k++){for(j = n; j >= i; j--)A[k][j] -= A[k][i] / A[i][i] * A[i][j];}}for(i = n-1; i >= 0; i--){for(j = i+1; j < n; j++)A[i][n] -= A[j][n] * A[i][j];A[i][n] /= A[i][i];}
}int main()
{int T, i, j, k;scanf("%d", &T);Matrix a, A;while(T--){for(i = 0; i < 12; i++)for(j = 0; j < 12; j++)scanf("%lf", &a[i][j]);memset(A, 0, sizeof(A));for(i = 0; i < 11; i++){for(j = 0; j < 11 ; j++){A[i][j] = a[i][j] - a[i+1][j];}A[i][11] = a[i+1][11] * a[i+1][11] - a[i][11] * a[i][11];for(k = 0; k < 11; k++)A[i][11] += a[i][k] * a[i][k] - a[i+1][k] * a[i+1][k];A[i][11] /= 2;}gause_elimination(A, 11);for(i = 0;i < 10; i++)printf("%.2lf ", A[i][11]);printf("%.2lf\n", A[10][11]);}return 0;
}