Bzoj 1013 & Gaussian elimination element

Test instructions

Tells you a k+1 point on a sphere of k-dimensional spheres asking the spherical coordinates.

Sol

At first glance I thought it was a k-dimensional answer and judged the distance ... What a jerk. You look at the messy topics and then forget the most useful basic calculations ...

We can see that, assuming the center O, according to the formula that he tells us we can get an equation of any two points and a center point, the equation has K unknowns, then we randomly construct K equations and then run a Gaussian elimination.

The mechanical industry's line is still quite clear and understandable ... Each time you enumerate to one of the main rows, the following each one is eliminated ... Well... More intuitive ...

Because the last line of change also PE ... A little tipsy = =

CODE:

　　

/*==========================================================================# last modified:2016-03-03 15:29# filename:1013.cpp# Description: ==========================================================================*/# Define me Acrossthesky #include <cstdio> #include <cmath> #include <ctime> #include <string> #inc (in)  Lude <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <set> #include <map> #include <stack> #include <queue> #include <vector> #define LOWBIT (x) (x) & (-X ) #define for (I,A,B) for ((i) = (a);(i) <= (b);(i) + +) #define FORP (I,A,B) for (int i= (a); i<= (b); i++) #define FORM (i,a,b ) for (int i= (a); i>= (b); i--) #define LS (A, a) (((a) + (b)) << 1) #define RS (A, B) (((a) + (c)) >> 1) #define GETLC (a) ch[(a)][0] #define GETRC (a) ch[(a)][1] #define Delta 1e-6#define MAXN 100000 #define MAXM 100000 #define PI 3.14159265 35898 #define _E 2.718281828459 #define INF 10700000The using namespace Std; typedef long Long LL;  typedef unsigned long long ull;     Template<class t> inline void read (t& num) {bool start=false,neg=false;     char c;     num=0;         while ((C=getchar ())!=eof) {if (c== '-') start=neg=true;             else if (c>= ' 0 ' && c<= ' 9 ') {start=true;         num=num*10+c-' 0 ';     } else if (start) break; } if (neg) num=-num; }/*==================split line==================*/Double m[20][20],p[20][20],ans[20];int n;int judge (Double X, Double T) {if (Fabs (x-t) <=delta) return 0;else if (x>t) return 1;return-1;} void Gauss_elimination () {double t=1.0; Forp (i,1,n) {if (Judge (m[i][i],0)!=0) T=m[i][i];else continue; Forp (j,i,n+1) m[i][j]/=t; Forp (j,i+1,n) {t=m[j][i]; Forp (k,i,n+1) m[j][k]-=t*m[i][k];} Check ();} FORM (i,n,1) {Forp (j,i+1,n) m[i][n+1]-=m[i][j]*ans[j];ans[i]=m[i][n+1];}} int main () {read (n); Forp (i,1,n+1) Forp (j,1,n) {scanf ("%lf", &p[i][j]);} Forp (i,2,n+1) {Forp (j,1,n) {m[i-1][j]=p[i][j]-p[1][j]; m[i-1][n+1]+= ((P[i][j]*p[i][j]-p[1][j]*p[1][j])/2.0);}} Gauss_elimination (); Forp (i,1,n-1) printf ("%.3lf", Ans[i]); printf ("%.3lf", Ans[n]);}

Bzoj 1013 & Gaussian elimination element